# 拓撲基底基礎 · 脈衝 R03

本批已通過 Schema、去重、來源快照與獨立答案重算；仍未標記為人工審定。

本題庫已有對應的中英逐步詳解。原題的 candidate 與英文 ai_candidate 狀態保留；AI 覆核、有限計算與 Lean 形式證明分開標示。

batch-20261014-advance-20261107-r03-topological-basis-foundations · 2026-10-14 · 100 題

欄序為來源綁定的呈現補充；原始 JSONL 與數學答案不變。

[欄序與來源綁定資料](https://ocme.evemisslab.com/data/i18n/question-output-protocols.json)



## 1. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-basis-qualification；資料={"n":5,"basis":[[0],[1],[2],[3],[4]]}。請依 [covers_carrier, local_refinement, first_failure, is_basis] 欄序回答。

查看答案與解釋(是, 是, null, 是)

Wave6根層核對=(是, 是, null, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0001/)

拓撲基底 · qf-mod38-topbasis-basis-qualification · candidate





## 2. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-basis-qualification；資料={"n":5,"basis":[[0],[0,1],[1],[2],[3],[4]]}。請依 [covers_carrier, local_refinement, first_failure, is_basis] 欄序回答。

查看答案與解釋(是, 是, null, 是)

Wave6根層核對=(是, 是, null, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0002/)

拓撲基底 · qf-mod38-topbasis-basis-qualification · candidate





## 3. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-basis-qualification；資料={"n":5,"basis":[[0],[0,2],[1],[2],[3],[4]]}。請依 [covers_carrier, local_refinement, first_failure, is_basis] 欄序回答。

查看答案與解釋(是, 是, null, 是)

Wave6根層核對=(是, 是, null, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0003/)

拓撲基底 · qf-mod38-topbasis-basis-qualification · candidate





## 4. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-basis-qualification；資料={"n":5,"basis":[[0],[0,1],[0,2],[1],[2],[3],[4]]}。請依 [covers_carrier, local_refinement, first_failure, is_basis] 欄序回答。

查看答案與解釋(是, 是, null, 是)

Wave6根層核對=(是, 是, null, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0004/)

拓撲基底 · qf-mod38-topbasis-basis-qualification · candidate





## 5. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-basis-qualification；資料={"n":5,"basis":[[0],[1],[1,2],[2],[3],[4]]}。請依 [covers_carrier, local_refinement, first_failure, is_basis] 欄序回答。

查看答案與解釋(是, 是, null, 是)

Wave6根層核對=(是, 是, null, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0005/)

拓撲基底 · qf-mod38-topbasis-basis-qualification · candidate





## 6. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-basis-qualification；資料={"n":5,"basis":[[0],[0,1],[1],[1,2],[2],[3],[4]]}。請依 [covers_carrier, local_refinement, first_failure, is_basis] 欄序回答。

查看答案與解釋(是, 是, null, 是)

Wave6根層核對=(是, 是, null, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0006/)

拓撲基底 · qf-mod38-topbasis-basis-qualification · candidate





## 7. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-basis-qualification；資料={"n":5,"basis":[[0],[0,2],[1],[1,2],[2],[3],[4]]}。請依 [covers_carrier, local_refinement, first_failure, is_basis] 欄序回答。

查看答案與解釋(是, 是, null, 是)

Wave6根層核對=(是, 是, null, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0007/)

拓撲基底 · qf-mod38-topbasis-basis-qualification · candidate





## 8. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-basis-qualification；資料={"n":5,"basis":[[0],[0,1],[0,2],[1],[1,2],[2],[3],[4]]}。請依 [covers_carrier, local_refinement, first_failure, is_basis] 欄序回答。

查看答案與解釋(是, 是, null, 是)

Wave6根層核對=(是, 是, null, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0008/)

拓撲基底 · qf-mod38-topbasis-basis-qualification · candidate





## 9. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-basis-qualification；資料={"n":5,"basis":[[0],[1],[2],[2,3],[3],[4]]}。請依 [covers_carrier, local_refinement, first_failure, is_basis] 欄序回答。

查看答案與解釋(是, 是, null, 是)

Wave6根層核對=(是, 是, null, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0009/)

拓撲基底 · qf-mod38-topbasis-basis-qualification · candidate





## 10. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-basis-qualification；資料={"n":5,"basis":[[0],[0,1],[1],[2],[2,3],[3],[4]]}。請依 [covers_carrier, local_refinement, first_failure, is_basis] 欄序回答。

查看答案與解釋(是, 是, null, 是)

Wave6根層核對=(是, 是, null, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0010/)

拓撲基底 · qf-mod38-topbasis-basis-qualification · candidate





## 11. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-basis-qualification；資料={"n":5,"basis":[[0],[0,2],[1],[2],[2,3],[3],[4]]}。請依 [covers_carrier, local_refinement, first_failure, is_basis] 欄序回答。

查看答案與解釋(是, 是, null, 是)

Wave6根層核對=(是, 是, null, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0011/)

拓撲基底 · qf-mod38-topbasis-basis-qualification · candidate





## 12. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-basis-qualification；資料={"n":5,"basis":[[0],[0,1],[0,2],[1],[2],[2,3],[3],[4]]}。請依 [covers_carrier, local_refinement, first_failure, is_basis] 欄序回答。

查看答案與解釋(是, 是, null, 是)

Wave6根層核對=(是, 是, null, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0012/)

拓撲基底 · qf-mod38-topbasis-basis-qualification · candidate





## 13. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-basis-qualification；資料={"n":5,"basis":[[0],[1],[1,2],[2],[2,3],[3],[4]]}。請依 [covers_carrier, local_refinement, first_failure, is_basis] 欄序回答。

查看答案與解釋(是, 是, null, 是)

Wave6根層核對=(是, 是, null, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0013/)

拓撲基底 · qf-mod38-topbasis-basis-qualification · candidate





## 14. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-basis-qualification；資料={"n":5,"basis":[[0],[0,1],[1],[1,2],[2],[2,3],[3],[4]]}。請依 [covers_carrier, local_refinement, first_failure, is_basis] 欄序回答。

查看答案與解釋(是, 是, null, 是)

Wave6根層核對=(是, 是, null, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0014/)

拓撲基底 · qf-mod38-topbasis-basis-qualification · candidate





## 15. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-basis-qualification；資料={"n":5,"basis":[[0],[0,2],[1],[1,2],[2],[2,3],[3],[4]]}。請依 [covers_carrier, local_refinement, first_failure, is_basis] 欄序回答。

查看答案與解釋(是, 是, null, 是)

Wave6根層核對=(是, 是, null, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0015/)

拓撲基底 · qf-mod38-topbasis-basis-qualification · candidate





## 16. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-basis-qualification；資料={"n":5,"basis":[[0],[0,1],[0,2],[1],[1,2],[2],[2,3],[3],[4]]}。請依 [covers_carrier, local_refinement, first_failure, is_basis] 欄序回答。

查看答案與解釋(是, 是, null, 是)

Wave6根層核對=(是, 是, null, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0016/)

拓撲基底 · qf-mod38-topbasis-basis-qualification · candidate





## 17. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-basis-qualification；資料={"n":5,"basis":[[0],[1],[2],[3],[3,4],[4]]}。請依 [covers_carrier, local_refinement, first_failure, is_basis] 欄序回答。

查看答案與解釋(是, 是, null, 是)

Wave6根層核對=(是, 是, null, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0017/)

