{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0001","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"53dbcef575b5395f28087848deb321255e6fbf709b88ab94e84426dda6e6950c","family_id":"qf-mod33-compact-finite-space-certificate","seed":202611025001,"semantic_template":"qf-mod33-compact-finite-space-certificate/{\"n\":5,\"topology\":[[],[0,1,2,3,4],[0,1,2,3],[0,1,2],[0,1],[0]]}"},"fields":{"title":"Finiteness Supplies the Universal Compactness Certificate","stem":"Use the complete finite topology=[[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).","answer":{"kind":"expression","value":"(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)","explanation":"The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)."},"skills":["Finiteness Supplies the Universal Compactness Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0002","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0de99a3b4791060253e5fb7d6f6489ce1a660b9d900e8c68dd91eee4dbbda459","family_id":"qf-mod33-compact-finite-space-certificate","seed":202611025002,"semantic_template":"qf-mod33-compact-finite-space-certificate/{\"n\":5,\"topology\":[[],[0,1,2,3,4],[0,1,2,4],[0,1,2],[0,1],[0]]}"},"fields":{"title":"Finiteness Supplies the Universal Compactness Certificate","stem":"Use the complete finite topology=[[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).","answer":{"kind":"expression","value":"(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)","explanation":"The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)."},"skills":["Finiteness Supplies the Universal Compactness Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0003","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"df6f170f28425a93e58a9399de7ec09076debdd5b528474318cd3150ede2b951","family_id":"qf-mod33-compact-finite-space-certificate","seed":202611025003,"semantic_template":"qf-mod33-compact-finite-space-certificate/{\"n\":5,\"topology\":[[],[0,1,2,3,4],[0,1,2,3],[0,1,3],[0,1],[0]]}"},"fields":{"title":"Finiteness Supplies the Universal Compactness Certificate","stem":"Use the complete finite topology=[[], [0], [0, 1], [0, 1, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).","answer":{"kind":"expression","value":"(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)","explanation":"The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)."},"skills":["Finiteness Supplies the Universal Compactness Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0004","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5641b2553a939d2bcc7fa6b8f0f0a0926e3903bc2b03e8de7d048e13b9a9ec7f","family_id":"qf-mod33-compact-finite-space-certificate","seed":202611025004,"semantic_template":"qf-mod33-compact-finite-space-certificate/{\"n\":5,\"topology\":[[],[0,1,2,3,4],[0,1,3,4],[0,1,3],[0,1],[0]]}"},"fields":{"title":"Finiteness Supplies the Universal Compactness Certificate","stem":"Use the complete finite topology=[[], [0], [0, 1], [0, 1, 3], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).","answer":{"kind":"expression","value":"(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)","explanation":"The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)."},"skills":["Finiteness Supplies the Universal Compactness Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0005","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bc814bf366c4aac5010f3bf70655dea6da2b30c52b8d30931cd010529f222ce7","family_id":"qf-mod33-compact-finite-space-certificate","seed":202611025005,"semantic_template":"qf-mod33-compact-finite-space-certificate/{\"n\":5,\"topology\":[[],[0,1,2,3,4],[0,1,2,4],[0,1,4],[0,1],[0]]}"},"fields":{"title":"Finiteness Supplies the Universal Compactness Certificate","stem":"Use the complete finite topology=[[], [0], [0, 1], [0, 1, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).","answer":{"kind":"expression","value":"(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)","explanation":"The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)."},"skills":["Finiteness Supplies the Universal Compactness Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0006","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f0d42388b5e694f41a2e55875dbdd7c0c038156f410c58a7601f5e20e44418ff","family_id":"qf-mod33-compact-finite-space-certificate","seed":202611025006,"semantic_template":"qf-mod33-compact-finite-space-certificate/{\"n\":5,\"topology\":[[],[0,1,2,3,4],[0,1,3,4],[0,1,4],[0,1],[0]]}"},"fields":{"title":"Finiteness Supplies the Universal Compactness Certificate","stem":"Use the complete finite topology=[[], [0], [0, 1], [0, 1, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).","answer":{"kind":"expression","value":"(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)","explanation":"The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)."},"skills":["Finiteness Supplies the Universal Compactness Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0007","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"02a88bb2769d51fef4a714ee39c777c4fbd1193ea121521def242e912e7e3ddf","family_id":"qf-mod33-compact-finite-space-certificate","seed":202611025007,"semantic_template":"qf-mod33-compact-finite-space-certificate/{\"n\":5,\"topology\":[[],[0,1,2,3,4],[0,1,2,3],[0,1,2],[0,2],[0]]}"},"fields":{"title":"Finiteness Supplies the Universal Compactness Certificate","stem":"Use the complete finite topology=[[], [0], [0, 2], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).","answer":{"kind":"expression","value":"(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)","explanation":"The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)."},"skills":["Finiteness Supplies the Universal Compactness Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0008","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4c74ee5cb44c4df2a64e8ed6e8a2c3b41448ee879cd3f85bca7f9fcb5708f557","family_id":"qf-mod33-compact-finite-space-certificate","seed":202611025008,"semantic_template":"qf-mod33-compact-finite-space-certificate/{\"n\":5,\"topology\":[[],[0,1,2,3,4],[0,1,2,4],[0,1,2],[0,2],[0]]}"},"fields":{"title":"Finiteness Supplies the Universal Compactness Certificate","stem":"Use the complete finite topology=[[], [0], [0, 2], [0, 1, 2], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).","answer":{"kind":"expression","value":"(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)","explanation":"The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)."},"skills":["Finiteness Supplies the Universal Compactness Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0009","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"99da1e6a8890318be7a842c0a211e34548b884531081937aa917b797e66272fe","family_id":"qf-mod33-compact-finite-space-certificate","seed":202611025009,"semantic_template":"qf-mod33-compact-finite-space-certificate/{\"n\":5,\"topology\":[[],[0,1,2,3,4],[0,1,2,3],[0,2,3],[0,2],[0]]}"},"fields":{"title":"Finiteness Supplies the Universal Compactness Certificate","stem":"Use the complete finite topology=[[], [0], [0, 2], [0, 2, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).","answer":{"kind":"expression","value":"(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)","explanation":"The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)."},"skills":["Finiteness Supplies the Universal Compactness Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0010","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"353ebee3c906b69975d8ffde8e2a0894301275e38e34b0cd1b5b1ad12ca1fefb","family_id":"qf-mod33-compact-finite-space-certificate","seed":202611025010,"semantic_template":"qf-mod33-compact-finite-space-certificate/{\"n\":5,\"topology\":[[],[0,1,2,3,4],[0,2,3,4],[0,2,3],[0,2],[0]]}"},"fields":{"title":"Finiteness Supplies the Universal Compactness Certificate","stem":"Use the complete finite topology=[[], [0], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).","answer":{"kind":"expression","value":"(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)","explanation":"The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)."},"skills":["Finiteness Supplies the Universal Compactness Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0011","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6b699d372134e8d4081701778097891ac728e64a0f1a9a2b99c157bb95fbe9aa","family_id":"qf-mod33-compact-finite-space-certificate","seed":202611025011,"semantic_template":"qf-mod33-compact-finite-space-certificate/{\"n\":5,\"topology\":[[],[0,1,2,3,4],[0,1,2,4],[0,2,4],[0,2],[0]]}"},"fields":{"title":"Finiteness Supplies the Universal Compactness Certificate","stem":"Use the complete finite topology=[[], [0], [0, 2], [0, 2, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).","answer":{"kind":"expression","value":"(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)","explanation":"The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)."},"skills":["Finiteness Supplies the Universal Compactness Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0012","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"10798486318af11a0946aecd4497bbd4c720e8b4c45daff4c3e2fd783f166caf","family_id":"qf-mod33-compact-finite-space-certificate","seed":202611025012,"semantic_template":"qf-mod33-compact-finite-space-certificate/{\"n\":5,\"topology\":[[],[0,1,2,3,4],[0,2,3,4],[0,2,4],[0,2],[0]]}"},"fields":{"title":"Finiteness Supplies the Universal Compactness Certificate","stem":"Use the complete finite topology=[[], [0], [0, 2], [0, 2, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).","answer":{"kind":"expression","value":"(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)","explanation":"The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)."},"skills":["Finiteness Supplies the Universal Compactness Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0013","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d6308749898d75268235d7199c5150e1bdc6960497c722c9d1421bf20550b2fe","family_id":"qf-mod33-compact-finite-space-certificate","seed":202611025013,"semantic_template":"qf-mod33-compact-finite-space-certificate/{\"n\":5,\"topology\":[[],[0,1,2,3,4],[0,1,2,3],[0,1,3],[0,3],[0]]}"},"fields":{"title":"Finiteness Supplies the Universal Compactness Certificate","stem":"Use the complete finite topology=[[], [0], [0, 3], [0, 1, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).","answer":{"kind":"expression","value":"(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)","explanation":"The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)."},"skills":["Finiteness Supplies the Universal Compactness Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0014","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"833d835a5a75cf26ec02029fe82988893c6ac742ae9617e9f7218d5dead42f9d","family_id":"qf-mod33-compact-finite-space-certificate","seed":202611025014,"semantic_template":"qf-mod33-compact-finite-space-certificate/{\"n\":5,\"topology\":[[],[0,1,2,3,4],[0,1,3,4],[0,1,3],[0,3],[0]]}"},"fields":{"title":"Finiteness Supplies the Universal Compactness Certificate","stem":"Use the complete finite topology=[[], [0], [0, 3], [0, 1, 3], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).","answer":{"kind":"expression","value":"(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)","explanation":"The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)."},"skills":["Finiteness Supplies the Universal Compactness Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0015","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f31fd1b14a1d99e27172c76d1f58358c33471d39db41fe8c03eaefda65a3078a","family_id":"qf-mod33-compact-finite-space-certificate","seed":202611025015,"semantic_template":"qf-mod33-compact-finite-space-certificate/{\"n\":5,\"topology\":[[],[0,1,2,3,4],[0,1,2,3],[0,2,3],[0,3],[0]]}"},"fields":{"title":"Finiteness Supplies the Universal Compactness Certificate","stem":"Use the complete finite topology=[[], [0], [0, 3], [0, 2, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).","answer":{"kind":"expression","value":"(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)","explanation":"The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)."