{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0001","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9abc709c8f2e0480adca0c1cf8a13f9d76219b655615452cf478599a189002f0","family_id":"qf-mod43-compactclosed-hypotheses-qualification","seed":202611123001,"semantic_template":"qf-mod43-compactclosed-hypotheses-qualification/{\"compact_subset\":[\"a0\",\"b0\"],\"points\":[\"a0\",\"b0\",\"c0\"],\"topology\":[[\"a0\",\"b0\",\"c0\"],[\"a0\",\"b0\"],[\"a0\",\"c0\"],[\"a0\"],[\"b0\",\"c0\"],[\"b0\"],[\"c0\"],[]]}"},"fields":{"title":"Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses","stem":"Use complete finite labeled carrier=[a0, b0, c0], topology=[[a0, b0, c0], [a0, b0], [a0, c0], [a0], [b0, c0], [b0], [c0], []] in its original supplied open-set order, and compact_subset K=[a0, b0]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, pair_separation_rows, Hausdorff, compact, certificate).","answer":{"kind":"expression","value":"([a0, b0, c0], [a0, b0], [[a0, b0, [[a0, c0], [b0]]], [a0, c0, [[a0, b0], [c0]]], [b0, c0, [[a0, b0], [c0]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)","explanation":"Require every distinct-point pair to have disjoint open neighborhoods, retaining first witnesses in supplied topology order. Report Hausdorffness and finite-subset compactness separately; do not drop the Hausdorff hypothesis in the general theorem. The exact result, retaining the requested order, is ([a0, b0, c0], [a0, b0], [[a0, b0, [[a0, c0], [b0]]], [a0, c0, [[a0, b0], [c0]]], [b0, c0, [[a0, b0], [c0]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)."},"skills":["Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0002","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a0b22cbb5107268ad94de44f6302039f2080fd64ce95c9c6a16f086f464d8a03","family_id":"qf-mod43-compactclosed-hypotheses-qualification","seed":202611123002,"semantic_template":"qf-mod43-compactclosed-hypotheses-qualification/{\"compact_subset\":[\"a1\",\"b1\"],\"points\":[\"a1\",\"b1\",\"c1\"],\"topology\":[[\"a1\",\"b1\",\"c1\"],[\"a1\",\"b1\"],[\"a1\",\"c1\"],[\"a1\"],[\"b1\",\"c1\"],[\"b1\"],[\"c1\"],[]]}"},"fields":{"title":"Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses","stem":"Use complete finite labeled carrier=[a1, b1, c1], topology=[[a1, b1, c1], [a1, b1], [a1, c1], [a1], [b1, c1], [b1], [c1], []] in its original supplied open-set order, and compact_subset K=[a1, b1]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, pair_separation_rows, Hausdorff, compact, certificate).","answer":{"kind":"expression","value":"([a1, b1, c1], [a1, b1], [[a1, b1, [[a1, c1], [b1]]], [a1, c1, [[a1, b1], [c1]]], [b1, c1, [[a1, b1], [c1]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)","explanation":"Require every distinct-point pair to have disjoint open neighborhoods, retaining first witnesses in supplied topology order. Report Hausdorffness and finite-subset compactness separately; do not drop the Hausdorff hypothesis in the general theorem. The exact result, retaining the requested order, is ([a1, b1, c1], [a1, b1], [[a1, b1, [[a1, c1], [b1]]], [a1, c1, [[a1, b1], [c1]]], [b1, c1, [[a1, b1], [c1]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)."},"skills":["Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0003","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a0ab200f9a02d9ab4f7a1eae787813f218869e1f54924f8813edf377b0227d86","family_id":"qf-mod43-compactclosed-hypotheses-qualification","seed":202611123003,"semantic_template":"qf-mod43-compactclosed-hypotheses-qualification/{\"compact_subset\":[\"a2\",\"b2\"],\"points\":[\"a2\",\"b2\",\"c2\"],\"topology\":[[\"a2\",\"b2\",\"c2\"],[\"a2\",\"b2\"],[\"a2\",\"c2\"],[\"a2\"],[\"b2\",\"c2\"],[\"b2\"],[\"c2\"],[]]}"},"fields":{"title":"Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses","stem":"Use complete finite labeled carrier=[a2, b2, c2], topology=[[a2, b2, c2], [a2, b2], [a2, c2], [a2], [b2, c2], [b2], [c2], []] in its original supplied open-set order, and compact_subset K=[a2, b2]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, pair_separation_rows, Hausdorff, compact, certificate).","answer":{"kind":"expression","value":"([a2, b2, c2], [a2, b2], [[a2, b2, [[a2, c2], [b2]]], [a2, c2, [[a2, b2], [c2]]], [b2, c2, [[a2, b2], [c2]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)","explanation":"Require every distinct-point pair to have disjoint open neighborhoods, retaining first witnesses in supplied topology order. Report Hausdorffness and finite-subset compactness separately; do not drop the Hausdorff hypothesis in the general theorem. The exact result, retaining the requested order, is ([a2, b2, c2], [a2, b2], [[a2, b2, [[a2, c2], [b2]]], [a2, c2, [[a2, b2], [c2]]], [b2, c2, [[a2, b2], [c2]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)."},"skills":["Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0004","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f337bee755f220eec4ac38757513b0ed95cf1ea8deccab4da91dd0202c107681","family_id":"qf-mod43-compactclosed-hypotheses-qualification","seed":202611123004,"semantic_template":"qf-mod43-compactclosed-hypotheses-qualification/{\"compact_subset\":[\"a3\",\"b3\"],\"points\":[\"a3\",\"b3\",\"c3\"],\"topology\":[[\"a3\",\"b3\",\"c3\"],[\"a3\",\"b3\"],[\"a3\",\"c3\"],[\"a3\"],[\"b3\",\"c3\"],[\"b3\"],[\"c3\"],[]]}"},"fields":{"title":"Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses","stem":"Use complete finite labeled carrier=[a3, b3, c3], topology=[[a3, b3, c3], [a3, b3], [a3, c3], [a3], [b3, c3], [b3], [c3], []] in its original supplied open-set order, and compact_subset K=[a3, b3]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, pair_separation_rows, Hausdorff, compact, certificate).","answer":{"kind":"expression","value":"([a3, b3, c3], [a3, b3], [[a3, b3, [[a3, c3], [b3]]], [a3, c3, [[a3, b3], [c3]]], [b3, c3, [[a3, b3], [c3]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)","explanation":"Require every distinct-point pair to have disjoint open neighborhoods, retaining first witnesses in supplied topology order. Report Hausdorffness and finite-subset compactness separately; do not drop the Hausdorff hypothesis in the general theorem. The exact result, retaining the requested order, is ([a3, b3, c3], [a3, b3], [[a3, b3, [[a3, c3], [b3]]], [a3, c3, [[a3, b3], [c3]]], [b3, c3, [[a3, b3], [c3]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)."},"skills":["Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0005","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8431ac847d3e4e80ca5d1b6c0535854a88e3dee637bc55b98f60ee73b01d3f71","family_id":"qf-mod43-compactclosed-hypotheses-qualification","seed":202611123005,"semantic_template":"qf-mod43-compactclosed-hypotheses-qualification/{\"compact_subset\":[\"a4\",\"b4\"],\"points\":[\"a4\",\"b4\",\"c4\"],\"topology\":[[\"a4\",\"b4\",\"c4\"],[\"a4\",\"b4\"],[\"a4\",\"c4\"],[\"a4\"],[\"b4\",\"c4\"],[\"b4\"],[\"c4\"],[]]}"},"fields":{"title":"Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses","stem":"Use complete finite labeled carrier=[a4, b4, c4], topology=[[a4, b4, c4], [a4, b4], [a4, c4], [a4], [b4, c4], [b4], [c4], []] in its original supplied open-set order, and compact_subset K=[a4, b4]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, pair_separation_rows, Hausdorff, compact, certificate).","answer":{"kind":"expression","value":"([a4, b4, c4], [a4, b4], [[a4, b4, [[a4, c4], [b4]]], [a4, c4, [[a4, b4], [c4]]], [b4, c4, [[a4, b4], [c4]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)","explanation":"Require every distinct-point pair to have disjoint open neighborhoods, retaining first witnesses in supplied topology order. Report Hausdorffness and finite-subset compactness separately; do not drop the Hausdorff hypothesis in the general theorem. The exact result, retaining the requested order, is ([a4, b4, c4], [a4, b4], [[a4, b4, [[a4, c4], [b4]]], [a4, c4, [[a4, b4], [c4]]], [b4, c4, [[a4, b4], [c4]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)."},"skills":["Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0006","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c5d24c7a39bbd6e05b216e0840466654cba94d4578bd8c95f227deb1b28a1e8c","family_id":"qf-mod43-compactclosed-hypotheses-qualification","seed":202611123006,"semantic_template":"qf-mod43-compactclosed-hypotheses-qualification/{\"compact_subset\":[\"a5\",\"b5\"],\"points\":[\"a5\",\"b5\",\"c5\"],\"topology\":[[\"a5\",\"b5\",\"c5\"],[\"a5\",\"b5\"],[\"a5\",\"c5\"],[\"a5\"],[\"b5\",\"c5\"],[\"b5\"],[\"c5\"],[]]}"},"fields":{"title":"Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses","stem":"Use complete finite labeled carrier=[a5, b5, c5], topology=[[a5, b5, c5], [a5, b5], [a5, c5], [a5], [b5, c5], [b5], [c5], []] in its original supplied open-set order, and compact_subset K=[a5, b5]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, pair_separation_rows, Hausdorff, compact, certificate).","answer":{"kind":"expression","value":"([a5, b5, c5], [a5, b5], [[a5, b5, [[a5, c5], [b5]]], [a5, c5, [[a5, b5], [c5]]], [b5, c5, [[a5, b5], [c5]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)","explanation":"Require every distinct-point pair to have disjoint open neighborhoods, retaining first witnesses in supplied topology order. Report Hausdorffness and finite-subset compactness separately; do not drop the Hausdorff hypothesis in the general theorem. The exact result, retaining the requested order, is ([a5, b5, c5], [a5, b5], [[a5, b5, [[a5, c5], [b5]]], [a5, c5, [[a5, b5], [c5]]], [b5, c5, [[a5, b5], [c5]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)."},"skills":["Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0007","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"70aa65286e23d8f98e65aa9b42c67d51bf73f0c04b02c4e6235d2681cb8ac579","family_id":"qf-mod43-compactclosed-hypotheses-qualification","seed":202611123007,"semantic_template":"qf-mod43-compactclosed-hypotheses-qualification/{\"compact_subset\":[\"a6\",\"b6\"],\"points\":[\"a6\",\"b6\",\"c6\"],\"topology\":[[\"a6\",\"b6\",\"c6\"],[\"a6\",\"b6\"],[\"a6\",\"c6\"],[\"a6\"],[\"b6\",\"c6\"],[\"b6\"],[\"c6\"],[]]}"},"fields":{"title":"Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses","stem":"Use complete finite labeled carrier=[a6, b6, c6], topology=[[a6, b6, c6], [a6, b6], [a6, c6], [a6], [b6, c6], [b6], [c6], []] in its original supplied open-set order, and compact_subset K=[a6, b6]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, pair_separation_rows, Hausdorff, compact, certificate).","answer":{"kind":"expression","value":"([a6, b6, c6], [a6, b6], [[a6, b6, [[a6, c6], [b6]]], [a6, c6, [[a6, b6], [c6]]], [b6, c6, [[a6, b6], [c6]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)","explanation":"Require every distinct-point pair to have disjoint open neighborhoods, retaining first witnesses in supplied topology order. Report Hausdorffness and finite-subset compactness separately; do not drop the Hausdorff hypothesis in the general theorem. The exact result, retaining the requested order, is ([a6, b6, c6], [a6, b6], [[a6, b6, [[a6, c6], [b6]]], [a6, c6, [[a6, b6], [c6]]], [b6, c6, [[a6, b6], [c6]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)."},"skills":["Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0008","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"380f0e45c359c2b06ab064f0c0a97e489e3f863e5c836f3af04e4f227cf52727","family_id":"qf-mod43-compactclosed-hypotheses-qualification","seed":202611123008,"semantic_template":"qf-mod43-compactclosed-hypotheses-qualification/{\"compact_subset\":[\"a7\",\"b7\"],\"points\":[\"a7\",\"b7\",\"c7\"],\"topology\":[[\"a7\",\"b7\",\"c7\"],[\"a7\",\"b7\"],[\"a7\",\"c7\"],[\"a7\"],[\"b7\",\"c7\"],[\"b7\"],[\"c7\"],[]]}"},"fields":{"title":"Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses","stem":"Use complete finite labeled carrier=[a7, b7, c7], topology=[[a7, b7, c7], [a7, b7], [a7, c7], [a7], [b7, c7], [b7], [c7], []] in its original supplied open-set order, and compact_subset K=[a7, b7]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, pair_separation_rows, Hausdorff, compact, certificate).","answer":{"kind":"expression","value":"([a7, b7, c7], [a7, b7], [[a7, b7, [[a7, c7], [b7]]], [a7, c7, [[a7, b7], [c7]]], [b7, c7, [[a7, b7], [c7]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)","explanation":"Require every distinct-point pair to have disjoint open neighborhoods, retaining first witnesses in supplied topology order. Report Hausdorffness and finite-subset compactness separately; do not drop the Hausdorff hypothesis in the general theorem. The exact result, retaining the requested order, is ([a7, b7, c7], [a7, b7], [[a7, b7, [[a7, c7], [b7]]], [a7, c7, [[a7, b7], [c7]]], [b7, c7, [[a7, b7], [c7]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)."},"skills":["Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0009","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"60a74249176e216aac7a5e0038c9db496616f9566bf50ec8eb55c97204ae54fa","family_id":"qf-mod43-compactclosed-hypotheses-qualification","seed":202611123009,"semantic_template":"qf-mod43-compactclosed-hypotheses-qualification/{\"compact_subset\":[\"a8\",\"b8\"],\"points\":[\"a8\",\"b8\",\"c8\"],\"topology\":[[\"a8\",\"b8\",\"c8\"],[\"a8\",\"b8\"],[\"a8\",\"c8\"],[\"a8\"],[\"b8\",\"c8\"],[\"b8\"],[\"c8\"],[]]}"},"fields":{"title":"Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses","stem":"Use complete finite labeled carrier=[a8, b8, c8], topology=[[a8, b8, c8], [a8, b8], [a8, c8], [a8], [b8, c8], [b8], [c8], []] in its original supplied open-set order, and compact_subset K=[a8, b8]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, pair_separation_rows, Hausdorff, compact, certificate).","answer":{"kind":"expression","value":"([a8, b8, c8], [a8, b8], [[a8, b8, [[a8, c8], [b8]]], [a8, c8, [[a8, b8], [c8]]], [b8, c8, [[a8, b8], [c8]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)","explanation":"Require every distinct-point pair to have disjoint open neighborhoods, retaining first witnesses in supplied topology order. Report Hausdorffness and finite-subset compactness separately; do not drop the Hausdorff hypothesis in the general theorem. The exact result, retaining the requested order, is ([a8, b8, c8], [a8, b8], [[a8, b8, [[a8, c8], [b8]]], [a8, c8, [[a8, b8], [c8]]], [b8, c8, [[a8, b8], [c8]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)."},"skills":["Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0010","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2f32ef67a5be91bb3447ad48f6835ba800d9d78c5542b12bd06a218ade760447","family_id":"qf-mod43-compactclosed-hypotheses-qualification","seed":202611123010,"semantic_template":"qf-mod43-compactclosed-hypotheses-qualification/{\"compact_subset\":[\"a9\",\"b9\"],\"points\":[\"a9\",\"b9\",\"c9\"],\"topology\":[[\"a9\",\"b9\",\"c9\"],[\"a9\",\"b9\"],[\"a9\",\"c9\"],[\"a9\"],[\"b9\",\"c9\"],[\"b9\"],[\"c9\"],[]]}"},"fields":{"title":"Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses","stem":"Use complete finite labeled carrier=[a9, b9, c9], topology=[[a9, b9, c9], [a9, b9], [a9, c9], [a9], [b9, c9], [b9], [c9], []] in its original supplied open-set order, and compact_subset K=[a9, b9]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, pair_separation_rows, Hausdorff, compact, certificate).","answer":{"kind":"expression","value":"([a9, b9, c9], [a9, b9], [[a9, b9, [[a9, c9], [b9]]], [a9, c9, [[a9, b9], [c9]]], [b9, c9, [[a9, b9], [c9]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)","explanation":"Require every distinct-point pair to have disjoint open neighborhoods, retaining first witnesses in supplied topology order. Report Hausdorffness and finite-subset compactness separately; do not drop the Hausdorff hypothesis in the general theorem. The exact result, retaining the requested order, is ([a9, b9, c9], [a9, b9], [[a9, b9, [[a9, c9], [b9]]], [a9, c9, [[a9, b9], [c9]]], [b9, c9, [[a9, b9], [c9]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)."},"skills":["Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0011","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"71e734414cb5912f07f4e1bcf065c25ae2fbb97be46404a29b3733c279112a33","family_id":"qf-mod43-compactclosed-hypotheses-qualification","seed":202611123011,"semantic_template":"qf-mod43-compactclosed-hypotheses-qualification/{\"compact_subset\":[\"a10\",\"b10\"],\"points\":[\"a10\",\"b10\",\"c10\"],\"topology\":[[\"a10\",\"b10\",\"c10\"],[\"a10\",\"b10\"],[\"a10\",\"c10\"],[\"a10\"],[\"b10\",\"c10\"],[\"b10\"],[\"c10\"],[]]}"},"fields":{"title":"Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses","stem":"Use complete finite labeled carrier=[a10, b10, c10], topology=[[a10, b10, c10], [a10, b10], [a10, c10], [a10], [b10, c10], [b10], [c10], []] in its original supplied open-set order, and compact_subset K=[a10, b10]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, pair_separation_rows, Hausdorff, compact, certificate).","answer":{"kind":"expression","value":"([a10, b10, c10], [a10, b10], [[a10, b10, [[a10, c10], [b10]]], [a10, c10, [[a10, b10], [c10]]], [b10, c10, [[a10, b10], [c10]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)","explanation":"Require every distinct-point pair to have disjoint open neighborhoods, retaining first witnesses in supplied topology order. Report Hausdorffness and finite-subset compactness separately; do not drop the Hausdorff hypothesis in the general theorem. The exact result, retaining the requested order, is ([a10, b10, c10], [a10, b10], [[a10, b10, [[a10, c10], [b10]]], [a10, c10, [[a10, b10], [c10]]], [b10, c10, [[a10, b10], [c10]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)."},"skills":["Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0012","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ff0f693523e382d05d564e6f857c0e8c031c8dfa6afdac831d7b837963218df8","family_id":"qf-mod43-compactclosed-hypotheses-qualification","seed":202611123012,"semantic_template":"qf-mod43-compactclosed-hypotheses-qualification/{\"compact_subset\":[\"a11\",\"b11\"],\"points\":[\"a11\",\"b11\",\"c11\"],\"topology\":[[\"a11\",\"b11\",\"c11\"],[\"a11\",\"b11\"],[\"a11\",\"c11\"],[\"a11\"],[\"b11\",\"c11\"],[\"b11\"],[\"c11\"],[]]}"},"fields":{"title":"Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses","stem":"Use complete finite labeled carrier=[a11, b11, c11], topology=[[a11, b11, c11], [a11, b11], [a11, c11], [a11], [b11, c11], [b11], [c11], []] in its original supplied open-set order, and compact_subset K=[a11, b11]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, pair_separation_rows, Hausdorff, compact, certificate).","answer":{"kind":"expression","value":"([a11, b11, c11], [a11, b11], [[a11, b11, [[a11, c11], [b11]]], [a11, c11, [[a11, b11], [c11]]], [b11, c11, [[a11, b11], [c11]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)","explanation":"Require every distinct-point pair to have disjoint open neighborhoods, retaining first witnesses in supplied topology order. Report Hausdorffness and finite-subset compactness separately; do not drop the Hausdorff hypothesis in the general theorem. The exact result, retaining the requested order, is ([a11, b11, c11], [a11, b11], [[a11, b11, [[a11, c11], [b11]]], [a11, c11, [[a11, b11], [c11]]], [b11, c11, [[a11, b11], [c11]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)."