拓撲基底 · qf-mod38-topbasis-basis-qualification · candidate





## 18. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-basis-qualification；資料={"n":5,"basis":[[0],[0,1],[1],[2],[3],[3,4],[4]]}。請依 [covers_carrier, local_refinement, first_failure, is_basis] 欄序回答。

查看答案與解釋(是, 是, null, 是)

Wave6根層核對=(是, 是, null, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0018/)

拓撲基底 · qf-mod38-topbasis-basis-qualification · candidate





## 19. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-basis-qualification；資料={"n":5,"basis":[[0],[0,2],[1],[2],[3],[3,4],[4]]}。請依 [covers_carrier, local_refinement, first_failure, is_basis] 欄序回答。

查看答案與解釋(是, 是, null, 是)

Wave6根層核對=(是, 是, null, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0019/)

拓撲基底 · qf-mod38-topbasis-basis-qualification · candidate





## 20. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-basis-qualification；資料={"n":5,"basis":[[0],[0,1],[0,2],[1],[2],[3],[3,4],[4]]}。請依 [covers_carrier, local_refinement, first_failure, is_basis] 欄序回答。

查看答案與解釋(是, 是, null, 是)

Wave6根層核對=(是, 是, null, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0020/)

拓撲基底 · qf-mod38-topbasis-basis-qualification · candidate





## 21. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-basis-qualification；資料={"n":5,"basis":[[0],[1],[1,2],[2],[3],[3,4],[4]]}。請依 [covers_carrier, local_refinement, first_failure, is_basis] 欄序回答。

查看答案與解釋(是, 是, null, 是)

Wave6根層核對=(是, 是, null, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0021/)

拓撲基底 · qf-mod38-topbasis-basis-qualification · candidate





## 22. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-basis-qualification；資料={"n":5,"basis":[[0],[0,1],[1],[1,2],[2],[3],[3,4],[4]]}。請依 [covers_carrier, local_refinement, first_failure, is_basis] 欄序回答。

查看答案與解釋(是, 是, null, 是)

Wave6根層核對=(是, 是, null, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0022/)

拓撲基底 · qf-mod38-topbasis-basis-qualification · candidate





## 23. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-basis-qualification；資料={"n":5,"basis":[[0],[0,2],[1],[1,2],[2],[3],[3,4],[4]]}。請依 [covers_carrier, local_refinement, first_failure, is_basis] 欄序回答。

查看答案與解釋(是, 是, null, 是)

Wave6根層核對=(是, 是, null, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0023/)

拓撲基底 · qf-mod38-topbasis-basis-qualification · candidate





## 24. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-basis-qualification；資料={"n":5,"basis":[[0],[0,1],[0,2],[1],[1,2],[2],[3],[3,4],[4]]}。請依 [covers_carrier, local_refinement, first_failure, is_basis] 欄序回答。

查看答案與解釋(是, 是, null, 是)

Wave6根層核對=(是, 是, null, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0024/)

拓撲基底 · qf-mod38-topbasis-basis-qualification · candidate





## 25. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-basis-qualification；資料={"n":5,"basis":[[0],[1],[2],[2,3],[3],[3,4],[4]]}。請依 [covers_carrier, local_refinement, first_failure, is_basis] 欄序回答。

查看答案與解釋(是, 是, null, 是)

Wave6根層核對=(是, 是, null, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0025/)

拓撲基底 · qf-mod38-topbasis-basis-qualification · candidate





## 26. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-generated-topology；資料={"n":5,"basis":[[0],[1],[2],[3],[4]]}。請依 [topology, open_count, topology_certificate, certificate] 欄序回答。

查看答案與解釋([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)

Wave6根層核對=([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0026/)

拓撲基底 · qf-mod38-topbasis-generated-topology · candidate





## 27. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-generated-topology；資料={"n":5,"basis":[[0],[0,1],[1],[2],[3],[4]]}。請依 [topology, open_count, topology_certificate, certificate] 欄序回答。

查看答案與解釋([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)

Wave6根層核對=([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0027/)

拓撲基底 · qf-mod38-topbasis-generated-topology · candidate





## 28. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-generated-topology；資料={"n":5,"basis":[[0],[0,2],[1],[2],[3],[4]]}。請依 [topology, open_count, topology_certificate, certificate] 欄序回答。

查看答案與解釋([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)

Wave6根層核對=([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0028/)

拓撲基底 · qf-mod38-topbasis-generated-topology · candidate





## 29. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-generated-topology；資料={"n":5,"basis":[[0],[0,1],[0,2],[1],[2],[3],[4]]}。請依 [topology, open_count, topology_certificate, certificate] 欄序回答。

查看答案與解釋([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)

Wave6根層核對=([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0029/)

拓撲基底 · qf-mod38-topbasis-generated-topology · candidate





## 30. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-generated-topology；資料={"n":5,"basis":[[0],[1],[1,2],[2],[3],[4]]}。請依 [topology, open_count, topology_certificate, certificate] 欄序回答。

查看答案與解釋([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)

Wave6根層核對=([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0030/)

拓撲基底 · qf-mod38-topbasis-generated-topology · candidate





## 31. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-generated-topology；資料={"n":5,"basis":[[0],[0,1],[1],[1,2],[2],[3],[4]]}。請依 [topology, open_count, topology_certificate, certificate] 欄序回答。

查看答案與解釋([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)

Wave6根層核對=([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0031/)

拓撲基底 · qf-mod38-topbasis-generated-topology · candidate





## 32. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-generated-topology；資料={"n":5,"basis":[[0],[0,2],[1],[1,2],[2],[3],[4]]}。請依 [topology, open_count, topology_certificate, certificate] 欄序回答。

查看答案與解釋([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)

Wave6根層核對=([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0032/)

拓撲基底 · qf-mod38-topbasis-generated-topology · candidate





## 33. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-generated-topology；資料={"n":5,"basis":[[0],[0,1],[0,2],[1],[1,2],[2],[3],[4]]}。請依 [topology, open_count, topology_certificate, certificate] 欄序回答。

查看答案與解釋([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)

Wave6根層核對=([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0033/)

拓撲基底 · qf-mod38-topbasis-generated-topology · candidate





## 34. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-generated-topology；資料={"n":5,"basis":[[0],[1],[2],[2,3],[3],[4]]}。請依 [topology, open_count, topology_certificate, certificate] 欄序回答。

查看答案與解釋([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)

Wave6根層核對=([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0034/)

拓撲基底 · qf-mod38-topbasis-generated-topology · candidate





## 35. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-generated-topology；資料={"n":5,"basis":[[0],[0,1],[1],[2],[2,3],[3],[4]]}。請依 [topology, open_count, topology_certificate, certificate] 欄序回答。

查看答案與解釋([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)

Wave6根層核對=([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0035/)

拓撲基底 · qf-mod38-topbasis-generated-topology · candidate





## 36. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-generated-topology；資料={"n":5,"basis":[[0],[0,2],[1],[2],[2,3],[3],[4]]}。請依 [topology, open_count, topology_certificate, certificate] 欄序回答。

查看答案與解釋([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)

Wave6根層核對=([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0036/)

拓撲基底 · qf-mod38-topbasis-generated-topology · candidate





## 37. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-generated-topology；資料={"n":5,"basis":[[0],[0,1],[0,2],[1],[2],[2,3],[3],[4]]}。請依 [topology, open_count, topology_certificate, certificate] 欄序回答。

查看答案與解釋([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)

