{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261018-r01-topological-space-0001","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9193e80fb8d6f4bc91353b4ed4b705e0e5528e581277384f3d1c3b4721d8436c","family_id":"qf-ext18-topology-axiom-audit","seed":202610181001,"semantic_template":"qf-ext18-topology-axiom-audit/{\"n\":4,\"opens\":[[0],[0,1],[0,1,2],[0,1,2,3]]}"},"fields":{"title":"Complete Finite Topology Axiom Audit","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[0], [0, 1], [0, 1, 2], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).","answer":{"kind":"expression","value":"(4, [[0], [0, 1], [0, 1, 2], [0, 1, 2, 3]], No, Yes, Yes, Yes, No, none, none)","explanation":"Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[0], [0, 1], [0, 1, 2], [0, 1, 2, 3]], No, Yes, Yes, Yes, No, none, none)."},"skills":["Complete Finite Topology Axiom Audit"],"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-20261013-advance-20261018-r01-topological-space-0002","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e760ffb2b7e1b9985e2c721be87d6f8df2dcd905d76e1d8cf2a4dfef6bf35b8e","family_id":"qf-ext18-topology-axiom-audit","seed":202610181002,"semantic_template":"qf-ext18-topology-axiom-audit/{\"n\":4,\"opens\":[[],[1],[0,1],[0,1,2]]}"},"fields":{"title":"Complete Finite Topology Axiom Audit","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [0, 1], [0, 1, 2]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).","answer":{"kind":"expression","value":"(4, [[], [1], [0, 1], [0, 1, 2]], Yes, No, Yes, Yes, No, none, none)","explanation":"Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [0, 1], [0, 1, 2]], Yes, No, Yes, Yes, No, none, none)."},"skills":["Complete Finite Topology Axiom Audit"],"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-20261013-advance-20261018-r01-topological-space-0003","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e9e9a756c2f444e4b1dfcf55b61cd0a6b29be6a795fa5ec3527c65569ab3415f","family_id":"qf-ext18-topology-axiom-audit","seed":202610181003,"semantic_template":"qf-ext18-topology-axiom-audit/{\"n\":4,\"opens\":[[],[2],[0,2],[0,1,2],[3],[0,1,2,3]]}"},"fields":{"title":"Complete Finite Topology Axiom Audit","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [2], [0, 2], [0, 1, 2], [3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).","answer":{"kind":"expression","value":"(4, [[], [2], [0, 2], [0, 1, 2], [3], [0, 1, 2, 3]], Yes, Yes, No, Yes, No, [[2], [3], [2, 3]], none)","explanation":"Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [2], [0, 2], [0, 1, 2], [3], [0, 1, 2, 3]], Yes, Yes, No, Yes, No, [[2], [3], [2, 3]], none)."},"skills":["Complete Finite Topology Axiom Audit"],"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-20261013-advance-20261018-r01-topological-space-0004","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2d0a724cc9c6346b58e3587621d6924804af96135d7a81ca6552d52ce9d2901b","family_id":"qf-ext18-topology-axiom-audit","seed":202610181004,"semantic_template":"qf-ext18-topology-axiom-audit/{\"n\":4,\"opens\":[[],[3],[0,3],[0,1,3],[0,1,2,3]]}"},"fields":{"title":"Complete Finite Topology Axiom Audit","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [3], [0, 3], [0, 1, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).","answer":{"kind":"expression","value":"(4, [[], [3], [0, 3], [0, 1, 3], [0, 1, 2, 3]], Yes, Yes, Yes, Yes, Yes, none, none)","explanation":"Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [3], [0, 3], [0, 1, 3], [0, 1, 2, 3]], Yes, Yes, Yes, Yes, Yes, none, none)."},"skills":["Complete Finite Topology Axiom Audit"],"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-20261013-advance-20261018-r01-topological-space-0005","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"00c12cb1f4fd086ab268cc13a8734e50c520053286cb547f95c6791de8d42b66","family_id":"qf-ext18-topology-axiom-audit","seed":202610181005,"semantic_template":"qf-ext18-topology-axiom-audit/{\"n\":4,\"opens\":[[0],[0,2],[0,1,2],[0,1,2,3]]}"},"fields":{"title":"Complete Finite Topology Axiom Audit","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[0], [0, 2], [0, 1, 2], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).","answer":{"kind":"expression","value":"(4, [[0], [0, 2], [0, 1, 2], [0, 1, 2, 3]], No, Yes, Yes, Yes, No, none, none)","explanation":"Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[0], [0, 2], [0, 1, 2], [0, 1, 2, 3]], No, Yes, Yes, Yes, No, none, none)."},"skills":["Complete Finite Topology Axiom Audit"],"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-20261013-advance-20261018-r01-topological-space-0006","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fae7e29aed36be88513918aea540c1bbff3a82d0fdfa137dec7eeefbe5398054","family_id":"qf-ext18-topology-axiom-audit","seed":202610181006,"semantic_template":"qf-ext18-topology-axiom-audit/{\"n\":4,\"opens\":[[],[1],[1,2],[0,1,2]]}"},"fields":{"title":"Complete Finite Topology Axiom Audit","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [1, 2], [0, 1, 2]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).","answer":{"kind":"expression","value":"(4, [[], [1], [1, 2], [0, 1, 2]], Yes, No, Yes, Yes, No, none, none)","explanation":"Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [1, 2], [0, 1, 2]], Yes, No, Yes, Yes, No, none, none)."},"skills":["Complete Finite Topology Axiom Audit"],"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-20261013-advance-20261018-r01-topological-space-0007","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ec4b7daf9d2846451a28176320a393d072f9d9c64e0148f5853e98077426f91a","family_id":"qf-ext18-topology-axiom-audit","seed":202610181007,"semantic_template":"qf-ext18-topology-axiom-audit/{\"n\":4,\"opens\":[[],[2],[1,2],[0,1,2],[3],[0,1,2,3]]}"},"fields":{"title":"Complete Finite Topology Axiom Audit","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [2], [1, 2], [0, 1, 2], [3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).","answer":{"kind":"expression","value":"(4, [[], [2], [1, 2], [0, 1, 2], [3], [0, 1, 2, 3]], Yes, Yes, No, Yes, No, [[2], [3], [2, 3]], none)","explanation":"Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [2], [1, 2], [0, 1, 2], [3], [0, 1, 2, 3]], Yes, Yes, No, Yes, No, [[2], [3], [2, 3]], none)."},"skills":["Complete Finite Topology Axiom Audit"],"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-20261013-advance-20261018-r01-topological-space-0008","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"038a3d74ce0f54de605de20c2c995fe63695b1d2b07cab712c44a235c53c47fc","family_id":"qf-ext18-topology-axiom-audit","seed":202610181008,"semantic_template":"qf-ext18-topology-axiom-audit/{\"n\":4,\"opens\":[[],[3],[1,3],[0,1,3],[0,1,2,3]]}"},"fields":{"title":"Complete Finite Topology Axiom Audit","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [3], [1, 3], [0, 1, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).","answer":{"kind":"expression","value":"(4, [[], [3], [1, 3], [0, 1, 3], [0, 1, 2, 3]], Yes, Yes, Yes, Yes, Yes, none, none)","explanation":"Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [3], [1, 3], [0, 1, 3], [0, 1, 2, 3]], Yes, Yes, Yes, Yes, Yes, none, none)."},"skills":["Complete Finite Topology Axiom Audit"],"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-20261013-advance-20261018-r01-topological-space-0009","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"58d34da6243f91fc89ebad18c97fc5965f2cef3516afcfe93c1c02d15cd8f2f4","family_id":"qf-ext18-topology-axiom-audit","seed":202610181009,"semantic_template":"qf-ext18-topology-axiom-audit/{\"n\":4,\"opens\":[[0],[0,3],[0,1,3],[0,1,2,3]]}"},"fields":{"title":"Complete Finite Topology Axiom Audit","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[0], [0, 3], [0, 1, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).","answer":{"kind":"expression","value":"(4, [[0], [0, 3], [0, 1, 3], [0, 1, 2, 3]], No, Yes, Yes, Yes, No, none, none)","explanation":"Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[0], [0, 3], [0, 1, 3], [0, 1, 2, 3]], No, Yes, Yes, Yes, No, none, none)."},"skills":["Complete Finite Topology Axiom Audit"],"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-20261013-advance-20261018-r01-topological-space-0010","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3f853157e2b6eeaf7803d9428343874f0b38dcea6a58ad50b3f3482aadcada33","family_id":"qf-ext18-topology-axiom-audit","seed":202610181010,"semantic_template":"qf-ext18-topology-axiom-audit/{\"n\":4,\"opens\":[[],[1],[1,3],[0,1,3]]}"},"fields":{"title":"Complete Finite Topology Axiom Audit","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [1, 3], [0, 1, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).","answer":{"kind":"expression","value":"(4, [[], [1], [1, 3], [0, 1, 3]], Yes, No, Yes, Yes, No, none, none)","explanation":"Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [1, 3], [0, 1, 3]], Yes, No, Yes, Yes, No, none, none)."},"skills":["Complete Finite Topology Axiom Audit"],"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-20261013-advance-20261018-r01-topological-space-0011","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9e1099a35ffee1a0d0ba4fb97f1cfdcba2c90a2aac973719672ed605912d4806","family_id":"qf-ext18-topology-axiom-audit","seed":202610181011,"semantic_template":"qf-ext18-topology-axiom-audit/{\"n\":4,\"opens\":[[],[1],[2],[2,3],[0,2,3],[0,1,2,3]]}"},"fields":{"title":"Complete Finite Topology Axiom Audit","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [2], [2, 3], [0, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).","answer":{"kind":"expression","value":"(4, [[], [1], [2], [2, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, Yes, No, Yes, No, [[1], [2], [1, 2]], none)","explanation":"Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [2], [2, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, Yes, No, Yes, No, [[1], [2], [1, 2]], none)."},"skills":["Complete Finite Topology Axiom Audit"],"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-20261013-advance-20261018-r01-topological-space-0012","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"700f732e2f2e5bb5418f0f6ad5c28c24acfd8a23bd028c1e0df5f5105abeb6e3","family_id":"qf-ext18-topology-axiom-audit","seed":202610181012,"semantic_template":"qf-ext18-topology-axiom-audit/{\"n\":4,\"opens\":[[],[3],[2,3],[0,2,3],[0,1,2,3]]}"},"fields":{"title":"Complete Finite Topology Axiom Audit","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [3], [2, 3], [0, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).","answer":{"kind":"expression","value":"(4, [[], [3], [2, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, Yes, Yes, Yes, Yes, none, none)","explanation":"Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [3], [2, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, Yes, Yes, Yes, Yes, none, none)."},"skills":["Complete Finite Topology Axiom Audit"],"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-20261013-advance-20261018-r01-topological-space-0013","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3447f163821f595f6d611625c0059d5303708a8b64568a185cc2e8d84af3c918","family_id":"qf-ext18-topology-axiom-audit","seed":202610181013,"semantic_template":"qf-ext18-topology-axiom-audit/{\"n\":4,\"opens\":[[0],[0,1],[0,1,3],[0,1,2,3]]}"},"fields":{"title":"Complete Finite Topology Axiom Audit","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[0], [0, 1], [0, 1, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).","answer":{"kind":"expression","value":"(4, [[0], [0, 1], [0, 1, 3], [0, 1, 2, 3]], No, Yes, Yes, Yes, No, none, none)","explanation":"Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[0], [0, 1], [0, 1, 3], [0, 1, 2, 3]], No, Yes, Yes, Yes, No, none, none)."},"skills":["Complete Finite Topology Axiom Audit"],"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-20261013-advance-20261018-r01-topological-space-0014","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e279993acad8a9c8543cd1e90090d562960a6640135d543d4246118184e23137","family_id":"qf-ext18-topology-axiom-audit","seed":202610181014,"semantic_template":"qf-ext18-topology-axiom-audit/{\"n\":4,\"opens\":[[],[1],[0,1],[0,1,3]]}"},"fields":{"title":"Complete Finite Topology Axiom Audit","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [0, 1], [0, 1, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).","answer":{"kind":"expression","value":"(4, [[], [1], [0, 1], [0, 1, 3]], Yes, No, Yes, Yes, No, none, none)","explanation":"Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [0, 1], [0, 1, 3]], Yes, No, Yes, Yes, No, none, none)."