},"skills":["Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0013","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"15a7a9147c452563ff8620b3bf9053e0e97dc0a16365c7b7971904ffe8f093f1","family_id":"qf-mod43-compactclosed-hypotheses-qualification","seed":202611123013,"semantic_template":"qf-mod43-compactclosed-hypotheses-qualification/{\"compact_subset\":[\"a12\",\"b12\"],\"points\":[\"a12\",\"b12\",\"c12\"],\"topology\":[[\"a12\",\"b12\",\"c12\"],[\"a12\",\"b12\"],[\"a12\",\"c12\"],[\"a12\"],[\"b12\",\"c12\"],[\"b12\"],[\"c12\"],[]]}"},"fields":{"title":"Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses","stem":"Use complete finite labeled carrier=[a12, b12, c12], topology=[[a12, b12, c12], [a12, b12], [a12, c12], [a12], [b12, c12], [b12], [c12], []] in its original supplied open-set order, and compact_subset K=[a12, b12]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, pair_separation_rows, Hausdorff, compact, certificate).","answer":{"kind":"expression","value":"([a12, b12, c12], [a12, b12], [[a12, b12, [[a12, c12], [b12]]], [a12, c12, [[a12, b12], [c12]]], [b12, c12, [[a12, b12], [c12]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)","explanation":"Require every distinct-point pair to have disjoint open neighborhoods, retaining first witnesses in supplied topology order. Report Hausdorffness and finite-subset compactness separately; do not drop the Hausdorff hypothesis in the general theorem. The exact result, retaining the requested order, is ([a12, b12, c12], [a12, b12], [[a12, b12, [[a12, c12], [b12]]], [a12, c12, [[a12, b12], [c12]]], [b12, c12, [[a12, b12], [c12]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)."},"skills":["Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0014","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2990ed46e5ec2890c07b29240edcd15cb1439b08d9a888d0d62b7f3d30285352","family_id":"qf-mod43-compactclosed-hypotheses-qualification","seed":202611123014,"semantic_template":"qf-mod43-compactclosed-hypotheses-qualification/{\"compact_subset\":[\"a13\",\"b13\"],\"points\":[\"a13\",\"b13\",\"c13\"],\"topology\":[[\"a13\",\"b13\",\"c13\"],[\"a13\",\"b13\"],[\"a13\",\"c13\"],[\"a13\"],[\"b13\",\"c13\"],[\"b13\"],[\"c13\"],[]]}"},"fields":{"title":"Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses","stem":"Use complete finite labeled carrier=[a13, b13, c13], topology=[[a13, b13, c13], [a13, b13], [a13, c13], [a13], [b13, c13], [b13], [c13], []] in its original supplied open-set order, and compact_subset K=[a13, b13]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, pair_separation_rows, Hausdorff, compact, certificate).","answer":{"kind":"expression","value":"([a13, b13, c13], [a13, b13], [[a13, b13, [[a13, c13], [b13]]], [a13, c13, [[a13, b13], [c13]]], [b13, c13, [[a13, b13], [c13]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)","explanation":"Require every distinct-point pair to have disjoint open neighborhoods, retaining first witnesses in supplied topology order. Report Hausdorffness and finite-subset compactness separately; do not drop the Hausdorff hypothesis in the general theorem. The exact result, retaining the requested order, is ([a13, b13, c13], [a13, b13], [[a13, b13, [[a13, c13], [b13]]], [a13, c13, [[a13, b13], [c13]]], [b13, c13, [[a13, b13], [c13]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)."},"skills":["Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0015","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fe6ddbe28d85c27b20abe699884b452afff205f8919cbeee9d0f4590f043aafa","family_id":"qf-mod43-compactclosed-hypotheses-qualification","seed":202611123015,"semantic_template":"qf-mod43-compactclosed-hypotheses-qualification/{\"compact_subset\":[\"a14\",\"b14\"],\"points\":[\"a14\",\"b14\",\"c14\"],\"topology\":[[\"a14\",\"b14\",\"c14\"],[\"a14\",\"b14\"],[\"a14\",\"c14\"],[\"a14\"],[\"b14\",\"c14\"],[\"b14\"],[\"c14\"],[]]}"},"fields":{"title":"Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses","stem":"Use complete finite labeled carrier=[a14, b14, c14], topology=[[a14, b14, c14], [a14, b14], [a14, c14], [a14], [b14, c14], [b14], [c14], []] in its original supplied open-set order, and compact_subset K=[a14, b14]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, pair_separation_rows, Hausdorff, compact, certificate).","answer":{"kind":"expression","value":"([a14, b14, c14], [a14, b14], [[a14, b14, [[a14, c14], [b14]]], [a14, c14, [[a14, b14], [c14]]], [b14, c14, [[a14, b14], [c14]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)","explanation":"Require every distinct-point pair to have disjoint open neighborhoods, retaining first witnesses in supplied topology order. Report Hausdorffness and finite-subset compactness separately; do not drop the Hausdorff hypothesis in the general theorem. The exact result, retaining the requested order, is ([a14, b14, c14], [a14, b14], [[a14, b14, [[a14, c14], [b14]]], [a14, c14, [[a14, b14], [c14]]], [b14, c14, [[a14, b14], [c14]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)."},"skills":["Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0016","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"163afba725ce89fe9a600566279f866037cef5928daa2a0b1e851be5f85041f1","family_id":"qf-mod43-compactclosed-hypotheses-qualification","seed":202611123016,"semantic_template":"qf-mod43-compactclosed-hypotheses-qualification/{\"compact_subset\":[\"a15\",\"b15\"],\"points\":[\"a15\",\"b15\",\"c15\"],\"topology\":[[\"a15\",\"b15\",\"c15\"],[\"a15\",\"b15\"],[\"a15\",\"c15\"],[\"a15\"],[\"b15\",\"c15\"],[\"b15\"],[\"c15\"],[]]}"},"fields":{"title":"Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses","stem":"Use complete finite labeled carrier=[a15, b15, c15], topology=[[a15, b15, c15], [a15, b15], [a15, c15], [a15], [b15, c15], [b15], [c15], []] in its original supplied open-set order, and compact_subset K=[a15, b15]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, pair_separation_rows, Hausdorff, compact, certificate).","answer":{"kind":"expression","value":"([a15, b15, c15], [a15, b15], [[a15, b15, [[a15, c15], [b15]]], [a15, c15, [[a15, b15], [c15]]], [b15, c15, [[a15, b15], [c15]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)","explanation":"Require every distinct-point pair to have disjoint open neighborhoods, retaining first witnesses in supplied topology order. Report Hausdorffness and finite-subset compactness separately; do not drop the Hausdorff hypothesis in the general theorem. The exact result, retaining the requested order, is ([a15, b15, c15], [a15, b15], [[a15, b15, [[a15, c15], [b15]]], [a15, c15, [[a15, b15], [c15]]], [b15, c15, [[a15, b15], [c15]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)."},"skills":["Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0017","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"97fdfd66939789a95cfd3ef784472bfc417f9ccb6a591c280b60f773e3e15b4e","family_id":"qf-mod43-compactclosed-hypotheses-qualification","seed":202611123017,"semantic_template":"qf-mod43-compactclosed-hypotheses-qualification/{\"compact_subset\":[\"a16\",\"b16\"],\"points\":[\"a16\",\"b16\",\"c16\"],\"topology\":[[\"a16\",\"b16\",\"c16\"],[\"a16\",\"b16\"],[\"a16\",\"c16\"],[\"a16\"],[\"b16\",\"c16\"],[\"b16\"],[\"c16\"],[]]}"},"fields":{"title":"Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses","stem":"Use complete finite labeled carrier=[a16, b16, c16], topology=[[a16, b16, c16], [a16, b16], [a16, c16], [a16], [b16, c16], [b16], [c16], []] in its original supplied open-set order, and compact_subset K=[a16, b16]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, pair_separation_rows, Hausdorff, compact, certificate).","answer":{"kind":"expression","value":"([a16, b16, c16], [a16, b16], [[a16, b16, [[a16, c16], [b16]]], [a16, c16, [[a16, b16], [c16]]], [b16, c16, [[a16, b16], [c16]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)","explanation":"Require every distinct-point pair to have disjoint open neighborhoods, retaining first witnesses in supplied topology order. Report Hausdorffness and finite-subset compactness separately; do not drop the Hausdorff hypothesis in the general theorem. The exact result, retaining the requested order, is ([a16, b16, c16], [a16, b16], [[a16, b16, [[a16, c16], [b16]]], [a16, c16, [[a16, b16], [c16]]], [b16, c16, [[a16, b16], [c16]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)."},"skills":["Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0018","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cd2bae90c5345b2aa1a046b697255f1dfc26aaeb423c8bb58dbaa19c5f0c0ccb","family_id":"qf-mod43-compactclosed-hypotheses-qualification","seed":202611123018,"semantic_template":"qf-mod43-compactclosed-hypotheses-qualification/{\"compact_subset\":[\"a17\",\"b17\"],\"points\":[\"a17\",\"b17\",\"c17\"],\"topology\":[[\"a17\",\"b17\",\"c17\"],[\"a17\",\"b17\"],[\"a17\",\"c17\"],[\"a17\"],[\"b17\",\"c17\"],[\"b17\"],[\"c17\"],[]]}"},"fields":{"title":"Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses","stem":"Use complete finite labeled carrier=[a17, b17, c17], topology=[[a17, b17, c17], [a17, b17], [a17, c17], [a17], [b17, c17], [b17], [c17], []] in its original supplied open-set order, and compact_subset K=[a17, b17]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, pair_separation_rows, Hausdorff, compact, certificate).","answer":{"kind":"expression","value":"([a17, b17, c17], [a17, b17], [[a17, b17, [[a17, c17], [b17]]], [a17, c17, [[a17, b17], [c17]]], [b17, c17, [[a17, b17], [c17]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)","explanation":"Require every distinct-point pair to have disjoint open neighborhoods, retaining first witnesses in supplied topology order. Report Hausdorffness and finite-subset compactness separately; do not drop the Hausdorff hypothesis in the general theorem. The exact result, retaining the requested order, is ([a17, b17, c17], [a17, b17], [[a17, b17, [[a17, c17], [b17]]], [a17, c17, [[a17, b17], [c17]]], [b17, c17, [[a17, b17], [c17]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)."},"skills":["Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0019","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9d0839ebfec6ac5cac24a18436b580d567df09a0d9c021c783d04ae2008a53c3","family_id":"qf-mod43-compactclosed-hypotheses-qualification","seed":202611123019,"semantic_template":"qf-mod43-compactclosed-hypotheses-qualification/{\"compact_subset\":[\"a18\",\"b18\"],\"points\":[\"a18\",\"b18\",\"c18\"],\"topology\":[[\"a18\",\"b18\",\"c18\"],[\"a18\",\"b18\"],[\"a18\",\"c18\"],[\"a18\"],[\"b18\",\"c18\"],[\"b18\"],[\"c18\"],[]]}"},"fields":{"title":"Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses","stem":"Use complete finite labeled carrier=[a18, b18, c18], topology=[[a18, b18, c18], [a18, b18], [a18, c18], [a18], [b18, c18], [b18], [c18], []] in its original supplied open-set order, and compact_subset K=[a18, b18]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, pair_separation_rows, Hausdorff, compact, certificate).","answer":{"kind":"expression","value":"([a18, b18, c18], [a18, b18], [[a18, b18, [[a18, c18], [b18]]], [a18, c18, [[a18, b18], [c18]]], [b18, c18, [[a18, b18], [c18]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)","explanation":"Require every distinct-point pair to have disjoint open neighborhoods, retaining first witnesses in supplied topology order. Report Hausdorffness and finite-subset compactness separately; do not drop the Hausdorff hypothesis in the general theorem. The exact result, retaining the requested order, is ([a18, b18, c18], [a18, b18], [[a18, b18, [[a18, c18], [b18]]], [a18, c18, [[a18, b18], [c18]]], [b18, c18, [[a18, b18], [c18]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)."},"skills":["Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0020","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"be89e647ad381c0e32105e4827cb9cac6011363f0848b01ed21fb6656742774e","family_id":"qf-mod43-compactclosed-hypotheses-qualification","seed":202611123020,"semantic_template":"qf-mod43-compactclosed-hypotheses-qualification/{\"compact_subset\":[\"a19\",\"b19\"],\"points\":[\"a19\",\"b19\",\"c19\"],\"topology\":[[\"a19\",\"b19\",\"c19\"],[\"a19\",\"b19\"],[\"a19\",\"c19\"],[\"a19\"],[\"b19\",\"c19\"],[\"b19\"],[\"c19\"],[]]}"},"fields":{"title":"Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses","stem":"Use complete finite labeled carrier=[a19, b19, c19], topology=[[a19, b19, c19], [a19, b19], [a19, c19], [a19], [b19, c19], [b19], [c19], []] in its original supplied open-set order, and compact_subset K=[a19, b19]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, pair_separation_rows, Hausdorff, compact, certificate).","answer":{"kind":"expression","value":"([a19, b19, c19], [a19, b19], [[a19, b19, [[a19, c19], [b19]]], [a19, c19, [[a19, b19], [c19]]], [b19, c19, [[a19, b19], [c19]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)","explanation":"Require every distinct-point pair to have disjoint open neighborhoods, retaining first witnesses in supplied topology order. Report Hausdorffness and finite-subset compactness separately; do not drop the Hausdorff hypothesis in the general theorem. The exact result, retaining the requested order, is ([a19, b19, c19], [a19, b19], [[a19, b19, [[a19, c19], [b19]]], [a19, c19, [[a19, b19], [c19]]], [b19, c19, [[a19, b19], [c19]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)."},"skills":["Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0021","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7ffe51072b1d2a3fb7253b3bae112f87b489d63274c7b298704c85c6048bb26f","family_id":"qf-mod43-compactclosed-hypotheses-qualification","seed":202611123021,"semantic_template":"qf-mod43-compactclosed-hypotheses-qualification/{\"compact_subset\":[\"a20\",\"b20\"],\"points\":[\"a20\",\"b20\",\"c20\"],\"topology\":[[\"a20\",\"b20\",\"c20\"],[\"a20\",\"b20\"],[\"a20\",\"c20\"],[\"a20\"],[\"b20\",\"c20\"],[\"b20\"],[\"c20\"],[]]}"},"fields":{"title":"Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses","stem":"Use complete finite labeled carrier=[a20, b20, c20], topology=[[a20, b20, c20], [a20, b20], [a20, c20], [a20], [b20, c20], [b20], [c20], []] in its original supplied open-set order, and compact_subset K=[a20, b20]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, pair_separation_rows, Hausdorff, compact, certificate).","answer":{"kind":"expression","value":"([a20, b20, c20], [a20, b20], [[a20, b20, [[a20, c20], [b20]]], [a20, c20, [[a20, b20], [c20]]], [b20, c20, [[a20, b20], [c20]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)","explanation":"Require every distinct-point pair to have disjoint open neighborhoods, retaining first witnesses in supplied topology order. Report Hausdorffness and finite-subset compactness separately; do not drop the Hausdorff hypothesis in the general theorem. The exact result, retaining the requested order, is ([a20, b20, c20], [a20, b20], [[a20, b20, [[a20, c20], [b20]]], [a20, c20, [[a20, b20], [c20]]], [b20, c20, [[a20, b20], [c20]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)."},"skills":["Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0022","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b9020c1740f8cff28eda02b511b20720e00a1b05374e94c27b86a6b70fedb2ec","family_id":"qf-mod43-compactclosed-hypotheses-qualification","seed":202611123022,"semantic_template":"qf-mod43-compactclosed-hypotheses-qualification/{\"compact_subset\":[\"a21\",\"b21\"],\"points\":[\"a21\",\"b21\",\"c21\"],\"topology\":[[\"a21\",\"b21\",\"c21\"],[\"a21\",\"b21\"],[\"a21\",\"c21\"],[\"a21\"],[\"b21\",\"c21\"],[\"b21\"],[\"c21\"],[]]}"},"fields":{"title":"Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses","stem":"Use complete finite labeled carrier=[a21, b21, c21], topology=[[a21, b21, c21], [a21, b21], [a21, c21], [a21], [b21, c21], [b21], [c21], []] in its original supplied open-set order, and compact_subset K=[a21, b21]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, pair_separation_rows, Hausdorff, compact, certificate).","answer":{"kind":"expression","value":"([a21, b21, c21], [a21, b21], [[a21, b21, [[a21, c21], [b21]]], [a21, c21, [[a21, b21], [c21]]], [b21, c21, [[a21, b21], [c21]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)","explanation":"Require every distinct-point pair to have disjoint open neighborhoods, retaining first witnesses in supplied topology order. Report Hausdorffness and finite-subset compactness separately; do not drop the Hausdorff hypothesis in the general theorem. The exact result, retaining the requested order, is ([a21, b21, c21], [a21, b21], [[a21, b21, [[a21, c21], [b21]]], [a21, c21, [[a21, b21], [c21]]], [b21, c21, [[a21, b21], [c21]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)."},"skills":["Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0023","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5e3e18c20639b60cb5c5cfa03be5eff22bc92fdc3cef4684caf9bcfa59ab9587","family_id":"qf-mod43-compactclosed-hypotheses-qualification","seed":202611123023,"semantic_template":"qf-mod43-compactclosed-hypotheses-qualification/{\"compact_subset\":[\"a22\",\"b22\"],\"points\":[\"a22\",\"b22\",\"c22\"],\"topology\":[[\"a22\",\"b22\",\"c22\"],[\"a22\",\"b22\"],[\"a22\",\"c22\"],[\"a22\"],[\"b22\",\"c22\"],[\"b22\"],[\"c22\"],[]]}"},"fields":{"title":"Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses","stem":"Use complete finite labeled carrier=[a22, b22, c22], topology=[[a22, b22, c22], [a22, b22], [a22, c22], [a22], [b22, c22], [b22], [c22], []] in its original supplied open-set order, and compact_subset K=[a22, b22]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, pair_separation_rows, Hausdorff, compact, certificate).","answer":{"kind":"expression","value":"([a22, b22, c22], [a22, b22], [[a22, b22, [[a22, c22], [b22]]], [a22, c22, [[a22, b22], [c22]]], [b22, c22, [[a22, b22], [c22]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)","explanation":"Require every distinct-point pair to have disjoint open neighborhoods, retaining first witnesses in supplied topology order. Report Hausdorffness and finite-subset compactness separately; do not drop the Hausdorff hypothesis in the general theorem. The exact result, retaining the requested order, is ([a22, b22, c22], [a22, b22], [[a22, b22, [[a22, c22], [b22]]], [a22, c22, [[a22, b22], [c22]]], [b22, c22, [[a22, b22], [c22]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)."},"skills":["Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0024","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cf88f8dd7fe5453dc53250b4cb02eee09577cf08943558972334323bf80de979","family_id":"qf-mod43-compactclosed-hypotheses-qualification","seed":202611123024,"semantic_template":"qf-mod43-compactclosed-hypotheses-qualification/{\"compact_subset\":[\"a23\",\"b23\"],\"points\":[\"a23\",\"b23\",\"c23\"],\"topology\":[[\"a23\",\"b23\",\"c23\"],[\"a23\",\"b23\"],[\"a23\",\"c23\"],[\"a23\"],[\"b23\",\"c23\"],[\"b23\"],[\"c23\"],[]]}"},"fields":{"title":"Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses","stem":"Use complete finite labeled carrier=[a23, b23, c23], topology=[[a23, b23, c23], [a23, b23], [a23, c23], [a23], [b23, c23], [b23], [c23], []] in its original supplied open-set order, and compact_subset K=[a23, b23]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, pair_separation_rows, Hausdorff, compact, certificate).","answer":{"kind":"expression","value":"([a23, b23, c23], [a23, b23], [[a23, b23, [[a23, c23], [b23]]], [a23, c23, [[a23, b23], [c23]]], [b23, c23, [[a23, b23], [c23]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)","explanation":"Require every distinct-point pair to have disjoint open neighborhoods, retaining first witnesses in supplied topology order. Report Hausdorffness and finite-subset compactness separately; do not drop the Hausdorff hypothesis in the general theorem. The exact result, retaining the requested order, is ([a23, b23, c23], [a23, b23], [[a23, b23, [[a23, c23], [b23]]], [a23, c23, [[a23, b23], [c23]]], [b23, c23, [[a23, b23], [c23]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)."