},"skills":["Finiteness Supplies the Universal Compactness Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0016","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6a5b6c04a07202df80023229673935a754a4827430700474aefdc8f6b2bbc851","family_id":"qf-mod33-compact-finite-space-certificate","seed":202611025016,"semantic_template":"qf-mod33-compact-finite-space-certificate/{\"n\":5,\"topology\":[[],[0,1,2,3,4],[0,2,3,4],[0,2,3],[0,3],[0]]}"},"fields":{"title":"Finiteness Supplies the Universal Compactness Certificate","stem":"Use the complete finite topology=[[], [0], [0, 3], [0, 2, 3], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).","answer":{"kind":"expression","value":"(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)","explanation":"The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)."},"skills":["Finiteness Supplies the Universal Compactness Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0017","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d5d8799f8736fc5e5790005e92961de8af2b453037d346e41f58792ae6972882","family_id":"qf-mod33-compact-finite-space-certificate","seed":202611025017,"semantic_template":"qf-mod33-compact-finite-space-certificate/{\"n\":5,\"topology\":[[],[0,1,2,3,4],[0,1,3,4],[0,3,4],[0,3],[0]]}"},"fields":{"title":"Finiteness Supplies the Universal Compactness Certificate","stem":"Use the complete finite topology=[[], [0], [0, 3], [0, 3, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).","answer":{"kind":"expression","value":"(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)","explanation":"The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)."},"skills":["Finiteness Supplies the Universal Compactness Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0018","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c4f308b9823914a8c8e55d0032d39dd3274ee804288010f6151ff4026857786c","family_id":"qf-mod33-compact-finite-space-certificate","seed":202611025018,"semantic_template":"qf-mod33-compact-finite-space-certificate/{\"n\":5,\"topology\":[[],[0,1,2,3,4],[0,2,3,4],[0,3,4],[0,3],[0]]}"},"fields":{"title":"Finiteness Supplies the Universal Compactness Certificate","stem":"Use the complete finite topology=[[], [0], [0, 3], [0, 3, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).","answer":{"kind":"expression","value":"(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)","explanation":"The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)."},"skills":["Finiteness Supplies the Universal Compactness Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0019","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ae957b436a51d0eeaa71a62bac78273ba953d2122440fa81d9f07d71f3eae515","family_id":"qf-mod33-compact-finite-space-certificate","seed":202611025019,"semantic_template":"qf-mod33-compact-finite-space-certificate/{\"n\":5,\"topology\":[[],[0,1,2,3,4],[0,1,2,4],[0,1,4],[0,4],[0]]}"},"fields":{"title":"Finiteness Supplies the Universal Compactness Certificate","stem":"Use the complete finite topology=[[], [0], [0, 4], [0, 1, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).","answer":{"kind":"expression","value":"(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)","explanation":"The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)."},"skills":["Finiteness Supplies the Universal Compactness Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0020","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b1b0be9d36387cd17264dad13820dda8bdd87922e9296c3d257e045cd96070fa","family_id":"qf-mod33-compact-finite-space-certificate","seed":202611025020,"semantic_template":"qf-mod33-compact-finite-space-certificate/{\"n\":5,\"topology\":[[],[0,1,2,3,4],[0,1,3,4],[0,1,4],[0,4],[0]]}"},"fields":{"title":"Finiteness Supplies the Universal Compactness Certificate","stem":"Use the complete finite topology=[[], [0], [0, 4], [0, 1, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).","answer":{"kind":"expression","value":"(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)","explanation":"The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)."},"skills":["Finiteness Supplies the Universal Compactness Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0021","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a730c6b86902857b7e68c4951035bc629c718279168c390b6a79c6817545f756","family_id":"qf-mod33-compact-finite-space-certificate","seed":202611025021,"semantic_template":"qf-mod33-compact-finite-space-certificate/{\"n\":5,\"topology\":[[],[0,1,2,3,4],[0,1,2,4],[0,2,4],[0,4],[0]]}"},"fields":{"title":"Finiteness Supplies the Universal Compactness Certificate","stem":"Use the complete finite topology=[[], [0], [0, 4], [0, 2, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).","answer":{"kind":"expression","value":"(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)","explanation":"The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)."},"skills":["Finiteness Supplies the Universal Compactness Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0022","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8b1eb222d99b896b5062c30acc07968130ccc901e1104c2cf5a22c47113bb3f5","family_id":"qf-mod33-compact-finite-space-certificate","seed":202611025022,"semantic_template":"qf-mod33-compact-finite-space-certificate/{\"n\":5,\"topology\":[[],[0,1,2,3,4],[0,2,3,4],[0,2,4],[0,4],[0]]}"},"fields":{"title":"Finiteness Supplies the Universal Compactness Certificate","stem":"Use the complete finite topology=[[], [0], [0, 4], [0, 2, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).","answer":{"kind":"expression","value":"(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)","explanation":"The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)."},"skills":["Finiteness Supplies the Universal Compactness Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0023","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"03ef359e8150f7e5f7a935377a259cf641955066f706c5de831ca07f83043906","family_id":"qf-mod33-compact-finite-space-certificate","seed":202611025023,"semantic_template":"qf-mod33-compact-finite-space-certificate/{\"n\":5,\"topology\":[[],[0,1,2,3,4],[0,1,3,4],[0,3,4],[0,4],[0]]}"},"fields":{"title":"Finiteness Supplies the Universal Compactness Certificate","stem":"Use the complete finite topology=[[], [0], [0, 4], [0, 3, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).","answer":{"kind":"expression","value":"(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)","explanation":"The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)."},"skills":["Finiteness Supplies the Universal Compactness Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0024","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1292cd308927e4cac8ccf8eeca12939846f5d5ee3f5afec6669f92b28daf5fdf","family_id":"qf-mod33-compact-finite-space-certificate","seed":202611025024,"semantic_template":"qf-mod33-compact-finite-space-certificate/{\"n\":5,\"topology\":[[],[0,1,2,3,4],[0,2,3,4],[0,3,4],[0,4],[0]]}"},"fields":{"title":"Finiteness Supplies the Universal Compactness Certificate","stem":"Use the complete finite topology=[[], [0], [0, 4], [0, 3, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).","answer":{"kind":"expression","value":"(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)","explanation":"The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)."},"skills":["Finiteness Supplies the Universal Compactness Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0025","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a48e8d30b8833deb44497bf7f732f53a3b370908507e408ab63619210ae61be5","family_id":"qf-mod33-compact-finite-space-certificate","seed":202611025025,"semantic_template":"qf-mod33-compact-finite-space-certificate/{\"n\":5,\"topology\":[[],[0,1,2,3,4],[0,1,2,3],[0,1,2],[0,1],[1]]}"},"fields":{"title":"Finiteness Supplies the Universal Compactness Certificate","stem":"Use the complete finite topology=[[], [1], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).","answer":{"kind":"expression","value":"(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)","explanation":"The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)."},"skills":["Finiteness Supplies the Universal Compactness Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0026","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5008b53b82bc2e1f51f72ce994cb7a9ea3120040a3bdb705ade2dedee89c6dae","family_id":"qf-mod33-compact-cover-audit","seed":202611025026,"semantic_template":"qf-mod33-compact-cover-audit/{\"cover\":[[0,1],[0,1,2,3],[0,1,2,3,4]],\"n\":5,\"target\":[0,1,2,3],\"topology\":[[],[0,1,2,3,4],[0,1,2,3],[0,1,2],[0,1],[0]]}"},"fields":{"title":"An Invalid Supplied Cover Does Not Refute Finite-Space Compactness","stem":"Use the complete finite topology=[[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 1], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).","answer":{"kind":"expression","value":"([], [0, 1, 2, 3, 4], Yes, Yes)","explanation":"Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes)."},"skills":["An Invalid Supplied Cover Does Not Refute Finite-Space Compactness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0027","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"10b58abf0e9ffe7f8813081769ec4f04b7d89e1ebb94666e38def276f0d964cc","family_id":"qf-mod33-compact-cover-audit","seed":202611025027,"semantic_template":"qf-mod33-compact-cover-audit/{\"cover\":[[0,1],[0,1,2,4],[0,1,2,3,4]],\"n\":5,\"target\":[0,1,2,4],\"topology\":[[],[0,1,2,3,4],[0,1,2,4],[0,1,2],[0,1],[0]]}"},"fields":{"title":"An Invalid Supplied Cover Does Not Refute Finite-Space Compactness","stem":"Use the complete finite topology=[[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 4], original ordered cover=[[0, 1], [0, 1, 2, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).","answer":{"kind":"expression","value":"([], [0, 1, 2, 3, 4], Yes, Yes)","explanation":"Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes)."},"skills":["An Invalid Supplied Cover Does Not Refute Finite-Space Compactness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0028","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"22524f1fffead2e4ca65261ec8fc61b12e9564c7d7628303820951d5ff5fe759","family_id":"qf-mod33-compact-cover-audit","seed":202611025028,"semantic_template":"qf-mod33-compact-cover-audit/{\"cover\":[[0,1],[0,1,2,3],[0,1,2,3,4]],\"n\":5,\"target\":[0,1,2,3],\"topology\":[[],[0,1,2,3,4],[0,1,2,3],[0,1,3],[0,1],[0]]}"},"fields":{"title":"An Invalid Supplied Cover Does Not Refute Finite-Space Compactness","stem":"Use the complete finite topology=[[], [0], [0, 1], [0, 1, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 1], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).","answer":{"kind":"expression","value":"([], [0, 1, 2, 3, 4], Yes, Yes)","explanation":"Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes)."},"skills":["An Invalid Supplied Cover Does Not Refute Finite-Space Compactness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0029","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"acb91691cc1092321812d3c051859fb49017bdd90794af532ff9bf67e999107e","family_id":"qf-mod33-compact-cover-audit","seed":202611025029,"semantic_template":"qf-mod33-compact-cover-audit/{\"cover\":[[0,1],[0,1,3,4],[0,1,2,3,4]],\"n\":5,\"target\":[0,1,3,4],\"topology\":[[],[0,1,2,3,4],[0,1,3,4],[0,1,3],[0,1],[0]]}"},"fields":{"title":"An Invalid Supplied Cover Does Not Refute Finite-Space Compactness","stem":"Use the complete finite topology=[[], [0], [0, 1], [0, 1, 3], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 3, 4], original ordered cover=[[0, 1], [0, 1, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).","answer":{"kind":"expression","value":"([], [0, 1, 2, 3, 4], Yes, Yes)","explanation":"Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes)."