},"skills":["Complete Finite Topology Axiom Audit"],"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-20261013-advance-20261018-r01-topological-space-0015","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"81458bcbcab14f08f1bd96ed5bddaaa22423dde0273ce360517b0f2fa085ed69","family_id":"qf-ext18-topology-axiom-audit","seed":202610181015,"semantic_template":"qf-ext18-topology-axiom-audit/{\"n\":4,\"opens\":[[],[1],[2],[0,2],[0,2,3],[0,1,2,3]]}"},"fields":{"title":"Complete Finite Topology Axiom Audit","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [2], [0, 2], [0, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).","answer":{"kind":"expression","value":"(4, [[], [1], [2], [0, 2], [0, 2, 3], [0, 1, 2, 3]], Yes, Yes, No, Yes, No, [[1], [2], [1, 2]], none)","explanation":"Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [2], [0, 2], [0, 2, 3], [0, 1, 2, 3]], Yes, Yes, No, Yes, No, [[1], [2], [1, 2]], none)."},"skills":["Complete Finite Topology Axiom Audit"],"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-20261013-advance-20261018-r01-topological-space-0016","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2d4b219e5d6d2d2f0be1c0df607ba6dc043f16900252298c4ab6c6bcfc412413","family_id":"qf-ext18-topology-axiom-audit","seed":202610181016,"semantic_template":"qf-ext18-topology-axiom-audit/{\"n\":4,\"opens\":[[],[3],[0,3],[0,2,3],[0,1,2,3]]}"},"fields":{"title":"Complete Finite Topology Axiom Audit","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [3], [0, 3], [0, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).","answer":{"kind":"expression","value":"(4, [[], [3], [0, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, Yes, Yes, Yes, Yes, none, none)","explanation":"Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [3], [0, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, Yes, Yes, Yes, Yes, none, none)."},"skills":["Complete Finite Topology Axiom Audit"],"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-20261013-advance-20261018-r01-topological-space-0017","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8f265fda944cb6bde1af4d81d8cd5996df834db8110737bcd1cc8e7eb3b77d6c","family_id":"qf-ext18-topology-axiom-audit","seed":202610181017,"semantic_template":"qf-ext18-topology-axiom-audit/{\"n\":4,\"opens\":[[0],[0,2],[0,2,3],[0,1,2,3]]}"},"fields":{"title":"Complete Finite Topology Axiom Audit","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[0], [0, 2], [0, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).","answer":{"kind":"expression","value":"(4, [[0], [0, 2], [0, 2, 3], [0, 1, 2, 3]], No, Yes, Yes, Yes, No, none, none)","explanation":"Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[0], [0, 2], [0, 2, 3], [0, 1, 2, 3]], No, Yes, Yes, Yes, No, none, none)."},"skills":["Complete Finite Topology Axiom Audit"],"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-20261013-advance-20261018-r01-topological-space-0018","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3c1258b29e4d3a233a2fa816b33bf6bb6c970e27241412c6d1551a694ca798bb","family_id":"qf-ext18-topology-axiom-audit","seed":202610181018,"semantic_template":"qf-ext18-topology-axiom-audit/{\"n\":4,\"opens\":[[],[1],[1,2],[1,2,3]]}"},"fields":{"title":"Complete Finite Topology Axiom Audit","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [1, 2], [1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).","answer":{"kind":"expression","value":"(4, [[], [1], [1, 2], [1, 2, 3]], Yes, No, Yes, Yes, No, none, none)","explanation":"Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [1, 2], [1, 2, 3]], Yes, No, Yes, Yes, No, none, none)."},"skills":["Complete Finite Topology Axiom Audit"],"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-20261013-advance-20261018-r01-topological-space-0019","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b466e6687cdb27abb8bd2b3e7359dd61c44525221eb8f9fd7a1293d417e08df1","family_id":"qf-ext18-topology-axiom-audit","seed":202610181019,"semantic_template":"qf-ext18-topology-axiom-audit/{\"n\":4,\"opens\":[[],[0],[2],[1,2],[1,2,3],[0,1,2,3]]}"},"fields":{"title":"Complete Finite Topology Axiom Audit","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [0], [2], [1, 2], [1, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).","answer":{"kind":"expression","value":"(4, [[], [0], [2], [1, 2], [1, 2, 3], [0, 1, 2, 3]], Yes, Yes, No, Yes, No, [[0], [2], [0, 2]], none)","explanation":"Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0], [2], [1, 2], [1, 2, 3], [0, 1, 2, 3]], Yes, Yes, No, Yes, No, [[0], [2], [0, 2]], none)."},"skills":["Complete Finite Topology Axiom Audit"],"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-20261013-advance-20261018-r01-topological-space-0020","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a09517a93edfa2741fdb34f2594674d3c44c238275190d7e69572a6145d3b08e","family_id":"qf-ext18-topology-axiom-audit","seed":202610181020,"semantic_template":"qf-ext18-topology-axiom-audit/{\"n\":4,\"opens\":[[],[3],[1,3],[1,2,3],[0,1,2,3]]}"},"fields":{"title":"Complete Finite Topology Axiom Audit","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [3], [1, 3], [1, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).","answer":{"kind":"expression","value":"(4, [[], [3], [1, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, Yes, Yes, Yes, Yes, none, none)","explanation":"Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [3], [1, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, Yes, Yes, Yes, Yes, none, none)."},"skills":["Complete Finite Topology Axiom Audit"],"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-20261013-advance-20261018-r01-topological-space-0021","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7d1e09c970961ca9f1014df413732179ebb0cff14fdde6d4c8c275529ff8cdcc","family_id":"qf-ext18-topology-axiom-audit","seed":202610181021,"semantic_template":"qf-ext18-topology-axiom-audit/{\"n\":4,\"opens\":[[0],[0,3],[0,2,3],[0,1,2,3]]}"},"fields":{"title":"Complete Finite Topology Axiom Audit","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[0], [0, 3], [0, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).","answer":{"kind":"expression","value":"(4, [[0], [0, 3], [0, 2, 3], [0, 1, 2, 3]], No, Yes, Yes, Yes, No, none, none)","explanation":"Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[0], [0, 3], [0, 2, 3], [0, 1, 2, 3]], No, Yes, Yes, Yes, No, none, none)."},"skills":["Complete Finite Topology Axiom Audit"],"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-20261013-advance-20261018-r01-topological-space-0022","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"92cf647e8f5edcb2d7ec8fe674df1322989ee80c84adee36a961488444872dc7","family_id":"qf-ext18-topology-axiom-audit","seed":202610181022,"semantic_template":"qf-ext18-topology-axiom-audit/{\"n\":4,\"opens\":[[],[1],[1,3],[1,2,3]]}"},"fields":{"title":"Complete Finite Topology Axiom Audit","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [1, 3], [1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).","answer":{"kind":"expression","value":"(4, [[], [1], [1, 3], [1, 2, 3]], Yes, No, Yes, Yes, No, none, none)","explanation":"Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [1, 3], [1, 2, 3]], Yes, No, Yes, Yes, No, none, none)."},"skills":["Complete Finite Topology Axiom Audit"],"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-20261013-advance-20261018-r01-topological-space-0023","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cdce05e788a228bc8c3abb4d2351e74b33e86a2f5418d591d40b330434cfa2f4","family_id":"qf-ext18-topology-axiom-audit","seed":202610181023,"semantic_template":"qf-ext18-topology-axiom-audit/{\"n\":4,\"opens\":[[],[0],[2],[2,3],[1,2,3],[0,1,2,3]]}"},"fields":{"title":"Complete Finite Topology Axiom Audit","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [0], [2], [2, 3], [1, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).","answer":{"kind":"expression","value":"(4, [[], [0], [2], [2, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, Yes, No, Yes, No, [[0], [2], [0, 2]], none)","explanation":"Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0], [2], [2, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, Yes, No, Yes, No, [[0], [2], [0, 2]], none)."},"skills":["Complete Finite Topology Axiom Audit"],"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-20261013-advance-20261018-r01-topological-space-0024","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9136b38d6083a24d83e3d0ca2bf73db79046cb83c976b5ca337696248a32cc6c","family_id":"qf-ext18-topology-axiom-audit","seed":202610181024,"semantic_template":"qf-ext18-topology-axiom-audit/{\"n\":4,\"opens\":[[],[3],[2,3],[1,2,3],[0,1,2,3]]}"},"fields":{"title":"Complete Finite Topology Axiom Audit","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [3], [2, 3], [1, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).","answer":{"kind":"expression","value":"(4, [[], [3], [2, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, Yes, Yes, Yes, Yes, none, none)","explanation":"Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [3], [2, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, Yes, Yes, Yes, Yes, none, none)."},"skills":["Complete Finite Topology Axiom Audit"],"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-20261013-advance-20261018-r01-topological-space-0025","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"aa946bf318a3b1d373d232511fa7165bf452fe988f386f4d319d8ba0b1f3fad2","family_id":"qf-ext18-topology-axiom-audit","seed":202610181025,"semantic_template":"qf-ext18-topology-axiom-audit/{\"n\":3,\"opens\":[[],[0],[0,1],[0,1,2]]}"},"fields":{"title":"Complete Finite Topology Axiom Audit","stem":"Complete finite data: X=Fin(3)={0,…,2}. Candidate open family=[[], [0], [0, 1], [0, 1, 2]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).","answer":{"kind":"expression","value":"(3, [[], [0], [0, 1], [0, 1, 2]], Yes, Yes, Yes, Yes, Yes, none, none)","explanation":"Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (3, [[], [0], [0, 1], [0, 1, 2]], Yes, Yes, Yes, Yes, Yes, none, none)."},"skills":["Complete Finite Topology Axiom Audit"],"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-20261013-advance-20261018-r01-topological-space-0026","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"52205abe4a19a2803c6d16b0925ab3533948c3b7a857aa8b86afe1043bab8a86","family_id":"qf-ext18-topology-open-closed-inventory","seed":202610181026,"semantic_template":"qf-ext18-topology-open-closed-inventory/{\"n\":4,\"opens\":[[],[0],[0,1],[0,1,2],[0,1,2,3]]}"},"fields":{"title":"Every Open, Closed, and Clopen Set in a Finite Topology","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.","answer":{"kind":"expression","value":"(4, [[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3]], [[], [3], [2, 3], [1, 2, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])","explanation":"Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3]], [[], [3], [2, 3], [1, 2, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])."},"skills":["Every Open, Closed, and Clopen Set in a Finite Topology"],"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-20261013-advance-20261018-r01-topological-space-0027","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c63dbfd3154bb09fe4ee83c26ec7d80a7227fa9b52385eb5978ba810bbce9dfb","family_id":"qf-ext18-topology-open-closed-inventory","seed":202610181027,"semantic_template":"qf-ext18-topology-open-closed-inventory/{\"n\":4,\"opens\":[[],[1],[0,1],[0,1,2],[0,1,2,3]]}"},"fields":{"title":"Every Open, Closed, and Clopen Set in a Finite Topology","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [0, 1], [0, 1, 2], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.","answer":{"kind":"expression","value":"(4, [[], [1], [0, 1], [0, 1, 2], [0, 1, 2, 3]], [[], [3], [2, 3], [0, 2, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])","explanation":"Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [0, 1], [0, 1, 2], [0, 1, 2, 3]], [[], [3], [2, 3], [0, 2, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])."