},"skills":["Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0025","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"308989c9821c60ee43f1a1b1aa9f743172d821a016e341e44a9bf61439120eb7","family_id":"qf-mod43-compactclosed-hypotheses-qualification","seed":202611123025,"semantic_template":"qf-mod43-compactclosed-hypotheses-qualification/{\"compact_subset\":[\"a24\",\"b24\"],\"points\":[\"a24\",\"b24\",\"c24\"],\"topology\":[[\"a24\",\"b24\",\"c24\"],[\"a24\",\"b24\"],[\"a24\",\"c24\"],[\"a24\"],[\"b24\",\"c24\"],[\"b24\"],[\"c24\"],[]]}"},"fields":{"title":"Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses","stem":"Use complete finite labeled carrier=[a24, b24, c24], topology=[[a24, b24, c24], [a24, b24], [a24, c24], [a24], [b24, c24], [b24], [c24], []] in its original supplied open-set order, and compact_subset K=[a24, b24]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, pair_separation_rows, Hausdorff, compact, certificate).","answer":{"kind":"expression","value":"([a24, b24, c24], [a24, b24], [[a24, b24, [[a24, c24], [b24]]], [a24, c24, [[a24, b24], [c24]]], [b24, c24, [[a24, b24], [c24]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)","explanation":"Require every distinct-point pair to have disjoint open neighborhoods, retaining first witnesses in supplied topology order. Report Hausdorffness and finite-subset compactness separately; do not drop the Hausdorff hypothesis in the general theorem. The exact result, retaining the requested order, is ([a24, b24, c24], [a24, b24], [[a24, b24, [[a24, c24], [b24]]], [a24, c24, [[a24, b24], [c24]]], [b24, c24, [[a24, b24], [c24]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)."},"skills":["Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0026","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"291c61da57fc0c960f503135fce77dacaf9b5f1d35b39cf8cb8b12a72d683ac8","family_id":"qf-mod43-compactclosed-finite-closedness-audit","seed":202611123026,"semantic_template":"qf-mod43-compactclosed-finite-closedness-audit/{\"compact_subset\":[\"a0\",\"b0\"],\"points\":[\"a0\",\"b0\",\"c0\"],\"topology\":[[\"a0\",\"b0\",\"c0\"],[\"a0\",\"b0\"],[\"a0\",\"c0\"],[\"a0\"],[\"b0\",\"c0\"],[\"b0\"],[\"c0\"],[]]}"},"fields":{"title":"Closedness Uses the Open Complement in the Same Ambient Space","stem":"Use complete finite labeled carrier=[a0, b0, c0], topology=[[a0, b0, c0], [a0, b0], [a0, c0], [a0], [b0, c0], [b0], [c0], []] in its original supplied open-set order, and compact_subset K=[a0, b0]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (compact_subset, complement, compact, closed, certificate).","answer":{"kind":"expression","value":"([a0, b0], [c0], Yes, Yes, complement_is_open_so_compact_subset_is_closed)","explanation":"Require the positive Hausdorff instance and verify that X minus K is open. Empty K is also closed. Finite model verification does not prove the general theorem or its converse closed-implies-compact. The exact result, retaining the requested order, is ([a0, b0], [c0], Yes, Yes, complement_is_open_so_compact_subset_is_closed)."},"skills":["Closedness Uses the Open Complement in the Same Ambient Space"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0027","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"42603182e7c2aaeb60b2779b9d75214e618f514a29dfe3c9e72bf54d2683e45d","family_id":"qf-mod43-compactclosed-finite-closedness-audit","seed":202611123027,"semantic_template":"qf-mod43-compactclosed-finite-closedness-audit/{\"compact_subset\":[\"a1\",\"b1\"],\"points\":[\"a1\",\"b1\",\"c1\"],\"topology\":[[\"a1\",\"b1\",\"c1\"],[\"a1\",\"b1\"],[\"a1\",\"c1\"],[\"a1\"],[\"b1\",\"c1\"],[\"b1\"],[\"c1\"],[]]}"},"fields":{"title":"Closedness Uses the Open Complement in the Same Ambient Space","stem":"Use complete finite labeled carrier=[a1, b1, c1], topology=[[a1, b1, c1], [a1, b1], [a1, c1], [a1], [b1, c1], [b1], [c1], []] in its original supplied open-set order, and compact_subset K=[a1, b1]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (compact_subset, complement, compact, closed, certificate).","answer":{"kind":"expression","value":"([a1, b1], [c1], Yes, Yes, complement_is_open_so_compact_subset_is_closed)","explanation":"Require the positive Hausdorff instance and verify that X minus K is open. Empty K is also closed. Finite model verification does not prove the general theorem or its converse closed-implies-compact. The exact result, retaining the requested order, is ([a1, b1], [c1], Yes, Yes, complement_is_open_so_compact_subset_is_closed)."},"skills":["Closedness Uses the Open Complement in the Same Ambient Space"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0028","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4c5b45d317958e9972a0f6dad664c5ff35ae0332eb5ebabe6c74415354dd9355","family_id":"qf-mod43-compactclosed-finite-closedness-audit","seed":202611123028,"semantic_template":"qf-mod43-compactclosed-finite-closedness-audit/{\"compact_subset\":[\"a2\",\"b2\"],\"points\":[\"a2\",\"b2\",\"c2\"],\"topology\":[[\"a2\",\"b2\",\"c2\"],[\"a2\",\"b2\"],[\"a2\",\"c2\"],[\"a2\"],[\"b2\",\"c2\"],[\"b2\"],[\"c2\"],[]]}"},"fields":{"title":"Closedness Uses the Open Complement in the Same Ambient Space","stem":"Use complete finite labeled carrier=[a2, b2, c2], topology=[[a2, b2, c2], [a2, b2], [a2, c2], [a2], [b2, c2], [b2], [c2], []] in its original supplied open-set order, and compact_subset K=[a2, b2]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (compact_subset, complement, compact, closed, certificate).","answer":{"kind":"expression","value":"([a2, b2], [c2], Yes, Yes, complement_is_open_so_compact_subset_is_closed)","explanation":"Require the positive Hausdorff instance and verify that X minus K is open. Empty K is also closed. Finite model verification does not prove the general theorem or its converse closed-implies-compact. The exact result, retaining the requested order, is ([a2, b2], [c2], Yes, Yes, complement_is_open_so_compact_subset_is_closed)."},"skills":["Closedness Uses the Open Complement in the Same Ambient Space"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0029","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"987817a1c972fef0d834c7cac5029527f8bd134a7737abf1dd75a71156aa0e6e","family_id":"qf-mod43-compactclosed-finite-closedness-audit","seed":202611123029,"semantic_template":"qf-mod43-compactclosed-finite-closedness-audit/{\"compact_subset\":[\"a3\",\"b3\"],\"points\":[\"a3\",\"b3\",\"c3\"],\"topology\":[[\"a3\",\"b3\",\"c3\"],[\"a3\",\"b3\"],[\"a3\",\"c3\"],[\"a3\"],[\"b3\",\"c3\"],[\"b3\"],[\"c3\"],[]]}"},"fields":{"title":"Closedness Uses the Open Complement in the Same Ambient Space","stem":"Use complete finite labeled carrier=[a3, b3, c3], topology=[[a3, b3, c3], [a3, b3], [a3, c3], [a3], [b3, c3], [b3], [c3], []] in its original supplied open-set order, and compact_subset K=[a3, b3]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (compact_subset, complement, compact, closed, certificate).","answer":{"kind":"expression","value":"([a3, b3], [c3], Yes, Yes, complement_is_open_so_compact_subset_is_closed)","explanation":"Require the positive Hausdorff instance and verify that X minus K is open. Empty K is also closed. Finite model verification does not prove the general theorem or its converse closed-implies-compact. The exact result, retaining the requested order, is ([a3, b3], [c3], Yes, Yes, complement_is_open_so_compact_subset_is_closed)."},"skills":["Closedness Uses the Open Complement in the Same Ambient Space"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0030","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3f89926a53333e04baa6568ca72be6e04de7adc55d34f4f8d86cabcdef40f3e7","family_id":"qf-mod43-compactclosed-finite-closedness-audit","seed":202611123030,"semantic_template":"qf-mod43-compactclosed-finite-closedness-audit/{\"compact_subset\":[\"a4\",\"b4\"],\"points\":[\"a4\",\"b4\",\"c4\"],\"topology\":[[\"a4\",\"b4\",\"c4\"],[\"a4\",\"b4\"],[\"a4\",\"c4\"],[\"a4\"],[\"b4\",\"c4\"],[\"b4\"],[\"c4\"],[]]}"},"fields":{"title":"Closedness Uses the Open Complement in the Same Ambient Space","stem":"Use complete finite labeled carrier=[a4, b4, c4], topology=[[a4, b4, c4], [a4, b4], [a4, c4], [a4], [b4, c4], [b4], [c4], []] in its original supplied open-set order, and compact_subset K=[a4, b4]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (compact_subset, complement, compact, closed, certificate).","answer":{"kind":"expression","value":"([a4, b4], [c4], Yes, Yes, complement_is_open_so_compact_subset_is_closed)","explanation":"Require the positive Hausdorff instance and verify that X minus K is open. Empty K is also closed. Finite model verification does not prove the general theorem or its converse closed-implies-compact. The exact result, retaining the requested order, is ([a4, b4], [c4], Yes, Yes, complement_is_open_so_compact_subset_is_closed)."},"skills":["Closedness Uses the Open Complement in the Same Ambient Space"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0031","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"622a939743115fdc18a8ff05e5a988ff96000d106ac85feb84320c04c2046180","family_id":"qf-mod43-compactclosed-finite-closedness-audit","seed":202611123031,"semantic_template":"qf-mod43-compactclosed-finite-closedness-audit/{\"compact_subset\":[\"a5\",\"b5\"],\"points\":[\"a5\",\"b5\",\"c5\"],\"topology\":[[\"a5\",\"b5\",\"c5\"],[\"a5\",\"b5\"],[\"a5\",\"c5\"],[\"a5\"],[\"b5\",\"c5\"],[\"b5\"],[\"c5\"],[]]}"},"fields":{"title":"Closedness Uses the Open Complement in the Same Ambient Space","stem":"Use complete finite labeled carrier=[a5, b5, c5], topology=[[a5, b5, c5], [a5, b5], [a5, c5], [a5], [b5, c5], [b5], [c5], []] in its original supplied open-set order, and compact_subset K=[a5, b5]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (compact_subset, complement, compact, closed, certificate).","answer":{"kind":"expression","value":"([a5, b5], [c5], Yes, Yes, complement_is_open_so_compact_subset_is_closed)","explanation":"Require the positive Hausdorff instance and verify that X minus K is open. Empty K is also closed. Finite model verification does not prove the general theorem or its converse closed-implies-compact. The exact result, retaining the requested order, is ([a5, b5], [c5], Yes, Yes, complement_is_open_so_compact_subset_is_closed)."},"skills":["Closedness Uses the Open Complement in the Same Ambient Space"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0032","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"94caf5df81eabe5bc482a768cc334b180dc7880b91b7d3bd61ccfe37d8479e18","family_id":"qf-mod43-compactclosed-finite-closedness-audit","seed":202611123032,"semantic_template":"qf-mod43-compactclosed-finite-closedness-audit/{\"compact_subset\":[\"a6\",\"b6\"],\"points\":[\"a6\",\"b6\",\"c6\"],\"topology\":[[\"a6\",\"b6\",\"c6\"],[\"a6\",\"b6\"],[\"a6\",\"c6\"],[\"a6\"],[\"b6\",\"c6\"],[\"b6\"],[\"c6\"],[]]}"},"fields":{"title":"Closedness Uses the Open Complement in the Same Ambient Space","stem":"Use complete finite labeled carrier=[a6, b6, c6], topology=[[a6, b6, c6], [a6, b6], [a6, c6], [a6], [b6, c6], [b6], [c6], []] in its original supplied open-set order, and compact_subset K=[a6, b6]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (compact_subset, complement, compact, closed, certificate).","answer":{"kind":"expression","value":"([a6, b6], [c6], Yes, Yes, complement_is_open_so_compact_subset_is_closed)","explanation":"Require the positive Hausdorff instance and verify that X minus K is open. Empty K is also closed. Finite model verification does not prove the general theorem or its converse closed-implies-compact. The exact result, retaining the requested order, is ([a6, b6], [c6], Yes, Yes, complement_is_open_so_compact_subset_is_closed)."},"skills":["Closedness Uses the Open Complement in the Same Ambient Space"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0033","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"81d20058b5f179d81e4c45a8f713e5629c80a709b7aac7d24dfcbc7ec7e88ee6","family_id":"qf-mod43-compactclosed-finite-closedness-audit","seed":202611123033,"semantic_template":"qf-mod43-compactclosed-finite-closedness-audit/{\"compact_subset\":[\"a7\",\"b7\"],\"points\":[\"a7\",\"b7\",\"c7\"],\"topology\":[[\"a7\",\"b7\",\"c7\"],[\"a7\",\"b7\"],[\"a7\",\"c7\"],[\"a7\"],[\"b7\",\"c7\"],[\"b7\"],[\"c7\"],[]]}"},"fields":{"title":"Closedness Uses the Open Complement in the Same Ambient Space","stem":"Use complete finite labeled carrier=[a7, b7, c7], topology=[[a7, b7, c7], [a7, b7], [a7, c7], [a7], [b7, c7], [b7], [c7], []] in its original supplied open-set order, and compact_subset K=[a7, b7]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (compact_subset, complement, compact, closed, certificate).","answer":{"kind":"expression","value":"([a7, b7], [c7], Yes, Yes, complement_is_open_so_compact_subset_is_closed)","explanation":"Require the positive Hausdorff instance and verify that X minus K is open. Empty K is also closed. Finite model verification does not prove the general theorem or its converse closed-implies-compact. The exact result, retaining the requested order, is ([a7, b7], [c7], Yes, Yes, complement_is_open_so_compact_subset_is_closed)."},"skills":["Closedness Uses the Open Complement in the Same Ambient Space"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0034","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"50e0d17e51f7a0317905e16106d14bfa9e81a7a1da871ebaf248c8d4a5a0cc78","family_id":"qf-mod43-compactclosed-finite-closedness-audit","seed":202611123034,"semantic_template":"qf-mod43-compactclosed-finite-closedness-audit/{\"compact_subset\":[\"a8\",\"b8\"],\"points\":[\"a8\",\"b8\",\"c8\"],\"topology\":[[\"a8\",\"b8\",\"c8\"],[\"a8\",\"b8\"],[\"a8\",\"c8\"],[\"a8\"],[\"b8\",\"c8\"],[\"b8\"],[\"c8\"],[]]}"},"fields":{"title":"Closedness Uses the Open Complement in the Same Ambient Space","stem":"Use complete finite labeled carrier=[a8, b8, c8], topology=[[a8, b8, c8], [a8, b8], [a8, c8], [a8], [b8, c8], [b8], [c8], []] in its original supplied open-set order, and compact_subset K=[a8, b8]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (compact_subset, complement, compact, closed, certificate).","answer":{"kind":"expression","value":"([a8, b8], [c8], Yes, Yes, complement_is_open_so_compact_subset_is_closed)","explanation":"Require the positive Hausdorff instance and verify that X minus K is open. Empty K is also closed. Finite model verification does not prove the general theorem or its converse closed-implies-compact. The exact result, retaining the requested order, is ([a8, b8], [c8], Yes, Yes, complement_is_open_so_compact_subset_is_closed)."},"skills":["Closedness Uses the Open Complement in the Same Ambient Space"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0035","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"42a753177b7b87ab474969aa6c845445685f739e3d13fdfa8bfb6c630770cec5","family_id":"qf-mod43-compactclosed-finite-closedness-audit","seed":202611123035,"semantic_template":"qf-mod43-compactclosed-finite-closedness-audit/{\"compact_subset\":[\"a9\",\"b9\"],\"points\":[\"a9\",\"b9\",\"c9\"],\"topology\":[[\"a9\",\"b9\",\"c9\"],[\"a9\",\"b9\"],[\"a9\",\"c9\"],[\"a9\"],[\"b9\",\"c9\"],[\"b9\"],[\"c9\"],[]]}"},"fields":{"title":"Closedness Uses the Open Complement in the Same Ambient Space","stem":"Use complete finite labeled carrier=[a9, b9, c9], topology=[[a9, b9, c9], [a9, b9], [a9, c9], [a9], [b9, c9], [b9], [c9], []] in its original supplied open-set order, and compact_subset K=[a9, b9]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (compact_subset, complement, compact, closed, certificate).","answer":{"kind":"expression","value":"([a9, b9], [c9], Yes, Yes, complement_is_open_so_compact_subset_is_closed)","explanation":"Require the positive Hausdorff instance and verify that X minus K is open. Empty K is also closed. Finite model verification does not prove the general theorem or its converse closed-implies-compact. The exact result, retaining the requested order, is ([a9, b9], [c9], Yes, Yes, complement_is_open_so_compact_subset_is_closed)."},"skills":["Closedness Uses the Open Complement in the Same Ambient Space"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0036","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"67af22be52cf8f2d207b95e73d2ce7bd6b4d345deb5793f397f6c8944acac5a4","family_id":"qf-mod43-compactclosed-finite-closedness-audit","seed":202611123036,"semantic_template":"qf-mod43-compactclosed-finite-closedness-audit/{\"compact_subset\":[\"a10\",\"b10\"],\"points\":[\"a10\",\"b10\",\"c10\"],\"topology\":[[\"a10\",\"b10\",\"c10\"],[\"a10\",\"b10\"],[\"a10\",\"c10\"],[\"a10\"],[\"b10\",\"c10\"],[\"b10\"],[\"c10\"],[]]}"},"fields":{"title":"Closedness Uses the Open Complement in the Same Ambient Space","stem":"Use complete finite labeled carrier=[a10, b10, c10], topology=[[a10, b10, c10], [a10, b10], [a10, c10], [a10], [b10, c10], [b10], [c10], []] in its original supplied open-set order, and compact_subset K=[a10, b10]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (compact_subset, complement, compact, closed, certificate).","answer":{"kind":"expression","value":"([a10, b10], [c10], Yes, Yes, complement_is_open_so_compact_subset_is_closed)","explanation":"Require the positive Hausdorff instance and verify that X minus K is open. Empty K is also closed. Finite model verification does not prove the general theorem or its converse closed-implies-compact. The exact result, retaining the requested order, is ([a10, b10], [c10], Yes, Yes, complement_is_open_so_compact_subset_is_closed)."},"skills":["Closedness Uses the Open Complement in the Same Ambient Space"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0037","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b89c7c1babb8d88809c7672bb0d941b277ff562784e30f6a83227018b88553c0","family_id":"qf-mod43-compactclosed-finite-closedness-audit","seed":202611123037,"semantic_template":"qf-mod43-compactclosed-finite-closedness-audit/{\"compact_subset\":[\"a11\",\"b11\"],\"points\":[\"a11\",\"b11\",\"c11\"],\"topology\":[[\"a11\",\"b11\",\"c11\"],[\"a11\",\"b11\"],[\"a11\",\"c11\"],[\"a11\"],[\"b11\",\"c11\"],[\"b11\"],[\"c11\"],[]]}"},"fields":{"title":"Closedness Uses the Open Complement in the Same Ambient Space","stem":"Use complete finite labeled carrier=[a11, b11, c11], topology=[[a11, b11, c11], [a11, b11], [a11, c11], [a11], [b11, c11], [b11], [c11], []] in its original supplied open-set order, and compact_subset K=[a11, b11]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (compact_subset, complement, compact, closed, certificate).","answer":{"kind":"expression","value":"([a11, b11], [c11], Yes, Yes, complement_is_open_so_compact_subset_is_closed)","explanation":"Require the positive Hausdorff instance and verify that X minus K is open. Empty K is also closed. Finite model verification does not prove the general theorem or its converse closed-implies-compact. The exact result, retaining the requested order, is ([a11, b11], [c11], Yes, Yes, complement_is_open_so_compact_subset_is_closed)."},"skills":["Closedness Uses the Open Complement in the Same Ambient Space"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0038","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0b3fc4b183e7b8fc4952b52c9df67046662837376f0fc1c2301738dd9f93bc60","family_id":"qf-mod43-compactclosed-finite-closedness-audit","seed":202611123038,"semantic_template":"qf-mod43-compactclosed-finite-closedness-audit/{\"compact_subset\":[\"a12\",\"b12\"],\"points\":[\"a12\",\"b12\",\"c12\"],\"topology\":[[\"a12\",\"b12\",\"c12\"],[\"a12\",\"b12\"],[\"a12\",\"c12\"],[\"a12\"],[\"b12\",\"c12\"],[\"b12\"],[\"c12\"],[]]}"},"fields":{"title":"Closedness Uses the Open Complement in the Same Ambient Space","stem":"Use complete finite labeled carrier=[a12, b12, c12], topology=[[a12, b12, c12], [a12, b12], [a12, c12], [a12], [b12, c12], [b12], [c12], []] in its original supplied open-set order, and compact_subset K=[a12, b12]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (compact_subset, complement, compact, closed, certificate).","answer":{"kind":"expression","value":"([a12, b12], [c12], Yes, Yes, complement_is_open_so_compact_subset_is_closed)","explanation":"Require the positive Hausdorff instance and verify that X minus K is open. Empty K is also closed. Finite model verification does not prove the general theorem or its converse closed-implies-compact. The exact result, retaining the requested order, is ([a12, b12], [c12], Yes, Yes, complement_is_open_so_compact_subset_is_closed)."