{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-open-closed-set-01",
  "mko_id": "mko-open-closed-set",
  "title": {
    "zh-Hant": "開集與閉集：在四點拓撲中完整分類十六個子集",
    "en": "Open and closed sets: classifying all sixteen subsets of a four-point topology"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-open-closed-set-01",
    "source": {
      "mko_id": "mko-open-closed-set",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "78fdc40df7a4161259566db404d47fa805aa334d6f661420675bbc8c139c6508",
      "english_sha256": "54fd15465e9b24d23c770140d38653759ffbdbae2087c9777b70d3cf4211da31"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "set_operations_def",
      "set_extensionality",
      "subset_def",
      "finite_counting_basics",
      "nat_induction"
    ],
    "goal": {
      "zh-Hant": "設 X={1,2,3,4}，τ={∅,{1},{1,2},{3,4},{1,3,4},X}。先驗證 τ 是 X 上的拓撲，再列出全部閉集，並把 X 的十六個子集分成「既開又閉」「只開」「只閉」「既不開也不閉」四類；最後證明在任何拓撲空間中 ∅ 與 X 都既開又閉，以及兩個閉集的聯集仍是閉集。",
      "en": "Let X={1,2,3,4} and τ={∅,{1},{1,2},{3,4},{1,3,4},X}. First verify that τ is a topology on X, then list all closed sets and sort the sixteen subsets of X into four classes: clopen, open only, closed only, and neither; finally prove that in every topological space ∅ and X are clopen, and that the union of two closed sets is closed."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "底集合 X={1,2,3,4}（MKO 假設 a1）。",
          "en": "The underlying set X={1,2,3,4} (MKO assumption a1)."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "子集合族 τ={∅,{1},{1,2},{3,4},{1,3,4},X}，每個成員都是 X 的子集合。",
          "en": "The family τ={∅,{1},{1,2},{3,4},{1,3,4},X}; every member is a subset of X."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "所有補集都取相對於 X 的補集 X∖C，候選集合都先確認是 X 的子集合（MKO 假設 a2）。",
          "en": "Every complement is the complement X∖C relative to X, and every candidate set is first confirmed to be a subset of X (MKO assumption a2)."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2"
        ],
        "before_tex": "\\tau=\\{\\varnothing,\\{1\\},\\{1,2\\},\\{3,4\\},\\{1,3,4\\},X\\}",
        "after_tex": "(X,\\tau)\\ \\text{is a topological space}",
        "relation": "theorem_application",
        "rule": "L_finite_family",
        "conditions": [
          "\\tau\\ \\text{finite}",
          "\\varnothing,X\\in\\tau"
        ],
        "explanation": {
          "zh-Hant": "τ 的六個成員都是 X 的子集合，且 ∅∈τ、X∈τ。τ 是有限族，依引理 L_finite_family（在 S01.1、S01.2 證明），只要檢查任兩個成員的聯集與交集都在 τ，就得到對每個子族的聯集與每個有限子族的交集都封閉；這個兩兩檢查在 S01.3 完成。所以 (X,τ) 是拓撲空間，之後才能談開與閉。",
          "en": "All six members of τ are subsets of X, and ∅∈τ, X∈τ. τ is a finite family, so by Lemma L_finite_family (proved in S01.1 and S01.2) it suffices to check that the union and intersection of any two members lie in τ to get closure under unions of every subfamily and intersections of every finite subfamily; this pairwise check is done in S01.3. So (X,τ) is a topological space, and only then can open and closed be discussed."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G2",
            "L_finite_family",
            "mko_topological_space_def"
          ],
          "explanation": {
            "zh-Hant": "需要 τ 有限、含 ∅ 與 X、以及 S01.3 的兩兩檢查。引理只在有限族時成立；無限族只對兩兩聯集封閉並不足夠（見錯誤控制三）。",
            "en": "It needs τ finite, containing ∅ and X, and the pairwise check of S01.3. The lemma holds only for finite families; for an infinite family, closure under pairwise unions is not enough (see error control three)."
          }
        },
        "leaf_basis": "topology_def + set_operations_def + nat_induction"
      },
      {
        "id": "S01.1",
        "uses": [
          "nat_induction",
          "set_operations_def",
          "finite_counting_basics"
        ],
        "before_tex": "F\\subseteq\\tau,\\ |F|=k",
        "after_tex": "\\textstyle\\bigcup F\\in\\tau\\ \\text{for every}\\ k",
        "relation": "induction_step",
        "rule": "nat_induction",
        "conditions": [
          "\\varnothing\\in\\tau",
          "U,V\\in\\tau\\Rightarrow U\\cup V\\in\\tau"
        ],
        "explanation": {
          "zh-Hant": "τ 有限，所以每個子族 F 都有限，對其成員數 k 歸納。k=0：空族的聯集是 ∅，而 ∅∈τ。k→k+1：取 F 的一個成員 U，令 F′=F∖{U}，則 |F′|=k，∪F=(∪F′)∪U；依歸納假設 ∪F′∈τ，再依兩兩聯集封閉得 ∪F∈τ。這證得引理的聯集部分。",
          "en": "τ is finite, so every subfamily F is finite; induct on its number of members k. k=0: the union of the empty family is ∅, and ∅∈τ. k→k+1: pick a member U of F and let F′=F∖{U}, so |F′|=k and ∪F=(∪F′)∪U; by the induction hypothesis ∪F′∈τ, and closure under pairwise unions gives ∪F∈τ. This proves the union part of the lemma."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "nat_induction",
            "set_operations_def",
            "finite_counting_basics"
          ],
          "explanation": {
            "zh-Hant": "歸納需要每個子族都有有限個成員，這來自 τ 有限。基底需要 ∅∈τ，不能省略空族。",
            "en": "The induction needs every subfamily to have finitely many members, which comes from τ being finite. The base case needs ∅∈τ; the empty family may not be skipped."