},"skills":["An Invalid Supplied Cover Does Not Refute Finite-Space Compactness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0030","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"464ccf21dd0aa73c0a1ce646341c2c85dba3f27a7f7a2e2a54feb4b4c41eec3a","family_id":"qf-mod33-compact-cover-audit","seed":202611025030,"semantic_template":"qf-mod33-compact-cover-audit/{\"cover\":[[0,1],[0,1,2,4],[0,1,2,3,4]],\"n\":5,\"target\":[0,1,2,4],\"topology\":[[],[0,1,2,3,4],[0,1,2,4],[0,1,4],[0,1],[0]]}"},"fields":{"title":"An Invalid Supplied Cover Does Not Refute Finite-Space Compactness","stem":"Use the complete finite topology=[[], [0], [0, 1], [0, 1, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 4], original ordered cover=[[0, 1], [0, 1, 2, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).","answer":{"kind":"expression","value":"([], [0, 1, 2, 3, 4], Yes, Yes)","explanation":"Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes)."},"skills":["An Invalid Supplied Cover Does Not Refute Finite-Space Compactness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0031","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"17729c1dbe3f0e444bd06e9400eeaf965866b2454efc6acbc3cc0a970f774a32","family_id":"qf-mod33-compact-cover-audit","seed":202611025031,"semantic_template":"qf-mod33-compact-cover-audit/{\"cover\":[[0,1],[0,1,3,4],[0,1,2,3,4]],\"n\":5,\"target\":[0,1,3,4],\"topology\":[[],[0,1,2,3,4],[0,1,3,4],[0,1,4],[0,1],[0]]}"},"fields":{"title":"An Invalid Supplied Cover Does Not Refute Finite-Space Compactness","stem":"Use the complete finite topology=[[], [0], [0, 1], [0, 1, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 3, 4], original ordered cover=[[0, 1], [0, 1, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).","answer":{"kind":"expression","value":"([], [0, 1, 2, 3, 4], Yes, Yes)","explanation":"Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes)."},"skills":["An Invalid Supplied Cover Does Not Refute Finite-Space Compactness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0032","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a9e3616a6837396c04bd2560aea15627095768728b6bd3e7c0366cb2956c5eed","family_id":"qf-mod33-compact-cover-audit","seed":202611025032,"semantic_template":"qf-mod33-compact-cover-audit/{\"cover\":[[0,2],[0,1,2,3],[0,1,2,3,4]],\"n\":5,\"target\":[0,1,2,3],\"topology\":[[],[0,1,2,3,4],[0,1,2,3],[0,1,2],[0,2],[0]]}"},"fields":{"title":"An Invalid Supplied Cover Does Not Refute Finite-Space Compactness","stem":"Use the complete finite topology=[[], [0], [0, 2], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).","answer":{"kind":"expression","value":"([], [0, 1, 2, 3, 4], Yes, Yes)","explanation":"Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes)."},"skills":["An Invalid Supplied Cover Does Not Refute Finite-Space Compactness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0033","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e660021e6b6d98d3c80e52ed6720d634c09a87a4e06bf3b43b9fc7cfc848baaa","family_id":"qf-mod33-compact-cover-audit","seed":202611025033,"semantic_template":"qf-mod33-compact-cover-audit/{\"cover\":[[0,2],[0,1,2,4],[0,1,2,3,4]],\"n\":5,\"target\":[0,1,2,4],\"topology\":[[],[0,1,2,3,4],[0,1,2,4],[0,1,2],[0,2],[0]]}"},"fields":{"title":"An Invalid Supplied Cover Does Not Refute Finite-Space Compactness","stem":"Use the complete finite topology=[[], [0], [0, 2], [0, 1, 2], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 4], original ordered cover=[[0, 2], [0, 1, 2, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).","answer":{"kind":"expression","value":"([], [0, 1, 2, 3, 4], Yes, Yes)","explanation":"Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes)."},"skills":["An Invalid Supplied Cover Does Not Refute Finite-Space Compactness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0034","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"02fbb73d276504ef3619fb2b879a3b375f4b7a0eddf22f0eadd3ae3a30eaa504","family_id":"qf-mod33-compact-cover-audit","seed":202611025034,"semantic_template":"qf-mod33-compact-cover-audit/{\"cover\":[[0,2],[0,1,2,3],[0,1,2,3,4]],\"n\":5,\"target\":[0,1,2,3],\"topology\":[[],[0,1,2,3,4],[0,1,2,3],[0,2,3],[0,2],[0]]}"},"fields":{"title":"An Invalid Supplied Cover Does Not Refute Finite-Space Compactness","stem":"Use the complete finite topology=[[], [0], [0, 2], [0, 2, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).","answer":{"kind":"expression","value":"([], [0, 1, 2, 3, 4], Yes, Yes)","explanation":"Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes)."},"skills":["An Invalid Supplied Cover Does Not Refute Finite-Space Compactness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0035","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fa088008c57572643c1568225869c5d9637a0096e7640f232d9528a2dbe51e78","family_id":"qf-mod33-compact-cover-audit","seed":202611025035,"semantic_template":"qf-mod33-compact-cover-audit/{\"cover\":[[0,2],[0,2,3,4],[0,1,2,3,4]],\"n\":5,\"target\":[0,2,3,4],\"topology\":[[],[0,1,2,3,4],[0,2,3,4],[0,2,3],[0,2],[0]]}"},"fields":{"title":"An Invalid Supplied Cover Does Not Refute Finite-Space Compactness","stem":"Use the complete finite topology=[[], [0], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 2, 3, 4], original ordered cover=[[0, 2], [0, 2, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).","answer":{"kind":"expression","value":"([], [0, 1, 2, 3, 4], Yes, Yes)","explanation":"Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes)."},"skills":["An Invalid Supplied Cover Does Not Refute Finite-Space Compactness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0036","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b117407b8785572cb2f652ab08a0804d42bc7412f433933a2df868fa0d6e92a1","family_id":"qf-mod33-compact-cover-audit","seed":202611025036,"semantic_template":"qf-mod33-compact-cover-audit/{\"cover\":[[0,2],[0,1,2,4],[0,1,2,3,4]],\"n\":5,\"target\":[0,1,2,4],\"topology\":[[],[0,1,2,3,4],[0,1,2,4],[0,2,4],[0,2],[0]]}"},"fields":{"title":"An Invalid Supplied Cover Does Not Refute Finite-Space Compactness","stem":"Use the complete finite topology=[[], [0], [0, 2], [0, 2, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 4], original ordered cover=[[0, 2], [0, 1, 2, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).","answer":{"kind":"expression","value":"([], [0, 1, 2, 3, 4], Yes, Yes)","explanation":"Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes)."},"skills":["An Invalid Supplied Cover Does Not Refute Finite-Space Compactness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0037","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b1ae0e23da877a050d7ace7ae7a68315852fd09e4aed036e01ae18bfb1f9cc8d","family_id":"qf-mod33-compact-cover-audit","seed":202611025037,"semantic_template":"qf-mod33-compact-cover-audit/{\"cover\":[[0,2],[0,2,3,4],[0,1,2,3,4]],\"n\":5,\"target\":[0,2,3,4],\"topology\":[[],[0,1,2,3,4],[0,2,3,4],[0,2,4],[0,2],[0]]}"},"fields":{"title":"An Invalid Supplied Cover Does Not Refute Finite-Space Compactness","stem":"Use the complete finite topology=[[], [0], [0, 2], [0, 2, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 2, 3, 4], original ordered cover=[[0, 2], [0, 2, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).","answer":{"kind":"expression","value":"([], [0, 1, 2, 3, 4], Yes, Yes)","explanation":"Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes)."},"skills":["An Invalid Supplied Cover Does Not Refute Finite-Space Compactness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0038","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"66e84c82824c75cece39691dcd63e5035aad0f72456176f6c66daeca0c882a12","family_id":"qf-mod33-compact-cover-audit","seed":202611025038,"semantic_template":"qf-mod33-compact-cover-audit/{\"cover\":[[0,3],[0,1,2,3],[0,1,2,3,4]],\"n\":5,\"target\":[0,1,2,3],\"topology\":[[],[0,1,2,3,4],[0,1,2,3],[0,1,3],[0,3],[0]]}"},"fields":{"title":"An Invalid Supplied Cover Does Not Refute Finite-Space Compactness","stem":"Use the complete finite topology=[[], [0], [0, 3], [0, 1, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).","answer":{"kind":"expression","value":"([], [0, 1, 2, 3, 4], Yes, Yes)","explanation":"Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes)."},"skills":["An Invalid Supplied Cover Does Not Refute Finite-Space Compactness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0039","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0f0f2883b6887d843f8bacefcbdc57ea377e59cc5ddbcadddf51eba46fcc15f0","family_id":"qf-mod33-compact-cover-audit","seed":202611025039,"semantic_template":"qf-mod33-compact-cover-audit/{\"cover\":[[0,3],[0,1,3,4],[0,1,2,3,4]],\"n\":5,\"target\":[0,1,3,4],\"topology\":[[],[0,1,2,3,4],[0,1,3,4],[0,1,3],[0,3],[0]]}"},"fields":{"title":"An Invalid Supplied Cover Does Not Refute Finite-Space Compactness","stem":"Use the complete finite topology=[[], [0], [0, 3], [0, 1, 3], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 3, 4], original ordered cover=[[0, 3], [0, 1, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).","answer":{"kind":"expression","value":"([], [0, 1, 2, 3, 4], Yes, Yes)","explanation":"Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes)."},"skills":["An Invalid Supplied Cover Does Not Refute Finite-Space Compactness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0040","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e9bde67fdcac0db3651fd1ba091b6e86984d2be81ad5ea1e6880772a7050af1b","family_id":"qf-mod33-compact-cover-audit","seed":202611025040,"semantic_template":"qf-mod33-compact-cover-audit/{\"cover\":[[0,3],[0,1,2,3],[0,1,2,3,4]],\"n\":5,\"target\":[0,1,2,3],\"topology\":[[],[0,1,2,3,4],[0,1,2,3],[0,2,3],[0,3],[0]]}"},"fields":{"title":"An Invalid Supplied Cover Does Not Refute Finite-Space Compactness","stem":"Use the complete finite topology=[[], [0], [0, 3], [0, 2, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).","answer":{"kind":"expression","value":"([], [0, 1, 2, 3, 4], Yes, Yes)","explanation":"Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes)."},"skills":["An Invalid Supplied Cover Does Not Refute Finite-Space Compactness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0041","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5716d6aea41a0ba22748cb6ea70166eb6b14ecf98e6937261788a2a3e56f950c","family_id":"qf-mod33-compact-cover-audit","seed":202611025041,"semantic_template":"qf-mod33-compact-cover-audit/{\"cover\":[[0,3],[0,2,3,4],[0,1,2,3,4]],\"n\":5,\"target\":[0,2,3,4],\"topology\":[[],[0,1,2,3,4],[0,2,3,4],[0,2,3],[0,3],[0]]}"},"fields":{"title":"An Invalid Supplied Cover Does Not Refute Finite-Space Compactness","stem":"Use the complete finite topology=[[], [0], [0, 3], [0, 2, 3], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 2, 3, 4], original ordered cover=[[0, 3], [0, 2, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).","answer":{"kind":"expression","value":"([], [0, 1, 2, 3, 4], Yes, Yes)","explanation":"Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes)."