{
  "schema_version": "ocme-reviewed-question-reader-v1",
  "question_id": "q-20261014-advance-20261108-r03-subspace-topology-0086",
  "pilot_number": "R0041-21211",
  "number": 21211,
  "family_id": "qf-mod39-subtop-ambient-boundary",
  "title": {
    "zh-Hant": "第 21211 題：子空間拓撲",
    "en": "Question 21211: Relative-Open but Not Ambient-Open Is a Valid Boundary Case"
  },
  "original_question": {
    "schema_version": "ocme-question-v0.1",
    "id": "q-20261014-advance-20261108-r03-subspace-topology-0086",
    "batch_id": "batch-20261014-advance-20261108-r03-subspace-topology-foundations",
    "language": "zh-Hant",
    "source_mko_ids": [
      "mko-subspace-topology"
    ],
    "title_zh": "子空間拓撲",
    "stem_zh": "完整有限母拓撲、明示Y subset X與相對全集Y。操作=subtop-ambient-boundary；資料={\"n\":5,\"parent_topology\":[[],[0],[0,1,2,3,4],[0,1,2,4],[0,2],[0,2,4]],\"subset\":[1,2,3],\"candidate\":[1,2,3]}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。",
    "answer": {
      "kind": "expression",
      "value": "([1, 2, 3], 是, 否, 是, O_intersect_Y_not_O_subset_Y)",
      "explanation_zh": "Wave6第二層核對=([1, 2, 3], 是, 否, 是, O_intersect_Y_not_O_subset_Y)；依完整明示資料獨立重算；tau_Y恰為全部U intersect Y；relative open不必ambient open且不要求U subset Y；relative closed以Y內補集或F intersect Y判定；重複與空交集須正規化。"
    },
    "difficulty": {
      "band": "undergraduate_core",
      "level": 2,
      "skills": [
        "子空間拓撲"
      ]
    },
    "variant": {
      "family_id": "qf-mod39-subtop-ambient-boundary",
      "seed": 202611083086,
      "parameters": {
        "n": 5,
        "parent_topology": [
          [],
          [
            0
          ],
          [
            0,
            1,
            2,
            3,
            4
          ],
          [
            0,
            1,
            2,
            4
          ],
          [
            0,
            2
          ],
          [
            0,
            2,
            4
          ]
        ],
        "subset": [
          1,
          2,
          3
        ],
        "candidate": [
          1,
          2,
          3
        ],
        "domain": "Complete_finite_subspace_topology",
        "operation": "subtop-ambient-boundary",
        "semantic_template": "qf-mod39-subtop-ambient-boundary/{\"candidate\":[1,2,3],\"n\":5,\"parent_topology\":[[],[0,1,2,3,4],[0,1,2,4],[0,2,4],[0,2],[0]],\"subset\":[1,2,3]}"
      }
    },
    "verification": {
      "status": "candidate",
      "checks": [
        "producer_exact_finite_or_symbolic_model",
        "producer_source_snapshot_check",
        "producer_substantive_template_deduplication",
        "producer_json_schema"
      ],
      "warning_zh": "候選須主端重算；不構成人工審定、single-step orbit、finite-prefix monotonicity、ambient-relative topology混同、Chebyshev equality、pairwise-only容斥、無Euclid-lemma uniqueness、loopful digraph、Lean或Evidence。"
    },
    "provenance": {
      "generator_role": "question_producer",
      "generator_model": "model_unknown",
      "generated_at": "2026-10-14",
      "contract_version": "ocme-question-factory-v0.1"
    }
  },
  "original_english_edition": {
    "schema_version": "ocme-question-localization-v0.1",
    "question_id": "q-20261014-advance-20261108-r03-subspace-topology-0086",
    "batch_id": "batch-20261014-advance-20261108-r03-subspace-topology-foundations",
    "locale": "en",
    "translation_state": "ai_candidate",
    "source": {
      "question_sha256": "a8556a1f655bf7c238c3ba5f7399e0bb5f7c645239b4978d2efde09404abe1de",
      "family_id": "qf-mod39-subtop-ambient-boundary",
      "seed": 202611083086,