},"skills":["Every Open, Closed, and Clopen Set in a Finite Topology"],"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-20261013-advance-20261018-r01-topological-space-0028","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"89448e6b2f68c892e138f8ada9c3a3d3c64b9b7176ffe3cf6e05c7f33366f434","family_id":"qf-ext18-topology-open-closed-inventory","seed":202610181028,"semantic_template":"qf-ext18-topology-open-closed-inventory/{\"n\":4,\"opens\":[[],[2],[0,2],[0,1,2],[0,1,2,3]]}"},"fields":{"title":"Every Open, Closed, and Clopen Set in a Finite Topology","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [2], [0, 2], [0, 1, 2], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.","answer":{"kind":"expression","value":"(4, [[], [2], [0, 2], [0, 1, 2], [0, 1, 2, 3]], [[], [3], [1, 3], [0, 1, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])","explanation":"Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [2], [0, 2], [0, 1, 2], [0, 1, 2, 3]], [[], [3], [1, 3], [0, 1, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])."},"skills":["Every Open, Closed, and Clopen Set in a Finite Topology"],"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-20261013-advance-20261018-r01-topological-space-0029","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b15bcb7bbfa4ffe1a8ff0a30bf7b4de66d98c89e5f72ea54005fba7ed37b6f66","family_id":"qf-ext18-topology-open-closed-inventory","seed":202610181029,"semantic_template":"qf-ext18-topology-open-closed-inventory/{\"n\":4,\"opens\":[[],[3],[0,3],[0,1,3],[0,1,2,3]]}"},"fields":{"title":"Every Open, Closed, and Clopen Set in a Finite Topology","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [3], [0, 3], [0, 1, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.","answer":{"kind":"expression","value":"(4, [[], [3], [0, 3], [0, 1, 3], [0, 1, 2, 3]], [[], [2], [1, 2], [0, 1, 2], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])","explanation":"Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [3], [0, 3], [0, 1, 3], [0, 1, 2, 3]], [[], [2], [1, 2], [0, 1, 2], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])."},"skills":["Every Open, Closed, and Clopen Set in a Finite Topology"],"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-20261013-advance-20261018-r01-topological-space-0030","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f40bb3b645693873b855bd1c65eb4647a4b618aef54967e61a02a3dc1512dd4e","family_id":"qf-ext18-topology-open-closed-inventory","seed":202610181030,"semantic_template":"qf-ext18-topology-open-closed-inventory/{\"n\":4,\"opens\":[[],[0],[0,2],[0,1,2],[0,1,2,3]]}"},"fields":{"title":"Every Open, Closed, and Clopen Set in a Finite Topology","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [0], [0, 2], [0, 1, 2], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.","answer":{"kind":"expression","value":"(4, [[], [0], [0, 2], [0, 1, 2], [0, 1, 2, 3]], [[], [3], [1, 3], [1, 2, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])","explanation":"Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0], [0, 2], [0, 1, 2], [0, 1, 2, 3]], [[], [3], [1, 3], [1, 2, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])."},"skills":["Every Open, Closed, and Clopen Set in a Finite Topology"],"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-20261013-advance-20261018-r01-topological-space-0031","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"83b3836571fafb372aa141efad5e51a6a3c45fd1ac1e3e1329ff3b2852533c4c","family_id":"qf-ext18-topology-open-closed-inventory","seed":202610181031,"semantic_template":"qf-ext18-topology-open-closed-inventory/{\"n\":4,\"opens\":[[],[1],[1,2],[0,1,2],[0,1,2,3]]}"},"fields":{"title":"Every Open, Closed, and Clopen Set in a Finite Topology","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [1, 2], [0, 1, 2], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.","answer":{"kind":"expression","value":"(4, [[], [1], [1, 2], [0, 1, 2], [0, 1, 2, 3]], [[], [3], [0, 3], [0, 2, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])","explanation":"Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [1, 2], [0, 1, 2], [0, 1, 2, 3]], [[], [3], [0, 3], [0, 2, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])."},"skills":["Every Open, Closed, and Clopen Set in a Finite Topology"],"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-20261013-advance-20261018-r01-topological-space-0032","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b48209db4485721fbaea6174889280690168778290a82cf434edb11d76a3fba2","family_id":"qf-ext18-topology-open-closed-inventory","seed":202610181032,"semantic_template":"qf-ext18-topology-open-closed-inventory/{\"n\":4,\"opens\":[[],[2],[1,2],[0,1,2],[0,1,2,3]]}"},"fields":{"title":"Every Open, Closed, and Clopen Set in a Finite Topology","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [2], [1, 2], [0, 1, 2], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.","answer":{"kind":"expression","value":"(4, [[], [2], [1, 2], [0, 1, 2], [0, 1, 2, 3]], [[], [3], [0, 3], [0, 1, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])","explanation":"Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [2], [1, 2], [0, 1, 2], [0, 1, 2, 3]], [[], [3], [0, 3], [0, 1, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])."},"skills":["Every Open, Closed, and Clopen Set in a Finite Topology"],"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-20261013-advance-20261018-r01-topological-space-0033","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1edb8db80799380ca9e6e4ea314853064184421b4226eaf045f487a38b17b02e","family_id":"qf-ext18-topology-open-closed-inventory","seed":202610181033,"semantic_template":"qf-ext18-topology-open-closed-inventory/{\"n\":4,\"opens\":[[],[3],[1,3],[0,1,3],[0,1,2,3]]}"},"fields":{"title":"Every Open, Closed, and Clopen Set in a Finite Topology","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [3], [1, 3], [0, 1, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.","answer":{"kind":"expression","value":"(4, [[], [3], [1, 3], [0, 1, 3], [0, 1, 2, 3]], [[], [2], [0, 2], [0, 1, 2], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])","explanation":"Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [3], [1, 3], [0, 1, 3], [0, 1, 2, 3]], [[], [2], [0, 2], [0, 1, 2], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])."},"skills":["Every Open, Closed, and Clopen Set in a Finite Topology"],"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-20261013-advance-20261018-r01-topological-space-0034","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d181ae0d5e965016856ffd42363cf7f435159805ed288e5ca0574655b353c966","family_id":"qf-ext18-topology-open-closed-inventory","seed":202610181034,"semantic_template":"qf-ext18-topology-open-closed-inventory/{\"n\":4,\"opens\":[[],[0],[0,3],[0,1,3],[0,1,2,3]]}"},"fields":{"title":"Every Open, Closed, and Clopen Set in a Finite Topology","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [0], [0, 3], [0, 1, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.","answer":{"kind":"expression","value":"(4, [[], [0], [0, 3], [0, 1, 3], [0, 1, 2, 3]], [[], [2], [1, 2], [1, 2, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])","explanation":"Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0], [0, 3], [0, 1, 3], [0, 1, 2, 3]], [[], [2], [1, 2], [1, 2, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])."},"skills":["Every Open, Closed, and Clopen Set in a Finite Topology"],"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-20261013-advance-20261018-r01-topological-space-0035","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c883343cc5b4e19f17e5defdfada865c8294f66dcd48ca4375aa437218fb847e","family_id":"qf-ext18-topology-open-closed-inventory","seed":202610181035,"semantic_template":"qf-ext18-topology-open-closed-inventory/{\"n\":4,\"opens\":[[],[1],[1,3],[0,1,3],[0,1,2,3]]}"},"fields":{"title":"Every Open, Closed, and Clopen Set in a Finite Topology","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [1, 3], [0, 1, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.","answer":{"kind":"expression","value":"(4, [[], [1], [1, 3], [0, 1, 3], [0, 1, 2, 3]], [[], [2], [0, 2], [0, 2, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])","explanation":"Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [1, 3], [0, 1, 3], [0, 1, 2, 3]], [[], [2], [0, 2], [0, 2, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])."},"skills":["Every Open, Closed, and Clopen Set in a Finite Topology"],"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-20261013-advance-20261018-r01-topological-space-0036","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"00eec335462e7290f25213d5c60be91c258f7ddf8e3b3e12eae959ad00749bf2","family_id":"qf-ext18-topology-open-closed-inventory","seed":202610181036,"semantic_template":"qf-ext18-topology-open-closed-inventory/{\"n\":4,\"opens\":[[],[2],[2,3],[0,2,3],[0,1,2,3]]}"},"fields":{"title":"Every Open, Closed, and Clopen Set in a Finite Topology","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [2], [2, 3], [0, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.","answer":{"kind":"expression","value":"(4, [[], [2], [2, 3], [0, 2, 3], [0, 1, 2, 3]], [[], [1], [0, 1], [0, 1, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])","explanation":"Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [2], [2, 3], [0, 2, 3], [0, 1, 2, 3]], [[], [1], [0, 1], [0, 1, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])."},"skills":["Every Open, Closed, and Clopen Set in a Finite Topology"],"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-20261013-advance-20261018-r01-topological-space-0037","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c1f8b64d2f8334bbc208d3a7d9486c5f6a44c72aa929543f4991006ce0e40ff3","family_id":"qf-ext18-topology-open-closed-inventory","seed":202610181037,"semantic_template":"qf-ext18-topology-open-closed-inventory/{\"n\":4,\"opens\":[[],[3],[2,3],[0,2,3],[0,1,2,3]]}"},"fields":{"title":"Every Open, Closed, and Clopen Set in a Finite Topology","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [3], [2, 3], [0, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.","answer":{"kind":"expression","value":"(4, [[], [3], [2, 3], [0, 2, 3], [0, 1, 2, 3]], [[], [1], [0, 1], [0, 1, 2], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])","explanation":"Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [3], [2, 3], [0, 2, 3], [0, 1, 2, 3]], [[], [1], [0, 1], [0, 1, 2], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])."},"skills":["Every Open, Closed, and Clopen Set in a Finite Topology"],"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-20261013-advance-20261018-r01-topological-space-0038","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5463975667336d9e1ecf8ba5e0c58687b4409ce97d1bd71f897e26e1e429967e","family_id":"qf-ext18-topology-open-closed-inventory","seed":202610181038,"semantic_template":"qf-ext18-topology-open-closed-inventory/{\"n\":4,\"opens\":[[],[0],[0,1],[0,1,3],[0,1,2,3]]}"},"fields":{"title":"Every Open, Closed, and Clopen Set in a Finite Topology","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [0], [0, 1], [0, 1, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.","answer":{"kind":"expression","value":"(4, [[], [0], [0, 1], [0, 1, 3], [0, 1, 2, 3]], [[], [2], [2, 3], [1, 2, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])","explanation":"Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0], [0, 1], [0, 1, 3], [0, 1, 2, 3]], [[], [2], [2, 3], [1, 2, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])."