},"skills":["An Invalid Supplied Cover Does Not Refute Finite-Space Compactness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0042","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"388c808a77ff79f289dd040db4d38b255151fd5e3cbbe93a9c4ff338f880de95","family_id":"qf-mod33-compact-cover-audit","seed":202611025042,"semantic_template":"qf-mod33-compact-cover-audit/{\"cover\":[[0,3],[0,1,3,4],[0,1,2,3,4]],\"n\":5,\"target\":[0,1,3,4],\"topology\":[[],[0,1,2,3,4],[0,1,3,4],[0,3,4],[0,3],[0]]}"},"fields":{"title":"An Invalid Supplied Cover Does Not Refute Finite-Space Compactness","stem":"Use the complete finite topology=[[], [0], [0, 3], [0, 3, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 3, 4], original ordered cover=[[0, 3], [0, 1, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).","answer":{"kind":"expression","value":"([], [0, 1, 2, 3, 4], Yes, Yes)","explanation":"Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes)."},"skills":["An Invalid Supplied Cover Does Not Refute Finite-Space Compactness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0043","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"33c6f3be9f2a138e8d70abf5ab69070c45d869427fb838329cab865a7dfd9d04","family_id":"qf-mod33-compact-cover-audit","seed":202611025043,"semantic_template":"qf-mod33-compact-cover-audit/{\"cover\":[[0,3],[0,2,3,4],[0,1,2,3,4]],\"n\":5,\"target\":[0,2,3,4],\"topology\":[[],[0,1,2,3,4],[0,2,3,4],[0,3,4],[0,3],[0]]}"},"fields":{"title":"An Invalid Supplied Cover Does Not Refute Finite-Space Compactness","stem":"Use the complete finite topology=[[], [0], [0, 3], [0, 3, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 2, 3, 4], original ordered cover=[[0, 3], [0, 2, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).","answer":{"kind":"expression","value":"([], [0, 1, 2, 3, 4], Yes, Yes)","explanation":"Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes)."},"skills":["An Invalid Supplied Cover Does Not Refute Finite-Space Compactness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0044","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"878d055e182231849d4c2f65f9f851b32bc6f69a0b125b0d75c7bb9029c562c8","family_id":"qf-mod33-compact-cover-audit","seed":202611025044,"semantic_template":"qf-mod33-compact-cover-audit/{\"cover\":[[0,4],[0,1,2,4],[0,1,2,3,4]],\"n\":5,\"target\":[0,1,2,4],\"topology\":[[],[0,1,2,3,4],[0,1,2,4],[0,1,4],[0,4],[0]]}"},"fields":{"title":"An Invalid Supplied Cover Does Not Refute Finite-Space Compactness","stem":"Use the complete finite topology=[[], [0], [0, 4], [0, 1, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 4], original ordered cover=[[0, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).","answer":{"kind":"expression","value":"([], [0, 1, 2, 3, 4], Yes, Yes)","explanation":"Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes)."},"skills":["An Invalid Supplied Cover Does Not Refute Finite-Space Compactness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0045","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"38d54b0d805142b85323e4432bbe2a944ae0059142225d84ebf5c7cd99466e85","family_id":"qf-mod33-compact-cover-audit","seed":202611025045,"semantic_template":"qf-mod33-compact-cover-audit/{\"cover\":[[0,4],[0,1,3,4],[0,1,2,3,4]],\"n\":5,\"target\":[0,1,3,4],\"topology\":[[],[0,1,2,3,4],[0,1,3,4],[0,1,4],[0,4],[0]]}"},"fields":{"title":"An Invalid Supplied Cover Does Not Refute Finite-Space Compactness","stem":"Use the complete finite topology=[[], [0], [0, 4], [0, 1, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 3, 4], original ordered cover=[[0, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).","answer":{"kind":"expression","value":"([], [0, 1, 2, 3, 4], Yes, Yes)","explanation":"Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes)."},"skills":["An Invalid Supplied Cover Does Not Refute Finite-Space Compactness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0046","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cfd5170e13dc06774b333784388190721c378834bcccbecf5f1e0d220fcd8d07","family_id":"qf-mod33-compact-cover-audit","seed":202611025046,"semantic_template":"qf-mod33-compact-cover-audit/{\"cover\":[[0,4],[0,1,2,4],[0,1,2,3,4]],\"n\":5,\"target\":[0,1,2,4],\"topology\":[[],[0,1,2,3,4],[0,1,2,4],[0,2,4],[0,4],[0]]}"},"fields":{"title":"An Invalid Supplied Cover Does Not Refute Finite-Space Compactness","stem":"Use the complete finite topology=[[], [0], [0, 4], [0, 2, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 4], original ordered cover=[[0, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).","answer":{"kind":"expression","value":"([], [0, 1, 2, 3, 4], Yes, Yes)","explanation":"Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes)."},"skills":["An Invalid Supplied Cover Does Not Refute Finite-Space Compactness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0047","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"69dcee6878a9b03fb8590e4b2f79ea6b030c158b581c7d5ed109c6a7b08429fe","family_id":"qf-mod33-compact-cover-audit","seed":202611025047,"semantic_template":"qf-mod33-compact-cover-audit/{\"cover\":[[0,4],[0,2,3,4],[0,1,2,3,4]],\"n\":5,\"target\":[0,2,3,4],\"topology\":[[],[0,1,2,3,4],[0,2,3,4],[0,2,4],[0,4],[0]]}"},"fields":{"title":"An Invalid Supplied Cover Does Not Refute Finite-Space Compactness","stem":"Use the complete finite topology=[[], [0], [0, 4], [0, 2, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 2, 3, 4], original ordered cover=[[0, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).","answer":{"kind":"expression","value":"([], [0, 1, 2, 3, 4], Yes, Yes)","explanation":"Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes)."},"skills":["An Invalid Supplied Cover Does Not Refute Finite-Space Compactness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0048","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a6ab385ffdb9b61b9dc985975947be7b0e9b116dc40c5d7a4402b7bd8f6323a8","family_id":"qf-mod33-compact-cover-audit","seed":202611025048,"semantic_template":"qf-mod33-compact-cover-audit/{\"cover\":[[0,4],[0,1,3,4],[0,1,2,3,4]],\"n\":5,\"target\":[0,1,3,4],\"topology\":[[],[0,1,2,3,4],[0,1,3,4],[0,3,4],[0,4],[0]]}"},"fields":{"title":"An Invalid Supplied Cover Does Not Refute Finite-Space Compactness","stem":"Use the complete finite topology=[[], [0], [0, 4], [0, 3, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 3, 4], original ordered cover=[[0, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).","answer":{"kind":"expression","value":"([], [0, 1, 2, 3, 4], Yes, Yes)","explanation":"Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes)."},"skills":["An Invalid Supplied Cover Does Not Refute Finite-Space Compactness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0049","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2b572146d4ba0ffe5b6e6f1a7e0eb94aac1bd7b67e57e41b75de0ae6d35a3caa","family_id":"qf-mod33-compact-cover-audit","seed":202611025049,"semantic_template":"qf-mod33-compact-cover-audit/{\"cover\":[[0,4],[0,2,3,4],[0,1,2,3,4]],\"n\":5,\"target\":[0,2,3,4],\"topology\":[[],[0,1,2,3,4],[0,2,3,4],[0,3,4],[0,4],[0]]}"},"fields":{"title":"An Invalid Supplied Cover Does Not Refute Finite-Space Compactness","stem":"Use the complete finite topology=[[], [0], [0, 4], [0, 3, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 2, 3, 4], original ordered cover=[[0, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).","answer":{"kind":"expression","value":"([], [0, 1, 2, 3, 4], Yes, Yes)","explanation":"Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes)."},"skills":["An Invalid Supplied Cover Does Not Refute Finite-Space Compactness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0050","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dd97e60e964ac141d2a3d3fa0670724a15c1baa538c77fe3d1276539b236d8b5","family_id":"qf-mod33-compact-cover-audit","seed":202611025050,"semantic_template":"qf-mod33-compact-cover-audit/{\"cover\":[[0,1],[0,1,2,3],[0,1,2,3,4]],\"n\":5,\"target\":[0,1,2,3],\"topology\":[[],[0,1,2,3,4],[0,1,2,3],[0,1,2],[0,1],[1]]}"},"fields":{"title":"An Invalid Supplied Cover Does Not Refute Finite-Space Compactness","stem":"Use the complete finite topology=[[], [1], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 1], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).","answer":{"kind":"expression","value":"([], [0, 1, 2, 3, 4], Yes, Yes)","explanation":"Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes)."},"skills":["An Invalid Supplied Cover Does Not Refute Finite-Space Compactness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0051","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1780235ced9eade30fcb8adc8b0141fef77206b83abdc610201ebf6c1531b3a3","family_id":"qf-mod33-compact-subcover-witness","seed":202611025051,"semantic_template":"qf-mod33-compact-subcover-witness/{\"cover\":[[0,1],[0,1,2,3],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,1,2,3],\"topology\":[[],[0,1,2,3,4],[0,1,2,3],[0,1,2],[0,1],[0]]}"},"fields":{"title":"A Subcover Witness Is Bound to Its Original Supplied Cover","stem":"Use the complete finite topology=[[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 1], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).","answer":{"kind":"expression","value":"([1], [0, 1, 2, 3], Yes, finite_subcover_for_supplied_cover)","explanation":"Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 1, 2, 3], Yes, finite_subcover_for_supplied_cover)."},"skills":["A Subcover Witness Is Bound to Its Original Supplied Cover"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0052","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"eac272df43dce0e3cb0917eeebb5a07b90c9072fb7fb02eb619dc3b954effde2","family_id":"qf-mod33-compact-subcover-witness","seed":202611025052,"semantic_template":"qf-mod33-compact-subcover-witness/{\"cover\":[[0,1],[0,1,2,4],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,1,2,4],\"topology\":[[],[0,1,2,3,4],[0,1,2,4],[0,1,2],[0,1],[0]]}"},"fields":{"title":"A Subcover Witness Is Bound to Its Original Supplied Cover","stem":"Use the complete finite topology=[[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 4], original ordered cover=[[0, 1], [0, 1, 2, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).","answer":{"kind":"expression","value":"([1], [0, 1, 2, 4], Yes, finite_subcover_for_supplied_cover)","explanation":"Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 1, 2, 4], Yes, finite_subcover_for_supplied_cover)."},"skills":["A Subcover Witness Is Bound to Its Original Supplied Cover"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0053","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f85fe6340a0b6f6952f3f1ef8400f4d7ccda931e5bd5dffbecfe80f8b31c8dbc","family_id":"qf-mod33-compact-subcover-witness","seed":202611025053,"semantic_template":"qf-mod33-compact-subcover-witness/{\"cover\":[[0,1],[0,1,2,3],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,1,2,3],\"topology\":[[],[0,1,2,3,4],[0,1,2,3],[0,1,3],[0,1],[0]]}"},"fields":{"title":"A Subcover Witness Is Bound to Its Original Supplied Cover","stem":"Use the complete finite topology=[[], [0], [0, 1], [0, 1, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 1], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).","answer":{"kind":"expression","value":"([1], [0, 1, 2, 3], Yes, finite_subcover_for_supplied_cover)","explanation":"Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 1, 2, 3], Yes, finite_subcover_for_supplied_cover)."},"skills":["A Subcover Witness Is Bound to Its Original Supplied Cover"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0054","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f3792b3a4346502ead5ff1fb2bee795e7dcea2d9beaf40abe992406e26b04017","family_id":"qf-mod33-compact-subcover-witness","seed":202611025054,"semantic_template":"qf-mod33-compact-subcover-witness/{\"cover\":[[0,1],[0,1,3,4],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,1,3,4],\"topology\":[[],[0,1,2,3,4],[0,1,3,4],[0,1,3],[0,1],[0]]}"},"fields":{"title":"A Subcover Witness Is Bound to Its Original Supplied Cover","stem":"Use the complete finite topology=[[], [0], [0, 1], [0, 1, 3], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 3, 4], original ordered cover=[[0, 1], [0, 1, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).","answer":{"kind":"expression","value":"([1], [0, 1, 3, 4], Yes, finite_subcover_for_supplied_cover)","explanation":"Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 1, 3, 4], Yes, finite_subcover_for_supplied_cover)."