{ "schema_version": "ocme-reviewed-question-reader-v1", "question_id": "q-20261014-advance-20261101-r02-closure-and-interior-0055", "pilot_number": "R0032-16530", "number": 16530, "family_id": "qf-mod32-topology-closure-membership", "title": { "zh-Hant": "第 16530 題：閉包與內部", "en": "Question 16530: Closure Membership May Hold outside the Original Subset" }, "original_question": { "schema_version": "ocme-question-v0.1", "id": "q-20261014-advance-20261101-r02-closure-and-interior-0055", "batch_id": "batch-20261014-advance-20261101-r02-closure-and-interior-foundations", "language": "zh-Hant", "source_mko_ids": [ "mko-closure-and-interior" ], "title_zh": "閉包與內部", "stem_zh": "完整有限拓撲與同一載體子集。操作=topology-closure-membership；資料={\"n\":5,\"topology\":[[],[0],[0,1],[0,1,4],[0,1,2,4],[0,1,2,3,4]],\"subset\":[1,4],\"point\":3}。請依 [point,closure,in_closure,boundary] 欄序回答。", "answer": { "kind": "expression", "value": "(3, [1, 2, 3, 4], 是, topology_relative_closure)", "explanation_zh": "Wave5根節點核對=(3, [1, 2, 3, 4], 是, topology_relative_closure)；依完整明示資料重算；閉包取全部閉超集之交且內部取全部開子集之聯；結果相對指定拓撲；未供度量時不使用度量推論。" }, "difficulty": { "band": "undergraduate_core", "level": 1, "skills": [ "閉包與內部" ] }, "variant": { "family_id": "qf-mod32-topology-closure-membership", "seed": 202611012055, "parameters": { "n": 5, "topology": [ [], [ 0 ], [ 0, 1 ], [ 0, 1, 4 ], [ 0, 1, 2, 4 ], [ 0, 1, 2, 3, 4 ] ], "subset": [ 1, 4 ], "point": 3, "domain": "Complete_finite_topology_subset", "operation": "topology-closure-membership", "semantic_template": "qf-mod32-topology-closure-membership/{\"n\":5,\"point\":3,\"subset\":[1,4],\"topology\":[[],[0,1,2,3,4],[0,1,2,4],[0,1,4],[0,1],[0]]}" } }, "verification": { "status": "candidate", "checks": [ "producer_exact_finite_or_symbolic_model", "producer_source_snapshot_check", "producer_substantive_template_deduplication", "producer_json_schema" ], "warning_zh": "候選須由主端獨立重算；不構成人工審定、一般compactness／a.s.定理／maximum matching／語言等價、Lean或形式Evidence。" }, "provenance": { "generator_role": "question_producer", "generator_model": "model_unknown", "generated_at": "2026-10-14", "contract_version": "ocme-question-factory-v0.1" } }, "original_english_edition": { "schema_version": "ocme-question-localization-v0.1", "question_id": "q-20261014-advance-20261101-r02-closure-and-interior-0055", "batch_id": "batch-20261014-advance-20261101-r02-closure-and-interior-foundations", "locale": "en", "translation_state": "ai_candidate", "source": { "question_sha256": "aba918de903d9cbd79744d121626a557d8e59567f69888782c88815310d006d1", "family_id": "qf-mod32-topology-closure-membership", "seed": 202611012055, "semantic_template": "qf-mod32-topology-closure-membership/{\"n\":5,\"point\":3,\"subset\":[1,4],\"topology\":[[],[0,1,2,3,4],[0,1,2,4],[0,1,4],[0,1],[0]]}" }, "fields": { "title": "Closure Membership May Hold outside the Original Subset", "stem": "Use the complete finite topology=[[], [0], [0, 1], [0, 1, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Every subset is an increasing duplicate-free carrier list. First check the unique empty/whole opens and all union/intersection closure obligations; finite topology suffices for exhaustive closure checks. Complements refer to this carrier. No metric is silently supplied. Subset=[1, 4], point=3. Give (point,closure,in_closure?,topology_relative_closure).", "answer": { "kind": "expression", "value": "(3, [1, 2, 3, 4], Yes, topology_relative_closure)", "explanation": "Compute the full intersection of closed supersets first, then test this point against that closure. Do not replace closure membership with membership in the original subset. The exact result, retaining the requested order, is (3, [1, 2, 3, 4], Yes, topology_relative_closure)." }, "skills": [ "Closure Membership May Hold outside the Original 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." } }, "givens": [ { "id": "G001", "text": { "zh-Hant": "完整有限拓撲與同一載體子集。操作=topology-closure-membership；資料={\"n\":5,\"topology\":[[],[0],[0,1],[0,1,4],[0,1,2,4],[0,1,2,3,4]],\"subset\":[1,4],\"point\":3}。