      "semantic_template": "qf-mod39-subtop-ambient-boundary/{\"candidate\":[1,2,3],\"n\":5,\"parent_topology\":[[],[0,1,2,3,4],[0,1,2,4],[0,2,4],[0,2],[0]],\"subset\":[1,2,3]}"
    },
    "fields": {
      "title": "Relative-Open but Not Ambient-Open Is a Valid Boundary Case",
      "stem": "Use ambient carrier Fin(5)={0,…,4}, complete canonical parent_topology=[[], [0], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 2], [0, 2, 4]], and subset Y=[1, 2, 3]. Its relative topology is exactly every parent-open U intersected with Y, with duplicate intersections and empty intersections normalized. Complements of relative subsets are taken inside Y, not X. Candidate=[1, 2, 3] is a duplicate-free subset of Y. Give (candidate, relative_open, ambient_open, intersection_certificate, boundary).",
      "answer": {
        "kind": "expression",
        "value": "([1, 2, 3], Yes, No, Yes, O_intersect_Y_not_O_subset_Y)",
        "explanation": "Require the actual candidate to be relatively open but not ambient open. The parent-open intersection certificate does not require its ambient witness U to lie inside Y. The exact result, retaining the requested order, is ([1, 2, 3], Yes, No, Yes, O_intersect_Y_not_O_subset_Y)."
      },
      "skills": [
        "Relative-Open but Not Ambient-Open Is a Valid Boundary Case"
      ],
      "verification_warning": "This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."
    }
  },
  "givens": [
    {
      "id": "G001",
      "text": {
        "zh-Hant": "完整有限母拓撲、明示Y subset X與相對全集Y。操作=subtop-ambient-boundary；資料={\"n\":5,\"parent_topology\":[[],[0],[0,1,2,3,4],[0,1,2,4],[0,2],[0,2,4]],\"subset\":[1,2,3],\"candidate\":[1,2,3]}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。",
        "en": "Use ambient carrier Fin(5)={0,…,4}, complete canonical parent_topology=[[], [0], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 2], [0, 2, 4]], and subset Y=[1, 2, 3]. Its relative topology is exactly every parent-open U intersected with Y, with duplicate intersections and empty intersections normalized. Complements of relative subsets are taken inside Y, not X. Candidate=[1, 2, 3] is a duplicate-free subset of Y. Give (candidate, relative_open, ambient_open, intersection_certificate, boundary)."
      }
    }
  ],
  "steps": [
    {
      "id": "S001",
      "uses": [
        "G001"
      ],
      "before": "source operation=subtop-ambient-boundary;original snapshot and mathematics retained",
      "after": "explicit ordered source output fields=[candidate, relative_open, ambient_open, intersection_certificate, boundary]",
      "relation": "implies",
      "rule_id": "source_ordered_interface_explicit_without_javascript",
      "conditions": [],
      "explanation": {
        "zh-Hant": "原題歷史介面標記保留；此處直接列有序輸出欄位，不依賴JavaScript。欄位或tag不是額外數學證據，固定No保留原聲稱範圍。",
        "en": "Retain the historical interface marker while listing ordered output fields without JavaScript.Fields and tags are not extra mathematical evidence;fixed No markers retain their claim scope."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S002",
      "uses": [
        "G001",
        "S001"
      ],
      "before": "ORIGINAL parent Fin(5);parent open index=0,U=[];complete topology=[[], [0], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 2], [0, 2, 4]]",
      "after": "ALL [other_open_index,union,intersection] rows=[[0, [], []], [1, [0], []], [2, [0, 1, 2, 3, 4], []], [3, [0, 1, 2, 4], []], [4, [0, 2], []], [5, [0, 2, 4], []]];each belongs to the SAME parent family",
      "relation": "implies",