},"skills":["Every Open, Closed, and Clopen Set in a Finite Topology"],"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-20261013-advance-20261018-r01-topological-space-0039","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"24532b1147167975477b066e1ccb75021a9bef234ed8917253201aaca2cf6dfc","family_id":"qf-ext18-topology-open-closed-inventory","seed":202610181039,"semantic_template":"qf-ext18-topology-open-closed-inventory/{\"n\":4,\"opens\":[[],[1],[0,1],[0,1,3],[0,1,2,3]]}"},"fields":{"title":"Every Open, Closed, and Clopen Set in a Finite Topology","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [0, 1], [0, 1, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.","answer":{"kind":"expression","value":"(4, [[], [1], [0, 1], [0, 1, 3], [0, 1, 2, 3]], [[], [2], [2, 3], [0, 2, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])","explanation":"Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [0, 1], [0, 1, 3], [0, 1, 2, 3]], [[], [2], [2, 3], [0, 2, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])."},"skills":["Every Open, Closed, and Clopen Set in a Finite Topology"],"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-20261013-advance-20261018-r01-topological-space-0040","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b05a972e2ed73a6528b0897fbb614604a8a275040dfc7abf48fc33213e6ee145","family_id":"qf-ext18-topology-open-closed-inventory","seed":202610181040,"semantic_template":"qf-ext18-topology-open-closed-inventory/{\"n\":4,\"opens\":[[],[2],[0,2],[0,2,3],[0,1,2,3]]}"},"fields":{"title":"Every Open, Closed, and Clopen Set in a Finite Topology","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [2], [0, 2], [0, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.","answer":{"kind":"expression","value":"(4, [[], [2], [0, 2], [0, 2, 3], [0, 1, 2, 3]], [[], [1], [1, 3], [0, 1, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])","explanation":"Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [2], [0, 2], [0, 2, 3], [0, 1, 2, 3]], [[], [1], [1, 3], [0, 1, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])."},"skills":["Every Open, Closed, and Clopen Set in a Finite Topology"],"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-20261013-advance-20261018-r01-topological-space-0041","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dd76d51ccd9e35b8c1e54f31c5c5c6b4a779d758c600790c6e3490f85e531aa8","family_id":"qf-ext18-topology-open-closed-inventory","seed":202610181041,"semantic_template":"qf-ext18-topology-open-closed-inventory/{\"n\":4,\"opens\":[[],[3],[0,3],[0,2,3],[0,1,2,3]]}"},"fields":{"title":"Every Open, Closed, and Clopen Set in a Finite Topology","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [3], [0, 3], [0, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.","answer":{"kind":"expression","value":"(4, [[], [3], [0, 3], [0, 2, 3], [0, 1, 2, 3]], [[], [1], [1, 2], [0, 1, 2], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])","explanation":"Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [3], [0, 3], [0, 2, 3], [0, 1, 2, 3]], [[], [1], [1, 2], [0, 1, 2], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])."},"skills":["Every Open, Closed, and Clopen Set in a Finite Topology"],"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-20261013-advance-20261018-r01-topological-space-0042","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"337ac42de4cef64392b1db920aed67772f59d6fe340373ffac60a3d0d2adabfc","family_id":"qf-ext18-topology-open-closed-inventory","seed":202610181042,"semantic_template":"qf-ext18-topology-open-closed-inventory/{\"n\":4,\"opens\":[[],[0],[0,2],[0,2,3],[0,1,2,3]]}"},"fields":{"title":"Every Open, Closed, and Clopen Set in a Finite Topology","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [0], [0, 2], [0, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.","answer":{"kind":"expression","value":"(4, [[], [0], [0, 2], [0, 2, 3], [0, 1, 2, 3]], [[], [1], [1, 3], [1, 2, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])","explanation":"Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0], [0, 2], [0, 2, 3], [0, 1, 2, 3]], [[], [1], [1, 3], [1, 2, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])."},"skills":["Every Open, Closed, and Clopen Set in a Finite Topology"],"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-20261013-advance-20261018-r01-topological-space-0043","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"36bbd8ef4bf4d21cf04d77d3ba0a3d61bfab418647e529fc0de430baf392a243","family_id":"qf-ext18-topology-open-closed-inventory","seed":202610181043,"semantic_template":"qf-ext18-topology-open-closed-inventory/{\"n\":4,\"opens\":[[],[1],[1,2],[1,2,3],[0,1,2,3]]}"},"fields":{"title":"Every Open, Closed, and Clopen Set in a Finite Topology","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [1, 2], [1, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.","answer":{"kind":"expression","value":"(4, [[], [1], [1, 2], [1, 2, 3], [0, 1, 2, 3]], [[], [0], [0, 3], [0, 2, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])","explanation":"Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [1, 2], [1, 2, 3], [0, 1, 2, 3]], [[], [0], [0, 3], [0, 2, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])."},"skills":["Every Open, Closed, and Clopen Set in a Finite Topology"],"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-20261013-advance-20261018-r01-topological-space-0044","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6761c557d7d9247f87df834cfa9395397021f45c4d109ee849e47fe1f69b0352","family_id":"qf-ext18-topology-open-closed-inventory","seed":202610181044,"semantic_template":"qf-ext18-topology-open-closed-inventory/{\"n\":4,\"opens\":[[],[2],[1,2],[1,2,3],[0,1,2,3]]}"},"fields":{"title":"Every Open, Closed, and Clopen Set in a Finite Topology","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [2], [1, 2], [1, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.","answer":{"kind":"expression","value":"(4, [[], [2], [1, 2], [1, 2, 3], [0, 1, 2, 3]], [[], [0], [0, 3], [0, 1, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])","explanation":"Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [2], [1, 2], [1, 2, 3], [0, 1, 2, 3]], [[], [0], [0, 3], [0, 1, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])."},"skills":["Every Open, Closed, and Clopen Set in a Finite Topology"],"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-20261013-advance-20261018-r01-topological-space-0045","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"201abbd2885ed9b0a59d6da5506273ceb398ff07a70f501a681d378f85f2af9b","family_id":"qf-ext18-topology-open-closed-inventory","seed":202610181045,"semantic_template":"qf-ext18-topology-open-closed-inventory/{\"n\":4,\"opens\":[[],[3],[1,3],[1,2,3],[0,1,2,3]]}"},"fields":{"title":"Every Open, Closed, and Clopen Set in a Finite Topology","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [3], [1, 3], [1, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.","answer":{"kind":"expression","value":"(4, [[], [3], [1, 3], [1, 2, 3], [0, 1, 2, 3]], [[], [0], [0, 2], [0, 1, 2], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])","explanation":"Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [3], [1, 3], [1, 2, 3], [0, 1, 2, 3]], [[], [0], [0, 2], [0, 1, 2], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])."},"skills":["Every Open, Closed, and Clopen Set in a Finite Topology"],"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-20261013-advance-20261018-r01-topological-space-0046","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6b66a6f74bf9856f1953cf028d7d7567c3d0e9b62d8666edd1c63e62abb7d599","family_id":"qf-ext18-topology-open-closed-inventory","seed":202610181046,"semantic_template":"qf-ext18-topology-open-closed-inventory/{\"n\":4,\"opens\":[[],[0],[0,3],[0,2,3],[0,1,2,3]]}"},"fields":{"title":"Every Open, Closed, and Clopen Set in a Finite Topology","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [0], [0, 3], [0, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.","answer":{"kind":"expression","value":"(4, [[], [0], [0, 3], [0, 2, 3], [0, 1, 2, 3]], [[], [1], [1, 2], [1, 2, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])","explanation":"Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0], [0, 3], [0, 2, 3], [0, 1, 2, 3]], [[], [1], [1, 2], [1, 2, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])."},"skills":["Every Open, Closed, and Clopen Set in a Finite Topology"],"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-20261013-advance-20261018-r01-topological-space-0047","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5be79041ca61fc5640c367aabf00bed74232a3169c3b3b19c9aec4acaa7be539","family_id":"qf-ext18-topology-open-closed-inventory","seed":202610181047,"semantic_template":"qf-ext18-topology-open-closed-inventory/{\"n\":4,\"opens\":[[],[1],[1,3],[1,2,3],[0,1,2,3]]}"},"fields":{"title":"Every Open, Closed, and Clopen Set in a Finite Topology","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [1, 3], [1, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.","answer":{"kind":"expression","value":"(4, [[], [1], [1, 3], [1, 2, 3], [0, 1, 2, 3]], [[], [0], [0, 2], [0, 2, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])","explanation":"Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [1, 3], [1, 2, 3], [0, 1, 2, 3]], [[], [0], [0, 2], [0, 2, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])."},"skills":["Every Open, Closed, and Clopen Set in a Finite Topology"],"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-20261013-advance-20261018-r01-topological-space-0048","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"186b4b3a3a69b3c7c6bdfdd1446875405e6dee554426e3865ff9ed30a21f8be0","family_id":"qf-ext18-topology-open-closed-inventory","seed":202610181048,"semantic_template":"qf-ext18-topology-open-closed-inventory/{\"n\":4,\"opens\":[[],[2],[2,3],[1,2,3],[0,1,2,3]]}"},"fields":{"title":"Every Open, Closed, and Clopen Set in a Finite Topology","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [2], [2, 3], [1, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.","answer":{"kind":"expression","value":"(4, [[], [2], [2, 3], [1, 2, 3], [0, 1, 2, 3]], [[], [0], [0, 1], [0, 1, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])","explanation":"Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [2], [2, 3], [1, 2, 3], [0, 1, 2, 3]], [[], [0], [0, 1], [0, 1, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])."},"skills":["Every Open, Closed, and Clopen Set in a Finite Topology"],"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-20261013-advance-20261018-r01-topological-space-0049","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"495507ae5626fdf43d9c010c61dafb409e0acaf645c2da09f5575b063fec6ebf","family_id":"qf-ext18-topology-open-closed-inventory","seed":202610181049,"semantic_template":"qf-ext18-topology-open-closed-inventory/{\"n\":4,\"opens\":[[],[3],[2,3],[1,2,3],[0,1,2,3]]}"},"fields":{"title":"Every Open, Closed, and Clopen Set in a Finite Topology","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [3], [2, 3], [1, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.","answer":{"kind":"expression","value":"(4, [[], [3], [2, 3], [1, 2, 3], [0, 1, 2, 3]], [[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])","explanation":"Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [3], [2, 3], [1, 2, 3], [0, 1, 2, 3]], [[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3]], [[], [0, 1, 2, 3]])."},"skills":["Every Open, Closed, and Clopen Set in a Finite Topology"],"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-20261013-advance-20261018-r01-topological-space-0050","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ad259f4ae041721b03bcd46059e9c85883134970152707a02d1ad31b70595fa3","family_id":"qf-ext18-topology-open-closed-inventory","seed":202610181050,"semantic_template":"qf-ext18-topology-open-closed-inventory/{\"n\":3,\"opens\":[[],[0],[0,1],[0,1,2]]}"},"fields":{"title":"Every Open, Closed, and Clopen Set in a Finite Topology","stem":"Complete finite data: X=Fin(3)={0,…,2}. Candidate open family=[[], [0], [0, 1], [0, 1, 2]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.","answer":{"kind":"expression","value":"(3, [[], [0], [0, 1], [0, 1, 2]], [[], [2], [1, 2], [0, 1, 2]], [[], [0, 1, 2]])","explanation":"Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (3, [[], [0], [0, 1], [0, 1, 2]], [[], [2], [1, 2], [0, 1, 2]], [[], [0, 1, 2]])."