Wave6根層核對=([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0037/)

拓撲基底 · qf-mod38-topbasis-generated-topology · candidate





## 38. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-generated-topology；資料={"n":5,"basis":[[0],[1],[1,2],[2],[2,3],[3],[4]]}。請依 [topology, open_count, topology_certificate, certificate] 欄序回答。

查看答案與解釋([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)

Wave6根層核對=([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0038/)

拓撲基底 · qf-mod38-topbasis-generated-topology · candidate





## 39. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-generated-topology；資料={"n":5,"basis":[[0],[0,1],[1],[1,2],[2],[2,3],[3],[4]]}。請依 [topology, open_count, topology_certificate, certificate] 欄序回答。

查看答案與解釋([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)

Wave6根層核對=([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0039/)

拓撲基底 · qf-mod38-topbasis-generated-topology · candidate





## 40. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-generated-topology；資料={"n":5,"basis":[[0],[0,2],[1],[1,2],[2],[2,3],[3],[4]]}。請依 [topology, open_count, topology_certificate, certificate] 欄序回答。

查看答案與解釋([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)

Wave6根層核對=([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0040/)

拓撲基底 · qf-mod38-topbasis-generated-topology · candidate





## 41. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-generated-topology；資料={"n":5,"basis":[[0],[0,1],[0,2],[1],[1,2],[2],[2,3],[3],[4]]}。請依 [topology, open_count, topology_certificate, certificate] 欄序回答。

查看答案與解釋([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)

Wave6根層核對=([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0041/)

拓撲基底 · qf-mod38-topbasis-generated-topology · candidate





## 42. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-generated-topology；資料={"n":5,"basis":[[0],[1],[2],[3],[3,4],[4]]}。請依 [topology, open_count, topology_certificate, certificate] 欄序回答。

查看答案與解釋([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)

Wave6根層核對=([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0042/)

拓撲基底 · qf-mod38-topbasis-generated-topology · candidate





## 43. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-generated-topology；資料={"n":5,"basis":[[0],[0,1],[1],[2],[3],[3,4],[4]]}。請依 [topology, open_count, topology_certificate, certificate] 欄序回答。

查看答案與解釋([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)

Wave6根層核對=([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0043/)

拓撲基底 · qf-mod38-topbasis-generated-topology · candidate





## 44. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-generated-topology；資料={"n":5,"basis":[[0],[0,2],[1],[2],[3],[3,4],[4]]}。請依 [topology, open_count, topology_certificate, certificate] 欄序回答。

查看答案與解釋([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)

Wave6根層核對=([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0044/)

拓撲基底 · qf-mod38-topbasis-generated-topology · candidate





## 45. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-generated-topology；資料={"n":5,"basis":[[0],[0,1],[0,2],[1],[2],[3],[3,4],[4]]}。請依 [topology, open_count, topology_certificate, certificate] 欄序回答。

查看答案與解釋([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)

Wave6根層核對=([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0045/)

拓撲基底 · qf-mod38-topbasis-generated-topology · candidate





## 46. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-generated-topology；資料={"n":5,"basis":[[0],[1],[1,2],[2],[3],[3,4],[4]]}。請依 [topology, open_count, topology_certificate, certificate] 欄序回答。

查看答案與解釋([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)

Wave6根層核對=([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0046/)

拓撲基底 · qf-mod38-topbasis-generated-topology · candidate





## 47. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-generated-topology；資料={"n":5,"basis":[[0],[0,1],[1],[1,2],[2],[3],[3,4],[4]]}。請依 [topology, open_count, topology_certificate, certificate] 欄序回答。

查看答案與解釋([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)

Wave6根層核對=([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0047/)

拓撲基底 · qf-mod38-topbasis-generated-topology · candidate





## 48. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-generated-topology；資料={"n":5,"basis":[[0],[0,2],[1],[1,2],[2],[3],[3,4],[4]]}。請依 [topology, open_count, topology_certificate, certificate] 欄序回答。

查看答案與解釋([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)

Wave6根層核對=([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0048/)

拓撲基底 · qf-mod38-topbasis-generated-topology · candidate





## 49. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-generated-topology；資料={"n":5,"basis":[[0],[0,1],[0,2],[1],[1,2],[2],[3],[3,4],[4]]}。請依 [topology, open_count, topology_certificate, certificate] 欄序回答。

查看答案與解釋([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)

Wave6根層核對=([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0049/)

拓撲基底 · qf-mod38-topbasis-generated-topology · candidate





## 50. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-generated-topology；資料={"n":5,"basis":[[0],[1],[2],[2,3],[3],[3,4],[4]]}。請依 [topology, open_count, topology_certificate, certificate] 欄序回答。

查看答案與解釋([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)

Wave6根層核對=([[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 1, 3], [0, 1, 3, 4], [0, 1, 4], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 2, 4], [0, 3], [0, 3, 4], [0, 4], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]], 32, 是, all_unions_of_basis_elements)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0050/)

拓撲基底 · qf-mod38-topbasis-generated-topology · candidate





## 51. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-existing-topology-audit；資料={"n":5,"basis":[[0],[1],[2],[3],[4]],"topology":[[],[0],[0,1],[0,1,2],[0,1,2,3],[0,1,2,3,4],[0,1,2,4],[0,1,3],[0,1,3,4],[0,1,4],[0,2],[0,2,3],[0,2,3,4],[0,2,4],[0,3],[0,3,4],[0,4],[1],[1,2],[1,2,3],[1,2,3,4],[1,2,4],[1,3],[1,3,4],[1,4],[2],[2,3],[2,3,4],[2,4],[3],[3,4],[4]]}。請依 [topology_valid, abstract_basis_valid, basis_elements_open, every_open_is_basis_union, basis_for_given_topology] 欄序回答。

查看答案與解釋(是, 是, 是, 是, 是)

Wave6根層核對=(是, 是, 是, 是, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0051/)

拓撲基底 · qf-mod38-topbasis-existing-topology-audit · candidate





## 52. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-existing-topology-audit；資料={"n":5,"basis":[[0],[0,1],[1],[2],[3],[4]],"topology":[[],[0],[0,1],[0,1,2],[0,1,2,3],[0,1,2,3,4],[0,1,2,4],[0,1,3],[0,1,3,4],[0,1,4],[0,2],[0,2,3],[0,2,3,4],[0,2,4],[0,3],[0,3,4],[0,4],[1],[1,2],[1,2,3],[1,2,3,4],[1,2,4],[1,3],[1,3,4],[1,4],[2],[2,3],[2,3,4],[2,4],[3],[3,4],[4]]}。請依 [topology_valid, abstract_basis_valid, basis_elements_open, every_open_is_basis_union, basis_for_given_topology] 欄序回答。

查看答案與解釋(是, 是, 是, 是, 是)

Wave6根層核對=(是, 是, 是, 是, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0052/)

拓撲基底 · qf-mod38-topbasis-existing-topology-audit · candidate





## 53. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-existing-topology-audit；資料={"n":5,"basis":[[0],[0,2],[1],[2],[3],[4]],"topology":[[],[0],[0,1],[0,1,2],[0,1,2,3],[0,1,2,3,4],[0,1,2,4],[0,1,3],[0,1,3,4],[0,1,4],[0,2],[0,2,3],[0,2,3,4],[0,2,4],[0,3],[0,3,4],[0,4],[1],[1,2],[1,2,3],[1,2,3,4],[1,2,4],[1,3],[1,3,4],[1,4],[2],[2,3],[2,3,4],[2,4],[3],[3,4],[4]]}。請依 [topology_valid, abstract_basis_valid, basis_elements_open, every_open_is_basis_union, basis_for_given_topology] 欄序回答。