          }
        },
        "leaf_basis": "nat_induction + set_operations_def + finite_counting_basics"
      },
      {
        "id": "S01.2",
        "uses": [
          "nat_induction",
          "set_operations_def",
          "finite_counting_basics"
        ],
        "before_tex": "F\\subseteq\\tau,\\ |F|=k",
        "after_tex": "\\textstyle\\bigcap F\\in\\tau\\ \\text{for every}\\ k\\ (\\bigcap\\varnothing=X)",
        "relation": "induction_step",
        "rule": "nat_induction",
        "conditions": [
          "X\\in\\tau",
          "U,V\\in\\tau\\Rightarrow U\\cap V\\in\\tau"
        ],
        "explanation": {
          "zh-Hant": "同樣對 k 歸納。k=0：空族的交集按 X 解讀，X∈τ。k→k+1：∩F=(∩F′)∩U，依歸納假設與兩兩交集封閉得 ∩F∈τ。S01.1 與 S01.2 合起來證得 L_finite_family。",
          "en": "Induct on k in the same way. k=0: the intersection of the empty family is read as X, and X∈τ. k→k+1: ∩F=(∩F′)∩U, and the induction hypothesis with closure under pairwise intersections gives ∩F∈τ. S01.1 and S01.2 together prove L_finite_family."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "nat_induction",
            "set_operations_def"
          ],
          "explanation": {
            "zh-Hant": "需要 X∈τ 作為空族交集的基底，與兩兩交集封閉。",
            "en": "It needs X∈τ as the base for the empty intersection, and closure under pairwise intersections."
          }
        },
        "leaf_basis": "nat_induction + set_operations_def"
      },
      {
        "id": "S01.3",
        "uses": [
          "G2"
        ],
        "before_tex": "U,V\\in\\tau",
        "after_tex": "U\\cup V\\in\\tau,\\ U\\cap V\\in\\tau",
        "relation": "implies",
        "rule": "set_operations_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "含 ∅ 或 X 的組合都直接回到 τ（∅∪A=A、∅∩A=∅、X∪A=X、X∩A=A）。其餘六組：{1}∪{1,2}={1,2}、交集 {1}；{1}∪{3,4}={1,3,4}、交集 ∅；{1}∪{1,3,4}={1,3,4}、交集 {1}；{1,2}∪{3,4}=X、交集 ∅；{1,2}∪{1,3,4}=X、交集 {1}；{3,4}∪{1,3,4}={1,3,4}、交集 {3,4}。每個結果都在 τ 中。",
          "en": "Any pair involving ∅ or X returns to τ directly (∅∪A=A, ∅∩A=∅, X∪A=X, X∩A=A). The remaining six pairs: {1}∪{1,2}={1,2} with intersection {1}; {1}∪{3,4}={1,3,4} with intersection ∅; {1}∪{1,3,4}={1,3,4} with intersection {1}; {1,2}∪{3,4}=X with intersection ∅; {1,2}∪{1,3,4}=X with intersection {1}; {3,4}∪{1,3,4}={1,3,4} with intersection {3,4}. Every result is in τ."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G2",
            "set_operations_def"
          ],
          "explanation": {
            "zh-Hant": "四個非平凡成員兩兩配對共有 comb(4,2)=6 組，全部列出（N1），沒有遺漏。",
            "en": "The four nontrivial members form comb(4,2)=6 pairs, all listed (N1), with none omitted."
          }
        },
        "leaf_basis": "set_operations_def + finite_counting_basics"
      },
      {
        "id": "S02",
        "uses": [
          "S01",
          "G3"
        ],
        "before_tex": "C\\subseteq X\\ \\text{closed}\\iff X\\setminus C\\in\\tau",
        "after_tex": "C\\ \\text{closed}\\iff C=X\\setminus U\\ \\text{for some}\\ U\\in\\tau",
        "relation": "equivalent",
        "rule": "L_double_complement",
        "conditions": [
          "C\\subseteq X"
        ],
        "explanation": {
          "zh-Hant": "依 MKO 定義，C 是閉集當且僅當 X∖C∈τ。(⇒) 令 U=X∖C∈τ，依 L_double_complement 得 C=X∖(X∖C)=X∖U。(⇐) 若 C=X∖U 且 U∈τ，則 X∖C=X∖(X∖U)=U∈τ。所以閉集恰好是開集的補集，閉集族可由逐一取補集算出。",
          "en": "By the MKO definition, C is closed exactly when X∖C∈τ. (⇒) Let U=X∖C∈τ; by L_double_complement, C=X∖(X∖C)=X∖U. (⇐) If C=X∖U with U∈τ, then X∖C=X∖(X∖U)=U∈τ. So the closed sets are exactly the complements of open sets, and the closed family can be computed by taking complements one by one."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_double_complement",
            "mko_open_closed_def"
          ],
          "explanation": {
            "zh-Hant": "兩個方向都已寫出理由，都只用到雙重補集與定義。補集必須相對同一個 X，否則雙重補集公式不成立。",
            "en": "Both directions have reasons, using only the double complement and the definition. The complement must be relative to the same X, or the double complement formula fails."