},"skills":["Every Open, Closed, and Clopen Set in a Finite Topology"],"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-20261013-advance-20261018-r01-topological-space-0051","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8c04e4134e118df32162c72dac7dc7d4ceb60324a6e8da6c257758d7d349af97","family_id":"qf-ext18-topology-discrete-indiscrete","seed":202610181051,"semantic_template":"qf-ext18-topology-discrete-indiscrete/{\"n\":4,\"opens\":[[],[0],[1],[0,1],[2],[0,2],[1,2],[0,1,2],[3],[0,3],[1,3],[0,1,3],[2,3],[0,2,3],[1,2,3],[0,1,2,3]]}"},"fields":{"title":"Topology Qualification before Discrete or Indiscrete Classification","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [0], [1], [0, 1], [2], [0, 2], [1, 2], [0, 1, 2], [3], [0, 3], [1, 3], [0, 1, 3], [2, 3], [0, 2, 3], [1, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.","answer":{"kind":"expression","value":"(4, [[], [0], [1], [0, 1], [2], [0, 2], [1, 2], [0, 1, 2], [3], [0, 3], [1, 3], [0, 1, 3], [2, 3], [0, 2, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, discrete)","explanation":"Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0], [1], [0, 1], [2], [0, 2], [1, 2], [0, 1, 2], [3], [0, 3], [1, 3], [0, 1, 3], [2, 3], [0, 2, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, discrete)."},"skills":["Topology Qualification before Discrete or Indiscrete Classification"],"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-20261013-advance-20261018-r01-topological-space-0052","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"eecafc0b5a948dc0bbd2a8b153194f28d5b72ce98eab2c1deb4a24a83d4a41fe","family_id":"qf-ext18-topology-discrete-indiscrete","seed":202610181052,"semantic_template":"qf-ext18-topology-discrete-indiscrete/{\"n\":4,\"opens\":[[],[0,1,2,3]]}"},"fields":{"title":"Topology Qualification before Discrete or Indiscrete Classification","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.","answer":{"kind":"expression","value":"(4, [[], [0, 1, 2, 3]], Yes, indiscrete)","explanation":"Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], Yes, indiscrete)."},"skills":["Topology Qualification before Discrete or Indiscrete Classification"],"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-20261013-advance-20261018-r01-topological-space-0053","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"128ad22cb200c94fe506229c403fa06f095b9e7dc7b1dd7ef16d3cacf3435c26","family_id":"qf-ext18-topology-discrete-indiscrete","seed":202610181053,"semantic_template":"qf-ext18-topology-discrete-indiscrete/{\"n\":4,\"opens\":[[],[0],[0,1],[0,1,2],[0,1,2,3]]}"},"fields":{"title":"Topology Qualification before Discrete or Indiscrete Classification","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.","answer":{"kind":"expression","value":"(4, [[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3]], Yes, other-topology)","explanation":"Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3]], Yes, other-topology)."},"skills":["Topology Qualification before Discrete or Indiscrete Classification"],"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-20261013-advance-20261018-r01-topological-space-0054","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a1b36cb6dafe0bbfdea2eb4a19241a0bc0276bfb5b98d5fc763e1be3398f52d6","family_id":"qf-ext18-topology-discrete-indiscrete","seed":202610181054,"semantic_template":"qf-ext18-topology-discrete-indiscrete/{\"n\":4,\"opens\":[[],[1],[0,1],[0,1,2],[0,1,2,3]]}"},"fields":{"title":"Topology Qualification before Discrete or Indiscrete Classification","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [0, 1], [0, 1, 2], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.","answer":{"kind":"expression","value":"(4, [[], [1], [0, 1], [0, 1, 2], [0, 1, 2, 3]], Yes, other-topology)","explanation":"Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [0, 1], [0, 1, 2], [0, 1, 2, 3]], Yes, other-topology)."},"skills":["Topology Qualification before Discrete or Indiscrete Classification"],"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-20261013-advance-20261018-r01-topological-space-0055","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2552ebc68d6ce793618bad71b46a034fc6669bb6f8f09d2c9d531c9ca3ebe5b4","family_id":"qf-ext18-topology-discrete-indiscrete","seed":202610181055,"semantic_template":"qf-ext18-topology-discrete-indiscrete/{\"n\":4,\"opens\":[[],[2],[0,2],[0,1,2],[0,1,2,3]]}"},"fields":{"title":"Topology Qualification before Discrete or Indiscrete Classification","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [2], [0, 2], [0, 1, 2], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.","answer":{"kind":"expression","value":"(4, [[], [2], [0, 2], [0, 1, 2], [0, 1, 2, 3]], Yes, other-topology)","explanation":"Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [2], [0, 2], [0, 1, 2], [0, 1, 2, 3]], Yes, other-topology)."},"skills":["Topology Qualification before Discrete or Indiscrete Classification"],"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-20261013-advance-20261018-r01-topological-space-0056","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ac3692ab5ba36c9c5dbd28696b494d7ce158e3dd6aa1c3fca6fc2b77c2e28363","family_id":"qf-ext18-topology-discrete-indiscrete","seed":202610181056,"semantic_template":"qf-ext18-topology-discrete-indiscrete/{\"n\":4,\"opens\":[[],[3],[0,3],[0,1,3],[0,1,2,3]]}"},"fields":{"title":"Topology Qualification before Discrete or Indiscrete Classification","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [3], [0, 3], [0, 1, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.","answer":{"kind":"expression","value":"(4, [[], [3], [0, 3], [0, 1, 3], [0, 1, 2, 3]], Yes, other-topology)","explanation":"Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [3], [0, 3], [0, 1, 3], [0, 1, 2, 3]], Yes, other-topology)."},"skills":["Topology Qualification before Discrete or Indiscrete Classification"],"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-20261013-advance-20261018-r01-topological-space-0057","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5c145009dac96999ae82671f93e38b20964122f6898c1bfb1bd1b8f5055663b2","family_id":"qf-ext18-topology-discrete-indiscrete","seed":202610181057,"semantic_template":"qf-ext18-topology-discrete-indiscrete/{\"n\":4,\"opens\":[[],[0],[0,2],[0,1,2],[0,1,2,3]]}"},"fields":{"title":"Topology Qualification before Discrete or Indiscrete Classification","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [0], [0, 2], [0, 1, 2], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.","answer":{"kind":"expression","value":"(4, [[], [0], [0, 2], [0, 1, 2], [0, 1, 2, 3]], Yes, other-topology)","explanation":"Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0], [0, 2], [0, 1, 2], [0, 1, 2, 3]], Yes, other-topology)."},"skills":["Topology Qualification before Discrete or Indiscrete Classification"],"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-20261013-advance-20261018-r01-topological-space-0058","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"147a2a4258d9adc3381eee37ce8a535b5117e63aeaff4b5ead06c055f15d048b","family_id":"qf-ext18-topology-discrete-indiscrete","seed":202610181058,"semantic_template":"qf-ext18-topology-discrete-indiscrete/{\"n\":4,\"opens\":[[],[1],[1,2],[0,1,2],[0,1,2,3]]}"},"fields":{"title":"Topology Qualification before Discrete or Indiscrete Classification","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [1, 2], [0, 1, 2], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.","answer":{"kind":"expression","value":"(4, [[], [1], [1, 2], [0, 1, 2], [0, 1, 2, 3]], Yes, other-topology)","explanation":"Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [1, 2], [0, 1, 2], [0, 1, 2, 3]], Yes, other-topology)."},"skills":["Topology Qualification before Discrete or Indiscrete Classification"],"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-20261013-advance-20261018-r01-topological-space-0059","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9ae64919a54735c566abd45a4cd4944519b9bfdfe4141a1515a89311a068d576","family_id":"qf-ext18-topology-discrete-indiscrete","seed":202610181059,"semantic_template":"qf-ext18-topology-discrete-indiscrete/{\"n\":4,\"opens\":[[],[2],[1,2],[0,1,2],[0,1,2,3]]}"},"fields":{"title":"Topology Qualification before Discrete or Indiscrete Classification","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [2], [1, 2], [0, 1, 2], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.","answer":{"kind":"expression","value":"(4, [[], [2], [1, 2], [0, 1, 2], [0, 1, 2, 3]], Yes, other-topology)","explanation":"Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [2], [1, 2], [0, 1, 2], [0, 1, 2, 3]], Yes, other-topology)."},"skills":["Topology Qualification before Discrete or Indiscrete Classification"],"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-20261013-advance-20261018-r01-topological-space-0060","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"081beeba5f0ea5450b4b88cb900da2139a6e04c47355e8581e1a1e49361670e0","family_id":"qf-ext18-topology-discrete-indiscrete","seed":202610181060,"semantic_template":"qf-ext18-topology-discrete-indiscrete/{\"n\":4,\"opens\":[[],[3],[1,3],[0,1,3],[0,1,2,3]]}"},"fields":{"title":"Topology Qualification before Discrete or Indiscrete Classification","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [3], [1, 3], [0, 1, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.","answer":{"kind":"expression","value":"(4, [[], [3], [1, 3], [0, 1, 3], [0, 1, 2, 3]], Yes, other-topology)","explanation":"Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [3], [1, 3], [0, 1, 3], [0, 1, 2, 3]], Yes, other-topology)."},"skills":["Topology Qualification before Discrete or Indiscrete Classification"],"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-20261013-advance-20261018-r01-topological-space-0061","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8d7e452e22c6cb1a963e2ce3ecbb21de6d5bb02db572611c34dc307bd01f6ce2","family_id":"qf-ext18-topology-discrete-indiscrete","seed":202610181061,"semantic_template":"qf-ext18-topology-discrete-indiscrete/{\"n\":4,\"opens\":[[],[0],[0,3],[0,1,3],[0,1,2,3]]}"},"fields":{"title":"Topology Qualification before Discrete or Indiscrete Classification","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [0], [0, 3], [0, 1, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.","answer":{"kind":"expression","value":"(4, [[], [0], [0, 3], [0, 1, 3], [0, 1, 2, 3]], Yes, other-topology)","explanation":"Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0], [0, 3], [0, 1, 3], [0, 1, 2, 3]], Yes, other-topology)."},"skills":["Topology Qualification before Discrete or Indiscrete Classification"],"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-20261013-advance-20261018-r01-topological-space-0062","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9bb30af287805ec3f2fc4e41fbc161897087f050b4ad4a95610ae42a5ec3b7be","family_id":"qf-ext18-topology-discrete-indiscrete","seed":202610181062,"semantic_template":"qf-ext18-topology-discrete-indiscrete/{\"n\":4,\"opens\":[[],[1],[1,3],[0,1,3],[0,1,2,3]]}"},"fields":{"title":"Topology Qualification before Discrete or Indiscrete Classification","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [1, 3], [0, 1, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.","answer":{"kind":"expression","value":"(4, [[], [1], [1, 3], [0, 1, 3], [0, 1, 2, 3]], Yes, other-topology)","explanation":"Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [1, 3], [0, 1, 3], [0, 1, 2, 3]], Yes, other-topology)."},"skills":["Topology Qualification before Discrete or Indiscrete Classification"],"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-20261013-advance-20261018-r01-topological-space-0063","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"562b191f9cafe2fd78a6afa09aaa7ab081c688b43e36e3b0a547aef4a93c97a3","family_id":"qf-ext18-topology-discrete-indiscrete","seed":202610181063,"semantic_template":"qf-ext18-topology-discrete-indiscrete/{\"n\":4,\"opens\":[[],[2],[2,3],[0,2,3],[0,1,2,3]]}"},"fields":{"title":"Topology Qualification before Discrete or Indiscrete Classification","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [2], [2, 3], [0, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.","answer":{"kind":"expression","value":"(4, [[], [2], [2, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, other-topology)","explanation":"Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [2], [2, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, other-topology)."