查看答案與解釋(是, 是, 是, 是, 是)

Wave6根層核對=(是, 是, 是, 是, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0053/)

拓撲基底 · qf-mod38-topbasis-existing-topology-audit · candidate





## 54. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-existing-topology-audit；資料={"n":5,"basis":[[0],[0,1],[0,2],[1],[2],[3],[4]],"topology":[[],[0],[0,1],[0,1,2],[0,1,2,3],[0,1,2,3,4],[0,1,2,4],[0,1,3],[0,1,3,4],[0,1,4],[0,2],[0,2,3],[0,2,3,4],[0,2,4],[0,3],[0,3,4],[0,4],[1],[1,2],[1,2,3],[1,2,3,4],[1,2,4],[1,3],[1,3,4],[1,4],[2],[2,3],[2,3,4],[2,4],[3],[3,4],[4]]}。請依 [topology_valid, abstract_basis_valid, basis_elements_open, every_open_is_basis_union, basis_for_given_topology] 欄序回答。

查看答案與解釋(是, 是, 是, 是, 是)

Wave6根層核對=(是, 是, 是, 是, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0054/)

拓撲基底 · qf-mod38-topbasis-existing-topology-audit · candidate





## 55. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-existing-topology-audit；資料={"n":5,"basis":[[0],[1],[1,2],[2],[3],[4]],"topology":[[],[0],[0,1],[0,1,2],[0,1,2,3],[0,1,2,3,4],[0,1,2,4],[0,1,3],[0,1,3,4],[0,1,4],[0,2],[0,2,3],[0,2,3,4],[0,2,4],[0,3],[0,3,4],[0,4],[1],[1,2],[1,2,3],[1,2,3,4],[1,2,4],[1,3],[1,3,4],[1,4],[2],[2,3],[2,3,4],[2,4],[3],[3,4],[4]]}。請依 [topology_valid, abstract_basis_valid, basis_elements_open, every_open_is_basis_union, basis_for_given_topology] 欄序回答。

查看答案與解釋(是, 是, 是, 是, 是)

Wave6根層核對=(是, 是, 是, 是, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0055/)

拓撲基底 · qf-mod38-topbasis-existing-topology-audit · candidate





## 56. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-existing-topology-audit；資料={"n":5,"basis":[[0],[0,1],[1],[1,2],[2],[3],[4]],"topology":[[],[0],[0,1],[0,1,2],[0,1,2,3],[0,1,2,3,4],[0,1,2,4],[0,1,3],[0,1,3,4],[0,1,4],[0,2],[0,2,3],[0,2,3,4],[0,2,4],[0,3],[0,3,4],[0,4],[1],[1,2],[1,2,3],[1,2,3,4],[1,2,4],[1,3],[1,3,4],[1,4],[2],[2,3],[2,3,4],[2,4],[3],[3,4],[4]]}。請依 [topology_valid, abstract_basis_valid, basis_elements_open, every_open_is_basis_union, basis_for_given_topology] 欄序回答。

查看答案與解釋(是, 是, 是, 是, 是)

Wave6根層核對=(是, 是, 是, 是, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0056/)

拓撲基底 · qf-mod38-topbasis-existing-topology-audit · candidate





## 57. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-existing-topology-audit；資料={"n":5,"basis":[[0],[0,2],[1],[1,2],[2],[3],[4]],"topology":[[],[0],[0,1],[0,1,2],[0,1,2,3],[0,1,2,3,4],[0,1,2,4],[0,1,3],[0,1,3,4],[0,1,4],[0,2],[0,2,3],[0,2,3,4],[0,2,4],[0,3],[0,3,4],[0,4],[1],[1,2],[1,2,3],[1,2,3,4],[1,2,4],[1,3],[1,3,4],[1,4],[2],[2,3],[2,3,4],[2,4],[3],[3,4],[4]]}。請依 [topology_valid, abstract_basis_valid, basis_elements_open, every_open_is_basis_union, basis_for_given_topology] 欄序回答。

查看答案與解釋(是, 是, 是, 是, 是)

Wave6根層核對=(是, 是, 是, 是, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0057/)

拓撲基底 · qf-mod38-topbasis-existing-topology-audit · candidate





## 58. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-existing-topology-audit；資料={"n":5,"basis":[[0],[0,1],[0,2],[1],[1,2],[2],[3],[4]],"topology":[[],[0],[0,1],[0,1,2],[0,1,2,3],[0,1,2,3,4],[0,1,2,4],[0,1,3],[0,1,3,4],[0,1,4],[0,2],[0,2,3],[0,2,3,4],[0,2,4],[0,3],[0,3,4],[0,4],[1],[1,2],[1,2,3],[1,2,3,4],[1,2,4],[1,3],[1,3,4],[1,4],[2],[2,3],[2,3,4],[2,4],[3],[3,4],[4]]}。請依 [topology_valid, abstract_basis_valid, basis_elements_open, every_open_is_basis_union, basis_for_given_topology] 欄序回答。

查看答案與解釋(是, 是, 是, 是, 是)

Wave6根層核對=(是, 是, 是, 是, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0058/)

拓撲基底 · qf-mod38-topbasis-existing-topology-audit · candidate





## 59. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-existing-topology-audit；資料={"n":5,"basis":[[0],[1],[2],[2,3],[3],[4]],"topology":[[],[0],[0,1],[0,1,2],[0,1,2,3],[0,1,2,3,4],[0,1,2,4],[0,1,3],[0,1,3,4],[0,1,4],[0,2],[0,2,3],[0,2,3,4],[0,2,4],[0,3],[0,3,4],[0,4],[1],[1,2],[1,2,3],[1,2,3,4],[1,2,4],[1,3],[1,3,4],[1,4],[2],[2,3],[2,3,4],[2,4],[3],[3,4],[4]]}。請依 [topology_valid, abstract_basis_valid, basis_elements_open, every_open_is_basis_union, basis_for_given_topology] 欄序回答。

查看答案與解釋(是, 是, 是, 是, 是)

Wave6根層核對=(是, 是, 是, 是, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0059/)

拓撲基底 · qf-mod38-topbasis-existing-topology-audit · candidate





## 60. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-existing-topology-audit；資料={"n":5,"basis":[[0],[0,1],[1],[2],[2,3],[3],[4]],"topology":[[],[0],[0,1],[0,1,2],[0,1,2,3],[0,1,2,3,4],[0,1,2,4],[0,1,3],[0,1,3,4],[0,1,4],[0,2],[0,2,3],[0,2,3,4],[0,2,4],[0,3],[0,3,4],[0,4],[1],[1,2],[1,2,3],[1,2,3,4],[1,2,4],[1,3],[1,3,4],[1,4],[2],[2,3],[2,3,4],[2,4],[3],[3,4],[4]]}。請依 [topology_valid, abstract_basis_valid, basis_elements_open, every_open_is_basis_union, basis_for_given_topology] 欄序回答。

查看答案與解釋(是, 是, 是, 是, 是)

Wave6根層核對=(是, 是, 是, 是, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0060/)

