第 16847 題：緊緻拓撲空間

原題與已知條件

完整有限carrier與完整拓撲或明示supplied cover。操作=compact-finite-space-certificate；資料={"n":5,"topology":[[],[0],[0,4],[0,2,4],[0,2,3,4],[0,1,2,3,4]]}。請依 [carrier_size,open_set_count,compact,finite_topology_certificate] 欄序回答。

• G001 — 完整有限carrier與完整拓撲或明示supplied cover。操作=compact-finite-space-certificate；資料={"n":5,"topology":[[],[0],[0,4],[0,2,4],[0,2,3,4],[0,1,2,3,4]]}。請依 [carrier_size,open_set_count,compact,finite_topology_certificate] 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。

• mko-compact-topological-space

逐步推導

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

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

complete finite topology contains exactly6 distinct opens on Fin(5)

for ANY set-family open cover U,each member lies in this finite topology;U is finite and covers by itself

same finite topology with6 distinct opens;an arbitrary INDEXED cover may repeat opens

select one original index per distinct open used;only finitely many choices;the representative family retains the same union

逐步反向核對

• 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

(5, 6, 是, every_open_cover_is_a_subfamily_of_a_finite_topology)

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。