      "rule_id": "subtop_qualified_parent_all_pairs_union_intersection",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先驗母拓撲含空集與全集，再核全部有序open pair的union／intersection閉合。有限family的任意union可去重成有限union，故二元閉合足夠；不能把不合格母資料當True。",
        "en": "First require the empty/full opens,then check unions and intersections for every ordered pair in the same parent family.Any union in a finite family reduces to finitely many distinct opens,so binary closure suffices;an invalid parent is rejected."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S003",
      "uses": [
        "G001",
        "S002"
      ],
      "before": "ORIGINAL parent Fin(5);parent open index=1,U=[0];complete topology=[[], [0], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 2], [0, 2, 4]]",
      "after": "ALL [other_open_index,union,intersection] rows=[[0, [0], []], [1, [0], [0]], [2, [0, 1, 2, 3, 4], [0]], [3, [0, 1, 2, 4], [0]], [4, [0, 2], [0]], [5, [0, 2, 4], [0]]];each belongs to the SAME parent family",
      "relation": "implies",
      "rule_id": "subtop_qualified_parent_all_pairs_union_intersection",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先驗母拓撲含空集與全集，再核全部有序open pair的union／intersection閉合。有限family的任意union可去重成有限union，故二元閉合足夠；不能把不合格母資料當True。",
        "en": "First require the empty/full opens,then check unions and intersections for every ordered pair in the same parent family.Any union in a finite family reduces to finitely many distinct opens,so binary closure suffices;an invalid parent is rejected."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S004",
      "uses": [
        "G001",
        "S003"
      ],
      "before": "ORIGINAL parent Fin(5);parent open index=2,U=[0, 1, 2, 3, 4];complete topology=[[], [0], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 2], [0, 2, 4]]",
      "after": "ALL [other_open_index,union,intersection] rows=[[0, [0, 1, 2, 3, 4], []], [1, [0, 1, 2, 3, 4], [0]], [2, [0, 1, 2, 3, 4], [0, 1, 2, 3, 4]], [3, [0, 1, 2, 3, 4], [0, 1, 2, 4]], [4, [0, 1, 2, 3, 4], [0, 2]], [5, [0, 1, 2, 3, 4], [0, 2, 4]]];each belongs to the SAME parent family",
      "relation": "implies",
      "rule_id": "subtop_qualified_parent_all_pairs_union_intersection",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先驗母拓撲含空集與全集，再核全部有序open pair的union／intersection閉合。有限family的任意union可去重成有限union，故二元閉合足夠；不能把不合格母資料當True。",
        "en": "First require the empty/full opens,then check unions and intersections for every ordered pair in the same parent family.Any union in a finite family reduces to finitely many distinct opens,so binary closure suffices;an invalid parent is rejected."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S005",
      "uses": [
        "G001",
        "S004"
      ],
      "before": "ORIGINAL parent Fin(5);parent open index=3,U=[0, 1, 2, 4];complete topology=[[], [0], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 2], [0, 2, 4]]",
      "after": "ALL [other_open_index,union,intersection] rows=[[0, [0, 1, 2, 4], []], [1, [0, 1, 2, 4], [0]], [2, [0, 1, 2, 3, 4], [0, 1, 2, 4]], [3, [0, 1, 2, 4], [0, 1, 2, 4]], [4, [0, 1, 2, 4], [0, 2]], [5, [0, 1, 2, 4], [0, 2, 4]]];each belongs to the SAME parent family",
      "relation": "implies",
      "rule_id": "subtop_qualified_parent_all_pairs_union_intersection",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先驗母拓撲含空集與全集，再核全部有序open pair的union／intersection閉合。有限family的任意union可去重成有限union，故二元閉合足夠；不能把不合格母資料當True。",
        "en": "First require the empty/full opens,then check unions and intersections for every ordered pair in the same parent family.Any union in a finite family reduces to finitely many distinct opens,so binary closure suffices;an invalid parent is rejected."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S006",