},"skills":["Closedness Uses the Open Complement in the Same Ambient Space"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0039","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9202183b34bd88a8691afd157e21dbb94a1e11bcbe40bc8665bde54d22293f33","family_id":"qf-mod43-compactclosed-finite-closedness-audit","seed":202611123039,"semantic_template":"qf-mod43-compactclosed-finite-closedness-audit/{\"compact_subset\":[\"a13\",\"b13\"],\"points\":[\"a13\",\"b13\",\"c13\"],\"topology\":[[\"a13\",\"b13\",\"c13\"],[\"a13\",\"b13\"],[\"a13\",\"c13\"],[\"a13\"],[\"b13\",\"c13\"],[\"b13\"],[\"c13\"],[]]}"},"fields":{"title":"Closedness Uses the Open Complement in the Same Ambient Space","stem":"Use complete finite labeled carrier=[a13, b13, c13], topology=[[a13, b13, c13], [a13, b13], [a13, c13], [a13], [b13, c13], [b13], [c13], []] in its original supplied open-set order, and compact_subset K=[a13, b13]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (compact_subset, complement, compact, closed, certificate).","answer":{"kind":"expression","value":"([a13, b13], [c13], Yes, Yes, complement_is_open_so_compact_subset_is_closed)","explanation":"Require the positive Hausdorff instance and verify that X minus K is open. Empty K is also closed. Finite model verification does not prove the general theorem or its converse closed-implies-compact. The exact result, retaining the requested order, is ([a13, b13], [c13], Yes, Yes, complement_is_open_so_compact_subset_is_closed)."},"skills":["Closedness Uses the Open Complement in the Same Ambient Space"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0040","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c80a9b582f4af603bc6f598e68b4cbf7b522d5afb88472831d14cbca85c2d0e7","family_id":"qf-mod43-compactclosed-finite-closedness-audit","seed":202611123040,"semantic_template":"qf-mod43-compactclosed-finite-closedness-audit/{\"compact_subset\":[\"a14\",\"b14\"],\"points\":[\"a14\",\"b14\",\"c14\"],\"topology\":[[\"a14\",\"b14\",\"c14\"],[\"a14\",\"b14\"],[\"a14\",\"c14\"],[\"a14\"],[\"b14\",\"c14\"],[\"b14\"],[\"c14\"],[]]}"},"fields":{"title":"Closedness Uses the Open Complement in the Same Ambient Space","stem":"Use complete finite labeled carrier=[a14, b14, c14], topology=[[a14, b14, c14], [a14, b14], [a14, c14], [a14], [b14, c14], [b14], [c14], []] in its original supplied open-set order, and compact_subset K=[a14, b14]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (compact_subset, complement, compact, closed, certificate).","answer":{"kind":"expression","value":"([a14, b14], [c14], Yes, Yes, complement_is_open_so_compact_subset_is_closed)","explanation":"Require the positive Hausdorff instance and verify that X minus K is open. Empty K is also closed. Finite model verification does not prove the general theorem or its converse closed-implies-compact. The exact result, retaining the requested order, is ([a14, b14], [c14], Yes, Yes, complement_is_open_so_compact_subset_is_closed)."},"skills":["Closedness Uses the Open Complement in the Same Ambient Space"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0041","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cb8f4227f71adebc58a88da269f1c83dd40ad3c25eb385984ed30f4867ae008d","family_id":"qf-mod43-compactclosed-finite-closedness-audit","seed":202611123041,"semantic_template":"qf-mod43-compactclosed-finite-closedness-audit/{\"compact_subset\":[\"a15\",\"b15\"],\"points\":[\"a15\",\"b15\",\"c15\"],\"topology\":[[\"a15\",\"b15\",\"c15\"],[\"a15\",\"b15\"],[\"a15\",\"c15\"],[\"a15\"],[\"b15\",\"c15\"],[\"b15\"],[\"c15\"],[]]}"},"fields":{"title":"Closedness Uses the Open Complement in the Same Ambient Space","stem":"Use complete finite labeled carrier=[a15, b15, c15], topology=[[a15, b15, c15], [a15, b15], [a15, c15], [a15], [b15, c15], [b15], [c15], []] in its original supplied open-set order, and compact_subset K=[a15, b15]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (compact_subset, complement, compact, closed, certificate).","answer":{"kind":"expression","value":"([a15, b15], [c15], Yes, Yes, complement_is_open_so_compact_subset_is_closed)","explanation":"Require the positive Hausdorff instance and verify that X minus K is open. Empty K is also closed. Finite model verification does not prove the general theorem or its converse closed-implies-compact. The exact result, retaining the requested order, is ([a15, b15], [c15], Yes, Yes, complement_is_open_so_compact_subset_is_closed)."},"skills":["Closedness Uses the Open Complement in the Same Ambient Space"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0042","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"966899963404f213cd4092cc59938cb918c8f5ef5836051b34afc51b3371bb6f","family_id":"qf-mod43-compactclosed-finite-closedness-audit","seed":202611123042,"semantic_template":"qf-mod43-compactclosed-finite-closedness-audit/{\"compact_subset\":[\"a16\",\"b16\"],\"points\":[\"a16\",\"b16\",\"c16\"],\"topology\":[[\"a16\",\"b16\",\"c16\"],[\"a16\",\"b16\"],[\"a16\",\"c16\"],[\"a16\"],[\"b16\",\"c16\"],[\"b16\"],[\"c16\"],[]]}"},"fields":{"title":"Closedness Uses the Open Complement in the Same Ambient Space","stem":"Use complete finite labeled carrier=[a16, b16, c16], topology=[[a16, b16, c16], [a16, b16], [a16, c16], [a16], [b16, c16], [b16], [c16], []] in its original supplied open-set order, and compact_subset K=[a16, b16]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (compact_subset, complement, compact, closed, certificate).","answer":{"kind":"expression","value":"([a16, b16], [c16], Yes, Yes, complement_is_open_so_compact_subset_is_closed)","explanation":"Require the positive Hausdorff instance and verify that X minus K is open. Empty K is also closed. Finite model verification does not prove the general theorem or its converse closed-implies-compact. The exact result, retaining the requested order, is ([a16, b16], [c16], Yes, Yes, complement_is_open_so_compact_subset_is_closed)."},"skills":["Closedness Uses the Open Complement in the Same Ambient Space"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0043","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0a8760b36f3278fb579d9fd71e9e30a93df732a74934319cf5db918acaae9103","family_id":"qf-mod43-compactclosed-finite-closedness-audit","seed":202611123043,"semantic_template":"qf-mod43-compactclosed-finite-closedness-audit/{\"compact_subset\":[\"a17\",\"b17\"],\"points\":[\"a17\",\"b17\",\"c17\"],\"topology\":[[\"a17\",\"b17\",\"c17\"],[\"a17\",\"b17\"],[\"a17\",\"c17\"],[\"a17\"],[\"b17\",\"c17\"],[\"b17\"],[\"c17\"],[]]}"},"fields":{"title":"Closedness Uses the Open Complement in the Same Ambient Space","stem":"Use complete finite labeled carrier=[a17, b17, c17], topology=[[a17, b17, c17], [a17, b17], [a17, c17], [a17], [b17, c17], [b17], [c17], []] in its original supplied open-set order, and compact_subset K=[a17, b17]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (compact_subset, complement, compact, closed, certificate).","answer":{"kind":"expression","value":"([a17, b17], [c17], Yes, Yes, complement_is_open_so_compact_subset_is_closed)","explanation":"Require the positive Hausdorff instance and verify that X minus K is open. Empty K is also closed. Finite model verification does not prove the general theorem or its converse closed-implies-compact. The exact result, retaining the requested order, is ([a17, b17], [c17], Yes, Yes, complement_is_open_so_compact_subset_is_closed)."},"skills":["Closedness Uses the Open Complement in the Same Ambient Space"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0044","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b87632fc5e47ddde5887efab78698a1f3ec44e83fdba9bed44bc9c9e5b4017ba","family_id":"qf-mod43-compactclosed-finite-closedness-audit","seed":202611123044,"semantic_template":"qf-mod43-compactclosed-finite-closedness-audit/{\"compact_subset\":[\"a18\",\"b18\"],\"points\":[\"a18\",\"b18\",\"c18\"],\"topology\":[[\"a18\",\"b18\",\"c18\"],[\"a18\",\"b18\"],[\"a18\",\"c18\"],[\"a18\"],[\"b18\",\"c18\"],[\"b18\"],[\"c18\"],[]]}"},"fields":{"title":"Closedness Uses the Open Complement in the Same Ambient Space","stem":"Use complete finite labeled carrier=[a18, b18, c18], topology=[[a18, b18, c18], [a18, b18], [a18, c18], [a18], [b18, c18], [b18], [c18], []] in its original supplied open-set order, and compact_subset K=[a18, b18]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (compact_subset, complement, compact, closed, certificate).","answer":{"kind":"expression","value":"([a18, b18], [c18], Yes, Yes, complement_is_open_so_compact_subset_is_closed)","explanation":"Require the positive Hausdorff instance and verify that X minus K is open. Empty K is also closed. Finite model verification does not prove the general theorem or its converse closed-implies-compact. The exact result, retaining the requested order, is ([a18, b18], [c18], Yes, Yes, complement_is_open_so_compact_subset_is_closed)."},"skills":["Closedness Uses the Open Complement in the Same Ambient Space"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0045","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"82749f3ae3a5da70e32ae7c07eae39aa5e33d8ce5352c60d0e194cea4e18ca4b","family_id":"qf-mod43-compactclosed-finite-closedness-audit","seed":202611123045,"semantic_template":"qf-mod43-compactclosed-finite-closedness-audit/{\"compact_subset\":[\"a19\",\"b19\"],\"points\":[\"a19\",\"b19\",\"c19\"],\"topology\":[[\"a19\",\"b19\",\"c19\"],[\"a19\",\"b19\"],[\"a19\",\"c19\"],[\"a19\"],[\"b19\",\"c19\"],[\"b19\"],[\"c19\"],[]]}"},"fields":{"title":"Closedness Uses the Open Complement in the Same Ambient Space","stem":"Use complete finite labeled carrier=[a19, b19, c19], topology=[[a19, b19, c19], [a19, b19], [a19, c19], [a19], [b19, c19], [b19], [c19], []] in its original supplied open-set order, and compact_subset K=[a19, b19]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (compact_subset, complement, compact, closed, certificate).","answer":{"kind":"expression","value":"([a19, b19], [c19], Yes, Yes, complement_is_open_so_compact_subset_is_closed)","explanation":"Require the positive Hausdorff instance and verify that X minus K is open. Empty K is also closed. Finite model verification does not prove the general theorem or its converse closed-implies-compact. The exact result, retaining the requested order, is ([a19, b19], [c19], Yes, Yes, complement_is_open_so_compact_subset_is_closed)."},"skills":["Closedness Uses the Open Complement in the Same Ambient Space"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0046","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cb12cef5838c87d00ae3aece7e9475dd489eb0c20fd8f1999f22aec8d4872d4f","family_id":"qf-mod43-compactclosed-finite-closedness-audit","seed":202611123046,"semantic_template":"qf-mod43-compactclosed-finite-closedness-audit/{\"compact_subset\":[\"a20\",\"b20\"],\"points\":[\"a20\",\"b20\",\"c20\"],\"topology\":[[\"a20\",\"b20\",\"c20\"],[\"a20\",\"b20\"],[\"a20\",\"c20\"],[\"a20\"],[\"b20\",\"c20\"],[\"b20\"],[\"c20\"],[]]}"},"fields":{"title":"Closedness Uses the Open Complement in the Same Ambient Space","stem":"Use complete finite labeled carrier=[a20, b20, c20], topology=[[a20, b20, c20], [a20, b20], [a20, c20], [a20], [b20, c20], [b20], [c20], []] in its original supplied open-set order, and compact_subset K=[a20, b20]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (compact_subset, complement, compact, closed, certificate).","answer":{"kind":"expression","value":"([a20, b20], [c20], Yes, Yes, complement_is_open_so_compact_subset_is_closed)","explanation":"Require the positive Hausdorff instance and verify that X minus K is open. Empty K is also closed. Finite model verification does not prove the general theorem or its converse closed-implies-compact. The exact result, retaining the requested order, is ([a20, b20], [c20], Yes, Yes, complement_is_open_so_compact_subset_is_closed)."},"skills":["Closedness Uses the Open Complement in the Same Ambient Space"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0047","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fa7c5f808ba61efbaa8ff4f804266a6bffcf99f1d3e361340a1ef64c62cfd999","family_id":"qf-mod43-compactclosed-finite-closedness-audit","seed":202611123047,"semantic_template":"qf-mod43-compactclosed-finite-closedness-audit/{\"compact_subset\":[\"a21\",\"b21\"],\"points\":[\"a21\",\"b21\",\"c21\"],\"topology\":[[\"a21\",\"b21\",\"c21\"],[\"a21\",\"b21\"],[\"a21\",\"c21\"],[\"a21\"],[\"b21\",\"c21\"],[\"b21\"],[\"c21\"],[]]}"},"fields":{"title":"Closedness Uses the Open Complement in the Same Ambient Space","stem":"Use complete finite labeled carrier=[a21, b21, c21], topology=[[a21, b21, c21], [a21, b21], [a21, c21], [a21], [b21, c21], [b21], [c21], []] in its original supplied open-set order, and compact_subset K=[a21, b21]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (compact_subset, complement, compact, closed, certificate).","answer":{"kind":"expression","value":"([a21, b21], [c21], Yes, Yes, complement_is_open_so_compact_subset_is_closed)","explanation":"Require the positive Hausdorff instance and verify that X minus K is open. Empty K is also closed. Finite model verification does not prove the general theorem or its converse closed-implies-compact. The exact result, retaining the requested order, is ([a21, b21], [c21], Yes, Yes, complement_is_open_so_compact_subset_is_closed)."},"skills":["Closedness Uses the Open Complement in the Same Ambient Space"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0048","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"81ba5a80ae41e78fb918401cda5aff4bdb49c445135d02698df97cfb6b7a95ca","family_id":"qf-mod43-compactclosed-finite-closedness-audit","seed":202611123048,"semantic_template":"qf-mod43-compactclosed-finite-closedness-audit/{\"compact_subset\":[\"a22\",\"b22\"],\"points\":[\"a22\",\"b22\",\"c22\"],\"topology\":[[\"a22\",\"b22\",\"c22\"],[\"a22\",\"b22\"],[\"a22\",\"c22\"],[\"a22\"],[\"b22\",\"c22\"],[\"b22\"],[\"c22\"],[]]}"},"fields":{"title":"Closedness Uses the Open Complement in the Same Ambient Space","stem":"Use complete finite labeled carrier=[a22, b22, c22], topology=[[a22, b22, c22], [a22, b22], [a22, c22], [a22], [b22, c22], [b22], [c22], []] in its original supplied open-set order, and compact_subset K=[a22, b22]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (compact_subset, complement, compact, closed, certificate).","answer":{"kind":"expression","value":"([a22, b22], [c22], Yes, Yes, complement_is_open_so_compact_subset_is_closed)","explanation":"Require the positive Hausdorff instance and verify that X minus K is open. Empty K is also closed. Finite model verification does not prove the general theorem or its converse closed-implies-compact. The exact result, retaining the requested order, is ([a22, b22], [c22], Yes, Yes, complement_is_open_so_compact_subset_is_closed)."},"skills":["Closedness Uses the Open Complement in the Same Ambient Space"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0049","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e933946c53c48a77ed5b8fe02915e99042941ea3f88bf2a67087a6d824e0a357","family_id":"qf-mod43-compactclosed-finite-closedness-audit","seed":202611123049,"semantic_template":"qf-mod43-compactclosed-finite-closedness-audit/{\"compact_subset\":[\"a23\",\"b23\"],\"points\":[\"a23\",\"b23\",\"c23\"],\"topology\":[[\"a23\",\"b23\",\"c23\"],[\"a23\",\"b23\"],[\"a23\",\"c23\"],[\"a23\"],[\"b23\",\"c23\"],[\"b23\"],[\"c23\"],[]]}"},"fields":{"title":"Closedness Uses the Open Complement in the Same Ambient Space","stem":"Use complete finite labeled carrier=[a23, b23, c23], topology=[[a23, b23, c23], [a23, b23], [a23, c23], [a23], [b23, c23], [b23], [c23], []] in its original supplied open-set order, and compact_subset K=[a23, b23]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (compact_subset, complement, compact, closed, certificate).","answer":{"kind":"expression","value":"([a23, b23], [c23], Yes, Yes, complement_is_open_so_compact_subset_is_closed)","explanation":"Require the positive Hausdorff instance and verify that X minus K is open. Empty K is also closed. Finite model verification does not prove the general theorem or its converse closed-implies-compact. The exact result, retaining the requested order, is ([a23, b23], [c23], Yes, Yes, complement_is_open_so_compact_subset_is_closed)."},"skills":["Closedness Uses the Open Complement in the Same Ambient Space"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0050","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2bd44bbb1681c60d209ce308155b7568d9fd815f9186139538584960d597fb46","family_id":"qf-mod43-compactclosed-finite-closedness-audit","seed":202611123050,"semantic_template":"qf-mod43-compactclosed-finite-closedness-audit/{\"compact_subset\":[\"a24\",\"b24\"],\"points\":[\"a24\",\"b24\",\"c24\"],\"topology\":[[\"a24\",\"b24\",\"c24\"],[\"a24\",\"b24\"],[\"a24\",\"c24\"],[\"a24\"],[\"b24\",\"c24\"],[\"b24\"],[\"c24\"],[]]}"},"fields":{"title":"Closedness Uses the Open Complement in the Same Ambient Space","stem":"Use complete finite labeled carrier=[a24, b24, c24], topology=[[a24, b24, c24], [a24, b24], [a24, c24], [a24], [b24, c24], [b24], [c24], []] in its original supplied open-set order, and compact_subset K=[a24, b24]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (compact_subset, complement, compact, closed, certificate).","answer":{"kind":"expression","value":"([a24, b24], [c24], Yes, Yes, complement_is_open_so_compact_subset_is_closed)","explanation":"Require the positive Hausdorff instance and verify that X minus K is open. Empty K is also closed. Finite model verification does not prove the general theorem or its converse closed-implies-compact. The exact result, retaining the requested order, is ([a24, b24], [c24], Yes, Yes, complement_is_open_so_compact_subset_is_closed)."},"skills":["Closedness Uses the Open Complement in the Same Ambient Space"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0051","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2b648389f92e68ab3a2a0854973fbfe70fc789fe70e87d45c9dbee33abdb0067","family_id":"qf-mod43-compactclosed-neighborhood-subcover-certificate","seed":202611123051,"semantic_template":"qf-mod43-compactclosed-neighborhood-subcover-certificate/{\"compact_subset\":[\"a0\",\"b0\"],\"exterior_point\":\"c0\",\"points\":[\"a0\",\"b0\",\"c0\"],\"topology\":[[\"a0\",\"b0\",\"c0\"],[\"a0\",\"b0\"],[\"a0\",\"c0\"],[\"a0\"],[\"b0\",\"c0\"],[\"b0\"],[\"c0\"],[]]}"},"fields":{"title":"Finite Intersections Separate an Exterior Point from the Whole Compact Subset","stem":"Use complete finite labeled carrier=[a0, b0, c0], topology=[[a0, b0, c0], [a0, b0], [a0, c0], [a0], [b0, c0], [b0], [c0], []] in its original supplied open-set order, and compact_subset K=[a0, b0]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Exterior point x=c0 must lie outside K. Give (exterior_point, separation_rows, exterior_neighborhood, cover_union, covers_K, neighborhood_disjoint_from_K, neighborhood_open, certificate).","answer":{"kind":"expression","value":"(c0, [[a0, [[b0, c0], [a0]]], [b0, [[a0, c0], [b0]]]], [c0], [a0, b0], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)","explanation":"For each y∈K choose disjoint U_y containing x and V_y containing y in original open-list order. The V_y union covers K but may exceed K; intersect the finitely many U_y to obtain an open neighborhood disjoint from K. For empty K the empty intersection is X. No minimum or unique subcover is asserted; the general proof uses compactness to obtain a finite subcover. The exact result, retaining the requested order, is (c0, [[a0, [[b0, c0], [a0]]], [b0, [[a0, c0], [b0]]]], [c0], [a0, b0], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)."