          }
        },
        "leaf_basis": "def_unfold + set_operations_def + set_extensionality"
      },
      {
        "id": "S02.1",
        "uses": [
          "set_operations_def",
          "set_extensionality"
        ],
        "before_tex": "x\\in X\\setminus(X\\setminus C)",
        "after_tex": "x\\in C",
        "relation": "equivalent",
        "rule": "set_operations_def",
        "conditions": [
          "C\\subseteq X"
        ],
        "explanation": {
          "zh-Hant": "x∈X∖(X∖C) ⇔ x∈X 且不是（x∈X 且 x∉C）⇔ x∈X 且 x∈C ⇔ x∈C；最後一步用到 C⊆X。依集合外延性，X∖(X∖C)=C，證得 L_double_complement。",
          "en": "x∈X∖(X∖C) ⇔ x∈X and not (x∈X and x∉C) ⇔ x∈X and x∈C ⇔ x∈C; the last step uses C⊆X. By extensionality, X∖(X∖C)=C, proving L_double_complement."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "set_operations_def",
            "set_extensionality",
            "de_morgan_logic"
          ],
          "explanation": {
            "zh-Hant": "每一步都是邏輯等價（De Morgan 與 x∈X 已知），所以雙向成立。若 C 不是 X 的子集，最後一步只得 x∈C∩X，公式失效。",
            "en": "Each move is a logical equivalence (De Morgan, with x∈X known), so both directions hold. If C were not a subset of X, the last move would give only x∈C∩X and the formula would fail."
          }
        },
        "leaf_basis": "set_operations_def + de_morgan_logic + set_extensionality"
      },
      {
        "id": "S03",
        "uses": [
          "S02",
          "G2"
        ],
        "before_tex": "\\{X\\setminus U\\mid U\\in\\tau\\}",
        "after_tex": "\\mathcal{C}=\\{X,\\{2,3,4\\},\\{3,4\\},\\{1,2\\},\\{2\\},\\varnothing\\}",
        "relation": "implies",
        "rule": "set_operations_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐一取補集：X∖∅=X，X∖{1}={2,3,4}，X∖{1,2}={3,4}，X∖{3,4}={1,2}，X∖{1,3,4}={2}，X∖X=∅。依 S02，這六個集合就是全部閉集。因為取補集可由自身逆回（L_double_complement），不同開集給出不同閉集，所以恰有 6 個閉集。",
          "en": "Take complements one by one: X∖∅=X, X∖{1}={2,3,4}, X∖{1,2}={3,4}, X∖{3,4}={1,2}, X∖{1,3,4}={2}, X∖X=∅. By S02 these six sets are all the closed sets. Since taking complements undoes itself (L_double_complement), distinct open sets give distinct closed sets, so there are exactly 6 closed sets."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "set_operations_def",
            "L_double_complement"
          ],
          "explanation": {
            "zh-Hant": "需要 S02 的等價與 τ 的完整列表；每個補集都在 X 內計算。",
            "en": "It needs the equivalence of S02 and the complete list of τ; every complement is computed inside X."
          }
        },
        "leaf_basis": "set_operations_def + injective_surjective_def"
      },
      {
        "id": "S04",
        "uses": [
          "G2",
          "S03"
        ],
        "before_tex": "\\tau,\\ \\mathcal{C}",
        "after_tex": "\\tau\\cap\\mathcal{C}=\\{\\varnothing,\\{1,2\\},\\{3,4\\},X\\}",
        "relation": "implies",
        "rule": "set_operations_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "比對兩張表，同時出現在 τ 與 𝒞 中的是 ∅、{1,2}、{3,4}、X，這四個集合既開又閉。除了必然的 ∅ 與 X，{1,2} 與 {3,4} 也是 clopen：它們互為補集，而且都是開集。",
          "en": "Comparing the two lists, the sets in both τ and 𝒞 are ∅, {1,2}, {3,4} and X; these four are clopen. Besides the inevitable ∅ and X, {1,2} and {3,4} are clopen too: they are complements of each other and both are open."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G2",
            "S03",
            "set_operations_def"
          ],
          "explanation": {
            "zh-Hant": "交集需要兩張完整列表；結論是集合族的等式。",
            "en": "The intersection needs both complete lists; the conclusion is an equality of families."
          }
        },
        "leaf_basis": "set_operations_def"
      },
      {
        "id": "S05",
        "uses": [
          "G2",
          "S03"
        ],
        "before_tex": "\\tau,\\ \\mathcal{C}",
        "after_tex": "\\tau\\setminus\\mathcal{C}=\\{\\{1\\},\\{1,3,4\\}\\}",
        "relation": "implies",
        "rule": "set_operations_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "在 τ 中但不在 𝒞 中的是 {1} 與 {1,3,4}：它們是開集，但補集 {2,3,4} 與 {2} 不在 τ，所以不是閉集。",
          "en": "The members of τ not in 𝒞 are {1} and {1,3,4}: they are open, but their complements {2,3,4} and {2} are not in τ, so they are not closed."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "set_operations_def"
          ],
          "explanation": {
            "zh-Hant": "每個「不是閉集」都由補集不在 τ 這個具體檢查支持。",
            "en": "Each 'not closed' rests on the concrete check that the complement is not in τ."