      "uses": [
        "G001",
        "S005"
      ],
      "before": "ORIGINAL parent Fin(5);parent open index=4,U=[0, 2];complete topology=[[], [0], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 2], [0, 2, 4]]",
      "after": "ALL [other_open_index,union,intersection] rows=[[0, [0, 2], []], [1, [0, 2], [0]], [2, [0, 1, 2, 3, 4], [0, 2]], [3, [0, 1, 2, 4], [0, 2]], [4, [0, 2], [0, 2]], [5, [0, 2, 4], [0, 2]]];each belongs to the SAME parent family",
      "relation": "implies",
      "rule_id": "subtop_qualified_parent_all_pairs_union_intersection",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先驗母拓撲含空集與全集，再核全部有序open pair的union／intersection閉合。有限family的任意union可去重成有限union，故二元閉合足夠；不能把不合格母資料當True。",
        "en": "First require the empty/full opens,then check unions and intersections for every ordered pair in the same parent family.Any union in a finite family reduces to finitely many distinct opens,so binary closure suffices;an invalid parent is rejected."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S007",
      "uses": [
        "G001",
        "S006"
      ],
      "before": "ORIGINAL parent Fin(5);parent open index=5,U=[0, 2, 4];complete topology=[[], [0], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 2], [0, 2, 4]]",
      "after": "ALL [other_open_index,union,intersection] rows=[[0, [0, 2, 4], []], [1, [0, 2, 4], [0]], [2, [0, 1, 2, 3, 4], [0, 2, 4]], [3, [0, 1, 2, 4], [0, 2, 4]], [4, [0, 2, 4], [0, 2]], [5, [0, 2, 4], [0, 2, 4]]];each belongs to the SAME parent family",
      "relation": "implies",
      "rule_id": "subtop_qualified_parent_all_pairs_union_intersection",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先驗母拓撲含空集與全集，再核全部有序open pair的union／intersection閉合。有限family的任意union可去重成有限union，故二元閉合足夠；不能把不合格母資料當True。",
        "en": "First require the empty/full opens,then check unions and intersections for every ordered pair in the same parent family.Any union in a finite family reduces to finitely many distinct opens,so binary closure suffices;an invalid parent is rejected."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S008",
      "uses": [
        "G001",
        "S007"
      ],
      "before": "ORIGINAL parent open index=0,U=[];ORIGINAL subset Y=[1, 2, 3]",
      "after": "actual U intersect Y=[];retain parent index even if another open has the same intersection",
      "relation": "implies",
      "rule_id": "subtop_every_parent_open_intersection_with_original_Y",
      "conditions": [],
      "explanation": {
        "zh-Hant": "對每個母open逐點交Y；相同交集只在誘導family去重，原母witness不能丟。U可含Y外的點，不強迫U subset Y。",
        "en": "Intersect every parent open with Y pointwise.Deduplicate only the induced family,not its distinct ambient witnesses.U may contain points outside Y;do not require U to be a subset of Y."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S009",
      "uses": [
        "G001",
        "S008"
      ],
      "before": "ORIGINAL parent open index=1,U=[0];ORIGINAL subset Y=[1, 2, 3]",
      "after": "actual U intersect Y=[];retain parent index even if another open has the same intersection",
      "relation": "implies",
      "rule_id": "subtop_every_parent_open_intersection_with_original_Y",
      "conditions": [],
      "explanation": {
        "zh-Hant": "對每個母open逐點交Y；相同交集只在誘導family去重，原母witness不能丟。U可含Y外的點，不強迫U subset Y。",
        "en": "Intersect every parent open with Y pointwise.Deduplicate only the induced family,not its distinct ambient witnesses.U may contain points outside Y;do not require U to be a subset of Y."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S010",