拓撲基底 · qf-mod38-topbasis-existing-topology-audit · candidate





## 61. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-existing-topology-audit；資料={"n":5,"basis":[[0],[0,2],[1],[2],[2,3],[3],[4]],"topology":[[],[0],[0,1],[0,1,2],[0,1,2,3],[0,1,2,3,4],[0,1,2,4],[0,1,3],[0,1,3,4],[0,1,4],[0,2],[0,2,3],[0,2,3,4],[0,2,4],[0,3],[0,3,4],[0,4],[1],[1,2],[1,2,3],[1,2,3,4],[1,2,4],[1,3],[1,3,4],[1,4],[2],[2,3],[2,3,4],[2,4],[3],[3,4],[4]]}。請依 [topology_valid, abstract_basis_valid, basis_elements_open, every_open_is_basis_union, basis_for_given_topology] 欄序回答。

查看答案與解釋(是, 是, 是, 是, 是)

Wave6根層核對=(是, 是, 是, 是, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0061/)

拓撲基底 · qf-mod38-topbasis-existing-topology-audit · candidate





## 62. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-existing-topology-audit；資料={"n":5,"basis":[[0],[0,1],[0,2],[1],[2],[2,3],[3],[4]],"topology":[[],[0],[0,1],[0,1,2],[0,1,2,3],[0,1,2,3,4],[0,1,2,4],[0,1,3],[0,1,3,4],[0,1,4],[0,2],[0,2,3],[0,2,3,4],[0,2,4],[0,3],[0,3,4],[0,4],[1],[1,2],[1,2,3],[1,2,3,4],[1,2,4],[1,3],[1,3,4],[1,4],[2],[2,3],[2,3,4],[2,4],[3],[3,4],[4]]}。請依 [topology_valid, abstract_basis_valid, basis_elements_open, every_open_is_basis_union, basis_for_given_topology] 欄序回答。

查看答案與解釋(是, 是, 是, 是, 是)

Wave6根層核對=(是, 是, 是, 是, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0062/)

拓撲基底 · qf-mod38-topbasis-existing-topology-audit · candidate





## 63. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-existing-topology-audit；資料={"n":5,"basis":[[0],[1],[1,2],[2],[2,3],[3],[4]],"topology":[[],[0],[0,1],[0,1,2],[0,1,2,3],[0,1,2,3,4],[0,1,2,4],[0,1,3],[0,1,3,4],[0,1,4],[0,2],[0,2,3],[0,2,3,4],[0,2,4],[0,3],[0,3,4],[0,4],[1],[1,2],[1,2,3],[1,2,3,4],[1,2,4],[1,3],[1,3,4],[1,4],[2],[2,3],[2,3,4],[2,4],[3],[3,4],[4]]}。請依 [topology_valid, abstract_basis_valid, basis_elements_open, every_open_is_basis_union, basis_for_given_topology] 欄序回答。

查看答案與解釋(是, 是, 是, 是, 是)

Wave6根層核對=(是, 是, 是, 是, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0063/)

拓撲基底 · qf-mod38-topbasis-existing-topology-audit · candidate





## 64. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-existing-topology-audit；資料={"n":5,"basis":[[0],[0,1],[1],[1,2],[2],[2,3],[3],[4]],"topology":[[],[0],[0,1],[0,1,2],[0,1,2,3],[0,1,2,3,4],[0,1,2,4],[0,1,3],[0,1,3,4],[0,1,4],[0,2],[0,2,3],[0,2,3,4],[0,2,4],[0,3],[0,3,4],[0,4],[1],[1,2],[1,2,3],[1,2,3,4],[1,2,4],[1,3],[1,3,4],[1,4],[2],[2,3],[2,3,4],[2,4],[3],[3,4],[4]]}。請依 [topology_valid, abstract_basis_valid, basis_elements_open, every_open_is_basis_union, basis_for_given_topology] 欄序回答。

查看答案與解釋(是, 是, 是, 是, 是)

Wave6根層核對=(是, 是, 是, 是, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0064/)

拓撲基底 · qf-mod38-topbasis-existing-topology-audit · candidate





## 65. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-existing-topology-audit；資料={"n":5,"basis":[[0],[0,2],[1],[1,2],[2],[2,3],[3],[4]],"topology":[[],[0],[0,1],[0,1,2],[0,1,2,3],[0,1,2,3,4],[0,1,2,4],[0,1,3],[0,1,3,4],[0,1,4],[0,2],[0,2,3],[0,2,3,4],[0,2,4],[0,3],[0,3,4],[0,4],[1],[1,2],[1,2,3],[1,2,3,4],[1,2,4],[1,3],[1,3,4],[1,4],[2],[2,3],[2,3,4],[2,4],[3],[3,4],[4]]}。請依 [topology_valid, abstract_basis_valid, basis_elements_open, every_open_is_basis_union, basis_for_given_topology] 欄序回答。

查看答案與解釋(是, 是, 是, 是, 是)

Wave6根層核對=(是, 是, 是, 是, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0065/)

拓撲基底 · qf-mod38-topbasis-existing-topology-audit · candidate





## 66. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-existing-topology-audit；資料={"n":5,"basis":[[0],[0,1],[0,2],[1],[1,2],[2],[2,3],[3],[4]],"topology":[[],[0],[0,1],[0,1,2],[0,1,2,3],[0,1,2,3,4],[0,1,2,4],[0,1,3],[0,1,3,4],[0,1,4],[0,2],[0,2,3],[0,2,3,4],[0,2,4],[0,3],[0,3,4],[0,4],[1],[1,2],[1,2,3],[1,2,3,4],[1,2,4],[1,3],[1,3,4],[1,4],[2],[2,3],[2,3,4],[2,4],[3],[3,4],[4]]}。請依 [topology_valid, abstract_basis_valid, basis_elements_open, every_open_is_basis_union, basis_for_given_topology] 欄序回答。

查看答案與解釋(是, 是, 是, 是, 是)

Wave6根層核對=(是, 是, 是, 是, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0066/)

拓撲基底 · qf-mod38-topbasis-existing-topology-audit · candidate





## 67. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-existing-topology-audit；資料={"n":5,"basis":[[0],[1],[2],[3],[3,4],[4]],"topology":[[],[0],[0,1],[0,1,2],[0,1,2,3],[0,1,2,3,4],[0,1,2,4],[0,1,3],[0,1,3,4],[0,1,4],[0,2],[0,2,3],[0,2,3,4],[0,2,4],[0,3],[0,3,4],[0,4],[1],[1,2],[1,2,3],[1,2,3,4],[1,2,4],[1,3],[1,3,4],[1,4],[2],[2,3],[2,3,4],[2,4],[3],[3,4],[4]]}。請依 [topology_valid, abstract_basis_valid, basis_elements_open, every_open_is_basis_union, basis_for_given_topology] 欄序回答。

查看答案與解釋(是, 是, 是, 是, 是)

Wave6根層核對=(是, 是, 是, 是, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0067/)

拓撲基底 · qf-mod38-topbasis-existing-topology-audit · candidate





## 68. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-existing-topology-audit；資料={"n":5,"basis":[[0],[0,1],[1],[2],[3],[3,4],[4]],"topology":[[],[0],[0,1],[0,1,2],[0,1,2,3],[0,1,2,3,4],[0,1,2,4],[0,1,3],[0,1,3,4],[0,1,4],[0,2],[0,2,3],[0,2,3,4],[0,2,4],[0,3],[0,3,4],[0,4],[1],[1,2],[1,2,3],[1,2,3,4],[1,2,4],[1,3],[1,3,4],[1,4],[2],[2,3],[2,3,4],[2,4],[3],[3,4],[4]]}。請依 [topology_valid, abstract_basis_valid, basis_elements_open, every_open_is_basis_union, basis_for_given_topology] 欄序回答。