          }
        },
        "leaf_basis": "set_operations_def + mko_open_closed_def"
      },
      {
        "id": "S06",
        "uses": [
          "G2",
          "S03"
        ],
        "before_tex": "\\tau,\\ \\mathcal{C}",
        "after_tex": "\\mathcal{C}\\setminus\\tau=\\{\\{2\\},\\{2,3,4\\}\\}",
        "relation": "implies",
        "rule": "set_operations_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "在 𝒞 中但不在 τ 中的是 {2} 與 {2,3,4}：它們是閉集，但不是開集。",
          "en": "The members of 𝒞 not in τ are {2} and {2,3,4}: they are closed but not open."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "G2"
          ],
          "explanation": {
            "zh-Hant": "「不是開集」由與 τ 的六個成員逐一比對得到。",
            "en": "'Not open' comes from comparing with each of the six members of τ."
          }
        },
        "leaf_basis": "set_operations_def + mko_open_closed_def"
      },
      {
        "id": "S07",
        "uses": [
          "S04",
          "S05",
          "S06"
        ],
        "before_tex": "|\\mathcal{P}(X)|=2^4=16,\\ |\\tau\\cup\\mathcal{C}|=6+6-4=8",
        "after_tex": "\\mathcal{P}(X)\\setminus(\\tau\\cup\\mathcal{C})=\\{\\{3\\},\\{4\\},\\{1,3\\},\\{1,4\\},\\{2,3\\},\\{2,4\\},\\{1,2,3\\},\\{1,2,4\\}\\}",
        "relation": "implies",
        "rule": "finite_counting_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "X 有 2^4=16 個子集。τ∪𝒞 有 6+6−4=8 個成員（重疊的是 S04 的 4 個 clopen 集），所以恰有 16−8=8 個子集既不開也不閉，列出如上。核對：4 個 clopen、2 個只開、2 個只閉、8 個兩者皆非，合計 16。子步 S07.1 以 {3} 為例逐項核對。",
          "en": "X has 2^4=16 subsets. τ∪𝒞 has 6+6−4=8 members (the overlap is the 4 clopen sets of S04), so exactly 16−8=8 subsets are neither open nor closed, listed above. Check: 4 clopen, 2 open only, 2 closed only, 8 neither, total 16. Sub-step S07.1 checks {3} item by item."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "S05",
            "S06",
            "finite_counting_basics"
          ],
          "explanation": {
            "zh-Hant": "計數用到聯集的容斥（N2、N3、N4 可重算）。列出的八個集合都不在兩張表中，與計數一致。",
            "en": "The count uses inclusion and exclusion for the union (recomputable as N2, N3, N4). The eight listed sets appear in neither list, consistent with the count."
          }
        },
        "leaf_basis": "finite_counting_basics + numeric_evaluation + set_operations_def"
      },
      {
        "id": "S07.1",
        "uses": [
          "G2",
          "S03"
        ],
        "before_tex": "\\{3\\}",
        "after_tex": "\\{3\\}\\notin\\tau,\\ X\\setminus\\{3\\}=\\{1,2,4\\}\\notin\\tau",
        "relation": "implies",
        "rule": "mko_open_closed_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "{3} 不在 τ 的六個成員中，所以不是開集；它的補集 {1,2,4} 也不在 τ，所以 {3} 不是閉集。{3} 既不開也不閉，這直接否定「不開就是閉」。",
          "en": "{3} is not among the six members of τ, so it is not open; its complement {1,2,4} is not in τ either, so {3} is not closed. {3} is neither open nor closed, which directly refutes 'not open means closed'."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G2",
            "mko_open_closed_def"
          ],
          "explanation": {
            "zh-Hant": "兩個判定都只用定義與 τ 的完整列表；補集在 X 中計算。",
            "en": "Both verdicts use only the definition and the complete list of τ; the complement is computed in X."
          }
        },
        "leaf_basis": "mko_open_closed_def + set_operations_def"
      },
      {
        "id": "S08",
        "uses": [
          "mko_topological_space_def",
          "mko_open_closed_def"
        ],
        "before_tex": "(Y,\\sigma)\\ \\text{any topological space}",
        "after_tex": "\\varnothing\\ \\text{and}\\ Y\\ \\text{are clopen in}\\ (Y,\\sigma)",
        "relation": "implies",
        "rule": "forall_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "取任意拓撲空間 (Y,σ)。依拓撲定義 ∅∈σ、Y∈σ，所以兩者都是開集。又 Y∖∅=Y∈σ，故 ∅ 是閉集；Y∖Y=∅∈σ，故 Y 是閉集。證明沒有用到 Y 或 σ 的任何特殊性質，依全稱推廣對所有拓撲空間成立；本例中即 S04 的 ∅ 與 X。",
          "en": "Take any topological space (Y,σ). By the definition of a topology, ∅∈σ and Y∈σ, so both are open. Also Y∖∅=Y∈σ, so ∅ is closed; Y∖Y=∅∈σ, so Y is closed. The proof uses no special property of Y or σ, so by generalization it holds for every topological space; in our example these are ∅ and X in S04."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_topological_space_def",
            "mko_open_closed_def",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "只用到兩條拓撲公理與閉集定義；不宣稱其他集合也一定 clopen（例如 {1} 不是）。",
            "en": "Only two axioms of a topology and the definition of closed are used; it is not claimed that any other set must be clopen (for example {1} is not)."