},"skills":["Topology Qualification before Discrete or Indiscrete Classification"],"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-20261013-advance-20261018-r01-topological-space-0064","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a897a8c835dfd4712049484eb5bc3daaea67ef5b55649d16e0e9cc1141bed123","family_id":"qf-ext18-topology-discrete-indiscrete","seed":202610181064,"semantic_template":"qf-ext18-topology-discrete-indiscrete/{\"n\":4,\"opens\":[[],[3],[2,3],[0,2,3],[0,1,2,3]]}"},"fields":{"title":"Topology Qualification before Discrete or Indiscrete Classification","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [3], [2, 3], [0, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.","answer":{"kind":"expression","value":"(4, [[], [3], [2, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, other-topology)","explanation":"Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [3], [2, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, other-topology)."},"skills":["Topology Qualification before Discrete or Indiscrete Classification"],"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-20261013-advance-20261018-r01-topological-space-0065","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8ac9f864257d5eb8834c101c37c6a8957df225d5121f9558f4359b91d828a48f","family_id":"qf-ext18-topology-discrete-indiscrete","seed":202610181065,"semantic_template":"qf-ext18-topology-discrete-indiscrete/{\"n\":4,\"opens\":[[],[0],[0,1],[0,1,3],[0,1,2,3]]}"},"fields":{"title":"Topology Qualification before Discrete or Indiscrete Classification","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [0], [0, 1], [0, 1, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.","answer":{"kind":"expression","value":"(4, [[], [0], [0, 1], [0, 1, 3], [0, 1, 2, 3]], Yes, other-topology)","explanation":"Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0], [0, 1], [0, 1, 3], [0, 1, 2, 3]], Yes, other-topology)."},"skills":["Topology Qualification before Discrete or Indiscrete Classification"],"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-20261013-advance-20261018-r01-topological-space-0066","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bdaecb1468584ecb9ab1423460df918c8bbf42e3c86abe206af2ba270e45693e","family_id":"qf-ext18-topology-discrete-indiscrete","seed":202610181066,"semantic_template":"qf-ext18-topology-discrete-indiscrete/{\"n\":4,\"opens\":[[],[1],[0,1],[0,1,3],[0,1,2,3]]}"},"fields":{"title":"Topology Qualification before Discrete or Indiscrete Classification","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [0, 1], [0, 1, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.","answer":{"kind":"expression","value":"(4, [[], [1], [0, 1], [0, 1, 3], [0, 1, 2, 3]], Yes, other-topology)","explanation":"Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [0, 1], [0, 1, 3], [0, 1, 2, 3]], Yes, other-topology)."},"skills":["Topology Qualification before Discrete or Indiscrete Classification"],"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-20261013-advance-20261018-r01-topological-space-0067","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0578fc996059c17c2b4271341d5ed27c3006b78807ec88fdd909b19f1676bf0f","family_id":"qf-ext18-topology-discrete-indiscrete","seed":202610181067,"semantic_template":"qf-ext18-topology-discrete-indiscrete/{\"n\":4,\"opens\":[[],[2],[0,2],[0,2,3],[0,1,2,3]]}"},"fields":{"title":"Topology Qualification before Discrete or Indiscrete Classification","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [2], [0, 2], [0, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.","answer":{"kind":"expression","value":"(4, [[], [2], [0, 2], [0, 2, 3], [0, 1, 2, 3]], Yes, other-topology)","explanation":"Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [2], [0, 2], [0, 2, 3], [0, 1, 2, 3]], Yes, other-topology)."},"skills":["Topology Qualification before Discrete or Indiscrete Classification"],"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-20261013-advance-20261018-r01-topological-space-0068","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7c21b2bc9baa2b39ad0921d84a13da6697882c1356301d8e62ab1d054addaea6","family_id":"qf-ext18-topology-discrete-indiscrete","seed":202610181068,"semantic_template":"qf-ext18-topology-discrete-indiscrete/{\"n\":4,\"opens\":[[],[3],[0,3],[0,2,3],[0,1,2,3]]}"},"fields":{"title":"Topology Qualification before Discrete or Indiscrete Classification","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [3], [0, 3], [0, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.","answer":{"kind":"expression","value":"(4, [[], [3], [0, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, other-topology)","explanation":"Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [3], [0, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, other-topology)."},"skills":["Topology Qualification before Discrete or Indiscrete Classification"],"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-20261013-advance-20261018-r01-topological-space-0069","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"01c465a7d87ee82abc2f07e76a6da86eae6d99295bce5725429bc67f553a9398","family_id":"qf-ext18-topology-discrete-indiscrete","seed":202610181069,"semantic_template":"qf-ext18-topology-discrete-indiscrete/{\"n\":4,\"opens\":[[],[0],[0,2],[0,2,3],[0,1,2,3]]}"},"fields":{"title":"Topology Qualification before Discrete or Indiscrete Classification","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [0], [0, 2], [0, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.","answer":{"kind":"expression","value":"(4, [[], [0], [0, 2], [0, 2, 3], [0, 1, 2, 3]], Yes, other-topology)","explanation":"Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0], [0, 2], [0, 2, 3], [0, 1, 2, 3]], Yes, other-topology)."},"skills":["Topology Qualification before Discrete or Indiscrete Classification"],"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-20261013-advance-20261018-r01-topological-space-0070","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2a603d67bc0be48517683ff860e55d449c9194d91d3670e4a70a9169c8059976","family_id":"qf-ext18-topology-discrete-indiscrete","seed":202610181070,"semantic_template":"qf-ext18-topology-discrete-indiscrete/{\"n\":4,\"opens\":[[],[1],[1,2],[1,2,3],[0,1,2,3]]}"},"fields":{"title":"Topology Qualification before Discrete or Indiscrete Classification","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [1, 2], [1, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.","answer":{"kind":"expression","value":"(4, [[], [1], [1, 2], [1, 2, 3], [0, 1, 2, 3]], Yes, other-topology)","explanation":"Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [1, 2], [1, 2, 3], [0, 1, 2, 3]], Yes, other-topology)."},"skills":["Topology Qualification before Discrete or Indiscrete Classification"],"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-20261013-advance-20261018-r01-topological-space-0071","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"89c589617221295cef42706553b29e460de80a7806d37c51b9004e7d2660a4de","family_id":"qf-ext18-topology-discrete-indiscrete","seed":202610181071,"semantic_template":"qf-ext18-topology-discrete-indiscrete/{\"n\":4,\"opens\":[[],[2],[1,2],[1,2,3],[0,1,2,3]]}"},"fields":{"title":"Topology Qualification before Discrete or Indiscrete Classification","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [2], [1, 2], [1, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.","answer":{"kind":"expression","value":"(4, [[], [2], [1, 2], [1, 2, 3], [0, 1, 2, 3]], Yes, other-topology)","explanation":"Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [2], [1, 2], [1, 2, 3], [0, 1, 2, 3]], Yes, other-topology)."},"skills":["Topology Qualification before Discrete or Indiscrete Classification"],"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-20261013-advance-20261018-r01-topological-space-0072","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bbc6b2f25f2baed4b880e9565ba2d68eace17e592559063066304b2124c47f33","family_id":"qf-ext18-topology-discrete-indiscrete","seed":202610181072,"semantic_template":"qf-ext18-topology-discrete-indiscrete/{\"n\":4,\"opens\":[[],[3],[1,3],[1,2,3],[0,1,2,3]]}"},"fields":{"title":"Topology Qualification before Discrete or Indiscrete Classification","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [3], [1, 3], [1, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.","answer":{"kind":"expression","value":"(4, [[], [3], [1, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, other-topology)","explanation":"Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [3], [1, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, other-topology)."},"skills":["Topology Qualification before Discrete or Indiscrete Classification"],"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-20261013-advance-20261018-r01-topological-space-0073","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"07e2d749a2cb272e02664f9f74dca25737f4348b7796d665bdf61b8c07224c75","family_id":"qf-ext18-topology-discrete-indiscrete","seed":202610181073,"semantic_template":"qf-ext18-topology-discrete-indiscrete/{\"n\":4,\"opens\":[[],[0],[0,3],[0,2,3],[0,1,2,3]]}"},"fields":{"title":"Topology Qualification before Discrete or Indiscrete Classification","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [0], [0, 3], [0, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.","answer":{"kind":"expression","value":"(4, [[], [0], [0, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, other-topology)","explanation":"Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0], [0, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, other-topology)."},"skills":["Topology Qualification before Discrete or Indiscrete Classification"],"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-20261013-advance-20261018-r01-topological-space-0074","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4d337fa979ec1d1b4aa6315a05efa35368bb741ffefc3a153c3b5e4951580a1c","family_id":"qf-ext18-topology-discrete-indiscrete","seed":202610181074,"semantic_template":"qf-ext18-topology-discrete-indiscrete/{\"n\":4,\"opens\":[[],[1],[1,3],[1,2,3],[0,1,2,3]]}"},"fields":{"title":"Topology Qualification before Discrete or Indiscrete Classification","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [1, 3], [1, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.","answer":{"kind":"expression","value":"(4, [[], [1], [1, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, other-topology)","explanation":"Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [1, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, other-topology)."},"skills":["Topology Qualification before Discrete or Indiscrete Classification"],"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-20261013-advance-20261018-r01-topological-space-0075","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e1b968e40dd66db88131219d6b99f7cc6ff46d5c52da5629ecf5de92db496c43","family_id":"qf-ext18-topology-discrete-indiscrete","seed":202610181075,"semantic_template":"qf-ext18-topology-discrete-indiscrete/{\"n\":4,\"opens\":[[],[2],[2,3],[1,2,3],[0,1,2,3]]}"},"fields":{"title":"Topology Qualification before Discrete or Indiscrete Classification","stem":"Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [2], [2, 3], [1, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.","answer":{"kind":"expression","value":"(4, [[], [2], [2, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, other-topology)","explanation":"Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [2], [2, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, other-topology)."},"skills":["Topology Qualification before Discrete or Indiscrete Classification"],"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-20261013-advance-20261018-r01-topological-space-0076","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6339cd4fc82d5101495f76b9a3371eb79f7420c5717303ef58a57cf8da6f2b22","family_id":"qf-ext18-topology-finer-coarser","seed":202610181076,"semantic_template":"qf-ext18-topology-finer-coarser/{\"n\":4,\"opens_a\":[[],[0,1,2,3]],\"opens_b\":[[],[0],[0,1],[0,1,2],[0,1,2,3]]}"},"fields":{"title":"Compare Two Qualified Topologies on the Same Finite Set","stem":"Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.","answer":{"kind":"expression","value":"(4, [[], [0, 1, 2, 3]], [[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3]], Yes, No, A-coarser)","explanation":"Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3]], Yes, No, A-coarser)."