      "uses": [
        "G001",
        "S009"
      ],
      "before": "ORIGINAL parent open index=2,U=[0, 1, 2, 3, 4];ORIGINAL subset Y=[1, 2, 3]",
      "after": "actual U intersect Y=[1, 2, 3];retain parent index even if another open has the same intersection",
      "relation": "implies",
      "rule_id": "subtop_every_parent_open_intersection_with_original_Y",
      "conditions": [],
      "explanation": {
        "zh-Hant": "對每個母open逐點交Y；相同交集只在誘導family去重，原母witness不能丟。U可含Y外的點，不強迫U subset Y。",
        "en": "Intersect every parent open with Y pointwise.Deduplicate only the induced family,not its distinct ambient witnesses.U may contain points outside Y;do not require U to be a subset of Y."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S011",
      "uses": [
        "G001",
        "S010"
      ],
      "before": "ORIGINAL parent open index=3,U=[0, 1, 2, 4];ORIGINAL subset Y=[1, 2, 3]",
      "after": "actual U intersect Y=[1, 2];retain parent index even if another open has the same intersection",
      "relation": "implies",
      "rule_id": "subtop_every_parent_open_intersection_with_original_Y",
      "conditions": [],
      "explanation": {
        "zh-Hant": "對每個母open逐點交Y；相同交集只在誘導family去重，原母witness不能丟。U可含Y外的點，不強迫U subset Y。",
        "en": "Intersect every parent open with Y pointwise.Deduplicate only the induced family,not its distinct ambient witnesses.U may contain points outside Y;do not require U to be a subset of Y."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S012",
      "uses": [
        "G001",
        "S011"
      ],
      "before": "ORIGINAL parent open index=4,U=[0, 2];ORIGINAL subset Y=[1, 2, 3]",
      "after": "actual U intersect Y=[2];retain parent index even if another open has the same intersection",
      "relation": "implies",
      "rule_id": "subtop_every_parent_open_intersection_with_original_Y",
      "conditions": [],
      "explanation": {
        "zh-Hant": "對每個母open逐點交Y；相同交集只在誘導family去重，原母witness不能丟。U可含Y外的點，不強迫U subset Y。",
        "en": "Intersect every parent open with Y pointwise.Deduplicate only the induced family,not its distinct ambient witnesses.U may contain points outside Y;do not require U to be a subset of Y."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S013",
      "uses": [
        "G001",
        "S012"
      ],
      "before": "ORIGINAL parent open index=5,U=[0, 2, 4];ORIGINAL subset Y=[1, 2, 3]",
      "after": "actual U intersect Y=[2];retain parent index even if another open has the same intersection",
      "relation": "implies",
      "rule_id": "subtop_every_parent_open_intersection_with_original_Y",
      "conditions": [],
      "explanation": {
        "zh-Hant": "對每個母open逐點交Y；相同交集只在誘導family去重，原母witness不能丟。U可含Y外的點，不強迫U subset Y。",
        "en": "Intersect every parent open with Y pointwise.Deduplicate only the induced family,not its distinct ambient witnesses.U may contain points outside Y;do not require U to be a subset of Y."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S014",
      "uses": [
        "G001",
        "S013"
      ],
      "before": "subtop distinct relative open count := sum(1, 1, 1, 1)",
      "after": "subtop distinct relative open count = 4",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "依原題角色與已核中間量精確代值，保留原順序、正負號及量詞條件。",
        "en": "Substitute the original roles and established intermediates exactly,retaining order,signs and quantified conditions."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 subtop distinct relative open count，不以小數近似替代。",
          "en": "Recompute subtop distinct relative open count with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "1",