          }
        },
        "leaf_basis": "forall_intro + set_operations_def + mko_topological_space_def"
      },
      {
        "id": "S09",
        "uses": [
          "S01",
          "S02"
        ],
        "before_tex": "C_1,C_2\\in\\mathcal{C}",
        "after_tex": "X\\setminus(C_1\\cup C_2)=(X\\setminus C_1)\\cap(X\\setminus C_2)\\in\\tau,\\ \\text{so}\\ C_1\\cup C_2\\in\\mathcal{C}",
        "relation": "theorem_application",
        "rule": "L_demorgan",
        "conditions": [
          "C_1,C_2\\subseteq X"
        ],
        "explanation": {
          "zh-Hant": "設 C1、C2 是閉集，即 X∖C1、X∖C2∈τ。依 De Morgan 引理 L_demorgan（在 S09.1、S09.2 證明），X∖(C1∪C2)=(X∖C1)∩(X∖C2)，這是兩個開集的交集，依拓撲定義仍在 τ。所以 C1∪C2 是閉集。這個論證適用於任何拓撲空間，再對個數歸納可推廣到有限多個閉集。",
          "en": "Let C1 and C2 be closed, so X∖C1 and X∖C2 are in τ. By the De Morgan lemma L_demorgan (proved in S09.1 and S09.2), X∖(C1∪C2)=(X∖C1)∩(X∖C2), an intersection of two open sets, which is in τ by the definition of a topology. So C1∪C2 is closed. The argument works in any topological space, and induction on the number of sets extends it to finitely many closed sets."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_demorgan",
            "mko_topological_space_def",
            "mko_open_closed_def"
          ],
          "explanation": {
            "zh-Hant": "需要有限交集封閉；對無限多個閉集的聯集不宣稱成立（一般拓撲中可能不閉）。",
            "en": "It needs closure under finite intersections; nothing is claimed for unions of infinitely many closed sets (in general topologies they may fail to be closed)."
          }
        },
        "leaf_basis": "set_operations_def + de_morgan_logic + topology_def"
      },
      {
        "id": "S09.1",
        "uses": [
          "set_operations_def",
          "de_morgan_logic"
        ],
        "before_tex": "x\\in X\\setminus(A\\cup B)",
        "after_tex": "x\\in(X\\setminus A)\\cap(X\\setminus B)",
        "relation": "implies",
        "rule": "set_operations_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "(⊆) x∈X∖(A∪B) 表示 x∈X 且不是（x∈A 或 x∈B）；依邏輯 De Morgan，等於 x∈X、x∉A 且 x∉B，所以 x∈X∖A 且 x∈X∖B。",
          "en": "(⊆) x∈X∖(A∪B) means x∈X and not (x∈A or x∈B); by De Morgan in logic this is x∈X, x∉A and x∉B, so x∈X∖A and x∈X∖B."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "set_operations_def",
            "de_morgan_logic"
          ],
          "explanation": {
            "zh-Hant": "此子步只證一個包含方向；另一方向在 S09.2。",
            "en": "This sub-step proves one inclusion only; the other is S09.2."
          }
        },
        "leaf_basis": "set_operations_def + de_morgan_logic"
      },
      {
        "id": "S09.2",
        "uses": [
          "S09.1",
          "set_extensionality"
        ],
        "before_tex": "x\\in(X\\setminus A)\\cap(X\\setminus B)",
        "after_tex": "x\\in X\\setminus(A\\cup B);\\ \\text{hence}\\ X\\setminus(A\\cup B)=(X\\setminus A)\\cap(X\\setminus B)",
        "relation": "implies",
        "rule": "set_extensionality",
        "conditions": [],
        "explanation": {
          "zh-Hant": "(⊇) 若 x∈X、x∉A 且 x∉B，則 x 不在 A∪B，所以 x∈X∖(A∪B)。兩個包含合起來，依集合外延性得等式，證得 L_demorgan。",
          "en": "(⊇) If x∈X, x∉A and x∉B, then x is not in A∪B, so x∈X∖(A∪B). The two inclusions together give the equality by extensionality, proving L_demorgan."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S09.1",
            "set_extensionality",
            "set_operations_def"
          ],
          "explanation": {
            "zh-Hant": "集合等式需要兩個方向的包含，都已證明。",
            "en": "A set equality needs both inclusions, and both are proved."
          }
        },
        "leaf_basis": "set_operations_def + set_extensionality"
      },
      {
        "id": "S10",
        "uses": [
          "S09",
          "S03"
        ],
        "before_tex": "\\{2\\},\\ \\{3,4\\}\\in\\mathcal{C}",
        "after_tex": "\\{2\\}\\cup\\{3,4\\}=\\{2,3,4\\}\\in\\mathcal{C},\\quad X\\setminus\\{2,3,4\\}=\\{1\\}=\\{1,3,4\\}\\cap\\{1,2\\}",
        "relation": "implies",
        "rule": "forall_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "把 S09 用在 C1={2}、C2={3,4}：聯集 {2,3,4} 應是閉集。直接核對：X∖{2,3,4}={1}∈τ；而依 De Morgan，它也等於 (X∖{2})∩(X∖{3,4})={1,3,4}∩{1,2}={1}。兩種算法一致。",
          "en": "Apply S09 to C1={2} and C2={3,4}: the union {2,3,4} should be closed. Direct check: X∖{2,3,4}={1}∈τ; by De Morgan it also equals (X∖{2})∩(X∖{3,4})={1,3,4}∩{1,2}={1}. The two computations agree."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S09",
            "S03",
            "forall_elim"
          ],
          "explanation": {
            "zh-Hant": "這是一般結論的一個實例與交叉核對，不能反過來證明一般結論。",
            "en": "This is one instance of the general result and a cross-check; it cannot prove the general result in reverse."