},"skills":["Finite Intersections Separate an Exterior Point from the Whole Compact Subset"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0052","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7b750b4be50b14270edd31ab4c888689009cbf293f307af6254f49b3110a16e8","family_id":"qf-mod43-compactclosed-neighborhood-subcover-certificate","seed":202611123052,"semantic_template":"qf-mod43-compactclosed-neighborhood-subcover-certificate/{\"compact_subset\":[\"a1\",\"b1\"],\"exterior_point\":\"c1\",\"points\":[\"a1\",\"b1\",\"c1\"],\"topology\":[[\"a1\",\"b1\",\"c1\"],[\"a1\",\"b1\"],[\"a1\",\"c1\"],[\"a1\"],[\"b1\",\"c1\"],[\"b1\"],[\"c1\"],[]]}"},"fields":{"title":"Finite Intersections Separate an Exterior Point from the Whole Compact Subset","stem":"Use complete finite labeled carrier=[a1, b1, c1], topology=[[a1, b1, c1], [a1, b1], [a1, c1], [a1], [b1, c1], [b1], [c1], []] in its original supplied open-set order, and compact_subset K=[a1, b1]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Exterior point x=c1 must lie outside K. Give (exterior_point, separation_rows, exterior_neighborhood, cover_union, covers_K, neighborhood_disjoint_from_K, neighborhood_open, certificate).","answer":{"kind":"expression","value":"(c1, [[a1, [[b1, c1], [a1]]], [b1, [[a1, c1], [b1]]]], [c1], [a1, b1], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)","explanation":"For each y∈K choose disjoint U_y containing x and V_y containing y in original open-list order. The V_y union covers K but may exceed K; intersect the finitely many U_y to obtain an open neighborhood disjoint from K. For empty K the empty intersection is X. No minimum or unique subcover is asserted; the general proof uses compactness to obtain a finite subcover. The exact result, retaining the requested order, is (c1, [[a1, [[b1, c1], [a1]]], [b1, [[a1, c1], [b1]]]], [c1], [a1, b1], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)."},"skills":["Finite Intersections Separate an Exterior Point from the Whole Compact Subset"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0053","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f6b9bf538ac42d2cedac8419281762adf288df878486a7a48248d9b541d627e0","family_id":"qf-mod43-compactclosed-neighborhood-subcover-certificate","seed":202611123053,"semantic_template":"qf-mod43-compactclosed-neighborhood-subcover-certificate/{\"compact_subset\":[\"a2\",\"b2\"],\"exterior_point\":\"c2\",\"points\":[\"a2\",\"b2\",\"c2\"],\"topology\":[[\"a2\",\"b2\",\"c2\"],[\"a2\",\"b2\"],[\"a2\",\"c2\"],[\"a2\"],[\"b2\",\"c2\"],[\"b2\"],[\"c2\"],[]]}"},"fields":{"title":"Finite Intersections Separate an Exterior Point from the Whole Compact Subset","stem":"Use complete finite labeled carrier=[a2, b2, c2], topology=[[a2, b2, c2], [a2, b2], [a2, c2], [a2], [b2, c2], [b2], [c2], []] in its original supplied open-set order, and compact_subset K=[a2, b2]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Exterior point x=c2 must lie outside K. Give (exterior_point, separation_rows, exterior_neighborhood, cover_union, covers_K, neighborhood_disjoint_from_K, neighborhood_open, certificate).","answer":{"kind":"expression","value":"(c2, [[a2, [[b2, c2], [a2]]], [b2, [[a2, c2], [b2]]]], [c2], [a2, b2], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)","explanation":"For each y∈K choose disjoint U_y containing x and V_y containing y in original open-list order. The V_y union covers K but may exceed K; intersect the finitely many U_y to obtain an open neighborhood disjoint from K. For empty K the empty intersection is X. No minimum or unique subcover is asserted; the general proof uses compactness to obtain a finite subcover. The exact result, retaining the requested order, is (c2, [[a2, [[b2, c2], [a2]]], [b2, [[a2, c2], [b2]]]], [c2], [a2, b2], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)."},"skills":["Finite Intersections Separate an Exterior Point from the Whole Compact Subset"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0054","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d521c88968775c5939db64e69369eee8e2112d46356c4129c652bfcb0485db13","family_id":"qf-mod43-compactclosed-neighborhood-subcover-certificate","seed":202611123054,"semantic_template":"qf-mod43-compactclosed-neighborhood-subcover-certificate/{\"compact_subset\":[\"a3\",\"b3\"],\"exterior_point\":\"c3\",\"points\":[\"a3\",\"b3\",\"c3\"],\"topology\":[[\"a3\",\"b3\",\"c3\"],[\"a3\",\"b3\"],[\"a3\",\"c3\"],[\"a3\"],[\"b3\",\"c3\"],[\"b3\"],[\"c3\"],[]]}"},"fields":{"title":"Finite Intersections Separate an Exterior Point from the Whole Compact Subset","stem":"Use complete finite labeled carrier=[a3, b3, c3], topology=[[a3, b3, c3], [a3, b3], [a3, c3], [a3], [b3, c3], [b3], [c3], []] in its original supplied open-set order, and compact_subset K=[a3, b3]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Exterior point x=c3 must lie outside K. Give (exterior_point, separation_rows, exterior_neighborhood, cover_union, covers_K, neighborhood_disjoint_from_K, neighborhood_open, certificate).","answer":{"kind":"expression","value":"(c3, [[a3, [[b3, c3], [a3]]], [b3, [[a3, c3], [b3]]]], [c3], [a3, b3], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)","explanation":"For each y∈K choose disjoint U_y containing x and V_y containing y in original open-list order. The V_y union covers K but may exceed K; intersect the finitely many U_y to obtain an open neighborhood disjoint from K. For empty K the empty intersection is X. No minimum or unique subcover is asserted; the general proof uses compactness to obtain a finite subcover. The exact result, retaining the requested order, is (c3, [[a3, [[b3, c3], [a3]]], [b3, [[a3, c3], [b3]]]], [c3], [a3, b3], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)."},"skills":["Finite Intersections Separate an Exterior Point from the Whole Compact Subset"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0055","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9f9cf558787a6082512536c6f6fcb63d0a227dc574f804c2c8595835c13bf840","family_id":"qf-mod43-compactclosed-neighborhood-subcover-certificate","seed":202611123055,"semantic_template":"qf-mod43-compactclosed-neighborhood-subcover-certificate/{\"compact_subset\":[\"a4\",\"b4\"],\"exterior_point\":\"c4\",\"points\":[\"a4\",\"b4\",\"c4\"],\"topology\":[[\"a4\",\"b4\",\"c4\"],[\"a4\",\"b4\"],[\"a4\",\"c4\"],[\"a4\"],[\"b4\",\"c4\"],[\"b4\"],[\"c4\"],[]]}"},"fields":{"title":"Finite Intersections Separate an Exterior Point from the Whole Compact Subset","stem":"Use complete finite labeled carrier=[a4, b4, c4], topology=[[a4, b4, c4], [a4, b4], [a4, c4], [a4], [b4, c4], [b4], [c4], []] in its original supplied open-set order, and compact_subset K=[a4, b4]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Exterior point x=c4 must lie outside K. Give (exterior_point, separation_rows, exterior_neighborhood, cover_union, covers_K, neighborhood_disjoint_from_K, neighborhood_open, certificate).","answer":{"kind":"expression","value":"(c4, [[a4, [[b4, c4], [a4]]], [b4, [[a4, c4], [b4]]]], [c4], [a4, b4], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)","explanation":"For each y∈K choose disjoint U_y containing x and V_y containing y in original open-list order. The V_y union covers K but may exceed K; intersect the finitely many U_y to obtain an open neighborhood disjoint from K. For empty K the empty intersection is X. No minimum or unique subcover is asserted; the general proof uses compactness to obtain a finite subcover. The exact result, retaining the requested order, is (c4, [[a4, [[b4, c4], [a4]]], [b4, [[a4, c4], [b4]]]], [c4], [a4, b4], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)."},"skills":["Finite Intersections Separate an Exterior Point from the Whole Compact Subset"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0056","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9ded039a1a3f03c9adbb259992ddd9fb8906b79648d41bd1e9c5ad663e61bd24","family_id":"qf-mod43-compactclosed-neighborhood-subcover-certificate","seed":202611123056,"semantic_template":"qf-mod43-compactclosed-neighborhood-subcover-certificate/{\"compact_subset\":[\"a5\",\"b5\"],\"exterior_point\":\"c5\",\"points\":[\"a5\",\"b5\",\"c5\"],\"topology\":[[\"a5\",\"b5\",\"c5\"],[\"a5\",\"b5\"],[\"a5\",\"c5\"],[\"a5\"],[\"b5\",\"c5\"],[\"b5\"],[\"c5\"],[]]}"},"fields":{"title":"Finite Intersections Separate an Exterior Point from the Whole Compact Subset","stem":"Use complete finite labeled carrier=[a5, b5, c5], topology=[[a5, b5, c5], [a5, b5], [a5, c5], [a5], [b5, c5], [b5], [c5], []] in its original supplied open-set order, and compact_subset K=[a5, b5]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Exterior point x=c5 must lie outside K. Give (exterior_point, separation_rows, exterior_neighborhood, cover_union, covers_K, neighborhood_disjoint_from_K, neighborhood_open, certificate).","answer":{"kind":"expression","value":"(c5, [[a5, [[b5, c5], [a5]]], [b5, [[a5, c5], [b5]]]], [c5], [a5, b5], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)","explanation":"For each y∈K choose disjoint U_y containing x and V_y containing y in original open-list order. The V_y union covers K but may exceed K; intersect the finitely many U_y to obtain an open neighborhood disjoint from K. For empty K the empty intersection is X. No minimum or unique subcover is asserted; the general proof uses compactness to obtain a finite subcover. The exact result, retaining the requested order, is (c5, [[a5, [[b5, c5], [a5]]], [b5, [[a5, c5], [b5]]]], [c5], [a5, b5], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)."},"skills":["Finite Intersections Separate an Exterior Point from the Whole Compact Subset"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0057","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8ee2636951a8fa3d47b329a5687e390f60d8e6c81c52e78015dc0e256b2e6669","family_id":"qf-mod43-compactclosed-neighborhood-subcover-certificate","seed":202611123057,"semantic_template":"qf-mod43-compactclosed-neighborhood-subcover-certificate/{\"compact_subset\":[\"a6\",\"b6\"],\"exterior_point\":\"c6\",\"points\":[\"a6\",\"b6\",\"c6\"],\"topology\":[[\"a6\",\"b6\",\"c6\"],[\"a6\",\"b6\"],[\"a6\",\"c6\"],[\"a6\"],[\"b6\",\"c6\"],[\"b6\"],[\"c6\"],[]]}"},"fields":{"title":"Finite Intersections Separate an Exterior Point from the Whole Compact Subset","stem":"Use complete finite labeled carrier=[a6, b6, c6], topology=[[a6, b6, c6], [a6, b6], [a6, c6], [a6], [b6, c6], [b6], [c6], []] in its original supplied open-set order, and compact_subset K=[a6, b6]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Exterior point x=c6 must lie outside K. Give (exterior_point, separation_rows, exterior_neighborhood, cover_union, covers_K, neighborhood_disjoint_from_K, neighborhood_open, certificate).","answer":{"kind":"expression","value":"(c6, [[a6, [[b6, c6], [a6]]], [b6, [[a6, c6], [b6]]]], [c6], [a6, b6], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)","explanation":"For each y∈K choose disjoint U_y containing x and V_y containing y in original open-list order. The V_y union covers K but may exceed K; intersect the finitely many U_y to obtain an open neighborhood disjoint from K. For empty K the empty intersection is X. No minimum or unique subcover is asserted; the general proof uses compactness to obtain a finite subcover. The exact result, retaining the requested order, is (c6, [[a6, [[b6, c6], [a6]]], [b6, [[a6, c6], [b6]]]], [c6], [a6, b6], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)."},"skills":["Finite Intersections Separate an Exterior Point from the Whole Compact Subset"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0058","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1befc6283d2ffa3ff9e1f106d75cd2bfe6fdaffd2ad13ae62ccb939da4a5d0fd","family_id":"qf-mod43-compactclosed-neighborhood-subcover-certificate","seed":202611123058,"semantic_template":"qf-mod43-compactclosed-neighborhood-subcover-certificate/{\"compact_subset\":[\"a7\",\"b7\"],\"exterior_point\":\"c7\",\"points\":[\"a7\",\"b7\",\"c7\"],\"topology\":[[\"a7\",\"b7\",\"c7\"],[\"a7\",\"b7\"],[\"a7\",\"c7\"],[\"a7\"],[\"b7\",\"c7\"],[\"b7\"],[\"c7\"],[]]}"},"fields":{"title":"Finite Intersections Separate an Exterior Point from the Whole Compact Subset","stem":"Use complete finite labeled carrier=[a7, b7, c7], topology=[[a7, b7, c7], [a7, b7], [a7, c7], [a7], [b7, c7], [b7], [c7], []] in its original supplied open-set order, and compact_subset K=[a7, b7]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Exterior point x=c7 must lie outside K. Give (exterior_point, separation_rows, exterior_neighborhood, cover_union, covers_K, neighborhood_disjoint_from_K, neighborhood_open, certificate).","answer":{"kind":"expression","value":"(c7, [[a7, [[b7, c7], [a7]]], [b7, [[a7, c7], [b7]]]], [c7], [a7, b7], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)","explanation":"For each y∈K choose disjoint U_y containing x and V_y containing y in original open-list order. The V_y union covers K but may exceed K; intersect the finitely many U_y to obtain an open neighborhood disjoint from K. For empty K the empty intersection is X. No minimum or unique subcover is asserted; the general proof uses compactness to obtain a finite subcover. The exact result, retaining the requested order, is (c7, [[a7, [[b7, c7], [a7]]], [b7, [[a7, c7], [b7]]]], [c7], [a7, b7], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)."},"skills":["Finite Intersections Separate an Exterior Point from the Whole Compact Subset"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0059","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3e9df48b8325d35ea21f2048580bdeb8678270363fbd2c206bc833df7f87406c","family_id":"qf-mod43-compactclosed-neighborhood-subcover-certificate","seed":202611123059,"semantic_template":"qf-mod43-compactclosed-neighborhood-subcover-certificate/{\"compact_subset\":[\"a8\",\"b8\"],\"exterior_point\":\"c8\",\"points\":[\"a8\",\"b8\",\"c8\"],\"topology\":[[\"a8\",\"b8\",\"c8\"],[\"a8\",\"b8\"],[\"a8\",\"c8\"],[\"a8\"],[\"b8\",\"c8\"],[\"b8\"],[\"c8\"],[]]}"},"fields":{"title":"Finite Intersections Separate an Exterior Point from the Whole Compact Subset","stem":"Use complete finite labeled carrier=[a8, b8, c8], topology=[[a8, b8, c8], [a8, b8], [a8, c8], [a8], [b8, c8], [b8], [c8], []] in its original supplied open-set order, and compact_subset K=[a8, b8]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Exterior point x=c8 must lie outside K. Give (exterior_point, separation_rows, exterior_neighborhood, cover_union, covers_K, neighborhood_disjoint_from_K, neighborhood_open, certificate).","answer":{"kind":"expression","value":"(c8, [[a8, [[b8, c8], [a8]]], [b8, [[a8, c8], [b8]]]], [c8], [a8, b8], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)","explanation":"For each y∈K choose disjoint U_y containing x and V_y containing y in original open-list order. The V_y union covers K but may exceed K; intersect the finitely many U_y to obtain an open neighborhood disjoint from K. For empty K the empty intersection is X. No minimum or unique subcover is asserted; the general proof uses compactness to obtain a finite subcover. The exact result, retaining the requested order, is (c8, [[a8, [[b8, c8], [a8]]], [b8, [[a8, c8], [b8]]]], [c8], [a8, b8], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)."},"skills":["Finite Intersections Separate an Exterior Point from the Whole Compact Subset"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0060","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b389c6ba483d6b3de6e489ab9ebb1dc5c8a70c6d2fbc1109a7c3e39d167ba225","family_id":"qf-mod43-compactclosed-neighborhood-subcover-certificate","seed":202611123060,"semantic_template":"qf-mod43-compactclosed-neighborhood-subcover-certificate/{\"compact_subset\":[\"a9\",\"b9\"],\"exterior_point\":\"c9\",\"points\":[\"a9\",\"b9\",\"c9\"],\"topology\":[[\"a9\",\"b9\",\"c9\"],[\"a9\",\"b9\"],[\"a9\",\"c9\"],[\"a9\"],[\"b9\",\"c9\"],[\"b9\"],[\"c9\"],[]]}"},"fields":{"title":"Finite Intersections Separate an Exterior Point from the Whole Compact Subset","stem":"Use complete finite labeled carrier=[a9, b9, c9], topology=[[a9, b9, c9], [a9, b9], [a9, c9], [a9], [b9, c9], [b9], [c9], []] in its original supplied open-set order, and compact_subset K=[a9, b9]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Exterior point x=c9 must lie outside K. Give (exterior_point, separation_rows, exterior_neighborhood, cover_union, covers_K, neighborhood_disjoint_from_K, neighborhood_open, certificate).","answer":{"kind":"expression","value":"(c9, [[a9, [[b9, c9], [a9]]], [b9, [[a9, c9], [b9]]]], [c9], [a9, b9], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)","explanation":"For each y∈K choose disjoint U_y containing x and V_y containing y in original open-list order. The V_y union covers K but may exceed K; intersect the finitely many U_y to obtain an open neighborhood disjoint from K. For empty K the empty intersection is X. No minimum or unique subcover is asserted; the general proof uses compactness to obtain a finite subcover. The exact result, retaining the requested order, is (c9, [[a9, [[b9, c9], [a9]]], [b9, [[a9, c9], [b9]]]], [c9], [a9, b9], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)."},"skills":["Finite Intersections Separate an Exterior Point from the Whole Compact Subset"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0061","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3a81b8582d311a6da6a35f9f027f0349c4b74f56d8f6684a01b27c82268c7f75","family_id":"qf-mod43-compactclosed-neighborhood-subcover-certificate","seed":202611123061,"semantic_template":"qf-mod43-compactclosed-neighborhood-subcover-certificate/{\"compact_subset\":[\"a10\",\"b10\"],\"exterior_point\":\"c10\",\"points\":[\"a10\",\"b10\",\"c10\"],\"topology\":[[\"a10\",\"b10\",\"c10\"],[\"a10\",\"b10\"],[\"a10\",\"c10\"],[\"a10\"],[\"b10\",\"c10\"],[\"b10\"],[\"c10\"],[]]}"},"fields":{"title":"Finite Intersections Separate an Exterior Point from the Whole Compact Subset","stem":"Use complete finite labeled carrier=[a10, b10, c10], topology=[[a10, b10, c10], [a10, b10], [a10, c10], [a10], [b10, c10], [b10], [c10], []] in its original supplied open-set order, and compact_subset K=[a10, b10]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Exterior point x=c10 must lie outside K. Give (exterior_point, separation_rows, exterior_neighborhood, cover_union, covers_K, neighborhood_disjoint_from_K, neighborhood_open, certificate).","answer":{"kind":"expression","value":"(c10, [[a10, [[b10, c10], [a10]]], [b10, [[a10, c10], [b10]]]], [c10], [a10, b10], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)","explanation":"For each y∈K choose disjoint U_y containing x and V_y containing y in original open-list order. The V_y union covers K but may exceed K; intersect the finitely many U_y to obtain an open neighborhood disjoint from K. For empty K the empty intersection is X. No minimum or unique subcover is asserted; the general proof uses compactness to obtain a finite subcover. The exact result, retaining the requested order, is (c10, [[a10, [[b10, c10], [a10]]], [b10, [[a10, c10], [b10]]]], [c10], [a10, b10], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)."