},"skills":["A Subcover Witness Is Bound to Its Original Supplied Cover"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0055","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6c4dbead04fa54968f0627762d66e3289c5b08ed0c00334a20eff7745a0d406e","family_id":"qf-mod33-compact-subcover-witness","seed":202611025055,"semantic_template":"qf-mod33-compact-subcover-witness/{\"cover\":[[0,1],[0,1,2,4],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,1,2,4],\"topology\":[[],[0,1,2,3,4],[0,1,2,4],[0,1,4],[0,1],[0]]}"},"fields":{"title":"A Subcover Witness Is Bound to Its Original Supplied Cover","stem":"Use the complete finite topology=[[], [0], [0, 1], [0, 1, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 4], original ordered cover=[[0, 1], [0, 1, 2, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).","answer":{"kind":"expression","value":"([1], [0, 1, 2, 4], Yes, finite_subcover_for_supplied_cover)","explanation":"Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 1, 2, 4], Yes, finite_subcover_for_supplied_cover)."},"skills":["A Subcover Witness Is Bound to Its Original Supplied Cover"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0056","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3a645e7c6dc4836dd5f929dd4c0e1270e3bfc5c899c1f7679f10284686eb190b","family_id":"qf-mod33-compact-subcover-witness","seed":202611025056,"semantic_template":"qf-mod33-compact-subcover-witness/{\"cover\":[[0,1],[0,1,3,4],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,1,3,4],\"topology\":[[],[0,1,2,3,4],[0,1,3,4],[0,1,4],[0,1],[0]]}"},"fields":{"title":"A Subcover Witness Is Bound to Its Original Supplied Cover","stem":"Use the complete finite topology=[[], [0], [0, 1], [0, 1, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 3, 4], original ordered cover=[[0, 1], [0, 1, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).","answer":{"kind":"expression","value":"([1], [0, 1, 3, 4], Yes, finite_subcover_for_supplied_cover)","explanation":"Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 1, 3, 4], Yes, finite_subcover_for_supplied_cover)."},"skills":["A Subcover Witness Is Bound to Its Original Supplied Cover"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0057","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ed4f0e3d857fac325fe66081ff942fb24006bd38f465408595c5f803df533aa3","family_id":"qf-mod33-compact-subcover-witness","seed":202611025057,"semantic_template":"qf-mod33-compact-subcover-witness/{\"cover\":[[0,2],[0,1,2,3],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,1,2,3],\"topology\":[[],[0,1,2,3,4],[0,1,2,3],[0,1,2],[0,2],[0]]}"},"fields":{"title":"A Subcover Witness Is Bound to Its Original Supplied Cover","stem":"Use the complete finite topology=[[], [0], [0, 2], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).","answer":{"kind":"expression","value":"([1], [0, 1, 2, 3], Yes, finite_subcover_for_supplied_cover)","explanation":"Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 1, 2, 3], Yes, finite_subcover_for_supplied_cover)."},"skills":["A Subcover Witness Is Bound to Its Original Supplied Cover"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0058","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c24769f77dda888fceb426454401479045c67aa87ed4fa0b952cf037c1d759a5","family_id":"qf-mod33-compact-subcover-witness","seed":202611025058,"semantic_template":"qf-mod33-compact-subcover-witness/{\"cover\":[[0,2],[0,1,2,4],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,1,2,4],\"topology\":[[],[0,1,2,3,4],[0,1,2,4],[0,1,2],[0,2],[0]]}"},"fields":{"title":"A Subcover Witness Is Bound to Its Original Supplied Cover","stem":"Use the complete finite topology=[[], [0], [0, 2], [0, 1, 2], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 4], original ordered cover=[[0, 2], [0, 1, 2, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).","answer":{"kind":"expression","value":"([1], [0, 1, 2, 4], Yes, finite_subcover_for_supplied_cover)","explanation":"Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 1, 2, 4], Yes, finite_subcover_for_supplied_cover)."},"skills":["A Subcover Witness Is Bound to Its Original Supplied Cover"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0059","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"72c4235bda715d0817666800026090426262702328d2b023adf8de679274df40","family_id":"qf-mod33-compact-subcover-witness","seed":202611025059,"semantic_template":"qf-mod33-compact-subcover-witness/{\"cover\":[[0,2],[0,1,2,3],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,1,2,3],\"topology\":[[],[0,1,2,3,4],[0,1,2,3],[0,2,3],[0,2],[0]]}"},"fields":{"title":"A Subcover Witness Is Bound to Its Original Supplied Cover","stem":"Use the complete finite topology=[[], [0], [0, 2], [0, 2, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).","answer":{"kind":"expression","value":"([1], [0, 1, 2, 3], Yes, finite_subcover_for_supplied_cover)","explanation":"Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 1, 2, 3], Yes, finite_subcover_for_supplied_cover)."},"skills":["A Subcover Witness Is Bound to Its Original Supplied Cover"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0060","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"210d700939b09f7d430930677dc96fd09327b5a413b17c03ae3908dfe27d4a0a","family_id":"qf-mod33-compact-subcover-witness","seed":202611025060,"semantic_template":"qf-mod33-compact-subcover-witness/{\"cover\":[[0,2],[0,2,3,4],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,2,3,4],\"topology\":[[],[0,1,2,3,4],[0,2,3,4],[0,2,3],[0,2],[0]]}"},"fields":{"title":"A Subcover Witness Is Bound to Its Original Supplied Cover","stem":"Use the complete finite topology=[[], [0], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 2, 3, 4], original ordered cover=[[0, 2], [0, 2, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).","answer":{"kind":"expression","value":"([1], [0, 2, 3, 4], Yes, finite_subcover_for_supplied_cover)","explanation":"Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 2, 3, 4], Yes, finite_subcover_for_supplied_cover)."},"skills":["A Subcover Witness Is Bound to Its Original Supplied Cover"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0061","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"814df8147a5225e5f42140cbe3eb1deae798dc5d1ff65ede77adac9c1dea2b26","family_id":"qf-mod33-compact-subcover-witness","seed":202611025061,"semantic_template":"qf-mod33-compact-subcover-witness/{\"cover\":[[0,2],[0,1,2,4],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,1,2,4],\"topology\":[[],[0,1,2,3,4],[0,1,2,4],[0,2,4],[0,2],[0]]}"},"fields":{"title":"A Subcover Witness Is Bound to Its Original Supplied Cover","stem":"Use the complete finite topology=[[], [0], [0, 2], [0, 2, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 4], original ordered cover=[[0, 2], [0, 1, 2, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).","answer":{"kind":"expression","value":"([1], [0, 1, 2, 4], Yes, finite_subcover_for_supplied_cover)","explanation":"Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 1, 2, 4], Yes, finite_subcover_for_supplied_cover)."},"skills":["A Subcover Witness Is Bound to Its Original Supplied Cover"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0062","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d14697d0c4ae042715ff9abbb7bd4890b9ab6fa939668f80f7f5ee9c370da722","family_id":"qf-mod33-compact-subcover-witness","seed":202611025062,"semantic_template":"qf-mod33-compact-subcover-witness/{\"cover\":[[0,2],[0,2,3,4],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,2,3,4],\"topology\":[[],[0,1,2,3,4],[0,2,3,4],[0,2,4],[0,2],[0]]}"},"fields":{"title":"A Subcover Witness Is Bound to Its Original Supplied Cover","stem":"Use the complete finite topology=[[], [0], [0, 2], [0, 2, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 2, 3, 4], original ordered cover=[[0, 2], [0, 2, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).","answer":{"kind":"expression","value":"([1], [0, 2, 3, 4], Yes, finite_subcover_for_supplied_cover)","explanation":"Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 2, 3, 4], Yes, finite_subcover_for_supplied_cover)."},"skills":["A Subcover Witness Is Bound to Its Original Supplied Cover"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0063","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ae64146954b2a09ce4931139b8bf931efa7d34ff3fa080d8f070f6d051c3baa4","family_id":"qf-mod33-compact-subcover-witness","seed":202611025063,"semantic_template":"qf-mod33-compact-subcover-witness/{\"cover\":[[0,3],[0,1,2,3],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,1,2,3],\"topology\":[[],[0,1,2,3,4],[0,1,2,3],[0,1,3],[0,3],[0]]}"},"fields":{"title":"A Subcover Witness Is Bound to Its Original Supplied Cover","stem":"Use the complete finite topology=[[], [0], [0, 3], [0, 1, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).","answer":{"kind":"expression","value":"([1], [0, 1, 2, 3], Yes, finite_subcover_for_supplied_cover)","explanation":"Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 1, 2, 3], Yes, finite_subcover_for_supplied_cover)."},"skills":["A Subcover Witness Is Bound to Its Original Supplied Cover"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0064","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"657356c74d72f7497a4cb674cb825b9df3439c67029009773d302506ebc3ff8a","family_id":"qf-mod33-compact-subcover-witness","seed":202611025064,"semantic_template":"qf-mod33-compact-subcover-witness/{\"cover\":[[0,3],[0,1,3,4],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,1,3,4],\"topology\":[[],[0,1,2,3,4],[0,1,3,4],[0,1,3],[0,3],[0]]}"},"fields":{"title":"A Subcover Witness Is Bound to Its Original Supplied Cover","stem":"Use the complete finite topology=[[], [0], [0, 3], [0, 1, 3], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 3, 4], original ordered cover=[[0, 3], [0, 1, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).","answer":{"kind":"expression","value":"([1], [0, 1, 3, 4], Yes, finite_subcover_for_supplied_cover)","explanation":"Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 1, 3, 4], Yes, finite_subcover_for_supplied_cover)."},"skills":["A Subcover Witness Is Bound to Its Original Supplied Cover"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0065","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"52cd7093864f8f764620494bfac3af47a0ee54fc5d15618326baffa1fa4859dc","family_id":"qf-mod33-compact-subcover-witness","seed":202611025065,"semantic_template":"qf-mod33-compact-subcover-witness/{\"cover\":[[0,3],[0,1,2,3],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,1,2,3],\"topology\":[[],[0,1,2,3,4],[0,1,2,3],[0,2,3],[0,3],[0]]}"},"fields":{"title":"A Subcover Witness Is Bound to Its Original Supplied Cover","stem":"Use the complete finite topology=[[], [0], [0, 3], [0, 2, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).","answer":{"kind":"expression","value":"([1], [0, 1, 2, 3], Yes, finite_subcover_for_supplied_cover)","explanation":"Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 1, 2, 3], Yes, finite_subcover_for_supplied_cover)."