          }
        },
        "leaf_basis": "forall_elim + set_operations_def"
      },
      {
        "id": "S11",
        "uses": [
          "S04",
          "S07.1"
        ],
        "before_tex": "\\text{claim:}\\ C\\ \\text{closed}\\iff C\\notin\\tau",
        "after_tex": "\\{1,2\\}\\in\\tau\\cap\\mathcal{C},\\ \\{3\\}\\notin\\tau\\cup\\mathcal{C}\\ \\Rightarrow\\ \\text{claim false in both directions}",
        "relation": "implies",
        "rule": "quantifier_negation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "檢驗 MKO 指出的誤解「閉就是不開」。(⇒) 方向的反例：{1,2} 是閉集，卻也是開集。(⇐) 方向的反例：{3} 不是開集，卻也不是閉集（S07.1）。兩個方向都有反例，所以開與閉是兩個獨立的性質，四類都可能出現。",
          "en": "Test the misconception identified in the MKO 'closed means not open'. Counterexample to (⇒): {1,2} is closed and also open. Counterexample to (⇐): {3} is not open, and it is not closed either (S07.1). Both directions have counterexamples, so open and closed are independent properties and all four classes can occur."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "S07.1",
            "quantifier_negation"
          ],
          "explanation": {
            "zh-Hant": "此處要分別否定兩個方向，所以每個方向各給一個反例，兩者都由前面的具體計算提供；若只需否定整個全稱等價，任一方向的一個反例已足夠。",
            "en": "Here each direction is refuted separately by a counterexample from the computations above; a counterexample to either one direction alone already suffices to refute the universal equivalence."
          }
        },
        "leaf_basis": "quantifier_negation + iff_intro"
      }
    ],
    "result": {
      "zh-Hant": "τ 是 X={1,2,3,4} 上的拓撲；閉集族為 {X,{2,3,4},{3,4},{1,2},{2},∅}。十六個子集分為：既開又閉 ∅、{1,2}、{3,4}、X；只開 {1}、{1,3,4}；只閉 {2}、{2,3,4}；其餘八個既不開也不閉。一般地，∅ 與 X 在任何拓撲空間中都既開又閉，兩個閉集的聯集仍是閉集。",
      "en": "τ is a topology on X={1,2,3,4}; the closed family is {X,{2,3,4},{3,4},{1,2},{2},∅}. The sixteen subsets split into: clopen ∅, {1,2}, {3,4}, X; open only {1}, {1,3,4}; closed only {2}, {2,3,4}; the other eight are neither open nor closed. In general, ∅ and X are clopen in every topological space, and the union of two closed sets is closed."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "{3} 不是開集，所以 {3} 是閉集",
          "en": "{3} is not open, so {3} is closed"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "X∖{3}={1,2,4}∉τ，所以 {3} 不是閉集；它既不開也不閉。",
          "en": "X∖{3}={1,2,4}∉τ, so {3} is not closed; it is neither open nor closed."
        }
      },
      {
        "claim": {
          "zh-Hant": "{1,2} 是閉集，所以它不是開集",
          "en": "{1,2} is closed, so it is not open"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "{1,2}∈τ，它同時開且閉。",
          "en": "{1,2}∈τ, so it is both open and closed."
        }
      },
      {
        "claim": {
          "zh-Hant": "在 {1,2,3,4,5} 中 {3,4} 的補集是 {1,2,5}∉τ，所以 {3,4} 不是閉集",
          "en": "In {1,2,3,4,5} the complement of {3,4} is {1,2,5}∉τ, so {3,4} is not closed"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "補集必須相對 X：X∖{3,4}={1,2}∈τ，{3,4} 是閉集。",
          "en": "The complement must be taken in X: X∖{3,4}={1,2}∈τ, so {3,4} is closed."
        }
      },
      {
        "claim": {
          "zh-Hant": "閉集恰好是不在 τ 中的 10 個子集",
          "en": "The closed sets are exactly the 10 subsets not in τ"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "閉集只有 6 個（S03），其中 4 個也在 τ 中；不在 τ 中的 10 個子集裡有 8 個不是閉集。",
          "en": "There are only 6 closed sets (S03), and 4 of them are in τ; of the 10 subsets not in τ, 8 are not closed."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "set_operations_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "集合運算：x∈A∪B ⇔ x∈A 或 x∈B；x∈A∩B ⇔ x∈A 且 x∈B；x∈A∖B ⇔ x∈A 且 x∉B；在全集 U 中 Aᶜ=U∖A。",
          "en": "Set operations: x∈A∪B ⇔ x∈A or x∈B; x∈A∩B ⇔ x∈A and x∈B; x∈A∖B ⇔ x∈A and x∉B; in a universe U, Aᶜ=U∖A."
        }
      },
      {
        "id": "set_extensionality",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "集合相等：A=B 當且僅當 A⊆B 且 B⊆A，即兩者有相同元素。",
          "en": "Set equality: A=B if and only if A⊆B and B⊆A, i.e. they have the same elements."
        }
      },
      {
        "id": "subset_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "子集：A⊆B 表示 A 的每個元素都是 B 的元素，即 ∀x (x∈A⇒x∈B)。",
          "en": "Subset: A⊆B means every element of A is an element of B, i.e. ∀x (x∈A⇒x∈B)."
        }
      },
      {
        "id": "finite_counting_basics",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "有限計數：互斥有限集合 A、B 有 |A∪B|=|A|+|B|；|A×B|=|A|·|B|；有限集之間存在雙射 ⇔ 元素個數相同。",
          "en": "Finite counting: for disjoint finite sets A, B, |A∪B|=|A|+|B|; |A×B|=|A|·|B|; there is a bijection between finite sets ⇔ they have the same number of elements."