查看答案與解釋(是, 是, 是, 是, 是)

Wave6根層核對=(是, 是, 是, 是, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0068/)

拓撲基底 · qf-mod38-topbasis-existing-topology-audit · candidate





## 69. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-existing-topology-audit；資料={"n":5,"basis":[[0],[0,2],[1],[2],[3],[3,4],[4]],"topology":[[],[0],[0,1],[0,1,2],[0,1,2,3],[0,1,2,3,4],[0,1,2,4],[0,1,3],[0,1,3,4],[0,1,4],[0,2],[0,2,3],[0,2,3,4],[0,2,4],[0,3],[0,3,4],[0,4],[1],[1,2],[1,2,3],[1,2,3,4],[1,2,4],[1,3],[1,3,4],[1,4],[2],[2,3],[2,3,4],[2,4],[3],[3,4],[4]]}。請依 [topology_valid, abstract_basis_valid, basis_elements_open, every_open_is_basis_union, basis_for_given_topology] 欄序回答。

查看答案與解釋(是, 是, 是, 是, 是)

Wave6根層核對=(是, 是, 是, 是, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0069/)

拓撲基底 · qf-mod38-topbasis-existing-topology-audit · candidate





## 70. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-existing-topology-audit；資料={"n":5,"basis":[[0],[0,1],[0,2],[1],[2],[3],[3,4],[4]],"topology":[[],[0],[0,1],[0,1,2],[0,1,2,3],[0,1,2,3,4],[0,1,2,4],[0,1,3],[0,1,3,4],[0,1,4],[0,2],[0,2,3],[0,2,3,4],[0,2,4],[0,3],[0,3,4],[0,4],[1],[1,2],[1,2,3],[1,2,3,4],[1,2,4],[1,3],[1,3,4],[1,4],[2],[2,3],[2,3,4],[2,4],[3],[3,4],[4]]}。請依 [topology_valid, abstract_basis_valid, basis_elements_open, every_open_is_basis_union, basis_for_given_topology] 欄序回答。

查看答案與解釋(是, 是, 是, 是, 是)

Wave6根層核對=(是, 是, 是, 是, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0070/)

拓撲基底 · qf-mod38-topbasis-existing-topology-audit · candidate





## 71. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-existing-topology-audit；資料={"n":5,"basis":[[0],[1],[1,2],[2],[3],[3,4],[4]],"topology":[[],[0],[0,1],[0,1,2],[0,1,2,3],[0,1,2,3,4],[0,1,2,4],[0,1,3],[0,1,3,4],[0,1,4],[0,2],[0,2,3],[0,2,3,4],[0,2,4],[0,3],[0,3,4],[0,4],[1],[1,2],[1,2,3],[1,2,3,4],[1,2,4],[1,3],[1,3,4],[1,4],[2],[2,3],[2,3,4],[2,4],[3],[3,4],[4]]}。請依 [topology_valid, abstract_basis_valid, basis_elements_open, every_open_is_basis_union, basis_for_given_topology] 欄序回答。

查看答案與解釋(是, 是, 是, 是, 是)

Wave6根層核對=(是, 是, 是, 是, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0071/)

拓撲基底 · qf-mod38-topbasis-existing-topology-audit · candidate





## 72. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-existing-topology-audit；資料={"n":5,"basis":[[0],[0,1],[1],[1,2],[2],[3],[3,4],[4]],"topology":[[],[0],[0,1],[0,1,2],[0,1,2,3],[0,1,2,3,4],[0,1,2,4],[0,1,3],[0,1,3,4],[0,1,4],[0,2],[0,2,3],[0,2,3,4],[0,2,4],[0,3],[0,3,4],[0,4],[1],[1,2],[1,2,3],[1,2,3,4],[1,2,4],[1,3],[1,3,4],[1,4],[2],[2,3],[2,3,4],[2,4],[3],[3,4],[4]]}。請依 [topology_valid, abstract_basis_valid, basis_elements_open, every_open_is_basis_union, basis_for_given_topology] 欄序回答。

查看答案與解釋(是, 是, 是, 是, 是)

Wave6根層核對=(是, 是, 是, 是, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0072/)

拓撲基底 · qf-mod38-topbasis-existing-topology-audit · candidate





## 73. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-existing-topology-audit；資料={"n":5,"basis":[[0],[0,2],[1],[1,2],[2],[3],[3,4],[4]],"topology":[[],[0],[0,1],[0,1,2],[0,1,2,3],[0,1,2,3,4],[0,1,2,4],[0,1,3],[0,1,3,4],[0,1,4],[0,2],[0,2,3],[0,2,3,4],[0,2,4],[0,3],[0,3,4],[0,4],[1],[1,2],[1,2,3],[1,2,3,4],[1,2,4],[1,3],[1,3,4],[1,4],[2],[2,3],[2,3,4],[2,4],[3],[3,4],[4]]}。請依 [topology_valid, abstract_basis_valid, basis_elements_open, every_open_is_basis_union, basis_for_given_topology] 欄序回答。

查看答案與解釋(是, 是, 是, 是, 是)

Wave6根層核對=(是, 是, 是, 是, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0073/)

拓撲基底 · qf-mod38-topbasis-existing-topology-audit · candidate





## 74. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-existing-topology-audit；資料={"n":5,"basis":[[0],[0,1],[0,2],[1],[1,2],[2],[3],[3,4],[4]],"topology":[[],[0],[0,1],[0,1,2],[0,1,2,3],[0,1,2,3,4],[0,1,2,4],[0,1,3],[0,1,3,4],[0,1,4],[0,2],[0,2,3],[0,2,3,4],[0,2,4],[0,3],[0,3,4],[0,4],[1],[1,2],[1,2,3],[1,2,3,4],[1,2,4],[1,3],[1,3,4],[1,4],[2],[2,3],[2,3,4],[2,4],[3],[3,4],[4]]}。請依 [topology_valid, abstract_basis_valid, basis_elements_open, every_open_is_basis_union, basis_for_given_topology] 欄序回答。

查看答案與解釋(是, 是, 是, 是, 是)

Wave6根層核對=(是, 是, 是, 是, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0074/)

拓撲基底 · qf-mod38-topbasis-existing-topology-audit · candidate





## 75. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-existing-topology-audit；資料={"n":5,"basis":[[0],[1],[2],[2,3],[3],[3,4],[4]],"topology":[[],[0],[0,1],[0,1,2],[0,1,2,3],[0,1,2,3,4],[0,1,2,4],[0,1,3],[0,1,3,4],[0,1,4],[0,2],[0,2,3],[0,2,3,4],[0,2,4],[0,3],[0,3,4],[0,4],[1],[1,2],[1,2,3],[1,2,3,4],[1,2,4],[1,3],[1,3,4],[1,4],[2],[2,3],[2,3,4],[2,4],[3],[3,4],[4]]}。請依 [topology_valid, abstract_basis_valid, basis_elements_open, every_open_is_basis_union, basis_for_given_topology] 欄序回答。

查看答案與解釋(是, 是, 是, 是, 是)

Wave6根層核對=(是, 是, 是, 是, 是)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0075/)

拓撲基底 · qf-mod38-topbasis-existing-topology-audit · candidate





## 76. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-subbasis-boundary；資料={"n":5,"basis":[[0,1],[1,2],[2,3],[3,4]]}。請依 [covers_carrier, local_refinement, is_basis, cover_alone_suffices_general_claim, boundary] 欄序回答。