},"skills":["Finite Intersections Separate an Exterior Point from the Whole Compact Subset"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0062","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5255cb68831e52ca1c46ea8628af73002e1e58be74a74f5de0b13dcc9f652eab","family_id":"qf-mod43-compactclosed-neighborhood-subcover-certificate","seed":202611123062,"semantic_template":"qf-mod43-compactclosed-neighborhood-subcover-certificate/{\"compact_subset\":[\"a11\",\"b11\"],\"exterior_point\":\"c11\",\"points\":[\"a11\",\"b11\",\"c11\"],\"topology\":[[\"a11\",\"b11\",\"c11\"],[\"a11\",\"b11\"],[\"a11\",\"c11\"],[\"a11\"],[\"b11\",\"c11\"],[\"b11\"],[\"c11\"],[]]}"},"fields":{"title":"Finite Intersections Separate an Exterior Point from the Whole Compact Subset","stem":"Use complete finite labeled carrier=[a11, b11, c11], topology=[[a11, b11, c11], [a11, b11], [a11, c11], [a11], [b11, c11], [b11], [c11], []] in its original supplied open-set order, and compact_subset K=[a11, b11]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Exterior point x=c11 must lie outside K. Give (exterior_point, separation_rows, exterior_neighborhood, cover_union, covers_K, neighborhood_disjoint_from_K, neighborhood_open, certificate).","answer":{"kind":"expression","value":"(c11, [[a11, [[b11, c11], [a11]]], [b11, [[a11, c11], [b11]]]], [c11], [a11, b11], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)","explanation":"For each y∈K choose disjoint U_y containing x and V_y containing y in original open-list order. The V_y union covers K but may exceed K; intersect the finitely many U_y to obtain an open neighborhood disjoint from K. For empty K the empty intersection is X. No minimum or unique subcover is asserted; the general proof uses compactness to obtain a finite subcover. The exact result, retaining the requested order, is (c11, [[a11, [[b11, c11], [a11]]], [b11, [[a11, c11], [b11]]]], [c11], [a11, b11], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)."},"skills":["Finite Intersections Separate an Exterior Point from the Whole Compact Subset"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0063","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"eae96a7c95a2d535a39d31adb440aecaa9fdf77cbef4b021523c38b9e1f1cf37","family_id":"qf-mod43-compactclosed-neighborhood-subcover-certificate","seed":202611123063,"semantic_template":"qf-mod43-compactclosed-neighborhood-subcover-certificate/{\"compact_subset\":[\"a12\",\"b12\"],\"exterior_point\":\"c12\",\"points\":[\"a12\",\"b12\",\"c12\"],\"topology\":[[\"a12\",\"b12\",\"c12\"],[\"a12\",\"b12\"],[\"a12\",\"c12\"],[\"a12\"],[\"b12\",\"c12\"],[\"b12\"],[\"c12\"],[]]}"},"fields":{"title":"Finite Intersections Separate an Exterior Point from the Whole Compact Subset","stem":"Use complete finite labeled carrier=[a12, b12, c12], topology=[[a12, b12, c12], [a12, b12], [a12, c12], [a12], [b12, c12], [b12], [c12], []] in its original supplied open-set order, and compact_subset K=[a12, b12]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Exterior point x=c12 must lie outside K. Give (exterior_point, separation_rows, exterior_neighborhood, cover_union, covers_K, neighborhood_disjoint_from_K, neighborhood_open, certificate).","answer":{"kind":"expression","value":"(c12, [[a12, [[b12, c12], [a12]]], [b12, [[a12, c12], [b12]]]], [c12], [a12, b12], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)","explanation":"For each y∈K choose disjoint U_y containing x and V_y containing y in original open-list order. The V_y union covers K but may exceed K; intersect the finitely many U_y to obtain an open neighborhood disjoint from K. For empty K the empty intersection is X. No minimum or unique subcover is asserted; the general proof uses compactness to obtain a finite subcover. The exact result, retaining the requested order, is (c12, [[a12, [[b12, c12], [a12]]], [b12, [[a12, c12], [b12]]]], [c12], [a12, b12], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)."},"skills":["Finite Intersections Separate an Exterior Point from the Whole Compact Subset"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0064","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a9bb7cce86962934364e36b8899e28a15062c374edd28351239144a69e5691af","family_id":"qf-mod43-compactclosed-neighborhood-subcover-certificate","seed":202611123064,"semantic_template":"qf-mod43-compactclosed-neighborhood-subcover-certificate/{\"compact_subset\":[\"a13\",\"b13\"],\"exterior_point\":\"c13\",\"points\":[\"a13\",\"b13\",\"c13\"],\"topology\":[[\"a13\",\"b13\",\"c13\"],[\"a13\",\"b13\"],[\"a13\",\"c13\"],[\"a13\"],[\"b13\",\"c13\"],[\"b13\"],[\"c13\"],[]]}"},"fields":{"title":"Finite Intersections Separate an Exterior Point from the Whole Compact Subset","stem":"Use complete finite labeled carrier=[a13, b13, c13], topology=[[a13, b13, c13], [a13, b13], [a13, c13], [a13], [b13, c13], [b13], [c13], []] in its original supplied open-set order, and compact_subset K=[a13, b13]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Exterior point x=c13 must lie outside K. Give (exterior_point, separation_rows, exterior_neighborhood, cover_union, covers_K, neighborhood_disjoint_from_K, neighborhood_open, certificate).","answer":{"kind":"expression","value":"(c13, [[a13, [[b13, c13], [a13]]], [b13, [[a13, c13], [b13]]]], [c13], [a13, b13], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)","explanation":"For each y∈K choose disjoint U_y containing x and V_y containing y in original open-list order. The V_y union covers K but may exceed K; intersect the finitely many U_y to obtain an open neighborhood disjoint from K. For empty K the empty intersection is X. No minimum or unique subcover is asserted; the general proof uses compactness to obtain a finite subcover. The exact result, retaining the requested order, is (c13, [[a13, [[b13, c13], [a13]]], [b13, [[a13, c13], [b13]]]], [c13], [a13, b13], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)."},"skills":["Finite Intersections Separate an Exterior Point from the Whole Compact Subset"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0065","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"81bf4c4a39ef3d5de586cf27536c6298fcd96a5d637d2530eccbcc97d2872b80","family_id":"qf-mod43-compactclosed-neighborhood-subcover-certificate","seed":202611123065,"semantic_template":"qf-mod43-compactclosed-neighborhood-subcover-certificate/{\"compact_subset\":[\"a14\",\"b14\"],\"exterior_point\":\"c14\",\"points\":[\"a14\",\"b14\",\"c14\"],\"topology\":[[\"a14\",\"b14\",\"c14\"],[\"a14\",\"b14\"],[\"a14\",\"c14\"],[\"a14\"],[\"b14\",\"c14\"],[\"b14\"],[\"c14\"],[]]}"},"fields":{"title":"Finite Intersections Separate an Exterior Point from the Whole Compact Subset","stem":"Use complete finite labeled carrier=[a14, b14, c14], topology=[[a14, b14, c14], [a14, b14], [a14, c14], [a14], [b14, c14], [b14], [c14], []] in its original supplied open-set order, and compact_subset K=[a14, b14]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Exterior point x=c14 must lie outside K. Give (exterior_point, separation_rows, exterior_neighborhood, cover_union, covers_K, neighborhood_disjoint_from_K, neighborhood_open, certificate).","answer":{"kind":"expression","value":"(c14, [[a14, [[b14, c14], [a14]]], [b14, [[a14, c14], [b14]]]], [c14], [a14, b14], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)","explanation":"For each y∈K choose disjoint U_y containing x and V_y containing y in original open-list order. The V_y union covers K but may exceed K; intersect the finitely many U_y to obtain an open neighborhood disjoint from K. For empty K the empty intersection is X. No minimum or unique subcover is asserted; the general proof uses compactness to obtain a finite subcover. The exact result, retaining the requested order, is (c14, [[a14, [[b14, c14], [a14]]], [b14, [[a14, c14], [b14]]]], [c14], [a14, b14], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)."},"skills":["Finite Intersections Separate an Exterior Point from the Whole Compact Subset"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0066","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e043cdae1a6b8b6649858088561668aed710477cdaabcdca76c6c6420228f919","family_id":"qf-mod43-compactclosed-neighborhood-subcover-certificate","seed":202611123066,"semantic_template":"qf-mod43-compactclosed-neighborhood-subcover-certificate/{\"compact_subset\":[\"a15\",\"b15\"],\"exterior_point\":\"c15\",\"points\":[\"a15\",\"b15\",\"c15\"],\"topology\":[[\"a15\",\"b15\",\"c15\"],[\"a15\",\"b15\"],[\"a15\",\"c15\"],[\"a15\"],[\"b15\",\"c15\"],[\"b15\"],[\"c15\"],[]]}"},"fields":{"title":"Finite Intersections Separate an Exterior Point from the Whole Compact Subset","stem":"Use complete finite labeled carrier=[a15, b15, c15], topology=[[a15, b15, c15], [a15, b15], [a15, c15], [a15], [b15, c15], [b15], [c15], []] in its original supplied open-set order, and compact_subset K=[a15, b15]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Exterior point x=c15 must lie outside K. Give (exterior_point, separation_rows, exterior_neighborhood, cover_union, covers_K, neighborhood_disjoint_from_K, neighborhood_open, certificate).","answer":{"kind":"expression","value":"(c15, [[a15, [[b15, c15], [a15]]], [b15, [[a15, c15], [b15]]]], [c15], [a15, b15], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)","explanation":"For each y∈K choose disjoint U_y containing x and V_y containing y in original open-list order. The V_y union covers K but may exceed K; intersect the finitely many U_y to obtain an open neighborhood disjoint from K. For empty K the empty intersection is X. No minimum or unique subcover is asserted; the general proof uses compactness to obtain a finite subcover. The exact result, retaining the requested order, is (c15, [[a15, [[b15, c15], [a15]]], [b15, [[a15, c15], [b15]]]], [c15], [a15, b15], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)."},"skills":["Finite Intersections Separate an Exterior Point from the Whole Compact Subset"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0067","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"90b9908e429cbb106bcd3a1bde2f56a3f08fff1f851b56385da31a1cf3f011fe","family_id":"qf-mod43-compactclosed-neighborhood-subcover-certificate","seed":202611123067,"semantic_template":"qf-mod43-compactclosed-neighborhood-subcover-certificate/{\"compact_subset\":[\"a16\",\"b16\"],\"exterior_point\":\"c16\",\"points\":[\"a16\",\"b16\",\"c16\"],\"topology\":[[\"a16\",\"b16\",\"c16\"],[\"a16\",\"b16\"],[\"a16\",\"c16\"],[\"a16\"],[\"b16\",\"c16\"],[\"b16\"],[\"c16\"],[]]}"},"fields":{"title":"Finite Intersections Separate an Exterior Point from the Whole Compact Subset","stem":"Use complete finite labeled carrier=[a16, b16, c16], topology=[[a16, b16, c16], [a16, b16], [a16, c16], [a16], [b16, c16], [b16], [c16], []] in its original supplied open-set order, and compact_subset K=[a16, b16]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Exterior point x=c16 must lie outside K. Give (exterior_point, separation_rows, exterior_neighborhood, cover_union, covers_K, neighborhood_disjoint_from_K, neighborhood_open, certificate).","answer":{"kind":"expression","value":"(c16, [[a16, [[b16, c16], [a16]]], [b16, [[a16, c16], [b16]]]], [c16], [a16, b16], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)","explanation":"For each y∈K choose disjoint U_y containing x and V_y containing y in original open-list order. The V_y union covers K but may exceed K; intersect the finitely many U_y to obtain an open neighborhood disjoint from K. For empty K the empty intersection is X. No minimum or unique subcover is asserted; the general proof uses compactness to obtain a finite subcover. The exact result, retaining the requested order, is (c16, [[a16, [[b16, c16], [a16]]], [b16, [[a16, c16], [b16]]]], [c16], [a16, b16], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)."},"skills":["Finite Intersections Separate an Exterior Point from the Whole Compact Subset"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0068","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e6f3ac9479fc4e0fbd46eccb5e2968bc7fa25f9eea51d244d25ca14a20b78083","family_id":"qf-mod43-compactclosed-neighborhood-subcover-certificate","seed":202611123068,"semantic_template":"qf-mod43-compactclosed-neighborhood-subcover-certificate/{\"compact_subset\":[\"a17\",\"b17\"],\"exterior_point\":\"c17\",\"points\":[\"a17\",\"b17\",\"c17\"],\"topology\":[[\"a17\",\"b17\",\"c17\"],[\"a17\",\"b17\"],[\"a17\",\"c17\"],[\"a17\"],[\"b17\",\"c17\"],[\"b17\"],[\"c17\"],[]]}"},"fields":{"title":"Finite Intersections Separate an Exterior Point from the Whole Compact Subset","stem":"Use complete finite labeled carrier=[a17, b17, c17], topology=[[a17, b17, c17], [a17, b17], [a17, c17], [a17], [b17, c17], [b17], [c17], []] in its original supplied open-set order, and compact_subset K=[a17, b17]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Exterior point x=c17 must lie outside K. Give (exterior_point, separation_rows, exterior_neighborhood, cover_union, covers_K, neighborhood_disjoint_from_K, neighborhood_open, certificate).","answer":{"kind":"expression","value":"(c17, [[a17, [[b17, c17], [a17]]], [b17, [[a17, c17], [b17]]]], [c17], [a17, b17], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)","explanation":"For each y∈K choose disjoint U_y containing x and V_y containing y in original open-list order. The V_y union covers K but may exceed K; intersect the finitely many U_y to obtain an open neighborhood disjoint from K. For empty K the empty intersection is X. No minimum or unique subcover is asserted; the general proof uses compactness to obtain a finite subcover. The exact result, retaining the requested order, is (c17, [[a17, [[b17, c17], [a17]]], [b17, [[a17, c17], [b17]]]], [c17], [a17, b17], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)."},"skills":["Finite Intersections Separate an Exterior Point from the Whole Compact Subset"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0069","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8de3581609c43757133ec4c99771d9eca61a8a6b5ce9a4c45469c77145d7f643","family_id":"qf-mod43-compactclosed-neighborhood-subcover-certificate","seed":202611123069,"semantic_template":"qf-mod43-compactclosed-neighborhood-subcover-certificate/{\"compact_subset\":[\"a18\",\"b18\"],\"exterior_point\":\"c18\",\"points\":[\"a18\",\"b18\",\"c18\"],\"topology\":[[\"a18\",\"b18\",\"c18\"],[\"a18\",\"b18\"],[\"a18\",\"c18\"],[\"a18\"],[\"b18\",\"c18\"],[\"b18\"],[\"c18\"],[]]}"},"fields":{"title":"Finite Intersections Separate an Exterior Point from the Whole Compact Subset","stem":"Use complete finite labeled carrier=[a18, b18, c18], topology=[[a18, b18, c18], [a18, b18], [a18, c18], [a18], [b18, c18], [b18], [c18], []] in its original supplied open-set order, and compact_subset K=[a18, b18]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Exterior point x=c18 must lie outside K. Give (exterior_point, separation_rows, exterior_neighborhood, cover_union, covers_K, neighborhood_disjoint_from_K, neighborhood_open, certificate).","answer":{"kind":"expression","value":"(c18, [[a18, [[b18, c18], [a18]]], [b18, [[a18, c18], [b18]]]], [c18], [a18, b18], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)","explanation":"For each y∈K choose disjoint U_y containing x and V_y containing y in original open-list order. The V_y union covers K but may exceed K; intersect the finitely many U_y to obtain an open neighborhood disjoint from K. For empty K the empty intersection is X. No minimum or unique subcover is asserted; the general proof uses compactness to obtain a finite subcover. The exact result, retaining the requested order, is (c18, [[a18, [[b18, c18], [a18]]], [b18, [[a18, c18], [b18]]]], [c18], [a18, b18], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)."},"skills":["Finite Intersections Separate an Exterior Point from the Whole Compact Subset"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0070","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"70a234e602b108dd90e6017c45b9f3b2840ccde0388b53b81fed67e6a5c985c7","family_id":"qf-mod43-compactclosed-neighborhood-subcover-certificate","seed":202611123070,"semantic_template":"qf-mod43-compactclosed-neighborhood-subcover-certificate/{\"compact_subset\":[\"a19\",\"b19\"],\"exterior_point\":\"c19\",\"points\":[\"a19\",\"b19\",\"c19\"],\"topology\":[[\"a19\",\"b19\",\"c19\"],[\"a19\",\"b19\"],[\"a19\",\"c19\"],[\"a19\"],[\"b19\",\"c19\"],[\"b19\"],[\"c19\"],[]]}"},"fields":{"title":"Finite Intersections Separate an Exterior Point from the Whole Compact Subset","stem":"Use complete finite labeled carrier=[a19, b19, c19], topology=[[a19, b19, c19], [a19, b19], [a19, c19], [a19], [b19, c19], [b19], [c19], []] in its original supplied open-set order, and compact_subset K=[a19, b19]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Exterior point x=c19 must lie outside K. Give (exterior_point, separation_rows, exterior_neighborhood, cover_union, covers_K, neighborhood_disjoint_from_K, neighborhood_open, certificate).","answer":{"kind":"expression","value":"(c19, [[a19, [[b19, c19], [a19]]], [b19, [[a19, c19], [b19]]]], [c19], [a19, b19], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)","explanation":"For each y∈K choose disjoint U_y containing x and V_y containing y in original open-list order. The V_y union covers K but may exceed K; intersect the finitely many U_y to obtain an open neighborhood disjoint from K. For empty K the empty intersection is X. No minimum or unique subcover is asserted; the general proof uses compactness to obtain a finite subcover. The exact result, retaining the requested order, is (c19, [[a19, [[b19, c19], [a19]]], [b19, [[a19, c19], [b19]]]], [c19], [a19, b19], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)."},"skills":["Finite Intersections Separate an Exterior Point from the Whole Compact Subset"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0071","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bc911a8cdc2b481e1c24de5a8c2bfa8b12c387dfea676a0693882d5524fd635a","family_id":"qf-mod43-compactclosed-neighborhood-subcover-certificate","seed":202611123071,"semantic_template":"qf-mod43-compactclosed-neighborhood-subcover-certificate/{\"compact_subset\":[\"a20\",\"b20\"],\"exterior_point\":\"c20\",\"points\":[\"a20\",\"b20\",\"c20\"],\"topology\":[[\"a20\",\"b20\",\"c20\"],[\"a20\",\"b20\"],[\"a20\",\"c20\"],[\"a20\"],[\"b20\",\"c20\"],[\"b20\"],[\"c20\"],[]]}"},"fields":{"title":"Finite Intersections Separate an Exterior Point from the Whole Compact Subset","stem":"Use complete finite labeled carrier=[a20, b20, c20], topology=[[a20, b20, c20], [a20, b20], [a20, c20], [a20], [b20, c20], [b20], [c20], []] in its original supplied open-set order, and compact_subset K=[a20, b20]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Exterior point x=c20 must lie outside K. Give (exterior_point, separation_rows, exterior_neighborhood, cover_union, covers_K, neighborhood_disjoint_from_K, neighborhood_open, certificate).","answer":{"kind":"expression","value":"(c20, [[a20, [[b20, c20], [a20]]], [b20, [[a20, c20], [b20]]]], [c20], [a20, b20], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)","explanation":"For each y∈K choose disjoint U_y containing x and V_y containing y in original open-list order. The V_y union covers K but may exceed K; intersect the finitely many U_y to obtain an open neighborhood disjoint from K. For empty K the empty intersection is X. No minimum or unique subcover is asserted; the general proof uses compactness to obtain a finite subcover. The exact result, retaining the requested order, is (c20, [[a20, [[b20, c20], [a20]]], [b20, [[a20, c20], [b20]]]], [c20], [a20, b20], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)."},"skills":["Finite Intersections Separate an Exterior Point from the Whole Compact Subset"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0072","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8bc8717808fd627d19d194d7f5f1cdccd5f2a37dff29a32d296b9acb8cb49447","family_id":"qf-mod43-compactclosed-neighborhood-subcover-certificate","seed":202611123072,"semantic_template":"qf-mod43-compactclosed-neighborhood-subcover-certificate/{\"compact_subset\":[\"a21\",\"b21\"],\"exterior_point\":\"c21\",\"points\":[\"a21\",\"b21\",\"c21\"],\"topology\":[[\"a21\",\"b21\",\"c21\"],[\"a21\",\"b21\"],[\"a21\",\"c21\"],[\"a21\"],[\"b21\",\"c21\"],[\"b21\"],[\"c21\"],[]]}"},"fields":{"title":"Finite Intersections Separate an Exterior Point from the Whole Compact Subset","stem":"Use complete finite labeled carrier=[a21, b21, c21], topology=[[a21, b21, c21], [a21, b21], [a21, c21], [a21], [b21, c21], [b21], [c21], []] in its original supplied open-set order, and compact_subset K=[a21, b21]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Exterior point x=c21 must lie outside K. Give (exterior_point, separation_rows, exterior_neighborhood, cover_union, covers_K, neighborhood_disjoint_from_K, neighborhood_open, certificate).","answer":{"kind":"expression","value":"(c21, [[a21, [[b21, c21], [a21]]], [b21, [[a21, c21], [b21]]]], [c21], [a21, b21], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)","explanation":"For each y∈K choose disjoint U_y containing x and V_y containing y in original open-list order. The V_y union covers K but may exceed K; intersect the finitely many U_y to obtain an open neighborhood disjoint from K. For empty K the empty intersection is X. No minimum or unique subcover is asserted; the general proof uses compactness to obtain a finite subcover. The exact result, retaining the requested order, is (c21, [[a21, [[b21, c21], [a21]]], [b21, [[a21, c21], [b21]]]], [c21], [a21, b21], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)."