},"skills":["A Subcover Witness Is Bound to Its Original Supplied Cover"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0066","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"49ae3b325a4daf8a5f2260b20860dec748f302ef73a665ad20e4182dfd5b11ca","family_id":"qf-mod33-compact-subcover-witness","seed":202611025066,"semantic_template":"qf-mod33-compact-subcover-witness/{\"cover\":[[0,3],[0,2,3,4],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,2,3,4],\"topology\":[[],[0,1,2,3,4],[0,2,3,4],[0,2,3],[0,3],[0]]}"},"fields":{"title":"A Subcover Witness Is Bound to Its Original Supplied Cover","stem":"Use the complete finite topology=[[], [0], [0, 3], [0, 2, 3], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 2, 3, 4], original ordered cover=[[0, 3], [0, 2, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).","answer":{"kind":"expression","value":"([1], [0, 2, 3, 4], Yes, finite_subcover_for_supplied_cover)","explanation":"Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 2, 3, 4], Yes, finite_subcover_for_supplied_cover)."},"skills":["A Subcover Witness Is Bound to Its Original Supplied Cover"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0067","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9c45f819957a64bbb6c67a0f1fe3f23ebc09a36d36693814d98fe0ca48e18c93","family_id":"qf-mod33-compact-subcover-witness","seed":202611025067,"semantic_template":"qf-mod33-compact-subcover-witness/{\"cover\":[[0,3],[0,1,3,4],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,1,3,4],\"topology\":[[],[0,1,2,3,4],[0,1,3,4],[0,3,4],[0,3],[0]]}"},"fields":{"title":"A Subcover Witness Is Bound to Its Original Supplied Cover","stem":"Use the complete finite topology=[[], [0], [0, 3], [0, 3, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 3, 4], original ordered cover=[[0, 3], [0, 1, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).","answer":{"kind":"expression","value":"([1], [0, 1, 3, 4], Yes, finite_subcover_for_supplied_cover)","explanation":"Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 1, 3, 4], Yes, finite_subcover_for_supplied_cover)."},"skills":["A Subcover Witness Is Bound to Its Original Supplied Cover"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0068","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a1db2e624d06444458a5ec5be55a452cd8ebf5cc681498cd8d5ee0ccc90a2125","family_id":"qf-mod33-compact-subcover-witness","seed":202611025068,"semantic_template":"qf-mod33-compact-subcover-witness/{\"cover\":[[0,3],[0,2,3,4],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,2,3,4],\"topology\":[[],[0,1,2,3,4],[0,2,3,4],[0,3,4],[0,3],[0]]}"},"fields":{"title":"A Subcover Witness Is Bound to Its Original Supplied Cover","stem":"Use the complete finite topology=[[], [0], [0, 3], [0, 3, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 2, 3, 4], original ordered cover=[[0, 3], [0, 2, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).","answer":{"kind":"expression","value":"([1], [0, 2, 3, 4], Yes, finite_subcover_for_supplied_cover)","explanation":"Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 2, 3, 4], Yes, finite_subcover_for_supplied_cover)."},"skills":["A Subcover Witness Is Bound to Its Original Supplied Cover"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0069","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"668c9e28d6139ae22ada39e9b7bc10c51dae015120a714277f3806efb16af939","family_id":"qf-mod33-compact-subcover-witness","seed":202611025069,"semantic_template":"qf-mod33-compact-subcover-witness/{\"cover\":[[0,4],[0,1,2,4],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,1,2,4],\"topology\":[[],[0,1,2,3,4],[0,1,2,4],[0,1,4],[0,4],[0]]}"},"fields":{"title":"A Subcover Witness Is Bound to Its Original Supplied Cover","stem":"Use the complete finite topology=[[], [0], [0, 4], [0, 1, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 4], original ordered cover=[[0, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).","answer":{"kind":"expression","value":"([1], [0, 1, 2, 4], Yes, finite_subcover_for_supplied_cover)","explanation":"Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 1, 2, 4], Yes, finite_subcover_for_supplied_cover)."},"skills":["A Subcover Witness Is Bound to Its Original Supplied Cover"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0070","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f4a106a702fe63146c840628122dd1b64576d32e12983197406f4a4eb44e6ff8","family_id":"qf-mod33-compact-subcover-witness","seed":202611025070,"semantic_template":"qf-mod33-compact-subcover-witness/{\"cover\":[[0,4],[0,1,3,4],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,1,3,4],\"topology\":[[],[0,1,2,3,4],[0,1,3,4],[0,1,4],[0,4],[0]]}"},"fields":{"title":"A Subcover Witness Is Bound to Its Original Supplied Cover","stem":"Use the complete finite topology=[[], [0], [0, 4], [0, 1, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 3, 4], original ordered cover=[[0, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).","answer":{"kind":"expression","value":"([1], [0, 1, 3, 4], Yes, finite_subcover_for_supplied_cover)","explanation":"Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 1, 3, 4], Yes, finite_subcover_for_supplied_cover)."},"skills":["A Subcover Witness Is Bound to Its Original Supplied Cover"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0071","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1df79eef2df32907ca4dc618556ff8082f2c905a5e5502b9a03055dd3f9da5bb","family_id":"qf-mod33-compact-subcover-witness","seed":202611025071,"semantic_template":"qf-mod33-compact-subcover-witness/{\"cover\":[[0,4],[0,1,2,4],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,1,2,4],\"topology\":[[],[0,1,2,3,4],[0,1,2,4],[0,2,4],[0,4],[0]]}"},"fields":{"title":"A Subcover Witness Is Bound to Its Original Supplied Cover","stem":"Use the complete finite topology=[[], [0], [0, 4], [0, 2, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 4], original ordered cover=[[0, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).","answer":{"kind":"expression","value":"([1], [0, 1, 2, 4], Yes, finite_subcover_for_supplied_cover)","explanation":"Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 1, 2, 4], Yes, finite_subcover_for_supplied_cover)."},"skills":["A Subcover Witness Is Bound to Its Original Supplied Cover"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0072","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8d09922430d8057b55eec6cc36a1b0ab627345f9be33b24250b94fd4f054a28c","family_id":"qf-mod33-compact-subcover-witness","seed":202611025072,"semantic_template":"qf-mod33-compact-subcover-witness/{\"cover\":[[0,4],[0,2,3,4],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,2,3,4],\"topology\":[[],[0,1,2,3,4],[0,2,3,4],[0,2,4],[0,4],[0]]}"},"fields":{"title":"A Subcover Witness Is Bound to Its Original Supplied Cover","stem":"Use the complete finite topology=[[], [0], [0, 4], [0, 2, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 2, 3, 4], original ordered cover=[[0, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).","answer":{"kind":"expression","value":"([1], [0, 2, 3, 4], Yes, finite_subcover_for_supplied_cover)","explanation":"Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 2, 3, 4], Yes, finite_subcover_for_supplied_cover)."},"skills":["A Subcover Witness Is Bound to Its Original Supplied Cover"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0073","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"26031d501f4eac8a8cc7d1c96515da75bc5a7a9f1c1f53097e4d77b9617aa42e","family_id":"qf-mod33-compact-subcover-witness","seed":202611025073,"semantic_template":"qf-mod33-compact-subcover-witness/{\"cover\":[[0,4],[0,1,3,4],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,1,3,4],\"topology\":[[],[0,1,2,3,4],[0,1,3,4],[0,3,4],[0,4],[0]]}"},"fields":{"title":"A Subcover Witness Is Bound to Its Original Supplied Cover","stem":"Use the complete finite topology=[[], [0], [0, 4], [0, 3, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 3, 4], original ordered cover=[[0, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).","answer":{"kind":"expression","value":"([1], [0, 1, 3, 4], Yes, finite_subcover_for_supplied_cover)","explanation":"Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 1, 3, 4], Yes, finite_subcover_for_supplied_cover)."},"skills":["A Subcover Witness Is Bound to Its Original Supplied Cover"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0074","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"97c724e49dc2e986b4fdc4e83d2f7dce98ce503637e5c651a5e1ad6945ee60af","family_id":"qf-mod33-compact-subcover-witness","seed":202611025074,"semantic_template":"qf-mod33-compact-subcover-witness/{\"cover\":[[0,4],[0,2,3,4],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,2,3,4],\"topology\":[[],[0,1,2,3,4],[0,2,3,4],[0,3,4],[0,4],[0]]}"},"fields":{"title":"A Subcover Witness Is Bound to Its Original Supplied Cover","stem":"Use the complete finite topology=[[], [0], [0, 4], [0, 3, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 2, 3, 4], original ordered cover=[[0, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).","answer":{"kind":"expression","value":"([1], [0, 2, 3, 4], Yes, finite_subcover_for_supplied_cover)","explanation":"Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 2, 3, 4], Yes, finite_subcover_for_supplied_cover)."},"skills":["A Subcover Witness Is Bound to Its Original Supplied Cover"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0075","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0ddc7717ed43109bfa605039044263614246e6c28d1e37a3bde351f8774e8e5a","family_id":"qf-mod33-compact-subcover-witness","seed":202611025075,"semantic_template":"qf-mod33-compact-subcover-witness/{\"cover\":[[0,1],[0,1,2,3],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,1,2,3],\"topology\":[[],[0,1,2,3,4],[0,1,2,3],[0,1,2],[0,1],[1]]}"},"fields":{"title":"A Subcover Witness Is Bound to Its Original Supplied Cover","stem":"Use the complete finite topology=[[], [1], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 1], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).","answer":{"kind":"expression","value":"([1], [0, 1, 2, 3], Yes, finite_subcover_for_supplied_cover)","explanation":"Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 1, 2, 3], Yes, finite_subcover_for_supplied_cover)."},"skills":["A Subcover Witness Is Bound to Its Original Supplied Cover"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0076","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d61b67ed2f58aa5370fc97a133d13cda7b69aba4c8249f37b8b14a7c110a4e18","family_id":"qf-mod33-compact-single-cover-boundary","seed":202611025076,"semantic_template":"qf-mod33-compact-single-cover-boundary/{\"cover\":[[0,1],[0,1,2,3],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,1,2,3],\"topology\":[[],[0,1,2,3,4],[0,1,2,3],[0,1,2],[0,1],[0]]}"},"fields":{"title":"Keep One-Cover Evidence Separate from the Finite Universal Theorem","stem":"Use the complete finite topology=[[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 1], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).","answer":{"kind":"expression","value":"([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)","explanation":"The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)."},"skills":["Keep One-Cover Evidence Separate from the Finite Universal Theorem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0077","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c28858977adbee950aee0c79699211f4d42f58db4cc2a39bb57e870ac74650d8","family_id":"qf-mod33-compact-single-cover-boundary","seed":202611025077,"semantic_template":"qf-mod33-compact-single-cover-boundary/{\"cover\":[[0,1],[0,1,2,4],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,1,2,4],\"topology\":[[],[0,1,2,3,4],[0,1,2,4],[0,1,2],[0,1],[0]]}"},"fields":{"title":"Keep One-Cover Evidence Separate from the Finite Universal Theorem","stem":"Use the complete finite topology=[[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 4], original ordered cover=[[0, 1], [0, 1, 2, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).","answer":{"kind":"expression","value":"([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)","explanation":"The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)."