{
  "schema_version": "ocme-reviewed-question-reader-v1",
  "question_id": "q-20261013-advance-20261019-r03-topological-continuity-0031",
  "pilot_number": "R0017-09431",
  "number": 9431,
  "family_id": "qf-ext19-continuity-full-audit",
  "title": {
    "zh-Hant": "第 9431 題：有限拓撲連續性",
    "en": "Question 9431: Every Target Open Set and Its Full Preimage"
  },
  "original_question": {
    "schema_version": "ocme-question-v0.1",
    "id": "q-20261013-advance-20261019-r03-topological-continuity-0031",
    "batch_id": "batch-20261013-advance-20261019-r03-topological-continuity-foundations",
    "language": "zh-Hant",
    "source_mko_ids": [
      "mko-topological-continuity"
    ],
    "title_zh": "有限拓撲連續性",
    "stem_zh": "完整有限拓撲與總映射。操作=continuity-full-audit；資料={\"n\":4,\"m\":4,\"opens_x\":[[],[1],[1,2],[0,1,2],[0,1,2,3]],\"opens_y\":[[],[1],[0,1],[0,1,2],[0,1,2,3]],\"images\":[1,2,0,3]}。請依 [images,all_target_open_preimage_rows,continuous] 欄序回答。",
    "answer": {
      "kind": "expression",
      "value": "([1, 2, 0, 3], [[[], [], 是], [[1], [0], 否], [[0, 1], [0, 2], 否], [[0, 1, 2], [0, 1, 2], 是], [[0, 1, 2, 3], [0, 1, 2, 3], 是]], 否)",
      "explanation_zh": "第二批擴展核對=([1, 2, 0, 3], [[[], [], 是], [[1], [0], 否], [[0, 1], [0, 2], 否], [[0, 1, 2], [0, 1, 2], 是], [[0, 1, 2, 3], [0, 1, 2, 3], 是]], 否)；依完整明示資料重算；有限／合成資料核對不是一般形式證明；逐一核對全部目標開集逆像；開映射與連續性分離。"
    },
    "difficulty": {
      "band": "undergraduate_core",
      "level": 1,
      "skills": [
        "有限拓撲連續性"
      ]
    },
    "variant": {
      "family_id": "qf-ext19-continuity-full-audit",
      "seed": 202610193031,
      "parameters": {
        "n": 4,
        "m": 4,
        "opens_x": [
          [],
          [
            1
          ],
          [
            1,
            2
          ],
          [
            0,
            1,
            2
          ],
          [
            0,
            1,
            2,
            3
          ]
        ],
        "opens_y": [
          [],
          [
            1
          ],
          [
            0,
            1
          ],
          [
            0,
            1,
            2
          ],
          [
            0,
            1,
            2,
            3
          ]
        ],
        "images": [
          1,
          2,
          0,
          3
        ],
        "domain": "Fin_topological_map",
        "operation": "continuity-full-audit",
        "semantic_template": "qf-ext19-continuity-full-audit/{\"images\":[1,2,0,3],\"m\":4,\"n\":4,\"opens_x\":[[],[1],[1,2],[0,1,2],[0,1,2,3]],\"opens_y\":[[],[1],[0,1],[0,1,2],[0,1,2,3]]}"
      }
    },
    "verification": {
      "status": "candidate",
      "checks": [
        "producer_exact_finite_model",
        "producer_source_snapshot_check",
        "producer_substantive_template_deduplication",
        "producer_json_schema"
      ],
      "warning_zh": "僅為產生端候選及自檢；主端須另驗，不構成人工審定、一般程式證明、Lean執行或形式Evidence。"
    },
    "provenance": {
      "generator_role": "question_producer",
      "generator_model": "model_unknown",
      "generated_at": "2026-10-13",
      "contract_version": "ocme-question-factory-v0.1"
    }
  },
  "original_english_edition": {
    "schema_version": "ocme-question-localization-v0.1",