},"skills":["Finite Intersections Separate an Exterior Point from the Whole Compact Subset"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0073","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"af5146f3fc57e12f1775f5bde3b1ac9109d9585753f5a1dde7101362b7ddbad0","family_id":"qf-mod43-compactclosed-neighborhood-subcover-certificate","seed":202611123073,"semantic_template":"qf-mod43-compactclosed-neighborhood-subcover-certificate/{\"compact_subset\":[\"a22\",\"b22\"],\"exterior_point\":\"c22\",\"points\":[\"a22\",\"b22\",\"c22\"],\"topology\":[[\"a22\",\"b22\",\"c22\"],[\"a22\",\"b22\"],[\"a22\",\"c22\"],[\"a22\"],[\"b22\",\"c22\"],[\"b22\"],[\"c22\"],[]]}"},"fields":{"title":"Finite Intersections Separate an Exterior Point from the Whole Compact Subset","stem":"Use complete finite labeled carrier=[a22, b22, c22], topology=[[a22, b22, c22], [a22, b22], [a22, c22], [a22], [b22, c22], [b22], [c22], []] in its original supplied open-set order, and compact_subset K=[a22, b22]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Exterior point x=c22 must lie outside K. Give (exterior_point, separation_rows, exterior_neighborhood, cover_union, covers_K, neighborhood_disjoint_from_K, neighborhood_open, certificate).","answer":{"kind":"expression","value":"(c22, [[a22, [[b22, c22], [a22]]], [b22, [[a22, c22], [b22]]]], [c22], [a22, b22], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)","explanation":"For each y∈K choose disjoint U_y containing x and V_y containing y in original open-list order. The V_y union covers K but may exceed K; intersect the finitely many U_y to obtain an open neighborhood disjoint from K. For empty K the empty intersection is X. No minimum or unique subcover is asserted; the general proof uses compactness to obtain a finite subcover. The exact result, retaining the requested order, is (c22, [[a22, [[b22, c22], [a22]]], [b22, [[a22, c22], [b22]]]], [c22], [a22, b22], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)."},"skills":["Finite Intersections Separate an Exterior Point from the Whole Compact Subset"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0074","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"817cc9fd7316cbb947c76bde76bb9b10d5f3bc37b98980d1a3304c977c43ba52","family_id":"qf-mod43-compactclosed-neighborhood-subcover-certificate","seed":202611123074,"semantic_template":"qf-mod43-compactclosed-neighborhood-subcover-certificate/{\"compact_subset\":[\"a23\",\"b23\"],\"exterior_point\":\"c23\",\"points\":[\"a23\",\"b23\",\"c23\"],\"topology\":[[\"a23\",\"b23\",\"c23\"],[\"a23\",\"b23\"],[\"a23\",\"c23\"],[\"a23\"],[\"b23\",\"c23\"],[\"b23\"],[\"c23\"],[]]}"},"fields":{"title":"Finite Intersections Separate an Exterior Point from the Whole Compact Subset","stem":"Use complete finite labeled carrier=[a23, b23, c23], topology=[[a23, b23, c23], [a23, b23], [a23, c23], [a23], [b23, c23], [b23], [c23], []] in its original supplied open-set order, and compact_subset K=[a23, b23]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Exterior point x=c23 must lie outside K. Give (exterior_point, separation_rows, exterior_neighborhood, cover_union, covers_K, neighborhood_disjoint_from_K, neighborhood_open, certificate).","answer":{"kind":"expression","value":"(c23, [[a23, [[b23, c23], [a23]]], [b23, [[a23, c23], [b23]]]], [c23], [a23, b23], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)","explanation":"For each y∈K choose disjoint U_y containing x and V_y containing y in original open-list order. The V_y union covers K but may exceed K; intersect the finitely many U_y to obtain an open neighborhood disjoint from K. For empty K the empty intersection is X. No minimum or unique subcover is asserted; the general proof uses compactness to obtain a finite subcover. The exact result, retaining the requested order, is (c23, [[a23, [[b23, c23], [a23]]], [b23, [[a23, c23], [b23]]]], [c23], [a23, b23], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)."},"skills":["Finite Intersections Separate an Exterior Point from the Whole Compact Subset"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0075","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9924147417b3a1549feec9d4167bdd5c91daa33309afc44ce313002de3cc4306","family_id":"qf-mod43-compactclosed-neighborhood-subcover-certificate","seed":202611123075,"semantic_template":"qf-mod43-compactclosed-neighborhood-subcover-certificate/{\"compact_subset\":[\"a24\",\"b24\"],\"exterior_point\":\"c24\",\"points\":[\"a24\",\"b24\",\"c24\"],\"topology\":[[\"a24\",\"b24\",\"c24\"],[\"a24\",\"b24\"],[\"a24\",\"c24\"],[\"a24\"],[\"b24\",\"c24\"],[\"b24\"],[\"c24\"],[]]}"},"fields":{"title":"Finite Intersections Separate an Exterior Point from the Whole Compact Subset","stem":"Use complete finite labeled carrier=[a24, b24, c24], topology=[[a24, b24, c24], [a24, b24], [a24, c24], [a24], [b24, c24], [b24], [c24], []] in its original supplied open-set order, and compact_subset K=[a24, b24]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Exterior point x=c24 must lie outside K. Give (exterior_point, separation_rows, exterior_neighborhood, cover_union, covers_K, neighborhood_disjoint_from_K, neighborhood_open, certificate).","answer":{"kind":"expression","value":"(c24, [[a24, [[b24, c24], [a24]]], [b24, [[a24, c24], [b24]]]], [c24], [a24, b24], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)","explanation":"For each y∈K choose disjoint U_y containing x and V_y containing y in original open-list order. The V_y union covers K but may exceed K; intersect the finitely many U_y to obtain an open neighborhood disjoint from K. For empty K the empty intersection is X. No minimum or unique subcover is asserted; the general proof uses compactness to obtain a finite subcover. The exact result, retaining the requested order, is (c24, [[a24, [[b24, c24], [a24]]], [b24, [[a24, c24], [b24]]]], [c24], [a24, b24], Yes, Yes, Yes, finite_subcover_and_finite_intersection_separate_x_from_K)."},"skills":["Finite Intersections Separate an Exterior Point from the Whole Compact Subset"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0076","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f0a8b03718c74c3aac3796142d92fe7889d398ee3b4a93da91be68631c68fe58","family_id":"qf-mod43-compactclosed-nonhausdorff-boundary","seed":202611123076,"semantic_template":"qf-mod43-compactclosed-nonhausdorff-boundary/{\"compact_subset\":[\"u0\"],\"points\":[\"o0\",\"u0\"],\"topology\":[[\"o0\",\"u0\"],[\"u0\"],[]]}"},"fields":{"title":"A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis","stem":"Use complete finite labeled carrier=[o0, u0], topology=[[o0, u0], [u0], []] in its original supplied open-set order, and compact_subset K=[u0]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, complement, Hausdorff, compact, closed, boundary).","answer":{"kind":"expression","value":"([o0, u0], [u0], [o0], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)","explanation":"Require this actual finite space to be non-Hausdorff and K not closed. K remains compact by finiteness. This is a counterexample to removing the Hausdorff hypothesis, not a claim that every non-Hausdorff compact subset is nonclosed or a converse counterexample. The exact result, retaining the requested order, is ([o0, u0], [u0], [o0], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)."},"skills":["A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0077","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4c53d1948e6033454d92ddde0c8501267ea67372f50421a6afee05ce17bbeef8","family_id":"qf-mod43-compactclosed-nonhausdorff-boundary","seed":202611123077,"semantic_template":"qf-mod43-compactclosed-nonhausdorff-boundary/{\"compact_subset\":[\"u1\"],\"points\":[\"o1\",\"u1\"],\"topology\":[[\"o1\",\"u1\"],[\"u1\"],[]]}"},"fields":{"title":"A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis","stem":"Use complete finite labeled carrier=[o1, u1], topology=[[o1, u1], [u1], []] in its original supplied open-set order, and compact_subset K=[u1]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, complement, Hausdorff, compact, closed, boundary).","answer":{"kind":"expression","value":"([o1, u1], [u1], [o1], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)","explanation":"Require this actual finite space to be non-Hausdorff and K not closed. K remains compact by finiteness. This is a counterexample to removing the Hausdorff hypothesis, not a claim that every non-Hausdorff compact subset is nonclosed or a converse counterexample. The exact result, retaining the requested order, is ([o1, u1], [u1], [o1], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)."},"skills":["A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0078","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9ff7f25d223867d100af7f4c727cd6a5474420e2bec32577598737b92b176c3c","family_id":"qf-mod43-compactclosed-nonhausdorff-boundary","seed":202611123078,"semantic_template":"qf-mod43-compactclosed-nonhausdorff-boundary/{\"compact_subset\":[\"u2\"],\"points\":[\"o2\",\"u2\"],\"topology\":[[\"o2\",\"u2\"],[\"u2\"],[]]}"},"fields":{"title":"A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis","stem":"Use complete finite labeled carrier=[o2, u2], topology=[[o2, u2], [u2], []] in its original supplied open-set order, and compact_subset K=[u2]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, complement, Hausdorff, compact, closed, boundary).","answer":{"kind":"expression","value":"([o2, u2], [u2], [o2], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)","explanation":"Require this actual finite space to be non-Hausdorff and K not closed. K remains compact by finiteness. This is a counterexample to removing the Hausdorff hypothesis, not a claim that every non-Hausdorff compact subset is nonclosed or a converse counterexample. The exact result, retaining the requested order, is ([o2, u2], [u2], [o2], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)."},"skills":["A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0079","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fa224a60a0ff0bc0d1eb151b95fac688eca939033ef73394739ab5efab4e4a0d","family_id":"qf-mod43-compactclosed-nonhausdorff-boundary","seed":202611123079,"semantic_template":"qf-mod43-compactclosed-nonhausdorff-boundary/{\"compact_subset\":[\"u3\"],\"points\":[\"o3\",\"u3\"],\"topology\":[[\"o3\",\"u3\"],[\"u3\"],[]]}"},"fields":{"title":"A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis","stem":"Use complete finite labeled carrier=[o3, u3], topology=[[o3, u3], [u3], []] in its original supplied open-set order, and compact_subset K=[u3]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, complement, Hausdorff, compact, closed, boundary).","answer":{"kind":"expression","value":"([o3, u3], [u3], [o3], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)","explanation":"Require this actual finite space to be non-Hausdorff and K not closed. K remains compact by finiteness. This is a counterexample to removing the Hausdorff hypothesis, not a claim that every non-Hausdorff compact subset is nonclosed or a converse counterexample. The exact result, retaining the requested order, is ([o3, u3], [u3], [o3], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)."},"skills":["A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0080","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a0ab20d2570a4c28337cc983f36cbb090121bed10e328bed5fc6723658b039cd","family_id":"qf-mod43-compactclosed-nonhausdorff-boundary","seed":202611123080,"semantic_template":"qf-mod43-compactclosed-nonhausdorff-boundary/{\"compact_subset\":[\"u4\"],\"points\":[\"o4\",\"u4\"],\"topology\":[[\"o4\",\"u4\"],[\"u4\"],[]]}"},"fields":{"title":"A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis","stem":"Use complete finite labeled carrier=[o4, u4], topology=[[o4, u4], [u4], []] in its original supplied open-set order, and compact_subset K=[u4]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, complement, Hausdorff, compact, closed, boundary).","answer":{"kind":"expression","value":"([o4, u4], [u4], [o4], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)","explanation":"Require this actual finite space to be non-Hausdorff and K not closed. K remains compact by finiteness. This is a counterexample to removing the Hausdorff hypothesis, not a claim that every non-Hausdorff compact subset is nonclosed or a converse counterexample. The exact result, retaining the requested order, is ([o4, u4], [u4], [o4], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)."},"skills":["A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0081","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"00b449e9e1f48359f08895ec7343f2058134509c74d8ceb3904a762f683429e7","family_id":"qf-mod43-compactclosed-nonhausdorff-boundary","seed":202611123081,"semantic_template":"qf-mod43-compactclosed-nonhausdorff-boundary/{\"compact_subset\":[\"u5\"],\"points\":[\"o5\",\"u5\"],\"topology\":[[\"o5\",\"u5\"],[\"u5\"],[]]}"},"fields":{"title":"A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis","stem":"Use complete finite labeled carrier=[o5, u5], topology=[[o5, u5], [u5], []] in its original supplied open-set order, and compact_subset K=[u5]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, complement, Hausdorff, compact, closed, boundary).","answer":{"kind":"expression","value":"([o5, u5], [u5], [o5], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)","explanation":"Require this actual finite space to be non-Hausdorff and K not closed. K remains compact by finiteness. This is a counterexample to removing the Hausdorff hypothesis, not a claim that every non-Hausdorff compact subset is nonclosed or a converse counterexample. The exact result, retaining the requested order, is ([o5, u5], [u5], [o5], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)."},"skills":["A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0082","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bd079bbb41ca8161f7e8586f975be133a2e8c2e27d3223965d916a268b70e111","family_id":"qf-mod43-compactclosed-nonhausdorff-boundary","seed":202611123082,"semantic_template":"qf-mod43-compactclosed-nonhausdorff-boundary/{\"compact_subset\":[\"u6\"],\"points\":[\"o6\",\"u6\"],\"topology\":[[\"o6\",\"u6\"],[\"u6\"],[]]}"},"fields":{"title":"A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis","stem":"Use complete finite labeled carrier=[o6, u6], topology=[[o6, u6], [u6], []] in its original supplied open-set order, and compact_subset K=[u6]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, complement, Hausdorff, compact, closed, boundary).","answer":{"kind":"expression","value":"([o6, u6], [u6], [o6], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)","explanation":"Require this actual finite space to be non-Hausdorff and K not closed. K remains compact by finiteness. This is a counterexample to removing the Hausdorff hypothesis, not a claim that every non-Hausdorff compact subset is nonclosed or a converse counterexample. The exact result, retaining the requested order, is ([o6, u6], [u6], [o6], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)."},"skills":["A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0083","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"418d91387e01734025ba1da29bad5f98fd91974062449b4effafc6c461d67208","family_id":"qf-mod43-compactclosed-nonhausdorff-boundary","seed":202611123083,"semantic_template":"qf-mod43-compactclosed-nonhausdorff-boundary/{\"compact_subset\":[\"u7\"],\"points\":[\"o7\",\"u7\"],\"topology\":[[\"o7\",\"u7\"],[\"u7\"],[]]}"},"fields":{"title":"A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis","stem":"Use complete finite labeled carrier=[o7, u7], topology=[[o7, u7], [u7], []] in its original supplied open-set order, and compact_subset K=[u7]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, complement, Hausdorff, compact, closed, boundary).","answer":{"kind":"expression","value":"([o7, u7], [u7], [o7], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)","explanation":"Require this actual finite space to be non-Hausdorff and K not closed. K remains compact by finiteness. This is a counterexample to removing the Hausdorff hypothesis, not a claim that every non-Hausdorff compact subset is nonclosed or a converse counterexample. The exact result, retaining the requested order, is ([o7, u7], [u7], [o7], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)."},"skills":["A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0084","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"31a6be55ee2fd59fead7cb4aa22ef1a0f070027272674f1644a709ab58bb7c1d","family_id":"qf-mod43-compactclosed-nonhausdorff-boundary","seed":202611123084,"semantic_template":"qf-mod43-compactclosed-nonhausdorff-boundary/{\"compact_subset\":[\"u8\"],\"points\":[\"o8\",\"u8\"],\"topology\":[[\"o8\",\"u8\"],[\"u8\"],[]]}"},"fields":{"title":"A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis","stem":"Use complete finite labeled carrier=[o8, u8], topology=[[o8, u8], [u8], []] in its original supplied open-set order, and compact_subset K=[u8]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, complement, Hausdorff, compact, closed, boundary).","answer":{"kind":"expression","value":"([o8, u8], [u8], [o8], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)","explanation":"Require this actual finite space to be non-Hausdorff and K not closed. K remains compact by finiteness. This is a counterexample to removing the Hausdorff hypothesis, not a claim that every non-Hausdorff compact subset is nonclosed or a converse counterexample. The exact result, retaining the requested order, is ([o8, u8], [u8], [o8], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)."},"skills":["A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0085","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2377916349dcc5387af7a45b5a39b343fa6f296eede11993e1a65c1149c2060f","family_id":"qf-mod43-compactclosed-nonhausdorff-boundary","seed":202611123085,"semantic_template":"qf-mod43-compactclosed-nonhausdorff-boundary/{\"compact_subset\":[\"u9\"],\"points\":[\"o9\",\"u9\"],\"topology\":[[\"o9\",\"u9\"],[\"u9\"],[]]}"},"fields":{"title":"A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis","stem":"Use complete finite labeled carrier=[o9, u9], topology=[[o9, u9], [u9], []] in its original supplied open-set order, and compact_subset K=[u9]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, complement, Hausdorff, compact, closed, boundary).","answer":{"kind":"expression","value":"([o9, u9], [u9], [o9], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)","explanation":"Require this actual finite space to be non-Hausdorff and K not closed. K remains compact by finiteness. This is a counterexample to removing the Hausdorff hypothesis, not a claim that every non-Hausdorff compact subset is nonclosed or a converse counterexample. The exact result, retaining the requested order, is ([o9, u9], [u9], [o9], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)."