        }
      },
      {
        "id": "nat_induction",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "數學歸納原理：對從 n₀ 起的命題族 P(n)，若 P(n₀) 成立，且對每個 n≥n₀ 都有 P(n)⇒P(n+1)，則所有 n≥n₀ 都有 P(n)。",
          "en": "Principle of induction: for a family P(n) starting at n₀, if P(n₀) holds and P(n)⇒P(n+1) for every n≥n₀, then P(n) holds for every n≥n₀."
        }
      },
      {
        "id": "topology_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "拓撲：X 上的開集族 τ 包含 ∅ 與 X，對任意聯集與有限交集封閉。",
          "en": "Topology: a family τ of open subsets of X containing ∅ and X, closed under arbitrary unions and finite intersections."
        }
      },
      {
        "id": "def_unfold",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "展開定義：把一個已定義的詞換成它的定義內容（或反向收回），命題的真假不變。",
          "en": "Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true."
        }
      },
      {
        "id": "de_morgan_logic",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "邏輯的德摩根律：¬(P∧Q) ⇔ ¬P∨¬Q；¬(P∨Q) ⇔ ¬P∧¬Q。",
          "en": "De Morgan's laws for logic: ¬(P∧Q) ⇔ ¬P∨¬Q and ¬(P∨Q) ⇔ ¬P∧¬Q."
        }
      },
      {
        "id": "injective_surjective_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "單射：f(a)=f(a′) ⇒ a=a′。滿射：每個 b∈B 都有 a∈A 使 f(a)=b。雙射：既單又滿，此時有唯一反函數 f⁻¹ 使 f⁻¹∘f=id_A、f∘f⁻¹=id_B。",
          "en": "Injective: f(a)=f(a′) ⇒ a=a′. Surjective: every b∈B has some a∈A with f(a)=b. Bijective: both; then there is a unique inverse f⁻¹ with f⁻¹∘f=id_A and f∘f⁻¹=id_B."
        }
      },
      {
        "id": "numeric_evaluation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "明確整數或有理數算式的精確求值；可以獨立重算。",
          "en": "Exact evaluation of an explicit integer or rational expression; independently recomputable."
        }
      },
      {
        "id": "forall_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∀引入（全稱推廣）：若某命題對一個「任意的」元素 x 證成，而證明沒有對 x 加任何額外假設，則它對論域中所有 x 成立。",
          "en": "∀-introduction (generalization): if a statement is proved for an arbitrary element x, with no extra assumption on x, it holds for every x in the domain."
        }
      },
      {
        "id": "forall_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∀消去（代入特例）：若 ∀x∈D, P(x) 成立且 a∈D，則 P(a) 成立。",
          "en": "∀-elimination (instantiation): if ∀x∈D, P(x) holds and a∈D, then P(a) holds."
        }
      },
      {
        "id": "quantifier_negation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "量詞否定：¬(∀x, P(x)) 等價於 ∃x, ¬P(x)；¬(∃x, P(x)) 等價於 ∀x, ¬P(x)。",
          "en": "Negating quantifiers: ¬(∀x, P(x)) is equivalent to ∃x, ¬P(x), and ¬(∃x, P(x)) is equivalent to ∀x, ¬P(x)."
        }
      },
      {
        "id": "iff_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "雙條件：P⇒Q 與 Q⇒P 都成立時，P⇔Q 成立；反之亦然。",
          "en": "Biconditional: P⇔Q holds exactly when both P⇒Q and Q⇒P hold."
        }
      },
      {
        "id": "mko_open_closed_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-open-closed-set",
          "sha256": "78fdc40df7a4161259566db404d47fa805aa334d6f661420675bbc8c139c6508",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "給定拓撲空間 (X,τ)，U⊆X 稱為開集當且僅當 U∈τ；C⊆X 稱為閉集當且僅當 X∖C∈τ。開與閉不是互斥分類，∅ 與 X 永遠同時開且閉。",
          "en": "In a topological space (X,τ), a subset U⊆X is open exactly when U∈τ, and C⊆X is closed exactly when X∖C∈τ. Open and closed are not mutually exclusive; ∅ and X are always both open and closed."
        }
      },
      {
        "id": "mko_topological_space_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-topological-space",
          "sha256": "95f78b4f323b3139dd05068842f7209ac33b39b3d86feb5b73e0eb8849cdb43d",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "X 上的拓撲是子集合族 τ⊆P(X)，滿足 ∅∈τ、X∈τ、τ 中任意子族的聯集仍在 τ，且 τ 中任意有限子族的交集仍在 τ；空族的聯集是 ∅，空族的交集按 X 解讀。",
          "en": "A topology on X is a family τ⊆P(X) with ∅∈τ, X∈τ, the union of any subfamily of τ in τ, and the intersection of any finite subfamily of τ in τ; the union of the empty family is ∅ and its intersection is read as X."
        }
      },
      {
        "id": "L_finite_family",
        "kind": "lemma",
        "proved_in": [
          "S01.1",
          "S01.2"
        ],
        "statement": {
          "zh-Hant": "設 τ 是 X 的有限子集合族。若 ∅∈τ 且 τ 對兩個成員的聯集封閉，則 τ 的每個子族的聯集都在 τ；若 X∈τ 且 τ 對兩個成員的交集封閉，則 τ 的每個子族的交集（空族的交集為 X）都在 τ。",
          "en": "Let τ be a finite family of subsets of X. If ∅∈τ and τ is closed under the union of two members, then the union of every subfamily of τ is in τ; if X∈τ and τ is closed under the intersection of two members, then the intersection of every subfamily of τ (the empty one read as X) is in τ."