          "1",
          "1",
          "1"
        ],
        "output": "4",
        "label": "subtop distinct relative open count"
      }
    },
    {
      "id": "S015",
      "uses": [
        "G001",
        "S014"
      ],
      "before": "ORIGINAL Y=[1, 2, 3];ALL parent intersections=[[], [], [1, 2, 3], [1, 2], [2], [2]]",
      "after": "normalized relative topology=[[], [1, 2], [1, 2, 3], [2]];distinct count=4;empty and Y included;unions and finite intersections commute with intersect Y",
      "relation": "implies",
      "rule_id": "subtop_induced_axioms_and_empty_Y",
      "conditions": [],
      "explanation": {
        "zh-Hant": "任意union與有限intersection都可把交Y提出，再由母拓撲封閉得到相對open。Y空時所有交集都是空集，誘導拓撲是只含空集的一個family，不是零個open。",
        "en": "Intersection with Y commutes with arbitrary unions and finite intersections;parent closure gives the induced axioms.For empty Y every intersection is empty,so the topology has one open,not zero."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S016",
      "uses": [
        "G001",
        "S015"
      ],
      "before": "ORIGINAL candidate=[1, 2, 3] subset Y=[1, 2, 3];ALL original parent opens=[[], [0], [0, 1, 2, 3, 4], [0, 1, 2, 4], [0, 2], [0, 2, 4]]",
      "after": "ALL ambient U with U intersect Y=candidate=[[0, 1, 2, 3, 4]];relative_open=是;ambient_open=否;Y minus candidate=[];relative_closed=是",
      "relation": "implies",
      "rule_id": "subtop_relative_open_witnesses_and_relative_complement",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相對open以母open交Y見證，不等於candidate自己在母拓撲。相對closed要查Y內補集open，並由De Morgan律看出相對closed的母closed交Y形式；不改用母空間內補集。",
        "en": "Relative openness is witnessed by ambient-open intersections,not ambient membership of the candidate.Relative closedness tests the complement inside Y.De Morgan gives the ambient-closed-intersection form;do not substitute X minus candidate for the relative complement."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S017",
      "uses": [
        "G001",
        "S016"
      ],
      "before": "ACTUAL candidate=[1, 2, 3];relative_open=是;ambient_open=否",
      "after": "relative-only certificate=是;O intersect Y definition does NOT require O subset Y",
      "relation": "implies",
      "rule_id": "subtop_boundary_requires_actual_relative_not_ambient",
      "conditions": [],
      "explanation": {
        "zh-Hant": "這題先有真relative-open且非ambient-open的資料，才給boundary證書；不是聲稱任何relative-open都非ambient-open。",
        "en": "This boundary requires an actually relative-open but non-ambient-open candidate;it does not say every relative open fails ambient openness."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    }
  ],
  "math_model": {
    "result": [
      [
        1,
        2,
        3
      ],
      true,
      false,
      true,
      "O_intersect_Y_not_O_subset_Y"
    ]
  },
  "conclusion": {
    "canonical_value": "([1, 2, 3], 是, 否, 是, O_intersect_Y_not_O_subset_Y)",
    "explanation": {
      "zh-Hant": "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。",
      "en": "The full substitutions and premise checks give the original answer; source identity,format and status are unchanged."
    }
  },
  "definition_note": {
    "zh-Hant": "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。",
    "en": "Use the original question domain, assumptions and cited MKO definitions. This item has second-context mathematical and bilingual review, with the model, answer and every arithmetic step independently recomputed from the original parameters; the source question remains a candidate."