},"skills":["Keep One-Cover Evidence Separate from the Finite Universal Theorem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0078","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7710a916b54e791fbabe74a92c0b48c8d6bbbe0ea43a39a876ae9c2cba73f318","family_id":"qf-mod33-compact-single-cover-boundary","seed":202611025078,"semantic_template":"qf-mod33-compact-single-cover-boundary/{\"cover\":[[0,1],[0,1,2,3],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,1,2,3],\"topology\":[[],[0,1,2,3,4],[0,1,2,3],[0,1,3],[0,1],[0]]}"},"fields":{"title":"Keep One-Cover Evidence Separate from the Finite Universal Theorem","stem":"Use the complete finite topology=[[], [0], [0, 1], [0, 1, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 1], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).","answer":{"kind":"expression","value":"([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)","explanation":"The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)."},"skills":["Keep One-Cover Evidence Separate from the Finite Universal Theorem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0079","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8283af0ff44a115d6d9f33b0342ef5760ea7e5794460438cc4d92cb0d645d350","family_id":"qf-mod33-compact-single-cover-boundary","seed":202611025079,"semantic_template":"qf-mod33-compact-single-cover-boundary/{\"cover\":[[0,1],[0,1,3,4],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,1,3,4],\"topology\":[[],[0,1,2,3,4],[0,1,3,4],[0,1,3],[0,1],[0]]}"},"fields":{"title":"Keep One-Cover Evidence Separate from the Finite Universal Theorem","stem":"Use the complete finite topology=[[], [0], [0, 1], [0, 1, 3], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 3, 4], original ordered cover=[[0, 1], [0, 1, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).","answer":{"kind":"expression","value":"([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)","explanation":"The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)."},"skills":["Keep One-Cover Evidence Separate from the Finite Universal Theorem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0080","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"19c6fb6c1e73899fc747f14e47702acadaaab5342ac15f65cd91b4f17a788345","family_id":"qf-mod33-compact-single-cover-boundary","seed":202611025080,"semantic_template":"qf-mod33-compact-single-cover-boundary/{\"cover\":[[0,1],[0,1,2,4],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,1,2,4],\"topology\":[[],[0,1,2,3,4],[0,1,2,4],[0,1,4],[0,1],[0]]}"},"fields":{"title":"Keep One-Cover Evidence Separate from the Finite Universal Theorem","stem":"Use the complete finite topology=[[], [0], [0, 1], [0, 1, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 4], original ordered cover=[[0, 1], [0, 1, 2, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).","answer":{"kind":"expression","value":"([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)","explanation":"The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)."},"skills":["Keep One-Cover Evidence Separate from the Finite Universal Theorem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0081","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0aa52770378c1a0a177ba2760e216ad39bdaab8ba5fe85115f989b1dbb8baf35","family_id":"qf-mod33-compact-single-cover-boundary","seed":202611025081,"semantic_template":"qf-mod33-compact-single-cover-boundary/{\"cover\":[[0,1],[0,1,3,4],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,1,3,4],\"topology\":[[],[0,1,2,3,4],[0,1,3,4],[0,1,4],[0,1],[0]]}"},"fields":{"title":"Keep One-Cover Evidence Separate from the Finite Universal Theorem","stem":"Use the complete finite topology=[[], [0], [0, 1], [0, 1, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 3, 4], original ordered cover=[[0, 1], [0, 1, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).","answer":{"kind":"expression","value":"([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)","explanation":"The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)."},"skills":["Keep One-Cover Evidence Separate from the Finite Universal Theorem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0082","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"98b4ad3419798c9e2a11d0e41bd139cd957d7a7a1eb46cd9a7b459e763f6109c","family_id":"qf-mod33-compact-single-cover-boundary","seed":202611025082,"semantic_template":"qf-mod33-compact-single-cover-boundary/{\"cover\":[[0,2],[0,1,2,3],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,1,2,3],\"topology\":[[],[0,1,2,3,4],[0,1,2,3],[0,1,2],[0,2],[0]]}"},"fields":{"title":"Keep One-Cover Evidence Separate from the Finite Universal Theorem","stem":"Use the complete finite topology=[[], [0], [0, 2], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).","answer":{"kind":"expression","value":"([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)","explanation":"The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)."},"skills":["Keep One-Cover Evidence Separate from the Finite Universal Theorem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0083","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"91fbec7dac197caeeba449062f5f602f16a8f729b3e86fb0a6517b498d937e97","family_id":"qf-mod33-compact-single-cover-boundary","seed":202611025083,"semantic_template":"qf-mod33-compact-single-cover-boundary/{\"cover\":[[0,2],[0,1,2,4],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,1,2,4],\"topology\":[[],[0,1,2,3,4],[0,1,2,4],[0,1,2],[0,2],[0]]}"},"fields":{"title":"Keep One-Cover Evidence Separate from the Finite Universal Theorem","stem":"Use the complete finite topology=[[], [0], [0, 2], [0, 1, 2], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 4], original ordered cover=[[0, 2], [0, 1, 2, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).","answer":{"kind":"expression","value":"([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)","explanation":"The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)."},"skills":["Keep One-Cover Evidence Separate from the Finite Universal Theorem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0084","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dc0cb202b1ad4875f4e0b4b6752486023e397204f18c8506590889bbf4d03b01","family_id":"qf-mod33-compact-single-cover-boundary","seed":202611025084,"semantic_template":"qf-mod33-compact-single-cover-boundary/{\"cover\":[[0,2],[0,1,2,3],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,1,2,3],\"topology\":[[],[0,1,2,3,4],[0,1,2,3],[0,2,3],[0,2],[0]]}"},"fields":{"title":"Keep One-Cover Evidence Separate from the Finite Universal Theorem","stem":"Use the complete finite topology=[[], [0], [0, 2], [0, 2, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).","answer":{"kind":"expression","value":"([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)","explanation":"The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)."},"skills":["Keep One-Cover Evidence Separate from the Finite Universal Theorem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0085","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2207cdd0919ea479607b9e28131365de056d5a6f68385508dacbe4082a60b6f2","family_id":"qf-mod33-compact-single-cover-boundary","seed":202611025085,"semantic_template":"qf-mod33-compact-single-cover-boundary/{\"cover\":[[0,2],[0,2,3,4],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,2,3,4],\"topology\":[[],[0,1,2,3,4],[0,2,3,4],[0,2,3],[0,2],[0]]}"},"fields":{"title":"Keep One-Cover Evidence Separate from the Finite Universal Theorem","stem":"Use the complete finite topology=[[], [0], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 2, 3, 4], original ordered cover=[[0, 2], [0, 2, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).","answer":{"kind":"expression","value":"([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)","explanation":"The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)."},"skills":["Keep One-Cover Evidence Separate from the Finite Universal Theorem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0086","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7a20dd1ba41884065d3598df151a14d16f2911509b9dddd75272dce5b8c888f7","family_id":"qf-mod33-compact-single-cover-boundary","seed":202611025086,"semantic_template":"qf-mod33-compact-single-cover-boundary/{\"cover\":[[0,2],[0,1,2,4],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,1,2,4],\"topology\":[[],[0,1,2,3,4],[0,1,2,4],[0,2,4],[0,2],[0]]}"},"fields":{"title":"Keep One-Cover Evidence Separate from the Finite Universal Theorem","stem":"Use the complete finite topology=[[], [0], [0, 2], [0, 2, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 4], original ordered cover=[[0, 2], [0, 1, 2, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).","answer":{"kind":"expression","value":"([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)","explanation":"The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)."},"skills":["Keep One-Cover Evidence Separate from the Finite Universal Theorem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0087","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b80d1b670427b3561ad6e8cdbb34d8322a9f46ae855a8032f30d7c1de814d9e0","family_id":"qf-mod33-compact-single-cover-boundary","seed":202611025087,"semantic_template":"qf-mod33-compact-single-cover-boundary/{\"cover\":[[0,2],[0,2,3,4],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,2,3,4],\"topology\":[[],[0,1,2,3,4],[0,2,3,4],[0,2,4],[0,2],[0]]}"},"fields":{"title":"Keep One-Cover Evidence Separate from the Finite Universal Theorem","stem":"Use the complete finite topology=[[], [0], [0, 2], [0, 2, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 2, 3, 4], original ordered cover=[[0, 2], [0, 2, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).","answer":{"kind":"expression","value":"([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)","explanation":"The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)."},"skills":["Keep One-Cover Evidence Separate from the Finite Universal Theorem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0088","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"20eb78437a64de4ef7cde2f9589e32f3127221fa528d4de4365129f8415098af","family_id":"qf-mod33-compact-single-cover-boundary","seed":202611025088,"semantic_template":"qf-mod33-compact-single-cover-boundary/{\"cover\":[[0,3],[0,1,2,3],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,1,2,3],\"topology\":[[],[0,1,2,3,4],[0,1,2,3],[0,1,3],[0,3],[0]]}"},"fields":{"title":"Keep One-Cover Evidence Separate from the Finite Universal Theorem","stem":"Use the complete finite topology=[[], [0], [0, 3], [0, 1, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).","answer":{"kind":"expression","value":"([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)","explanation":"The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)."