原題與已知條件
完整有限拓撲與同一載體子集。操作=topology-closure;資料={"n":5,"topology":[[],[0],[0,3],[0,3,4],[0,1,3,4],[0,1,2,3,4]],"subset":[3,4]}。請依 [subset,closed_supersets,closure] 欄序回答。
G001 — 完整有限拓撲與同一載體子集。操作=topology-closure;資料={"n":5,"topology":[[],[0],[0,3],[0,3,4],[0,1,3,4],[0,1,2,3,4]],"subset":[3,4]}。請依 [subset,closed_supersets,closure] 欄序回答。
逐步推導
S001 · implies
操作前
same complete carrier Fin(5);original ordered complete topology=[[], [0], [0, 3], [0, 3, 4], [0, 1, 3, 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。
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=[3, 4]
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=[3, 4]
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=[3, 4]
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, 3, 4];same-carrier closed complement=[1, 2];subset=[3, 4]
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, 3, 4];same-carrier closed complement=[2];subset=[3, 4]
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=[3, 4]
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
結論
([3, 4], [[0, 1, 2, 3, 4], [1, 2, 3, 4]], [1, 2, 3, 4])
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。