  },
  "source": {
    "input_commit": "0f7be61e2c22317a4075561dc822c87d881a50db",
    "question_binding": {
      "batch_id": "batch-20261014-advance-20261108-r03-subspace-topology-foundations",
      "batch_path": "public/data/questions/batches/batch-20261014-advance-20261108-r03-subspace-topology-foundations/questions.jsonl",
      "batch_sha256": "62dfefafdbcf5526248b18a3436d1a2574778407b0029f5e2f9543a338f82959",
      "original_line": 86,
      "original_line_sha256": "aaaf5f55c0c6788528164e249dee1c293527979bc0a4eb4d590ca34936c9bdd7",
      "question_compact_sha256": "a8556a1f655bf7c238c3ba5f7399e0bb5f7c645239b4978d2efde09404abe1de",
      "english_row_compact_sha256": "9503c3af261cdbfc80444f77eed827f75d0a26872d14ae2dae486b4014433590",
      "english_manifest_path": "public/data/i18n/editions/batch-20261014-advance-20261108-r03-subspace-topology-foundations.en.json",
      "english_rows_path": "public/data/i18n/questions/batch-20261014-advance-20261108-r03-subspace-topology-foundations.en.jsonl"
    },
    "mko_bindings": {
      "mko-subspace-topology": {
        "sha256": "84259c8587e1acd913189fcf6b17200828b88d36622591d8bc5a52b3455544e7",
        "english_sha256": "6f532d1a2d042136c6f62e7c763d13337ac407fea6f50219d476e51f4dea0198",
        "original_path": "public/data/mko/subspace-topology.json",
        "english_path": "public/data/i18n/editions/mko-subspace-topology.en.json"
      }
    },
    "selected_manifest_sha256": "5233edba9654a6cffd8b6b2d83c932f82ded15f9c0b7c733bd049943e672acff",
    "selected_part_sha256": "2bda8f45e99ed8658b697be77ed3e5ec90ceff1b4f72fa7aa6a47d1a0c6254cf",
    "original_candidate_sha256": "5a5c043c51145f89a9804fca023305f8e7c04bcca6f900f014a6f79dd06b45d2",
    "original_candidate_line_sha256": "5a5c043c51145f89a9804fca023305f8e7c04bcca6f900f014a6f79dd06b45d2"
  },
  "historical_source_state": {
    "review_state": {
      "derivation": "candidate_authored_and_locally_checked",
      "source_question_state": "candidate",
      "independent_AI_or_human_review": "NOT_PERFORMED",
      "formalization": "not_formalized",
      "Lean": "NOT_RUN"
    },
    "scope_note": {
      "zh-Hant": "來源綁定教學推導候選；精確核對不等於獨立AI／人工審查或一般形式證明。原題、答案、MKO、公式、AST與歷史不變。",
      "en": "Source-bound pedagogical derivation candidate.Exact checking is not independent AI/human review or general formal proof.Original questions,answers,MKOs,formulas,ASTs and history remain unchanged."
    },
    "label": "Frozen producer metadata; current per-question review is separate below."
  },
  "revision": {
    "id": "review-20261026-009-p42",
    "review_date": "2026-10-26",
    "owner_authorized_reader_publication": true,
    "scope": "One hundred sixty family prototypes and their mathematical/bilingual reasoning variants, stems, boundary statements and every distinct explanation shape were reviewed in a context independent of the producer; parameter substitutions, source bindings, full math models, both answer strings and every arithmetic step were independently reconstructed per question from the original parameters only, with negative controls. This grouped method does not claim a separate literal AI reading of every repeated sentence or an extra AI voter.",
    "review_receipts": {
      "semantic_review_sha256": "8f134403487694af4c70d2e00ec84bab542043ef4b3c54517c04f3c409c0262a",
      "question_first_witness_sha256": "ad772845088ed8e99f87f90d4264b05ea97be35775305a02011af5ee35f0f4ce"
    },
    "status": "second_context_reviewed",
    "mathematical_semantics": true,
    "bilingual_semantics": true,
    "steps_reviewed": 17,
    "method": "family_template_semantic_review_plus_independent_per_instance_reconstruction",
    "exact_question_first_check": {
      "model_sha256": "752debbb94faf2441b755d365a0306948d8e1a8c870f93b78d1f5e5b6ca5f67f",
      "arithmetic_steps": 1
    },
    "uses_answer_as_premise": false,
    "human_review": "NOT_PERFORMED",
    "formalization": "not_formalized",
    "Lean": "NOT_RUN",
    "Canonical_truth_promotion": false,
    "authority": "Owner-authorized reviewed reader publication; original question/candidate states remain historical and unchanged."
  }
}