},"skills":["Keep One-Cover Evidence Separate from the Finite Universal Theorem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0089","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"89bdbc74374986315da370910f8cff015573ebd1ecfa67cd863deb4d0f7f8174","family_id":"qf-mod33-compact-single-cover-boundary","seed":202611025089,"semantic_template":"qf-mod33-compact-single-cover-boundary/{\"cover\":[[0,3],[0,1,3,4],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,1,3,4],\"topology\":[[],[0,1,2,3,4],[0,1,3,4],[0,1,3],[0,3],[0]]}"},"fields":{"title":"Keep One-Cover Evidence Separate from the Finite Universal Theorem","stem":"Use the complete finite topology=[[], [0], [0, 3], [0, 1, 3], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 3, 4], original ordered cover=[[0, 3], [0, 1, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).","answer":{"kind":"expression","value":"([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)","explanation":"The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)."},"skills":["Keep One-Cover Evidence Separate from the Finite Universal Theorem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0090","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"244ff4782cb58fec1ab208d6d53fc3c99fb4005f9fc3480626374af42d5d29b5","family_id":"qf-mod33-compact-single-cover-boundary","seed":202611025090,"semantic_template":"qf-mod33-compact-single-cover-boundary/{\"cover\":[[0,3],[0,1,2,3],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,1,2,3],\"topology\":[[],[0,1,2,3,4],[0,1,2,3],[0,2,3],[0,3],[0]]}"},"fields":{"title":"Keep One-Cover Evidence Separate from the Finite Universal Theorem","stem":"Use the complete finite topology=[[], [0], [0, 3], [0, 2, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).","answer":{"kind":"expression","value":"([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)","explanation":"The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)."},"skills":["Keep One-Cover Evidence Separate from the Finite Universal Theorem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0091","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"504bcab91fe3223a2390ad42118cea80846775082fb7325093dace8d8420524f","family_id":"qf-mod33-compact-single-cover-boundary","seed":202611025091,"semantic_template":"qf-mod33-compact-single-cover-boundary/{\"cover\":[[0,3],[0,2,3,4],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,2,3,4],\"topology\":[[],[0,1,2,3,4],[0,2,3,4],[0,2,3],[0,3],[0]]}"},"fields":{"title":"Keep One-Cover Evidence Separate from the Finite Universal Theorem","stem":"Use the complete finite topology=[[], [0], [0, 3], [0, 2, 3], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 2, 3, 4], original ordered cover=[[0, 3], [0, 2, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).","answer":{"kind":"expression","value":"([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)","explanation":"The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)."},"skills":["Keep One-Cover Evidence Separate from the Finite Universal Theorem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0092","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"210a1a31c2d4ba104df93c76e850aa9b41b60444bf27ef2a715e383239864579","family_id":"qf-mod33-compact-single-cover-boundary","seed":202611025092,"semantic_template":"qf-mod33-compact-single-cover-boundary/{\"cover\":[[0,3],[0,1,3,4],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,1,3,4],\"topology\":[[],[0,1,2,3,4],[0,1,3,4],[0,3,4],[0,3],[0]]}"},"fields":{"title":"Keep One-Cover Evidence Separate from the Finite Universal Theorem","stem":"Use the complete finite topology=[[], [0], [0, 3], [0, 3, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 3, 4], original ordered cover=[[0, 3], [0, 1, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).","answer":{"kind":"expression","value":"([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)","explanation":"The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)."},"skills":["Keep One-Cover Evidence Separate from the Finite Universal Theorem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0093","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9b36d7f81dafa1bc3612bd0456aaaf29d4463caf9e9a467d78d1b7ab30b5f716","family_id":"qf-mod33-compact-single-cover-boundary","seed":202611025093,"semantic_template":"qf-mod33-compact-single-cover-boundary/{\"cover\":[[0,3],[0,2,3,4],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,2,3,4],\"topology\":[[],[0,1,2,3,4],[0,2,3,4],[0,3,4],[0,3],[0]]}"},"fields":{"title":"Keep One-Cover Evidence Separate from the Finite Universal Theorem","stem":"Use the complete finite topology=[[], [0], [0, 3], [0, 3, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 2, 3, 4], original ordered cover=[[0, 3], [0, 2, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).","answer":{"kind":"expression","value":"([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)","explanation":"The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)."},"skills":["Keep One-Cover Evidence Separate from the Finite Universal Theorem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0094","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1ff6caac6113acd7c075556ede4090a0083341ba3cd3508b76c85e13079c5722","family_id":"qf-mod33-compact-single-cover-boundary","seed":202611025094,"semantic_template":"qf-mod33-compact-single-cover-boundary/{\"cover\":[[0,4],[0,1,2,4],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,1,2,4],\"topology\":[[],[0,1,2,3,4],[0,1,2,4],[0,1,4],[0,4],[0]]}"},"fields":{"title":"Keep One-Cover Evidence Separate from the Finite Universal Theorem","stem":"Use the complete finite topology=[[], [0], [0, 4], [0, 1, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 4], original ordered cover=[[0, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).","answer":{"kind":"expression","value":"([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)","explanation":"The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)."},"skills":["Keep One-Cover Evidence Separate from the Finite Universal Theorem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0095","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c88077018bdfe3e00d2d7131d14536c9c04406f13aaaf56ef835308df46eeb51","family_id":"qf-mod33-compact-single-cover-boundary","seed":202611025095,"semantic_template":"qf-mod33-compact-single-cover-boundary/{\"cover\":[[0,4],[0,1,3,4],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,1,3,4],\"topology\":[[],[0,1,2,3,4],[0,1,3,4],[0,1,4],[0,4],[0]]}"},"fields":{"title":"Keep One-Cover Evidence Separate from the Finite Universal Theorem","stem":"Use the complete finite topology=[[], [0], [0, 4], [0, 1, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 3, 4], original ordered cover=[[0, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).","answer":{"kind":"expression","value":"([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)","explanation":"The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)."},"skills":["Keep One-Cover Evidence Separate from the Finite Universal Theorem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0096","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"052e96cc9046736fa664179186a55593725d72e1d78c9ff13511fcd9c96e2c1c","family_id":"qf-mod33-compact-single-cover-boundary","seed":202611025096,"semantic_template":"qf-mod33-compact-single-cover-boundary/{\"cover\":[[0,4],[0,1,2,4],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,1,2,4],\"topology\":[[],[0,1,2,3,4],[0,1,2,4],[0,2,4],[0,4],[0]]}"},"fields":{"title":"Keep One-Cover Evidence Separate from the Finite Universal Theorem","stem":"Use the complete finite topology=[[], [0], [0, 4], [0, 2, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 4], original ordered cover=[[0, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).","answer":{"kind":"expression","value":"([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)","explanation":"The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)."},"skills":["Keep One-Cover Evidence Separate from the Finite Universal Theorem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0097","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c12ebe2d757400456ad76b53c16c49b6603ce2c42cda39408521cf220b273696","family_id":"qf-mod33-compact-single-cover-boundary","seed":202611025097,"semantic_template":"qf-mod33-compact-single-cover-boundary/{\"cover\":[[0,4],[0,2,3,4],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,2,3,4],\"topology\":[[],[0,1,2,3,4],[0,2,3,4],[0,2,4],[0,4],[0]]}"},"fields":{"title":"Keep One-Cover Evidence Separate from the Finite Universal Theorem","stem":"Use the complete finite topology=[[], [0], [0, 4], [0, 2, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 2, 3, 4], original ordered cover=[[0, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).","answer":{"kind":"expression","value":"([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)","explanation":"The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)."},"skills":["Keep One-Cover Evidence Separate from the Finite Universal Theorem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0098","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4765e182843eb301261abb854046d95ecd39f6b9471da895fa29e1a2c0b20154","family_id":"qf-mod33-compact-single-cover-boundary","seed":202611025098,"semantic_template":"qf-mod33-compact-single-cover-boundary/{\"cover\":[[0,4],[0,1,3,4],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,1,3,4],\"topology\":[[],[0,1,2,3,4],[0,1,3,4],[0,3,4],[0,4],[0]]}"},"fields":{"title":"Keep One-Cover Evidence Separate from the Finite Universal Theorem","stem":"Use the complete finite topology=[[], [0], [0, 4], [0, 3, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 3, 4], original ordered cover=[[0, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).","answer":{"kind":"expression","value":"([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)","explanation":"The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)."},"skills":["Keep One-Cover Evidence Separate from the Finite Universal Theorem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0099","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f4e51a35afdf09c4845da8f359eacf43ca9df03da7d3d0628d9fc01c5f97bf85","family_id":"qf-mod33-compact-single-cover-boundary","seed":202611025099,"semantic_template":"qf-mod33-compact-single-cover-boundary/{\"cover\":[[0,4],[0,2,3,4],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,2,3,4],\"topology\":[[],[0,1,2,3,4],[0,2,3,4],[0,3,4],[0,4],[0]]}"},"fields":{"title":"Keep One-Cover Evidence Separate from the Finite Universal Theorem","stem":"Use the complete finite topology=[[], [0], [0, 4], [0, 3, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 2, 3, 4], original ordered cover=[[0, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).","answer":{"kind":"expression","value":"([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)","explanation":"The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)."},"skills":["Keep One-Cover Evidence Separate from the Finite Universal Theorem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261102-r05-compact-topological-space-0100","batch_id":"batch-20261014-advance-20261102-r05-compact-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7eb8fe8f05022aa9bc5e9af593ae5eae76f912d900de3d3ff73171b32244891c","family_id":"qf-mod33-compact-single-cover-boundary","seed":202611025100,"semantic_template":"qf-mod33-compact-single-cover-boundary/{\"cover\":[[0,1],[0,1,2,3],[0,1,2,3,4]],\"n\":5,\"selected\":[1],\"target\":[0,1,2,3],\"topology\":[[],[0,1,2,3,4],[0,1,2,3],[0,1,2],[0,1],[1]]}"},"fields":{"title":"Keep One-Cover Evidence Separate from the Finite Universal Theorem","stem":"Use the complete finite topology=[[], [1], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 1], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).","answer":{"kind":"expression","value":"([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)","explanation":"The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)."},"skills":["Keep One-Cover Evidence Separate from the Finite Universal Theorem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}