    "question_id": "q-20261013-advance-20261019-r03-topological-continuity-0031",
    "batch_id": "batch-20261013-advance-20261019-r03-topological-continuity-foundations",
    "locale": "en",
    "translation_state": "ai_candidate",
    "source": {
      "question_sha256": "bcf2f64c4cb029ebac77629b11254a7789c4a117d0518cfeb1d007f135ade695",
      "family_id": "qf-ext19-continuity-full-audit",
      "seed": 202610193031,
      "semantic_template": "qf-ext19-continuity-full-audit/{\"images\":[1,2,0,3],\"m\":4,\"n\":4,\"opens_x\":[[],[1],[1,2],[0,1,2],[0,1,2,3]],\"opens_y\":[[],[1],[0,1],[0,1,2],[0,1,2,3]]}"
    },
    "fields": {
      "title": "Every Target Open Set and Its Full Preimage",
      "stem": "Use complete finite topologies X=Fin(4)={0,…,3}, opens_X=[[], [1], [1, 2], [0, 1, 2], [0, 1, 2, 3]] and Y=Fin(4)={0,…,3}, opens_Y=[[], [1], [0, 1], [0, 1, 2], [0, 1, 2, 3]]. The complete total map has images=[1, 2, 0, 3] in source order. All entries must type-check; set preimages are not inverse functions. Give (images, every row [target open, full preimage, source-open?], continuity?). Inspect all target opens in subset-mask order.",
      "answer": {
        "kind": "expression",
        "value": "([1, 2, 0, 3], [[[], [], Yes], [[1], [0], No], [[0, 1], [0, 2], No], [[0, 1, 2], [0, 1, 2], Yes], [[0, 1, 2, 3], [0, 1, 2, 3], Yes]], No)",
        "explanation": "Continuity requires all the open-preimage rows; sampling points or one open set is insufficient. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 2, 0, 3], [[[], [], Yes], [[1], [0], No], [[0, 1], [0, 2], No], [[0, 1, 2], [0, 1, 2], Yes], [[0, 1, 2, 3], [0, 1, 2, 3], Yes]], No)."
      },
      "skills": [
        "Every Target Open Set and Its Full Preimage"
      ],
      "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": "完整有限拓撲與總映射。操作=continuity-full-audit；資料={\"n\":4,\"m\":4,\"opens_x\":[[],[1],[1,2],[0,1,2],[0,1,2,3]],\"opens_y\":[[],[1],[0,1],[0,1,2],[0,1,2,3]],\"images\":[1,2,0,3]}。請依 [images,all_target_open_preimage_rows,continuous] 欄序回答。",
        "en": "Use complete finite topologies X=Fin(4)={0,…,3}, opens_X=[[], [1], [1, 2], [0, 1, 2], [0, 1, 2, 3]] and Y=Fin(4)={0,…,3}, opens_Y=[[], [1], [0, 1], [0, 1, 2], [0, 1, 2, 3]]. The complete total map has images=[1, 2, 0, 3] in source order. All entries must type-check; set preimages are not inverse functions. Give (images, every row [target open, full preimage, source-open?], continuity?). Inspect all target opens in subset-mask order."
      }
    }
  ],
  "steps": [
    {
      "id": "S001",
      "uses": [
        "G001"
      ],
      "before": "space=X;n=4;raw family=[[], [1], [1, 2], [0, 1, 2], [0, 1, 2, 3]]",
      "after": "universe=[0, 1, 2, 3];canonical opens=[[], [1], [1, 2], [0, 1, 2], [0, 1, 2, 3]]",
      "relation": "implies",
      "rule_id": "finite_topology_normalize_relative_universe",
      "conditions": [],
      "explanation": {
        "zh-Hant": "每個候選先合型為同一Fin(n)的子集，去重元素及重複子集，再以二進位mask排序；開集族與底集合是不同層資料。",
        "en": "Type every candidate as a subset of the same Fin(n),deduplicate elements and members,and order by binary masks.The open family and underlying set are distinct data layers."