},"skills":["Compare Two Qualified Topologies on the Same Finite Set"],"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-20261013-advance-20261018-r01-topological-space-0077","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"01adfe5b2f5f28e7313bc68c47f9ddb709d21e9e7414fda2b2753b252493e43a","family_id":"qf-ext18-topology-finer-coarser","seed":202610181077,"semantic_template":"qf-ext18-topology-finer-coarser/{\"n\":4,\"opens_a\":[[],[0,1,2,3]],\"opens_b\":[[],[1],[0,1],[0,1,2],[0,1,2,3]]}"},"fields":{"title":"Compare Two Qualified Topologies on the Same Finite Set","stem":"Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [1], [0, 1], [0, 1, 2], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.","answer":{"kind":"expression","value":"(4, [[], [0, 1, 2, 3]], [[], [1], [0, 1], [0, 1, 2], [0, 1, 2, 3]], Yes, No, A-coarser)","explanation":"Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [1], [0, 1], [0, 1, 2], [0, 1, 2, 3]], Yes, No, A-coarser)."},"skills":["Compare Two Qualified Topologies on the Same Finite Set"],"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-20261013-advance-20261018-r01-topological-space-0078","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f777fe34ce65a60139815e9e88f668233821fa4936eb5ea1598823a78a97489b","family_id":"qf-ext18-topology-finer-coarser","seed":202610181078,"semantic_template":"qf-ext18-topology-finer-coarser/{\"n\":4,\"opens_a\":[[],[0,1,2,3]],\"opens_b\":[[],[2],[0,2],[0,1,2],[0,1,2,3]]}"},"fields":{"title":"Compare Two Qualified Topologies on the Same Finite Set","stem":"Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [2], [0, 2], [0, 1, 2], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.","answer":{"kind":"expression","value":"(4, [[], [0, 1, 2, 3]], [[], [2], [0, 2], [0, 1, 2], [0, 1, 2, 3]], Yes, No, A-coarser)","explanation":"Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [2], [0, 2], [0, 1, 2], [0, 1, 2, 3]], Yes, No, A-coarser)."},"skills":["Compare Two Qualified Topologies on the Same Finite Set"],"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-20261013-advance-20261018-r01-topological-space-0079","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ac1726ae19d7bd9978ec789bffc144aa4b374bff1307ccb99c23c420ed374899","family_id":"qf-ext18-topology-finer-coarser","seed":202610181079,"semantic_template":"qf-ext18-topology-finer-coarser/{\"n\":4,\"opens_a\":[[],[0,1,2,3]],\"opens_b\":[[],[3],[0,3],[0,1,3],[0,1,2,3]]}"},"fields":{"title":"Compare Two Qualified Topologies on the Same Finite Set","stem":"Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [3], [0, 3], [0, 1, 3], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.","answer":{"kind":"expression","value":"(4, [[], [0, 1, 2, 3]], [[], [3], [0, 3], [0, 1, 3], [0, 1, 2, 3]], Yes, No, A-coarser)","explanation":"Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [3], [0, 3], [0, 1, 3], [0, 1, 2, 3]], Yes, No, A-coarser)."},"skills":["Compare Two Qualified Topologies on the Same Finite Set"],"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-20261013-advance-20261018-r01-topological-space-0080","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f2529f160b3f97ff58c566269d0c3930db309e180a296ed0512f19b860689615","family_id":"qf-ext18-topology-finer-coarser","seed":202610181080,"semantic_template":"qf-ext18-topology-finer-coarser/{\"n\":4,\"opens_a\":[[],[0,1,2,3]],\"opens_b\":[[],[0],[0,2],[0,1,2],[0,1,2,3]]}"},"fields":{"title":"Compare Two Qualified Topologies on the Same Finite Set","stem":"Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [0], [0, 2], [0, 1, 2], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.","answer":{"kind":"expression","value":"(4, [[], [0, 1, 2, 3]], [[], [0], [0, 2], [0, 1, 2], [0, 1, 2, 3]], Yes, No, A-coarser)","explanation":"Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [0], [0, 2], [0, 1, 2], [0, 1, 2, 3]], Yes, No, A-coarser)."},"skills":["Compare Two Qualified Topologies on the Same Finite Set"],"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-20261013-advance-20261018-r01-topological-space-0081","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"de38c7d54a2cf79ebd0738c7a27f8f4d2029b1d4a216aff3da33781fbb6600f6","family_id":"qf-ext18-topology-finer-coarser","seed":202610181081,"semantic_template":"qf-ext18-topology-finer-coarser/{\"n\":4,\"opens_a\":[[],[0,1,2,3]],\"opens_b\":[[],[1],[1,2],[0,1,2],[0,1,2,3]]}"},"fields":{"title":"Compare Two Qualified Topologies on the Same Finite Set","stem":"Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [1], [1, 2], [0, 1, 2], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.","answer":{"kind":"expression","value":"(4, [[], [0, 1, 2, 3]], [[], [1], [1, 2], [0, 1, 2], [0, 1, 2, 3]], Yes, No, A-coarser)","explanation":"Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [1], [1, 2], [0, 1, 2], [0, 1, 2, 3]], Yes, No, A-coarser)."},"skills":["Compare Two Qualified Topologies on the Same Finite Set"],"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-20261013-advance-20261018-r01-topological-space-0082","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"82553d9ccfe55bcbe957beba0f88a8bf1d0f93c41c8adb84b3f635b98d72d771","family_id":"qf-ext18-topology-finer-coarser","seed":202610181082,"semantic_template":"qf-ext18-topology-finer-coarser/{\"n\":4,\"opens_a\":[[],[0,1,2,3]],\"opens_b\":[[],[2],[1,2],[0,1,2],[0,1,2,3]]}"},"fields":{"title":"Compare Two Qualified Topologies on the Same Finite Set","stem":"Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [2], [1, 2], [0, 1, 2], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.","answer":{"kind":"expression","value":"(4, [[], [0, 1, 2, 3]], [[], [2], [1, 2], [0, 1, 2], [0, 1, 2, 3]], Yes, No, A-coarser)","explanation":"Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [2], [1, 2], [0, 1, 2], [0, 1, 2, 3]], Yes, No, A-coarser)."},"skills":["Compare Two Qualified Topologies on the Same Finite Set"],"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-20261013-advance-20261018-r01-topological-space-0083","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d4ef8d6edef06acf5afbce19689487de0b6fe676751bb045a5bc0f6054543a12","family_id":"qf-ext18-topology-finer-coarser","seed":202610181083,"semantic_template":"qf-ext18-topology-finer-coarser/{\"n\":4,\"opens_a\":[[],[0,1,2,3]],\"opens_b\":[[],[3],[1,3],[0,1,3],[0,1,2,3]]}"},"fields":{"title":"Compare Two Qualified Topologies on the Same Finite Set","stem":"Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [3], [1, 3], [0, 1, 3], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.","answer":{"kind":"expression","value":"(4, [[], [0, 1, 2, 3]], [[], [3], [1, 3], [0, 1, 3], [0, 1, 2, 3]], Yes, No, A-coarser)","explanation":"Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [3], [1, 3], [0, 1, 3], [0, 1, 2, 3]], Yes, No, A-coarser)."},"skills":["Compare Two Qualified Topologies on the Same Finite Set"],"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-20261013-advance-20261018-r01-topological-space-0084","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"915a8d1350edac6ddfe983ecdd8b57f751639712d71fd34e79e2a2750395ab31","family_id":"qf-ext18-topology-finer-coarser","seed":202610181084,"semantic_template":"qf-ext18-topology-finer-coarser/{\"n\":4,\"opens_a\":[[],[0,1,2,3]],\"opens_b\":[[],[0],[0,3],[0,1,3],[0,1,2,3]]}"},"fields":{"title":"Compare Two Qualified Topologies on the Same Finite Set","stem":"Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [0], [0, 3], [0, 1, 3], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.","answer":{"kind":"expression","value":"(4, [[], [0, 1, 2, 3]], [[], [0], [0, 3], [0, 1, 3], [0, 1, 2, 3]], Yes, No, A-coarser)","explanation":"Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [0], [0, 3], [0, 1, 3], [0, 1, 2, 3]], Yes, No, A-coarser)."},"skills":["Compare Two Qualified Topologies on the Same Finite Set"],"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-20261013-advance-20261018-r01-topological-space-0085","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"58d5ea1592a386718bedf5292dd25b83d67bf9bbab14f98a1c798e25c07cd8d3","family_id":"qf-ext18-topology-finer-coarser","seed":202610181085,"semantic_template":"qf-ext18-topology-finer-coarser/{\"n\":4,\"opens_a\":[[],[0,1,2,3]],\"opens_b\":[[],[1],[1,3],[0,1,3],[0,1,2,3]]}"},"fields":{"title":"Compare Two Qualified Topologies on the Same Finite Set","stem":"Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [1], [1, 3], [0, 1, 3], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.","answer":{"kind":"expression","value":"(4, [[], [0, 1, 2, 3]], [[], [1], [1, 3], [0, 1, 3], [0, 1, 2, 3]], Yes, No, A-coarser)","explanation":"Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [1], [1, 3], [0, 1, 3], [0, 1, 2, 3]], Yes, No, A-coarser)."},"skills":["Compare Two Qualified Topologies on the Same Finite Set"],"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-20261013-advance-20261018-r01-topological-space-0086","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5cb0e24f0f0aa09fe9f9c75a6b648802500eed0d6e52ba4348379b464a5e5b0a","family_id":"qf-ext18-topology-finer-coarser","seed":202610181086,"semantic_template":"qf-ext18-topology-finer-coarser/{\"n\":4,\"opens_a\":[[],[0,1,2,3]],\"opens_b\":[[],[2],[2,3],[0,2,3],[0,1,2,3]]}"},"fields":{"title":"Compare Two Qualified Topologies on the Same Finite Set","stem":"Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [2], [2, 3], [0, 2, 3], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.","answer":{"kind":"expression","value":"(4, [[], [0, 1, 2, 3]], [[], [2], [2, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)","explanation":"Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [2], [2, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)."},"skills":["Compare Two Qualified Topologies on the Same Finite Set"],"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-20261013-advance-20261018-r01-topological-space-0087","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4d3d77c39c837e51e5246b7708338c16285832d5302837d1a4b44a2b70e679a4","family_id":"qf-ext18-topology-finer-coarser","seed":202610181087,"semantic_template":"qf-ext18-topology-finer-coarser/{\"n\":4,\"opens_a\":[[],[0,1,2,3]],\"opens_b\":[[],[3],[2,3],[0,2,3],[0,1,2,3]]}"},"fields":{"title":"Compare Two Qualified Topologies on the Same Finite Set","stem":"Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [3], [2, 3], [0, 2, 3], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.","answer":{"kind":"expression","value":"(4, [[], [0, 1, 2, 3]], [[], [3], [2, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)","explanation":"Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [3], [2, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)."},"skills":["Compare Two Qualified Topologies on the Same Finite Set"],"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-20261013-advance-20261018-r01-topological-space-0088","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6af9e6b1fc82d5a92dc2e0919e21c37831b70b4770b48298097b9caeea9746c3","family_id":"qf-ext18-topology-finer-coarser","seed":202610181088,"semantic_template":"qf-ext18-topology-finer-coarser/{\"n\":4,\"opens_a\":[[],[0,1,2,3]],\"opens_b\":[[],[0],[0,1],[0,1,3],[0,1,2,3]]}"},"fields":{"title":"Compare Two Qualified Topologies on the Same Finite Set","stem":"Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [0], [0, 1], [0, 1, 3], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.","answer":{"kind":"expression","value":"(4, [[], [0, 1, 2, 3]], [[], [0], [0, 1], [0, 1, 3], [0, 1, 2, 3]], Yes, No, A-coarser)","explanation":"Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [0], [0, 1], [0, 1, 3], [0, 1, 2, 3]], Yes, No, A-coarser)."