請依 [point,closure,in_closure,boundary] 欄序回答。", "en": "Use the complete finite topology=[[], [0], [0, 1], [0, 1, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Every subset is an increasing duplicate-free carrier list. First check the unique empty/whole opens and all union/intersection closure obligations; finite topology suffices for exhaustive closure checks. Complements refer to this carrier. No metric is silently supplied. Subset=[1, 4], point=3. Give (point,closure,in_closure?,topology_relative_closure)." } } ], "steps": [ { "id": "S001", "uses": [ "G001" ], "before": "same complete carrier Fin(5);original ordered complete topology=[[], [0], [0, 1], [0, 1, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]]", "after": "unique empty/whole opens=是;ordered open pairs checked=36;first union failure=null;first intersection failure=null", "relation": "implies", "rule_id": "topology_complete_finite_parent_all_pairs", "conditions": [], "explanation": { "zh-Hant": "個別子集須照原載體遞增且無重複，拓撲族須有唯一空集、全集並核全部二元聯／交。因不同開集有限，任意聯只涉及有限種開集，二元封閉加空集便足夠；不另供度量。保留拓撲原順序。", "en": "Each subset is increasing and duplicate-free.The topology has unique empty/whole opens and closure under every binary union/intersection.Finitely many distinct opens reduce arbitrary unions to finite ones,handled by binary closure and the empty set.No metric is supplied.Keep original topology order." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S002", "uses": [ "G001", "S001" ], "before": "selected subset:original open index=0;open=[];same-carrier closed complement=[0, 1, 2, 3, 4];subset=[1, 4]", "after": "closed set contains whole subset=是;running intersection of qualifying closed sets=[0, 1, 2, 3, 4]", "relation": "implies", "rule_id": "closure_every_closed_superset_original_open_order", "conditions": [], "explanation": { "zh-Hant": "逐每個原開集取同載體補集成閉集，只有包含整個A才保留；反覆交所有合格者，而非挑一個方便閉超集。即使某點不在A，仍可能在最後交集中。", "en": "Complement every original open in the same carrier to obtain all closed sets.Keep exactly those containing all of A,and intersect every qualifying set rather than one convenient superset.A point outside A can still lie in the final intersection." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S003", "uses": [ "G001", "S002" ], "before": "selected subset:original open index=1;open=[0];same-carrier closed complement=[1, 2, 3, 4];subset=[1, 4]", "after": "closed set contains whole subset=是;running intersection of qualifying closed sets=[1, 2, 3, 4]", "relation": "implies", "rule_id": "closure_every_closed_superset_original_open_order", "conditions": [], "explanation": { "zh-Hant": "逐每個原開集取同載體補集成閉集，只有包含整個A才保留；反覆交所有合格者，而非挑一個方便閉超集。即使某點不在A，仍可能在最後交集中。", "en": "Complement every original open in the same carrier to obtain all closed sets.Keep exactly those containing all of A,and intersect every qualifying set rather than one convenient superset.A point outside A can still lie in the final intersection." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S004", "uses": [ "G001", "S003" ], "before": "selected subset:original open index=2;open=[0, 1];same-carrier closed complement=[2, 3, 4];subset=[1, 4]", "after": "closed set contains whole subset=否;running intersection of qualifying closed sets=[1, 2, 3, 4]", "relation": "implies", "rule_id": "closure_every_closed_superset_original_open_order", "conditions": [], "explanation": { "zh-Hant": "逐每個原開集取同載體補集成閉集，只有包含整個A才保留；反覆交所有合格者，而非挑一個方便閉超集。即使某點不在A，仍可能在最後交集中。", "en": "Complement every original open in the same carrier to obtain all closed sets.Keep exactly those containing all of A,and intersect every qualifying set rather than one convenient superset.A point outside A can still lie in the final intersection." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S005", "uses": [ "G001", "S004" ], "before": "selected subset:original open index=3;open=[0, 1, 4];same-carrier closed complement=[2, 3];subset=[1, 4]", "after": "closed set contains whole subset=否;running intersection of qualifying closed sets=[1, 2, 3, 4]", "relation": "implies", "rule_id": "closure_every_closed_superset_original_open_order", "conditions": [], "explanation": { "zh-Hant": "逐每個原開集取同載體補集成閉集，只有包含整個A才保留；反覆交所有合格者，而非挑一個方便閉超集。即使某點不在A，仍可能在最後交集中。", "en": "Complement every original open in the same carrier to obtain all closed sets.Keep exactly those containing all of A,and intersect every qualifying set rather than one convenient superset.A point outside A can still lie in the final intersection." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S006", "uses": [ "G001", "S005" ], "before": "selected subset:original open index=4;open=[0, 1, 2, 4];same-carrier closed complement=[3];subset=[1, 4]", "after": "closed set contains whole subset=否;running intersection of qualifying closed sets=[1, 2, 3, 4]", "relation": "implies", "rule_id": "closure_every_closed_superset_original_open_order", "conditions": [], "explanation": { "zh-Hant": "逐每個原開集取同載體補集成閉集，只有包含整個A才保留；反覆交所有合格者，而非挑一個方便閉超集。即使某點不在A，仍可能在最後交集中。", "en": "Complement every original open in the same carrier to obtain all closed sets.Keep exactly those containing all of A,and intersect every qualifying set rather than one convenient superset.A point outside A can still lie in the final intersection." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S007", "uses": [ "G001", "S006" ], "before": "selected subset:original open index=5;open=[0, 1, 2, 3, 4];same-carrier closed complement=[];subset=[1, 4]", "after": "closed set contains whole subset=否;running intersection of qualifying closed sets=[1, 2, 3, 4]", "relation": "implies", "rule_id": "closure_every_closed_superset_original_open_order", "conditions": [], "explanation": { "zh-Hant": "逐每個原開集取同載體補集成閉集，只有包含整個A才保留；反覆交所有合格者，而非挑一個方便閉超集。即使某點不在A，仍可能在最後交集中。", "en": "Complement every original open in the same carrier to obtain all closed sets.Keep exactly those containing all of A,and intersect every qualifying set rather than one convenient superset.A point outside A can still lie in the final intersection." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S008", "uses": [ "G001", "S007" ], "before": "selected subset:ALL qualifying closed supersets in original complement order=[[0, 1, 2, 3, 4], [1, 2, 3, 4]]", "after": "topology-relative closure=[1, 2, 3, 4]", "relation": "implies", "rule_id": "closure_all_candidates_intersection_not_metric_guess", "conditions": [], "explanation": { "zh-Hant": "全集是閉超集，所以交的家族不缺；所得是相對這份拓撲最小閉超集，原序只影響候選列表，不改集合值。", "en": "The whole carrier is a closed superset,so the family is present.The intersection is the least closed superset in this topology.Original order affects the candidate list,not its set value." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S009", "uses": [ "G001", "S008" ], "before": "original point=3;original subset=[1, 4];complete computed set=[1, 2, 3, 4]", "after": "membership=是;scope=topology_relative_closure", "relation": "implies", "rule_id": "topology_point_test_computed_closure_or_interior", "conditions": [], "explanation": { "zh-Hant": "先算完整閉包／內部才查指定點；不能拿原子集的成員資格或未指定的距離替代。", "en": "Compute the entire closure/interior before testing the original point.Do not substitute original-subset membership or an unspecified distance." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null } ], "math_model": { "result": [ 3, [ 1, 2, 3, 4 ], true, "topology_relative_closure" ] }, "conclusion": { "canonical_value": "(3, [1, 2, 3, 4], 是, topology_relative_closure)", "explanation": { "zh-Hant": "依上列完整代值及條件核對得到原題答案；原題編號、格式與狀態保持不變。", "en": "The full substitutions and premise checks give the original answer; source identity,format and status are unchanged." } }, "definition_note": { "zh-Hant": "沿用原題明示的論域、假設與來源 MKO 定義。