      },
      "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": "space=X;empty=是;full=是;union closed=是;intersection closed=是",
      "after": "topology=是;first union failure=無;first intersection failure=無",
      "relation": "implies",
      "rule_id": "finite_topology_axioms_joint_gate",
      "conditions": [],
      "explanation": {
        "zh-Hant": "四條件同時成立才是拓撲。底集合有限，所以不同開集也只有有限多個；任意聯集可約成有限不同成員的聯集，由二元閉合與空聯集得到。空交集需全集，無成員時二元條件真也不能略過空集／全集。",
        "en": "All four conditions are required.On a finite underlying set there are finitely many distinct opens,so an arbitrary union reduces to a finite distinct-member union,covered by binary closure and the empty union.The empty intersection needs the whole set.Vacuous binary closure does not waive empty/whole-set membership."
      },
      "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": "space=Y;n=4;raw family=[[], [1], [0, 1], [0, 1, 2], [0, 1, 2, 3]]",
      "after": "universe=[0, 1, 2, 3];canonical opens=[[], [1], [0, 1], [0, 1, 2], [0, 1, 2, 3]]",
      "relation": "implies",
      "rule_id": "finite_topology_normalize_relative_universe",
      "conditions": [],
      "explanation": {
        "zh-Hant": "每個候選先合型為同一Fin(n)的子集，去重元素及重複子集，再以二進位mask排序；開集族與底集合是不同層資料。",
        "en": "Type every candidate as a subset of the same Fin(n),deduplicate elements and members,and order by binary masks.The open family and underlying set are distinct data layers."
      },
      "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": "space=Y;empty=是;full=是;union closed=是;intersection closed=是",
      "after": "topology=是;first union failure=無;first intersection failure=無",
      "relation": "implies",
      "rule_id": "finite_topology_axioms_joint_gate",
      "conditions": [],
      "explanation": {
        "zh-Hant": "四條件同時成立才是拓撲。底集合有限，所以不同開集也只有有限多個；任意聯集可約成有限不同成員的聯集，由二元閉合與空聯集得到。空交集需全集，無成員時二元條件真也不能略過空集／全集。",
        "en": "All four conditions are required.On a finite underlying set there are finitely many distinct opens,so an arbitrary union reduces to a finite distinct-member union,covered by binary closure and the empty union.The empty intersection needs the whole set.Vacuous binary closure does not waive empty/whole-set membership."
      },
      "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": "f:Fin(4)->Fin(4);whole images=[1, 2, 0, 3]",
      "after": "all original image entries typed;source labels retained",
      "relation": "implies",
      "rule_id": "whole_topological_total_map_guard",
      "conditions": [],
      "explanation": {
        "zh-Hant": "即使本題只問一個空集逆像，也先核整張總表每格合型；保留原來源索引，不依查詢需求丟棄非法格。",
        "en": "Validate every cell of the complete total table even for a query about the empty preimage.Retain original source indices;do not discard invalid cells because a query does not use them."
      },
      "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": "map=f;images=[1, 2, 0, 3];target open=[]",
      "after": "full preimage=[];source open=是",
      "relation": "implies",
      "rule_id": "complete_open_preimage_row",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐一收集所有f(x)在U的原來源標籤，再查此完整集合是否在來源開集族；集合逆像不是反函數，不要求映射單射或滿射。",
        "en": "Collect every original source label with f(x) in U,then test the complete set in the source open family.A set preimage is not an inverse function and requires neither injectivity nor surjectivity."
      },
      "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": "map=f;images=[1, 2, 0, 3];target open=[1]",
      "after": "full preimage=[0];source open=否",
      "relation": "implies",
      "rule_id": "complete_open_preimage_row",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐一收集所有f(x)在U的原來源標籤，再查此完整集合是否在來源開集族；集合逆像不是反函數，不要求映射單射或滿射。",
        "en": "Collect every original source label with f(x) in U,then test the complete set in the source open family.A set preimage is not an inverse function and requires neither injectivity nor surjectivity."