最後的否是不接受「任意覆蓋族都足以作basis」的一般推論，不否定前面獨立核得的is_basis，也不是在判subbasis為假。

查看答案與解釋(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)

Wave6根層核對=(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0076/)

拓撲基底 · qf-mod38-topbasis-subbasis-boundary · candidate





## 77. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-subbasis-boundary；資料={"n":5,"basis":[[0,1],[0,2],[1,2],[2,3],[3,4]]}。請依 [covers_carrier, local_refinement, is_basis, cover_alone_suffices_general_claim, boundary] 欄序回答。

最後的否是不接受「任意覆蓋族都足以作basis」的一般推論，不否定前面獨立核得的is_basis，也不是在判subbasis為假。

查看答案與解釋(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)

Wave6根層核對=(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0077/)

拓撲基底 · qf-mod38-topbasis-subbasis-boundary · candidate





## 78. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-subbasis-boundary；資料={"n":5,"basis":[[0,1],[1,2],[1,3],[2,3],[3,4]]}。請依 [covers_carrier, local_refinement, is_basis, cover_alone_suffices_general_claim, boundary] 欄序回答。

最後的否是不接受「任意覆蓋族都足以作basis」的一般推論，不否定前面獨立核得的is_basis，也不是在判subbasis為假。

查看答案與解釋(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)

Wave6根層核對=(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0078/)

拓撲基底 · qf-mod38-topbasis-subbasis-boundary · candidate





## 79. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-subbasis-boundary；資料={"n":5,"basis":[[0,1],[0,2],[1,2],[1,3],[2,3],[3,4]]}。請依 [covers_carrier, local_refinement, is_basis, cover_alone_suffices_general_claim, boundary] 欄序回答。

最後的否是不接受「任意覆蓋族都足以作basis」的一般推論，不否定前面獨立核得的is_basis，也不是在判subbasis為假。

查看答案與解釋(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)

Wave6根層核對=(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0079/)

拓撲基底 · qf-mod38-topbasis-subbasis-boundary · candidate





## 80. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-subbasis-boundary；資料={"n":5,"basis":[[0,1],[1,2],[2,3],[2,4],[3,4]]}。請依 [covers_carrier, local_refinement, is_basis, cover_alone_suffices_general_claim, boundary] 欄序回答。

最後的否是不接受「任意覆蓋族都足以作basis」的一般推論，不否定前面獨立核得的is_basis，也不是在判subbasis為假。

查看答案與解釋(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)

Wave6根層核對=(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0080/)

拓撲基底 · qf-mod38-topbasis-subbasis-boundary · candidate





## 81. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-subbasis-boundary；資料={"n":5,"basis":[[0,1],[0,2],[1,2],[2,3],[2,4],[3,4]]}。請依 [covers_carrier, local_refinement, is_basis, cover_alone_suffices_general_claim, boundary] 欄序回答。

最後的否是不接受「任意覆蓋族都足以作basis」的一般推論，不否定前面獨立核得的is_basis，也不是在判subbasis為假。

查看答案與解釋(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)

Wave6根層核對=(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0081/)

拓撲基底 · qf-mod38-topbasis-subbasis-boundary · candidate





## 82. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-subbasis-boundary；資料={"n":5,"basis":[[0,1],[1,2],[1,3],[2,3],[2,4],[3,4]]}。請依 [covers_carrier, local_refinement, is_basis, cover_alone_suffices_general_claim, boundary] 欄序回答。

最後的否是不接受「任意覆蓋族都足以作basis」的一般推論，不否定前面獨立核得的is_basis，也不是在判subbasis為假。

查看答案與解釋(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)

Wave6根層核對=(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0082/)

拓撲基底 · qf-mod38-topbasis-subbasis-boundary · candidate





## 83. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-subbasis-boundary；資料={"n":5,"basis":[[0,1],[0,2],[1,2],[1,3],[2,3],[2,4],[3,4]]}。請依 [covers_carrier, local_refinement, is_basis, cover_alone_suffices_general_claim, boundary] 欄序回答。

最後的否是不接受「任意覆蓋族都足以作basis」的一般推論，不否定前面獨立核得的is_basis，也不是在判subbasis為假。

查看答案與解釋(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)

Wave6根層核對=(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0083/)

拓撲基底 · qf-mod38-topbasis-subbasis-boundary · candidate





## 84. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-subbasis-boundary；資料={"n":5,"basis":[[0,1],[0,3],[1,2],[2,3],[3,4]]}。請依 [covers_carrier, local_refinement, is_basis, cover_alone_suffices_general_claim, boundary] 欄序回答。

最後的否是不接受「任意覆蓋族都足以作basis」的一般推論，不否定前面獨立核得的is_basis，也不是在判subbasis為假。

查看答案與解釋(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)

Wave6根層核對=(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0084/)

拓撲基底 · qf-mod38-topbasis-subbasis-boundary · candidate





## 85. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-subbasis-boundary；資料={"n":5,"basis":[[0,1],[0,2],[0,3],[1,2],[2,3],[3,4]]}。請依 [covers_carrier, local_refinement, is_basis, cover_alone_suffices_general_claim, boundary] 欄序回答。

最後的否是不接受「任意覆蓋族都足以作basis」的一般推論，不否定前面獨立核得的is_basis，也不是在判subbasis為假。

查看答案與解釋(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)

Wave6根層核對=(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0085/)

拓撲基底 · qf-mod38-topbasis-subbasis-boundary · candidate





## 86. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-subbasis-boundary；資料={"n":5,"basis":[[0,1],[0,3],[1,2],[1,3],[2,3],[3,4]]}。請依 [covers_carrier, local_refinement, is_basis, cover_alone_suffices_general_claim, boundary] 欄序回答。

最後的否是不接受「任意覆蓋族都足以作basis」的一般推論，不否定前面獨立核得的is_basis，也不是在判subbasis為假。

查看答案與解釋(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)

Wave6根層核對=(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0086/)

拓撲基底 · qf-mod38-topbasis-subbasis-boundary · candidate





## 87. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-subbasis-boundary；資料={"n":5,"basis":[[0,1],[0,2],[0,3],[1,2],[1,3],[2,3],[3,4]]}。請依 [covers_carrier, local_refinement, is_basis, cover_alone_suffices_general_claim, boundary] 欄序回答。

最後的否是不接受「任意覆蓋族都足以作basis」的一般推論，不否定前面獨立核得的is_basis，也不是在判subbasis為假。

查看答案與解釋(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)

Wave6根層核對=(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0087/)

拓撲基底 · qf-mod38-topbasis-subbasis-boundary · candidate





## 88. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-subbasis-boundary；資料={"n":5,"basis":[[0,1],[0,3],[1,2],[2,3],[2,4],[3,4]]}。請依 [covers_carrier, local_refinement, is_basis, cover_alone_suffices_general_claim, boundary] 欄序回答。

最後的否是不接受「任意覆蓋族都足以作basis」的一般推論，不否定前面獨立核得的is_basis，也不是在判subbasis為假。

查看答案與解釋(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)

Wave6根層核對=(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0088/)

拓撲基底 · qf-mod38-topbasis-subbasis-boundary · candidate





## 89. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-subbasis-boundary；資料={"n":5,"basis":[[0,1],[0,2],[0,3],[1,2],[2,3],[2,4],[3,4]]}。請依 [covers_carrier, local_refinement, is_basis, cover_alone_suffices_general_claim, boundary] 欄序回答。