本題已完成第二對話數理與中英語義覆核，並由原始參數逐題獨立重算模型、答案與每個算術步驟；原題仍為 candidate。", "en": "Use the original question domain, assumptions and cited MKO definitions. This item has second-context mathematical and bilingual review, with the model, answer and every arithmetic step independently recomputed from the original parameters; the source question remains a candidate." }, "source": { "input_commit": "0f7be61e2c22317a4075561dc822c87d881a50db", "question_binding": { "batch_id": "batch-20261014-advance-20261101-r02-closure-and-interior-foundations", "batch_path": "public/data/questions/batches/batch-20261014-advance-20261101-r02-closure-and-interior-foundations/questions.jsonl", "batch_sha256": "32539df9729bb1d6629f5ef91723b5f8bd61f670d31a9dd806d336ccb48d1e62", "original_line": 55, "original_line_sha256": "a2435328650b4ab0679fbdc08fac46b7cde5478591afdc6c71119c76e36913d4", "question_compact_sha256": "aba918de903d9cbd79744d121626a557d8e59567f69888782c88815310d006d1", "english_row_compact_sha256": "6d4dbce58792b66038e2284571970af6b89e58e6f70637a48602f21c6359b7f6", "english_manifest_path": "public/data/i18n/editions/batch-20261014-advance-20261101-r02-closure-and-interior-foundations.en.json", "english_rows_path": "public/data/i18n/questions/batch-20261014-advance-20261101-r02-closure-and-interior-foundations.en.jsonl" }, "mko_bindings": { "mko-closure-and-interior": { "sha256": "72aef7da5935f151015d5eecaa4d9cdde8226564638885f222596bf4ef2433e4", "english_sha256": "8dc1d3fac60ca206027663c117e374595aff835c923f99ec01bdb7f6451fd1d2", "original_path": "public/data/mko/closure-and-interior.json", "english_path": "public/data/i18n/editions/mko-closure-and-interior.en.json" } }, "selected_manifest_sha256": "5233edba9654a6cffd8b6b2d83c932f82ded15f9c0b7c733bd049943e672acff", "selected_part_sha256": "fe4c48735753bdd58c10db9b2ca3a2a3baeb31d0424c3fc09dea62317185ddae", "original_candidate_sha256": "aac0c433fe84ab1f8b957a5404f8d8c86b0598de823b66f06244849b473180e0", "original_candidate_line_sha256": "aac0c433fe84ab1f8b957a5404f8d8c86b0598de823b66f06244849b473180e0" }, "historical_source_state": { "review_state": { "derivation": "candidate_authored_and_locally_checked", "source_question_state": "candidate", "independent_AI_or_human_review": "NOT_PERFORMED", "formalization": "not_formalized", "Lean": "NOT_RUN" }, "scope_note": { "zh-Hant": "來源綁定教學推導候選；精確核對不等於獨立AI／人工審查或一般形式證明。原題、答案、MKO、公式、AST與歷史不變。", "en": "Source-bound pedagogical derivation candidate.Exact checking is not independent AI/human review or general formal proof.Original questions,answers,MKOs,formulas,ASTs and history remain unchanged." }, "label": "Frozen producer metadata; current per-question review is separate below." }, "revision": { "id": "review-20261026-007-p33", "review_date": "2026-10-26", "owner_authorized_reader_publication": true, "scope": "One hundred sixty family prototypes and their mathematical/bilingual reasoning variants, stems, boundary statements and every distinct explanation shape were reviewed in a context independent of the producer; parameter substitutions, source bindings, full math models, both answer strings and every arithmetic step were independently reconstructed per question from the original parameters only, with negative controls. This grouped method does not claim a separate literal AI reading of every repeated sentence or an extra AI voter.", "review_receipts": { "semantic_review_sha256": "d64b4e08bda1c25dd393b7c7e4140a8734610a9b84ae47e6f0adcff7f30a0b8e", "question_first_witness_sha256": "790fd6623c04ba52a6ffb0a95cea52b532a02b23bb9453926d2d9f7a825fdec8" }, "status": "second_context_reviewed", "mathematical_semantics": true, "bilingual_semantics": true, "steps_reviewed": 9, "method": "family_template_semantic_review_plus_independent_per_instance_reconstruction", "exact_question_first_check": { "model_sha256": "633c90ed61928564dc5c06526ed2a430b1db70ec19b5b5202fabb251d7d55e5a", "arithmetic_steps": 0 }, "uses_answer_as_premise": false, "human_review": "NOT_PERFORMED", "formalization": "not_formalized", "Lean": "NOT_RUN", "Canonical_truth_promotion": false, "authority": "Owner-authorized reviewed reader publication; original question/candidate states remain historical and unchanged." } }