},"skills":["A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0086","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"681d982f34e42e739511193ecd06be9d764d6be397174c1514c11ccc843382f3","family_id":"qf-mod43-compactclosed-nonhausdorff-boundary","seed":202611123086,"semantic_template":"qf-mod43-compactclosed-nonhausdorff-boundary/{\"compact_subset\":[\"u10\"],\"points\":[\"o10\",\"u10\"],\"topology\":[[\"o10\",\"u10\"],[\"u10\"],[]]}"},"fields":{"title":"A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis","stem":"Use complete finite labeled carrier=[o10, u10], topology=[[o10, u10], [u10], []] in its original supplied open-set order, and compact_subset K=[u10]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, complement, Hausdorff, compact, closed, boundary).","answer":{"kind":"expression","value":"([o10, u10], [u10], [o10], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)","explanation":"Require this actual finite space to be non-Hausdorff and K not closed. K remains compact by finiteness. This is a counterexample to removing the Hausdorff hypothesis, not a claim that every non-Hausdorff compact subset is nonclosed or a converse counterexample. The exact result, retaining the requested order, is ([o10, u10], [u10], [o10], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)."},"skills":["A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0087","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a5f9e2f141fd608da12c2ac5fefd3330d30fd4ef7f074301937fba95a7fa3bde","family_id":"qf-mod43-compactclosed-nonhausdorff-boundary","seed":202611123087,"semantic_template":"qf-mod43-compactclosed-nonhausdorff-boundary/{\"compact_subset\":[\"u11\"],\"points\":[\"o11\",\"u11\"],\"topology\":[[\"o11\",\"u11\"],[\"u11\"],[]]}"},"fields":{"title":"A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis","stem":"Use complete finite labeled carrier=[o11, u11], topology=[[o11, u11], [u11], []] in its original supplied open-set order, and compact_subset K=[u11]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, complement, Hausdorff, compact, closed, boundary).","answer":{"kind":"expression","value":"([o11, u11], [u11], [o11], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)","explanation":"Require this actual finite space to be non-Hausdorff and K not closed. K remains compact by finiteness. This is a counterexample to removing the Hausdorff hypothesis, not a claim that every non-Hausdorff compact subset is nonclosed or a converse counterexample. The exact result, retaining the requested order, is ([o11, u11], [u11], [o11], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)."},"skills":["A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0088","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5628221079247ce555d628d9581b4308d18b7323ba5651e392f3914a06276973","family_id":"qf-mod43-compactclosed-nonhausdorff-boundary","seed":202611123088,"semantic_template":"qf-mod43-compactclosed-nonhausdorff-boundary/{\"compact_subset\":[\"u12\"],\"points\":[\"o12\",\"u12\"],\"topology\":[[\"o12\",\"u12\"],[\"u12\"],[]]}"},"fields":{"title":"A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis","stem":"Use complete finite labeled carrier=[o12, u12], topology=[[o12, u12], [u12], []] in its original supplied open-set order, and compact_subset K=[u12]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, complement, Hausdorff, compact, closed, boundary).","answer":{"kind":"expression","value":"([o12, u12], [u12], [o12], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)","explanation":"Require this actual finite space to be non-Hausdorff and K not closed. K remains compact by finiteness. This is a counterexample to removing the Hausdorff hypothesis, not a claim that every non-Hausdorff compact subset is nonclosed or a converse counterexample. The exact result, retaining the requested order, is ([o12, u12], [u12], [o12], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)."},"skills":["A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0089","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b492851fc45b55bb8fb278d052bd1e7113d37e73e4249edb0924079023d65982","family_id":"qf-mod43-compactclosed-nonhausdorff-boundary","seed":202611123089,"semantic_template":"qf-mod43-compactclosed-nonhausdorff-boundary/{\"compact_subset\":[\"u13\"],\"points\":[\"o13\",\"u13\"],\"topology\":[[\"o13\",\"u13\"],[\"u13\"],[]]}"},"fields":{"title":"A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis","stem":"Use complete finite labeled carrier=[o13, u13], topology=[[o13, u13], [u13], []] in its original supplied open-set order, and compact_subset K=[u13]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, complement, Hausdorff, compact, closed, boundary).","answer":{"kind":"expression","value":"([o13, u13], [u13], [o13], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)","explanation":"Require this actual finite space to be non-Hausdorff and K not closed. K remains compact by finiteness. This is a counterexample to removing the Hausdorff hypothesis, not a claim that every non-Hausdorff compact subset is nonclosed or a converse counterexample. The exact result, retaining the requested order, is ([o13, u13], [u13], [o13], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)."},"skills":["A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0090","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5abae3a344d1235b1135fab35ab62374a42796a92fc3aec3c60680661b015e2a","family_id":"qf-mod43-compactclosed-nonhausdorff-boundary","seed":202611123090,"semantic_template":"qf-mod43-compactclosed-nonhausdorff-boundary/{\"compact_subset\":[\"u14\"],\"points\":[\"o14\",\"u14\"],\"topology\":[[\"o14\",\"u14\"],[\"u14\"],[]]}"},"fields":{"title":"A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis","stem":"Use complete finite labeled carrier=[o14, u14], topology=[[o14, u14], [u14], []] in its original supplied open-set order, and compact_subset K=[u14]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, complement, Hausdorff, compact, closed, boundary).","answer":{"kind":"expression","value":"([o14, u14], [u14], [o14], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)","explanation":"Require this actual finite space to be non-Hausdorff and K not closed. K remains compact by finiteness. This is a counterexample to removing the Hausdorff hypothesis, not a claim that every non-Hausdorff compact subset is nonclosed or a converse counterexample. The exact result, retaining the requested order, is ([o14, u14], [u14], [o14], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)."},"skills":["A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0091","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"80ec198e74b304509898e1d6f34a49e55cf3d6921fd7d80636360e3d855dd687","family_id":"qf-mod43-compactclosed-nonhausdorff-boundary","seed":202611123091,"semantic_template":"qf-mod43-compactclosed-nonhausdorff-boundary/{\"compact_subset\":[\"u15\"],\"points\":[\"o15\",\"u15\"],\"topology\":[[\"o15\",\"u15\"],[\"u15\"],[]]}"},"fields":{"title":"A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis","stem":"Use complete finite labeled carrier=[o15, u15], topology=[[o15, u15], [u15], []] in its original supplied open-set order, and compact_subset K=[u15]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, complement, Hausdorff, compact, closed, boundary).","answer":{"kind":"expression","value":"([o15, u15], [u15], [o15], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)","explanation":"Require this actual finite space to be non-Hausdorff and K not closed. K remains compact by finiteness. This is a counterexample to removing the Hausdorff hypothesis, not a claim that every non-Hausdorff compact subset is nonclosed or a converse counterexample. The exact result, retaining the requested order, is ([o15, u15], [u15], [o15], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)."},"skills":["A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0092","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d287a533c33e553144a702a84e012f0e96bb6d226448c33b9728311f3d3fbc8b","family_id":"qf-mod43-compactclosed-nonhausdorff-boundary","seed":202611123092,"semantic_template":"qf-mod43-compactclosed-nonhausdorff-boundary/{\"compact_subset\":[\"u16\"],\"points\":[\"o16\",\"u16\"],\"topology\":[[\"o16\",\"u16\"],[\"u16\"],[]]}"},"fields":{"title":"A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis","stem":"Use complete finite labeled carrier=[o16, u16], topology=[[o16, u16], [u16], []] in its original supplied open-set order, and compact_subset K=[u16]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, complement, Hausdorff, compact, closed, boundary).","answer":{"kind":"expression","value":"([o16, u16], [u16], [o16], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)","explanation":"Require this actual finite space to be non-Hausdorff and K not closed. K remains compact by finiteness. This is a counterexample to removing the Hausdorff hypothesis, not a claim that every non-Hausdorff compact subset is nonclosed or a converse counterexample. The exact result, retaining the requested order, is ([o16, u16], [u16], [o16], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)."},"skills":["A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0093","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bbb83bb16baa83aaac154e71432d4b26ddcc2de08431d6e6eca0a75250922ee3","family_id":"qf-mod43-compactclosed-nonhausdorff-boundary","seed":202611123093,"semantic_template":"qf-mod43-compactclosed-nonhausdorff-boundary/{\"compact_subset\":[\"u17\"],\"points\":[\"o17\",\"u17\"],\"topology\":[[\"o17\",\"u17\"],[\"u17\"],[]]}"},"fields":{"title":"A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis","stem":"Use complete finite labeled carrier=[o17, u17], topology=[[o17, u17], [u17], []] in its original supplied open-set order, and compact_subset K=[u17]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, complement, Hausdorff, compact, closed, boundary).","answer":{"kind":"expression","value":"([o17, u17], [u17], [o17], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)","explanation":"Require this actual finite space to be non-Hausdorff and K not closed. K remains compact by finiteness. This is a counterexample to removing the Hausdorff hypothesis, not a claim that every non-Hausdorff compact subset is nonclosed or a converse counterexample. The exact result, retaining the requested order, is ([o17, u17], [u17], [o17], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)."},"skills":["A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0094","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b916ad2f561da40cd5ec9ae9b596745215b5996486ed6a88286605a4c9340905","family_id":"qf-mod43-compactclosed-nonhausdorff-boundary","seed":202611123094,"semantic_template":"qf-mod43-compactclosed-nonhausdorff-boundary/{\"compact_subset\":[\"u18\"],\"points\":[\"o18\",\"u18\"],\"topology\":[[\"o18\",\"u18\"],[\"u18\"],[]]}"},"fields":{"title":"A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis","stem":"Use complete finite labeled carrier=[o18, u18], topology=[[o18, u18], [u18], []] in its original supplied open-set order, and compact_subset K=[u18]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, complement, Hausdorff, compact, closed, boundary).","answer":{"kind":"expression","value":"([o18, u18], [u18], [o18], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)","explanation":"Require this actual finite space to be non-Hausdorff and K not closed. K remains compact by finiteness. This is a counterexample to removing the Hausdorff hypothesis, not a claim that every non-Hausdorff compact subset is nonclosed or a converse counterexample. The exact result, retaining the requested order, is ([o18, u18], [u18], [o18], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)."},"skills":["A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0095","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e66a99f38737e618817caed8af2c176ab42eebc83cfdde8cd7b669335b5dc0c1","family_id":"qf-mod43-compactclosed-nonhausdorff-boundary","seed":202611123095,"semantic_template":"qf-mod43-compactclosed-nonhausdorff-boundary/{\"compact_subset\":[\"u19\"],\"points\":[\"o19\",\"u19\"],\"topology\":[[\"o19\",\"u19\"],[\"u19\"],[]]}"},"fields":{"title":"A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis","stem":"Use complete finite labeled carrier=[o19, u19], topology=[[o19, u19], [u19], []] in its original supplied open-set order, and compact_subset K=[u19]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, complement, Hausdorff, compact, closed, boundary).","answer":{"kind":"expression","value":"([o19, u19], [u19], [o19], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)","explanation":"Require this actual finite space to be non-Hausdorff and K not closed. K remains compact by finiteness. This is a counterexample to removing the Hausdorff hypothesis, not a claim that every non-Hausdorff compact subset is nonclosed or a converse counterexample. The exact result, retaining the requested order, is ([o19, u19], [u19], [o19], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)."},"skills":["A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0096","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"194f8ade972d03a4f697e021e9c2947b1a95dd1e80d2eca9fda37bbbb8198287","family_id":"qf-mod43-compactclosed-nonhausdorff-boundary","seed":202611123096,"semantic_template":"qf-mod43-compactclosed-nonhausdorff-boundary/{\"compact_subset\":[\"u20\"],\"points\":[\"o20\",\"u20\"],\"topology\":[[\"o20\",\"u20\"],[\"u20\"],[]]}"},"fields":{"title":"A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis","stem":"Use complete finite labeled carrier=[o20, u20], topology=[[o20, u20], [u20], []] in its original supplied open-set order, and compact_subset K=[u20]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, complement, Hausdorff, compact, closed, boundary).","answer":{"kind":"expression","value":"([o20, u20], [u20], [o20], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)","explanation":"Require this actual finite space to be non-Hausdorff and K not closed. K remains compact by finiteness. This is a counterexample to removing the Hausdorff hypothesis, not a claim that every non-Hausdorff compact subset is nonclosed or a converse counterexample. The exact result, retaining the requested order, is ([o20, u20], [u20], [o20], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)."},"skills":["A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0097","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4dab81235876c89b92a3146acb74a76e0f4c1ab396d97b97740c6f3284b021e5","family_id":"qf-mod43-compactclosed-nonhausdorff-boundary","seed":202611123097,"semantic_template":"qf-mod43-compactclosed-nonhausdorff-boundary/{\"compact_subset\":[\"u21\"],\"points\":[\"o21\",\"u21\"],\"topology\":[[\"o21\",\"u21\"],[\"u21\"],[]]}"},"fields":{"title":"A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis","stem":"Use complete finite labeled carrier=[o21, u21], topology=[[o21, u21], [u21], []] in its original supplied open-set order, and compact_subset K=[u21]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, complement, Hausdorff, compact, closed, boundary).","answer":{"kind":"expression","value":"([o21, u21], [u21], [o21], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)","explanation":"Require this actual finite space to be non-Hausdorff and K not closed. K remains compact by finiteness. This is a counterexample to removing the Hausdorff hypothesis, not a claim that every non-Hausdorff compact subset is nonclosed or a converse counterexample. The exact result, retaining the requested order, is ([o21, u21], [u21], [o21], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)."},"skills":["A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0098","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"56eebd6327b94e2f4d5bb1d0357aa5f72292db542e4e5847d04cf0f69307dedd","family_id":"qf-mod43-compactclosed-nonhausdorff-boundary","seed":202611123098,"semantic_template":"qf-mod43-compactclosed-nonhausdorff-boundary/{\"compact_subset\":[\"u22\"],\"points\":[\"o22\",\"u22\"],\"topology\":[[\"o22\",\"u22\"],[\"u22\"],[]]}"},"fields":{"title":"A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis","stem":"Use complete finite labeled carrier=[o22, u22], topology=[[o22, u22], [u22], []] in its original supplied open-set order, and compact_subset K=[u22]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, complement, Hausdorff, compact, closed, boundary).","answer":{"kind":"expression","value":"([o22, u22], [u22], [o22], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)","explanation":"Require this actual finite space to be non-Hausdorff and K not closed. K remains compact by finiteness. This is a counterexample to removing the Hausdorff hypothesis, not a claim that every non-Hausdorff compact subset is nonclosed or a converse counterexample. The exact result, retaining the requested order, is ([o22, u22], [u22], [o22], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)."},"skills":["A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0099","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"284fb7a72b90dded703176b032b03e19f172ad52d4fe5eaf5756517197201536","family_id":"qf-mod43-compactclosed-nonhausdorff-boundary","seed":202611123099,"semantic_template":"qf-mod43-compactclosed-nonhausdorff-boundary/{\"compact_subset\":[\"u23\"],\"points\":[\"o23\",\"u23\"],\"topology\":[[\"o23\",\"u23\"],[\"u23\"],[]]}"},"fields":{"title":"A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis","stem":"Use complete finite labeled carrier=[o23, u23], topology=[[o23, u23], [u23], []] in its original supplied open-set order, and compact_subset K=[u23]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, complement, Hausdorff, compact, closed, boundary).","answer":{"kind":"expression","value":"([o23, u23], [u23], [o23], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)","explanation":"Require this actual finite space to be non-Hausdorff and K not closed. K remains compact by finiteness. This is a counterexample to removing the Hausdorff hypothesis, not a claim that every non-Hausdorff compact subset is nonclosed or a converse counterexample. The exact result, retaining the requested order, is ([o23, u23], [u23], [o23], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)."},"skills":["A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0100","batch_id":"batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7750ce00f3498062691e5d995c56ceb3a6c0559a4feec9d1d317b72845919a5f","family_id":"qf-mod43-compactclosed-nonhausdorff-boundary","seed":202611123100,"semantic_template":"qf-mod43-compactclosed-nonhausdorff-boundary/{\"compact_subset\":[\"u24\"],\"points\":[\"o24\",\"u24\"],\"topology\":[[\"o24\",\"u24\"],[\"u24\"],[]]}"},"fields":{"title":"A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis","stem":"Use complete finite labeled carrier=[o24, u24], topology=[[o24, u24], [u24], []] in its original supplied open-set order, and compact_subset K=[u24]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, complement, Hausdorff, compact, closed, boundary).","answer":{"kind":"expression","value":"([o24, u24], [u24], [o24], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)","explanation":"Require this actual finite space to be non-Hausdorff and K not closed. K remains compact by finiteness. This is a counterexample to removing the Hausdorff hypothesis, not a claim that every non-Hausdorff compact subset is nonclosed or a converse counterexample. The exact result, retaining the requested order, is ([o24, u24], [u24], [o24], No, Yes, No, Sierpinski_type_space_shows_Hausdorff_hypothesis_cannot_be_removed_and_converse_is_not_claimed)."},"skills":["A Non-Hausdorff Compact Nonclosed Instance Blocks Deleting the Hypothesis"],"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."}}