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第 16540 題:閉包與內部

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  1. 原題與已知條件
  2. 定義與適用範圍
  3. 逐步推導
  4. 逐步反向核對
  5. 結論
  6. 本題覆核與來源綁定

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← 逐題詳解目錄 · 原題批次

R0032-16540 · q-20261014-advance-20261101-r02-closure-and-interior-0065

原題與已知條件

完整有限拓撲與同一載體子集。操作=topology-closure-membership;資料={"n":5,"topology":[[],[0],[0,3],[0,2,3],[0,1,2,3],[0,1,2,3,4]],"subset":[2,3],"point":4}。請依 [point,closure,in_closure,boundary] 欄序回答。

  • G001 — 完整有限拓撲與同一載體子集。操作=topology-closure-membership;資料={"n":5,"topology":[[],[0],[0,3],[0,2,3],[0,1,2,3],[0,1,2,3,4]],"subset":[2,3],"point":4}。請依 [point,closure,in_closure,boundary] 欄序回答。

定義與適用範圍

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。

逐步推導

S001 · implies

操作前

same complete carrier Fin(5);original ordered complete topology=[[], [0], [0, 3], [0, 2, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]]

操作後

unique empty/whole opens=是;ordered open pairs checked=36;first union failure=null;first intersection failure=null

條件:無

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。

步驟識別與依賴

規則:topology_complete_finite_parent_all_pairs · 依賴:G001

S002 · implies

操作前

selected subset:original open index=0;open=[];same-carrier closed complement=[0, 1, 2, 3, 4];subset=[2, 3]

操作後

closed set contains whole subset=是;running intersection of qualifying closed sets=[0, 1, 2, 3, 4]

條件:無

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。

步驟識別與依賴

規則:closure_every_closed_superset_original_open_order · 依賴:G001, S001

S003 · implies

操作前

selected subset:original open index=1;open=[0];same-carrier closed complement=[1, 2, 3, 4];subset=[2, 3]

操作後

closed set contains whole subset=是;running intersection of qualifying closed sets=[1, 2, 3, 4]

條件:無

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。

步驟識別與依賴

規則:closure_every_closed_superset_original_open_order · 依賴:G001, S002

S004 · implies

操作前

selected subset:original open index=2;open=[0, 3];same-carrier closed complement=[1, 2, 4];subset=[2, 3]

操作後

closed set contains whole subset=否;running intersection of qualifying closed sets=[1, 2, 3, 4]

條件:無

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。

步驟識別與依賴

規則:closure_every_closed_superset_original_open_order · 依賴:G001, S003

S005 · implies

操作前

selected subset:original open index=3;open=[0, 2, 3];same-carrier closed complement=[1, 4];subset=[2, 3]

操作後

closed set contains whole subset=否;running intersection of qualifying closed sets=[1, 2, 3, 4]

條件:無

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。

步驟識別與依賴

規則:closure_every_closed_superset_original_open_order · 依賴:G001, S004

S006 · implies

操作前

selected subset:original open index=4;open=[0, 1, 2, 3];same-carrier closed complement=[4];subset=[2, 3]

操作後

closed set contains whole subset=否;running intersection of qualifying closed sets=[1, 2, 3, 4]

條件:無

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。

步驟識別與依賴

規則:closure_every_closed_superset_original_open_order · 依賴:G001, S005

S007 · implies

操作前

selected subset:original open index=5;open=[0, 1, 2, 3, 4];same-carrier closed complement=[];subset=[2, 3]

操作後

closed set contains whole subset=否;running intersection of qualifying closed sets=[1, 2, 3, 4]

條件:無

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。

步驟識別與依賴

規則:closure_every_closed_superset_original_open_order · 依賴:G001, S006

S008 · implies

操作前

selected subset:ALL qualifying closed supersets in original complement order=[[0, 1, 2, 3, 4], [1, 2, 3, 4]]

操作後

topology-relative closure=[1, 2, 3, 4]

條件:無

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。

步驟識別與依賴

規則:closure_all_candidates_intersection_not_metric_guess · 依賴:G001, S007

S009 · implies

操作前

original point=4;original subset=[2, 3];complete computed set=[1, 2, 3, 4]

操作後

membership=是;scope=topology_relative_closure

條件:無

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。

步驟識別與依賴

規則:topology_point_test_computed_closure_or_interior · 依賴:G001, S008

逐步反向核對

  • S009 · premise_and_result_audit — 回到原始資料與本步條件核對,不把單向推論誤稱等價。
  • S008 · premise_and_result_audit — 回到原始資料與本步條件核對,不把單向推論誤稱等價。
  • S007 · premise_and_result_audit — 回到原始資料與本步條件核對,不把單向推論誤稱等價。
  • S006 · premise_and_result_audit — 回到原始資料與本步條件核對,不把單向推論誤稱等價。
  • S005 · premise_and_result_audit — 回到原始資料與本步條件核對,不把單向推論誤稱等價。
  • S004 · premise_and_result_audit — 回到原始資料與本步條件核對,不把單向推論誤稱等價。
  • S003 · premise_and_result_audit — 回到原始資料與本步條件核對,不把單向推論誤稱等價。
  • S002 · premise_and_result_audit — 回到原始資料與本步條件核對,不把單向推論誤稱等價。
  • S001 · premise_and_result_audit — 回到原始資料與本步條件核對,不把單向推論誤稱等價。

結論

(4, [1, 2, 3, 4], 是, topology_relative_closure)

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。

本題覆核與來源綁定

覆核日期:2026-10-26

原題、答案、步驟表達式、數值 AST 與既有歷史保持原樣;原來的候選狀態屬於來源歷史,不覆寫為形式化證明。

來源雜湊與覆核憑證
  • input commit: 0f7be61e2c22317a4075561dc822c87d881a50db
  • original question SHA-256: 492b3246a4874975241d887a0319938a1ea5bb68facf08db7ab8078c12cffe13
  • candidate record SHA-256: 8e6e919139a9d37a8094afa6fffbf48d5de05fa18869a60f453bdbc1de36b39c
  • semantic review receipt SHA-256: d64b4e08bda1c25dd393b7c7e4140a8734610a9b84ae47e6f0adcff7f30a0b8e
  • question-first witness SHA-256: 790fd6623c04ba52a6ffb0a95cea52b532a02b23bb9453926d2d9f7a825fdec8
  • Lean: NOT_RUN · human review: NOT_PERFORMED
同源完整 JSON