        }
      },
      {
        "id": "L_double_complement",
        "kind": "lemma",
        "proved_in": [
          "S02.1"
        ],
        "statement": {
          "zh-Hant": "對每個 C⊆X，X∖(X∖C)=C。",
          "en": "For every C⊆X, X∖(X∖C)=C."
        }
      },
      {
        "id": "L_demorgan",
        "kind": "lemma",
        "proved_in": [
          "S09.1",
          "S09.2"
        ],
        "statement": {
          "zh-Hant": "對 A,B⊆X，X∖(A∪B)=(X∖A)∩(X∖B)。",
          "en": "For A,B⊆X, X∖(A∪B)=(X∖A)∩(X∖B)."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S01.3",
        "lhs": "comb(4,2)",
        "rhs": "6"
      },
      {
        "id": "N2",
        "step": "S07",
        "lhs": "2**4",
        "rhs": "16"
      },
      {
        "id": "N3",
        "step": "S07",
        "lhs": "16-(6+6-4)",
        "rhs": "8"
      },
      {
        "id": "N4",
        "step": "S07",
        "lhs": "4+2+2+8",
        "rhs": "16"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "在 MKO 範圍內工作：底集合 X 與拓撲 τ 固定（假設 a1），所有補集都相對同一個 X（假設 a2）。一般性結論只證明「∅ 與 X 永遠既開又閉」與「兩個閉集的聯集是閉集」；任意多個閉集之交的對偶敘述只說明不另證。不改動 canonical MKO 或英文 edition。",
      "en": "Works inside the MKO's scope: the underlying set X and topology τ are fixed (assumption a1), and every complement is taken relative to the same X (assumption a2). The general results proved are only '∅ and X are always clopen' and 'the union of two closed sets is closed'; the dual statement for intersections of arbitrarily many closed sets is mentioned but not proved separately. The canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "X",
        "底集合 {1,2,3,4}",
        "the underlying set {1,2,3,4}"
      ],
      [
        "\\tau",
        "X 上的拓撲（開集族）",
        "the topology on X (the family of open sets)"
      ],
      [
        "\\mathcal{C}",
        "閉集族 {C⊆X ∣ X∖C∈τ}",
        "the family of closed sets {C⊆X ∣ X∖C∈τ}"
      ],
      [
        "X\\setminus C",
        "C 相對於 X 的補集",
        "the complement of C relative to X"
      ],
      [
        "\\mathcal{P}(X)",
        "X 的全部子集",
        "all subsets of X"
      ]
    ],
    "md_error_controls": [
      {
        "title": "閉不是「不開」 / Closed does not mean 'not open'",
        "zh": "MKO 指出的誤解是把 closed 讀成 not open。本例中 {1,2} 同時開且閉，{3} 兩者皆非（S11）。判定閉集只能用定義：補集是否在 τ 中。",
        "en": "The misconception identified in the MKO is reading 'closed' as 'not open'. Here {1,2} is both open and closed, and {3} is neither (S11). Closedness can only be decided by the definition: is the complement in τ?"
      },
      {
        "title": "補集要相對同一個 X / Complements must be taken in the same X",
        "zh": "若把 {1,2} 的補集錯算在 {1,2,3} 中，會得到 {3}∉τ 而誤判 {1,2} 不閉；正確的 X∖{1,2}={3,4}∈τ。若在 {1,2,3,4,5} 中取 {3,4} 的補集，得到 {1,2,5}，它甚至不是 X 的子集，與 τ 的比對失去意義（MKO 假設 a2）。",
        "en": "If the complement of {1,2} is wrongly taken in {1,2,3}, one gets {3}∉τ and wrongly concludes {1,2} is not closed; the correct X∖{1,2}={3,4}∈τ. If the complement of {3,4} is taken in {1,2,3,4,5}, one gets {1,2,5}, not even a subset of X, so comparing with τ is meaningless (MKO assumption a2)."
      },
      {
        "title": "兩兩封閉只在有限族時足夠 / Pairwise closure suffices only for finite families",
        "zh": "S01 能只檢查兩兩聯集，是因為 τ 有限（L_finite_family）。反例：ℕ 的全部有限子集再加上 ℕ 本身，這個族含 ∅ 與 ℕ，對兩兩聯集與交集封閉，但所有 {2k} 的聯集是偶數集，它無限且不等於 ℕ，不在族中，所以不是拓撲。",
        "en": "S01 could check only pairwise unions because τ is finite (L_finite_family). Counterexample: the family of all finite subsets of ℕ together with ℕ itself contains ∅ and ℕ and is closed under pairwise unions and intersections, but the union of all {2k} is the set of even numbers, which is infinite and not equal to ℕ, so it is not in the family and the family is not a topology."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "MKO 第二段的對偶敘述「任意多個閉集之交仍閉」可由任意族的 De Morgan 定律推得；本稿只證明兩個閉集之聯集的情形，任意族情形未另寫子步。",
        "en": "The dual statement in MKO paragraph 2, that an intersection of arbitrarily many closed sets is closed, follows from De Morgan's law for arbitrary families; this guide proves only the case of the union of two closed sets and does not write sub-steps for arbitrary families."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "db4ff6acc8064d0f256eceeebb39a241b3986504cee1fd23ae8da1de54588f78",
    "original_source_sha256": "db4ff6acc8064d0f256eceeebb39a241b3986504cee1fd23ae8da1de54588f78",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}