最後的否是不接受「任意覆蓋族都足以作basis」的一般推論，不否定前面獨立核得的is_basis，也不是在判subbasis為假。

查看答案與解釋(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)

Wave6根層核對=(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0089/)

拓撲基底 · qf-mod38-topbasis-subbasis-boundary · candidate





## 90. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-subbasis-boundary；資料={"n":5,"basis":[[0,1],[0,3],[1,2],[1,3],[2,3],[2,4],[3,4]]}。請依 [covers_carrier, local_refinement, is_basis, cover_alone_suffices_general_claim, boundary] 欄序回答。

最後的否是不接受「任意覆蓋族都足以作basis」的一般推論，不否定前面獨立核得的is_basis，也不是在判subbasis為假。

查看答案與解釋(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)

Wave6根層核對=(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0090/)

拓撲基底 · qf-mod38-topbasis-subbasis-boundary · candidate





## 91. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-subbasis-boundary；資料={"n":5,"basis":[[0,1],[0,2],[0,3],[1,2],[1,3],[2,3],[2,4],[3,4]]}。請依 [covers_carrier, local_refinement, is_basis, cover_alone_suffices_general_claim, boundary] 欄序回答。

最後的否是不接受「任意覆蓋族都足以作basis」的一般推論，不否定前面獨立核得的is_basis，也不是在判subbasis為假。

查看答案與解釋(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)

Wave6根層核對=(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0091/)

拓撲基底 · qf-mod38-topbasis-subbasis-boundary · candidate





## 92. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-subbasis-boundary；資料={"n":5,"basis":[[0,1],[1,2],[1,4],[2,3],[3,4]]}。請依 [covers_carrier, local_refinement, is_basis, cover_alone_suffices_general_claim, boundary] 欄序回答。

最後的否是不接受「任意覆蓋族都足以作basis」的一般推論，不否定前面獨立核得的is_basis，也不是在判subbasis為假。

查看答案與解釋(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)

Wave6根層核對=(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0092/)

拓撲基底 · qf-mod38-topbasis-subbasis-boundary · candidate





## 93. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-subbasis-boundary；資料={"n":5,"basis":[[0,1],[0,2],[1,2],[1,4],[2,3],[3,4]]}。請依 [covers_carrier, local_refinement, is_basis, cover_alone_suffices_general_claim, boundary] 欄序回答。

最後的否是不接受「任意覆蓋族都足以作basis」的一般推論，不否定前面獨立核得的is_basis，也不是在判subbasis為假。

查看答案與解釋(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)

Wave6根層核對=(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0093/)

拓撲基底 · qf-mod38-topbasis-subbasis-boundary · candidate





## 94. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-subbasis-boundary；資料={"n":5,"basis":[[0,1],[1,2],[1,3],[1,4],[2,3],[3,4]]}。請依 [covers_carrier, local_refinement, is_basis, cover_alone_suffices_general_claim, boundary] 欄序回答。

最後的否是不接受「任意覆蓋族都足以作basis」的一般推論，不否定前面獨立核得的is_basis，也不是在判subbasis為假。

查看答案與解釋(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)

Wave6根層核對=(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0094/)

拓撲基底 · qf-mod38-topbasis-subbasis-boundary · candidate





## 95. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-subbasis-boundary；資料={"n":5,"basis":[[0,1],[0,2],[1,2],[1,3],[1,4],[2,3],[3,4]]}。請依 [covers_carrier, local_refinement, is_basis, cover_alone_suffices_general_claim, boundary] 欄序回答。

最後的否是不接受「任意覆蓋族都足以作basis」的一般推論，不否定前面獨立核得的is_basis，也不是在判subbasis為假。

查看答案與解釋(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)

Wave6根層核對=(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0095/)

拓撲基底 · qf-mod38-topbasis-subbasis-boundary · candidate





## 96. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-subbasis-boundary；資料={"n":5,"basis":[[0,1],[1,2],[1,4],[2,3],[2,4],[3,4]]}。請依 [covers_carrier, local_refinement, is_basis, cover_alone_suffices_general_claim, boundary] 欄序回答。

最後的否是不接受「任意覆蓋族都足以作basis」的一般推論，不否定前面獨立核得的is_basis，也不是在判subbasis為假。

查看答案與解釋(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)

Wave6根層核對=(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0096/)

拓撲基底 · qf-mod38-topbasis-subbasis-boundary · candidate





## 97. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-subbasis-boundary；資料={"n":5,"basis":[[0,1],[0,2],[1,2],[1,4],[2,3],[2,4],[3,4]]}。請依 [covers_carrier, local_refinement, is_basis, cover_alone_suffices_general_claim, boundary] 欄序回答。

最後的否是不接受「任意覆蓋族都足以作basis」的一般推論，不否定前面獨立核得的is_basis，也不是在判subbasis為假。

查看答案與解釋(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)

Wave6根層核對=(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0097/)

拓撲基底 · qf-mod38-topbasis-subbasis-boundary · candidate





## 98. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-subbasis-boundary；資料={"n":5,"basis":[[0,1],[1,2],[1,3],[1,4],[2,3],[2,4],[3,4]]}。請依 [covers_carrier, local_refinement, is_basis, cover_alone_suffices_general_claim, boundary] 欄序回答。

最後的否是不接受「任意覆蓋族都足以作basis」的一般推論，不否定前面獨立核得的is_basis，也不是在判subbasis為假。

查看答案與解釋(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)

Wave6根層核對=(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0098/)

拓撲基底 · qf-mod38-topbasis-subbasis-boundary · candidate





## 99. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-subbasis-boundary；資料={"n":5,"basis":[[0,1],[0,2],[1,2],[1,3],[1,4],[2,3],[2,4],[3,4]]}。請依 [covers_carrier, local_refinement, is_basis, cover_alone_suffices_general_claim, boundary] 欄序回答。

最後的否是不接受「任意覆蓋族都足以作basis」的一般推論，不否定前面獨立核得的is_basis，也不是在判subbasis為假。

查看答案與解釋(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)

Wave6根層核對=(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0099/)

拓撲基底 · qf-mod38-topbasis-subbasis-boundary · candidate





## 100. 拓撲基底

完整有限carrier與完整basis candidate。操作=topbasis-subbasis-boundary；資料={"n":5,"basis":[[0,1],[0,3],[1,2],[1,4],[2,3],[3,4]]}。請依 [covers_carrier, local_refinement, is_basis, cover_alone_suffices_general_claim, boundary] 欄序回答。

最後的否是不接受「任意覆蓋族都足以作basis」的一般推論，不否定前面獨立核得的is_basis，也不是在判subbasis為假。

查看答案與解釋(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)

Wave6根層核對=(是, 否, 否, 否, cover_without_local_intersection_refinement_is_not_a_basis_and_may_only_be_a_subbasis)；依完整明示資料獨立重算；basis須覆蓋X並逐點滿足局部交細化；basis elements皆為X子集；相對既有拓撲還須basis elements開且每個開集為basis聯集；basis不可混同subbasis。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261107-r03-topological-basis-0100/)

拓撲基底 · qf-mod38-topbasis-subbasis-boundary · candidate



[原始題目 JSONL](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261107-r03-topological-basis-foundations/questions.jsonl) · [驗收報告](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261107-r03-topological-basis-foundations/validation-report.json)