},"skills":["Compare Two Qualified Topologies on the Same Finite Set"],"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-20261013-advance-20261018-r01-topological-space-0089","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"73ad718d63c6a833da7748322b430230cf404dc326b6f22bcddfa0c55e14407d","family_id":"qf-ext18-topology-finer-coarser","seed":202610181089,"semantic_template":"qf-ext18-topology-finer-coarser/{\"n\":4,\"opens_a\":[[],[0,1,2,3]],\"opens_b\":[[],[1],[0,1],[0,1,3],[0,1,2,3]]}"},"fields":{"title":"Compare Two Qualified Topologies on the Same Finite Set","stem":"Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [1], [0, 1], [0, 1, 3], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.","answer":{"kind":"expression","value":"(4, [[], [0, 1, 2, 3]], [[], [1], [0, 1], [0, 1, 3], [0, 1, 2, 3]], Yes, No, A-coarser)","explanation":"Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [1], [0, 1], [0, 1, 3], [0, 1, 2, 3]], Yes, No, A-coarser)."},"skills":["Compare Two Qualified Topologies on the Same Finite Set"],"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-20261013-advance-20261018-r01-topological-space-0090","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5c4426e3f57487075443e9294605fcab271eeb7cdc924c9a611a6e9718869e3d","family_id":"qf-ext18-topology-finer-coarser","seed":202610181090,"semantic_template":"qf-ext18-topology-finer-coarser/{\"n\":4,\"opens_a\":[[],[0,1,2,3]],\"opens_b\":[[],[2],[0,2],[0,2,3],[0,1,2,3]]}"},"fields":{"title":"Compare Two Qualified Topologies on the Same Finite Set","stem":"Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [2], [0, 2], [0, 2, 3], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.","answer":{"kind":"expression","value":"(4, [[], [0, 1, 2, 3]], [[], [2], [0, 2], [0, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)","explanation":"Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [2], [0, 2], [0, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)."},"skills":["Compare Two Qualified Topologies on the Same Finite Set"],"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-20261013-advance-20261018-r01-topological-space-0091","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b35805abaf0e3e024fc5ed10ecd04da29a4be24f24c3a0d818320c1778482eb0","family_id":"qf-ext18-topology-finer-coarser","seed":202610181091,"semantic_template":"qf-ext18-topology-finer-coarser/{\"n\":4,\"opens_a\":[[],[0,1,2,3]],\"opens_b\":[[],[3],[0,3],[0,2,3],[0,1,2,3]]}"},"fields":{"title":"Compare Two Qualified Topologies on the Same Finite Set","stem":"Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [3], [0, 3], [0, 2, 3], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.","answer":{"kind":"expression","value":"(4, [[], [0, 1, 2, 3]], [[], [3], [0, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)","explanation":"Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [3], [0, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)."},"skills":["Compare Two Qualified Topologies on the Same Finite Set"],"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-20261013-advance-20261018-r01-topological-space-0092","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"63cc0bb3cddc2013a7b2d6d28a02b556d7901f5afe194bb0cf001592b64751cc","family_id":"qf-ext18-topology-finer-coarser","seed":202610181092,"semantic_template":"qf-ext18-topology-finer-coarser/{\"n\":4,\"opens_a\":[[],[0,1,2,3]],\"opens_b\":[[],[0],[0,2],[0,2,3],[0,1,2,3]]}"},"fields":{"title":"Compare Two Qualified Topologies on the Same Finite Set","stem":"Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [0], [0, 2], [0, 2, 3], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.","answer":{"kind":"expression","value":"(4, [[], [0, 1, 2, 3]], [[], [0], [0, 2], [0, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)","explanation":"Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [0], [0, 2], [0, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)."},"skills":["Compare Two Qualified Topologies on the Same Finite Set"],"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-20261013-advance-20261018-r01-topological-space-0093","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7fcb00c98f846231774c2f62a39c18495e0454df1e03c4c95989398ada110fab","family_id":"qf-ext18-topology-finer-coarser","seed":202610181093,"semantic_template":"qf-ext18-topology-finer-coarser/{\"n\":4,\"opens_a\":[[],[0,1,2,3]],\"opens_b\":[[],[1],[1,2],[1,2,3],[0,1,2,3]]}"},"fields":{"title":"Compare Two Qualified Topologies on the Same Finite Set","stem":"Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [1], [1, 2], [1, 2, 3], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.","answer":{"kind":"expression","value":"(4, [[], [0, 1, 2, 3]], [[], [1], [1, 2], [1, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)","explanation":"Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [1], [1, 2], [1, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)."},"skills":["Compare Two Qualified Topologies on the Same Finite Set"],"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-20261013-advance-20261018-r01-topological-space-0094","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e9ad47806193435831f62a342d615859895cadd8bc994ebec5db40c65a595555","family_id":"qf-ext18-topology-finer-coarser","seed":202610181094,"semantic_template":"qf-ext18-topology-finer-coarser/{\"n\":4,\"opens_a\":[[],[0,1,2,3]],\"opens_b\":[[],[2],[1,2],[1,2,3],[0,1,2,3]]}"},"fields":{"title":"Compare Two Qualified Topologies on the Same Finite Set","stem":"Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [2], [1, 2], [1, 2, 3], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.","answer":{"kind":"expression","value":"(4, [[], [0, 1, 2, 3]], [[], [2], [1, 2], [1, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)","explanation":"Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [2], [1, 2], [1, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)."},"skills":["Compare Two Qualified Topologies on the Same Finite Set"],"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-20261013-advance-20261018-r01-topological-space-0095","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fce890185738b1c4edc1c7482e02e7525d7de690c5b060da0ddc0a82d6acaf19","family_id":"qf-ext18-topology-finer-coarser","seed":202610181095,"semantic_template":"qf-ext18-topology-finer-coarser/{\"n\":4,\"opens_a\":[[],[0,1,2,3]],\"opens_b\":[[],[3],[1,3],[1,2,3],[0,1,2,3]]}"},"fields":{"title":"Compare Two Qualified Topologies on the Same Finite Set","stem":"Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [3], [1, 3], [1, 2, 3], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.","answer":{"kind":"expression","value":"(4, [[], [0, 1, 2, 3]], [[], [3], [1, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)","explanation":"Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [3], [1, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)."},"skills":["Compare Two Qualified Topologies on the Same Finite Set"],"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-20261013-advance-20261018-r01-topological-space-0096","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4fa49629417eabe0d3ea71c7e4a3b6d4dacdd9e43497a5ae26a9683fff5c3fcf","family_id":"qf-ext18-topology-finer-coarser","seed":202610181096,"semantic_template":"qf-ext18-topology-finer-coarser/{\"n\":4,\"opens_a\":[[],[0,1,2,3]],\"opens_b\":[[],[0],[0,3],[0,2,3],[0,1,2,3]]}"},"fields":{"title":"Compare Two Qualified Topologies on the Same Finite Set","stem":"Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [0], [0, 3], [0, 2, 3], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.","answer":{"kind":"expression","value":"(4, [[], [0, 1, 2, 3]], [[], [0], [0, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)","explanation":"Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [0], [0, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)."},"skills":["Compare Two Qualified Topologies on the Same Finite Set"],"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-20261013-advance-20261018-r01-topological-space-0097","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"850b9d5c1e7409d8be03c932725aed7c43e396e48c8c698aa92554b091338611","family_id":"qf-ext18-topology-finer-coarser","seed":202610181097,"semantic_template":"qf-ext18-topology-finer-coarser/{\"n\":4,\"opens_a\":[[],[0,1,2,3]],\"opens_b\":[[],[1],[1,3],[1,2,3],[0,1,2,3]]}"},"fields":{"title":"Compare Two Qualified Topologies on the Same Finite Set","stem":"Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [1], [1, 3], [1, 2, 3], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.","answer":{"kind":"expression","value":"(4, [[], [0, 1, 2, 3]], [[], [1], [1, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)","explanation":"Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [1], [1, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)."},"skills":["Compare Two Qualified Topologies on the Same Finite Set"],"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-20261013-advance-20261018-r01-topological-space-0098","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ae6f375cf005ef74048736865ed1750a2b846949960835aa4f002bf5232d4d09","family_id":"qf-ext18-topology-finer-coarser","seed":202610181098,"semantic_template":"qf-ext18-topology-finer-coarser/{\"n\":4,\"opens_a\":[[],[0,1,2,3]],\"opens_b\":[[],[2],[2,3],[1,2,3],[0,1,2,3]]}"},"fields":{"title":"Compare Two Qualified Topologies on the Same Finite Set","stem":"Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [2], [2, 3], [1, 2, 3], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.","answer":{"kind":"expression","value":"(4, [[], [0, 1, 2, 3]], [[], [2], [2, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)","explanation":"Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [2], [2, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)."},"skills":["Compare Two Qualified Topologies on the Same Finite Set"],"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-20261013-advance-20261018-r01-topological-space-0099","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"51d6de84e975bc674fbebe505707e69511b86570938036d586e4a90df4c69423","family_id":"qf-ext18-topology-finer-coarser","seed":202610181099,"semantic_template":"qf-ext18-topology-finer-coarser/{\"n\":4,\"opens_a\":[[],[0,1,2,3]],\"opens_b\":[[],[3],[2,3],[1,2,3],[0,1,2,3]]}"},"fields":{"title":"Compare Two Qualified Topologies on the Same Finite Set","stem":"Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [3], [2, 3], [1, 2, 3], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.","answer":{"kind":"expression","value":"(4, [[], [0, 1, 2, 3]], [[], [3], [2, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)","explanation":"Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [3], [2, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)."},"skills":["Compare Two Qualified Topologies on the Same Finite Set"],"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-20261013-advance-20261018-r01-topological-space-0100","batch_id":"batch-20261013-advance-20261018-r01-topological-space-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bb9c2856742b4d4aec5a370529d4f1dae2f85ed05fecbb3a6f3a17472a9bd9d4","family_id":"qf-ext18-topology-finer-coarser","seed":202610181100,"semantic_template":"qf-ext18-topology-finer-coarser/{\"n\":4,\"opens_a\":[[],[0],[0,1],[0,1,2],[0,1,2,3]],\"opens_b\":[[],[0],[1],[0,1],[2],[0,2],[1,2],[0,1,2],[3],[0,3],[1,3],[0,1,3],[2,3],[0,2,3],[1,2,3],[0,1,2,3]]}"},"fields":{"title":"Compare Two Qualified Topologies on the Same Finite Set","stem":"Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3]], B=[[], [0], [1], [0, 1], [2], [0, 2], [1, 2], [0, 1, 2], [3], [0, 3], [1, 3], [0, 1, 3], [2, 3], [0, 2, 3], [1, 2, 3], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.","answer":{"kind":"expression","value":"(4, [[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3]], [[], [0], [1], [0, 1], [2], [0, 2], [1, 2], [0, 1, 2], [3], [0, 3], [1, 3], [0, 1, 3], [2, 3], [0, 2, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)","explanation":"Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3]], [[], [0], [1], [0, 1], [2], [0, 2], [1, 2], [0, 1, 2], [3], [0, 3], [1, 3], [0, 1, 3], [2, 3], [0, 2, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)."},"skills":["Compare Two Qualified Topologies on the Same Finite Set"],"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."}}