R0032-16783 · q-20261014-advance-20261102-r04-topological-boundary-0033
完整有限拓撲與同一載體子集。操作=boundary-membership；資料={"n":5,"topology":[[],[0],[0,2],[0,1,2],[0,1,2,4],[0,1,2,3,4]],"subset":[1,2],"point":1}。請依 [point,boundary,in_boundary,scope] 欄序回答。
G001 — 完整有限拓撲與同一載體子集。操作=boundary-membership；資料={"n":5,"topology":[[],[0],[0,2],[0,1,2],[0,1,2,4],[0,1,2,3,4]],"subset":[1,2],"point":1}。請依 [point,boundary,in_boundary,scope] 欄序回答。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。
操作前
same complete carrier Fin(5);original ordered complete topology=[[], [0], [0, 2], [0, 1, 2], [0, 1, 2, 4], [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。
規則：topology_complete_finite_parent_all_pairs · 依賴：G001
操作前
boundary subset:original open index=0;open=[];same-carrier closed complement=[0, 1, 2, 3, 4];subset=[1, 2]
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。
規則：closure_every_closed_superset_original_open_order · 依賴：G001, S001
操作前
boundary subset:original open index=1;open=[0];same-carrier closed complement=[1, 2, 3, 4];subset=[1, 2]
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。
規則：closure_every_closed_superset_original_open_order · 依賴：G001, S002
操作前
boundary subset:original open index=2;open=[0, 2];same-carrier closed complement=[1, 3, 4];subset=[1, 2]
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。
規則：closure_every_closed_superset_original_open_order · 依賴：G001, S003
操作前
boundary subset:original open index=3;open=[0, 1, 2];same-carrier closed complement=[3, 4];subset=[1, 2]
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。
規則：closure_every_closed_superset_original_open_order · 依賴：G001, S004
操作前
boundary subset:original open index=4;open=[0, 1, 2, 4];same-carrier closed complement=[3];subset=[1, 2]
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。
規則：closure_every_closed_superset_original_open_order · 依賴：G001, S005
操作前
boundary subset:original open index=5;open=[0, 1, 2, 3, 4];same-carrier closed complement=[];subset=[1, 2]
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。
規則：closure_every_closed_superset_original_open_order · 依賴：G001, S006
操作前
boundary 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。
規則：closure_all_candidates_intersection_not_metric_guess · 依賴：G001, S007
操作前
boundary subset:original open index=0;open=[];subset=[1, 2]
操作後
whole open contained in subset=是;running union of qualifying opens=[]
條件：無
須整個開集包含在A才合格，不是只跟A相交。把全部合格開集聯起來，才有相對此拓撲的最大開子集。
規則：interior_every_contained_open_original_order · 依賴：G001, S008
操作前
boundary subset:original open index=1;open=[0];subset=[1, 2]
操作後
whole open contained in subset=否;running union of qualifying opens=[]
條件：無
須整個開集包含在A才合格，不是只跟A相交。把全部合格開集聯起來，才有相對此拓撲的最大開子集。
規則：interior_every_contained_open_original_order · 依賴：G001, S009
操作前
boundary subset:original open index=2;open=[0, 2];subset=[1, 2]
操作後
whole open contained in subset=否;running union of qualifying opens=[]
條件：無
須整個開集包含在A才合格，不是只跟A相交。把全部合格開集聯起來，才有相對此拓撲的最大開子集。
規則：interior_every_contained_open_original_order · 依賴：G001, S010
操作前
boundary subset:original open index=3;open=[0, 1, 2];subset=[1, 2]
操作後
whole open contained in subset=否;running union of qualifying opens=[]
條件：無
須整個開集包含在A才合格，不是只跟A相交。把全部合格開集聯起來，才有相對此拓撲的最大開子集。
規則：interior_every_contained_open_original_order · 依賴：G001, S011
操作前
boundary subset:original open index=4;open=[0, 1, 2, 4];subset=[1, 2]
操作後
whole open contained in subset=否;running union of qualifying opens=[]
條件：無
須整個開集包含在A才合格，不是只跟A相交。把全部合格開集聯起來，才有相對此拓撲的最大開子集。
規則：interior_every_contained_open_original_order · 依賴：G001, S012
操作前
boundary subset:original open index=5;open=[0, 1, 2, 3, 4];subset=[1, 2]
操作後
whole open contained in subset=否;running union of qualifying opens=[]
條件：無
須整個開集包含在A才合格，不是只跟A相交。把全部合格開集聯起來，才有相對此拓撲的最大開子集。
規則：interior_every_contained_open_original_order · 依賴：G001, S013
操作前
boundary subset:ALL contained opens in original topology order=[[]]
操作後
topology-relative interior=[]
條件：無
空開集永遠合格；結果只依原拓撲，不把A內每個點都自動當內點。
規則：interior_all_candidates_union_not_subset_membership · 依賴：G001, S014
操作前
boundary subset:same-topology closure=[1, 2, 3, 4];same-topology interior=[]
操作後
boundary=closure minus interior=[1, 2, 3, 4]
條件：無
把內部從閉包扣掉，不是從原A扣掉；非A成員仍可能是邊界點，結果沒有使用歐氏圖形或度量。
規則：boundary_same_topology_closure_minus_interior · 依賴：G001, S015
操作前
original point=1;original subset=[1, 2];complete relative boundary=[1, 2, 3, 4]
操作後
point in boundary=是;topology_relative_boundary
條件：無
只對已算出的完整邊界作成員測試，不按點是否在A或圖上看似邊緣來判。
規則：boundary_point_not_original_membership_or_picture · 依賴：G001, S016
S017 · 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。S016 · 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。S015 · 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。S014 · 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。S013 · 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。S012 · 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。S011 · 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。S010 · 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。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。(1, [1, 2, 3, 4], 是, topology_relative_boundary)
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。
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。
input commit: 0f7be61e2c22317a4075561dc822c87d881a50dboriginal question SHA-256: 4cb3110741ef2459175a961746fa4902857a11384a0d3fe1d50d65a9c0ea48bfcandidate record SHA-256: 56c38e59acac06697754a3679213178654b4ed4aa0852cf8aba312c47d551c36semantic review receipt SHA-256: d64b4e08bda1c25dd393b7c7e4140a8734610a9b84ae47e6f0adcff7f30a0b8equestion-first witness SHA-256: 790fd6623c04ba52a6ffb0a95cea52b532a02b23bb9453926d2d9f7a825fdec8Lean: NOT_RUN · human review: NOT_PERFORMED