      },
      "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": "map=f;images=[1, 2, 0, 3];target open=[0, 1]",
      "after": "full preimage=[0, 2];source open=否",
      "relation": "implies",
      "rule_id": "complete_open_preimage_row",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐一收集所有f(x)在U的原來源標籤，再查此完整集合是否在來源開集族；集合逆像不是反函數，不要求映射單射或滿射。",
        "en": "Collect every original source label with f(x) in U,then test the complete set in the source open family.A set preimage is not an inverse function and requires neither injectivity nor surjectivity."
      },
      "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": "map=f;images=[1, 2, 0, 3];target open=[0, 1, 2]",
      "after": "full preimage=[0, 1, 2];source open=是",
      "relation": "implies",
      "rule_id": "complete_open_preimage_row",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐一收集所有f(x)在U的原來源標籤，再查此完整集合是否在來源開集族；集合逆像不是反函數，不要求映射單射或滿射。",
        "en": "Collect every original source label with f(x) in U,then test the complete set in the source open family.A set preimage is not an inverse function and requires neither injectivity nor surjectivity."
      },
      "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": "map=f;images=[1, 2, 0, 3];target open=[0, 1, 2, 3]",
      "after": "full preimage=[0, 1, 2, 3];source open=是",
      "relation": "implies",
      "rule_id": "complete_open_preimage_row",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐一收集所有f(x)在U的原來源標籤，再查此完整集合是否在來源開集族；集合逆像不是反函數，不要求映射單射或滿射。",
        "en": "Collect every original source label with f(x) in U,then test the complete set in the source open family.A set preimage is not an inverse function and requires neither injectivity nor surjectivity."
      },
      "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": "map=f;all target opens=[[], [1], [0, 1], [0, 1, 2], [0, 1, 2, 3]];all preimage rows=[[[], [], 是], [[1], [0], 否], [[0, 1], [0, 2], 否], [[0, 1, 2], [0, 1, 2], 是], [[0, 1, 2, 3], [0, 1, 2, 3], 是]]",
      "after": "continuous=否",
      "relation": "implies",
      "rule_id": "continuity_all_target_opens",
      "conditions": [],
      "explanation": {
        "zh-Hant": "連續性要求每個目標開集的逆像都開；任一失敗即可否定，全數通過依完整有限拓撲資料才可肯定，不以單一成功列或點抽查代替。",
        "en": "Continuity requires an open preimage for every target open.A single failure refutes it;complete finite data permits affirmation only when all rows pass.One successful row or point sample is insufficient."
      },
      "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,
        0,
        3
      ],
      [
        [
          [],
          [],
          true
        ],
        [
          [
            1
          ],
          [
            0
          ],
          false
        ],
        [
          [
            0,
            1
          ],
          [
            0,
            2
          ],
          false
        ],
        [
          [
            0,
            1,
            2
          ],
          [
            0,
            1,
            2
          ],
          true
        ],
        [
          [
            0,
            1,
            2,
            3
          ],
          [
            0,
            1,
            2,
            3
          ],
          true
        ]
      ],
      false
    ]
  },
  "conclusion": {
    "canonical_value": "([1, 2, 0, 3], [[[], [], 是], [[1], [0], 否], [[0, 1], [0, 2], 否], [[0, 1, 2], [0, 1, 2], 是], [[0, 1, 2, 3], [0, 1, 2, 3], 是]], 否)",
    "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-20261013-advance-20261019-r03-topological-continuity-foundations",
      "batch_path": "public/data/questions/batches/batch-20261013-advance-20261019-r03-topological-continuity-foundations/questions.jsonl",
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      "english_manifest_path": "public/data/i18n/editions/batch-20261013-advance-20261019-r03-topological-continuity-foundations.en.json",
      "english_rows_path": "public/data/i18n/questions/batch-20261013-advance-20261019-r03-topological-continuity-foundations.en.jsonl"
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    "selected_part_sha256": "20e59f30c0b6218e6d9614c16170d770e56a2f1e4ae5ce39d0a8a3db206afa61",
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      "source_question_state": "candidate",
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      "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."
  },
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    "id": "review-20261026-006-p18",
    "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": {
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    "bilingual_semantics": true,
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    "authority": "Owner-authorized reviewed reader publication; original question/candidate states remain historical and unchanged."
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}