{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-topological-space-01",
  "mko_id": "mko-topological-space",
  "title": {
    "zh-Hant": "拓撲空間：完整驗證三點集合上的一個鏈狀開集族，並辨認非拓撲",
    "en": "Topological space: fully verifying a chain of open sets on a three-point set, and recognising non-topologies"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-topological-space-01",
    "source": {
      "mko_id": "mko-topological-space",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "95f78b4f323b3139dd05068842f7209ac33b39b3d86feb5b73e0eb8849cdb43d",
      "english_sha256": "d4be991db50b46add5676c87b921ccea446b23b0f93e496d3dac77e5f07c3d7c"
    },
    "mode": "definition_walkthrough",
    "audience_prerequisites": [
      "subset_def",
      "set_operations_def",
      "set_extensionality",
      "nat_induction",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "設 X={a,b,c}（a、b、c 相異）。(1) 證明 τ₁={∅,{a},{a,b},X} 是 X 上的拓撲，涵蓋全部 16 個子族（含空族）；(2) 證明 τ₂={∅,{a},{b},X} 不是拓撲，並指出它通過了哪些檢查；(3) 比較 τ₁ 與不可分拓撲、離散拓撲的粗細；(4) 在 ℕ 上，證明 τ_fin={ℕ 的有限子集}∪{ℕ} 對兩兩聯集與有限交集封閉，卻不是拓撲。",
      "en": "Let X={a,b,c} with a, b, c distinct. (1) Prove that τ₁={∅,{a},{a,b},X} is a topology on X, covering all 16 subfamilies (including the empty one); (2) prove that τ₂={∅,{a},{b},X} is not a topology, and say which checks it does pass; (3) compare τ₁ with the indiscrete and discrete topologies; (4) on ℕ, prove that τ_fin={finite subsets of ℕ}∪{ℕ} is closed under pairwise unions and finite intersections but is not a topology."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "X={a,b,c}，a、b、c 兩兩相異；τ₁={∅,{a},{a,b},X}；τ₂={∅,{a},{b},X}。",
          "en": "X={a,b,c} with a, b, c pairwise distinct; τ₁={∅,{a},{a,b},X}; τ₂={∅,{a},{b},X}."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "MKO 假設 a2：任意聯集包含空族情形，有限交包含空族情形；空族的交集按 X 解讀。",
          "en": "MKO assumption a2: arbitrary unions include the empty family and finite intersections include the empty family; the empty intersection is read as X."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "ℕ 包含 0；τ_fin={A⊆ℕ : A 有限}∪{ℕ}。",
          "en": "ℕ includes 0; τ_fin={A⊆ℕ : A finite}∪{ℕ}."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2",
          "mko_topology_def"
        ],
        "before_tex": "\\tau\\ \\text{is a topology on}\\ X",
        "after_tex": "\\tau\\subseteq\\mathcal{P}(X),\\ \\varnothing,X\\in\\tau,\\ \\forall\\mathcal{S}\\subseteq\\tau:\\ \\bigcup\\mathcal{S}\\in\\tau,\\ \\forall\\ \\text{finite}\\ \\mathcal{S}\\subseteq\\tau:\\ \\bigcap\\mathcal{S}\\in\\tau",
        "relation": "definition",
        "rule": "mko_topology_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO 的定義寫出要驗證的條件：每個成員是 X 的子集合；∅ 與 X 在 τ 中；每個子族（包括空族與無窮子族）的聯集在 τ 中；每個有限子族（包括空族）的交集在 τ 中，空族交集按 X 解讀（G2）。",
          "en": "Write out the conditions to verify from the MKO's definition: every member is a subset of X; ∅ and X are in τ; the union of every subfamily (including the empty one and infinite ones) is in τ; the intersection of every finite subfamily (including the empty one) is in τ, with the empty intersection read as X (G2)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "G2",
            "mko_topology_def"
          ],
          "explanation": {
            "zh-Hant": "此步只展開定義；任何一個條件失敗就不是拓撲。",
            "en": "This step only unfolds the definition; failure of any one condition means 'not a topology'."
          }
        },
        "leaf_basis": "mko_topology_def + def_unfold"
      },
      {
        "id": "S02",
        "uses": [
          "G1",
          "subset_def"
        ],
        "before_tex": "\\tau_1=\\{\\varnothing,\\{a\\},\\{a,b\\},X\\}",
        "after_tex": "\\tau_1\\subseteq\\mathcal{P}(X)",
        "relation": "implies",
        "rule": "subset_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "四個成員的元素都在 a、b、c 之中，所以每個成員都是 X 的子集合（MKO 假設 a1）。",
          "en": "The elements of all four members lie among a, b, c, so each member is a subset of X (MKO assumption a1)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "subset_def"
          ],
          "explanation": {
            "zh-Hant": "逐一檢查四個成員；不宣稱其他條件。",
            "en": "The four members are checked one by one; no other condition is claimed."
          }
        },
        "leaf_basis": "subset_def"
      },
      {
        "id": "S03",
        "uses": [
          "G2",
          "family_ops",
          "S02"
        ],
        "before_tex": "\\mathcal{S}=\\varnothing\\subseteq\\tau_1",
        "after_tex": "\\bigcup\\varnothing=\\varnothing\\in\\tau_1,\\quad \\bigcap\\varnothing=X\\in\\tau_1",
        "relation": "implies",
        "rule": "family_ops",
        "conditions": [],
        "explanation": {
          "zh-Hant": "空族沒有成員，沒有任何 x 屬於某個成員，所以 ⋃∅=∅；而「對每個 A∈∅ 都有 x∈A」對每個 x∈X 空真成立，所以 ⋂∅=X。∅ 與 X 都列在 τ₁ 中。可見公理「∅∈τ、X∈τ」正是聯集條件與交集條件的空族情形。",
          "en": "The empty family has no members, so no x lies in a member and ⋃∅=∅; and 'x∈A for every A∈∅' is vacuously true for every x∈X, so ⋂∅=X. Both ∅ and X are listed in τ₁. So the axioms '∅∈τ and X∈τ' are exactly the empty-family cases of the union and intersection conditions."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G2",
            "family_ops"
          ],
          "explanation": {
            "zh-Hant": "交集取 X 是相對於固定底集合的約定（MKO 假設 a2）；沒有這個約定，⋂∅ 沒有意義。",
            "en": "Taking X for the intersection is a convention relative to the fixed underlying set (MKO assumption a2); without it ⋂∅ would be meaningless."
          }
        },
        "leaf_basis": "family_ops + set_extensionality"
      },
      {
        "id": "S04",
        "uses": [
          "G1",
          "subset_def"
        ],
        "before_tex": "\\varnothing,\\ \\{a\\},\\ \\{a,b\\},\\ X",
        "after_tex": "\\varnothing\\subseteq\\{a\\}\\subseteq\\{a,b\\}\\subseteq X",
        "relation": "implies",
        "rule": "subset_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐一檢查包含：∅ 沒有元素；a∈{a,b}；a、b∈X。由 ⊆ 的遞移性，τ₁ 中任兩成員都可比較：τ₁ 是一條鏈。",
          "en": "Check each inclusion: ∅ has no elements; a∈{a,b}; a, b∈X. By transitivity of ⊆, any two members of τ₁ are comparable: τ₁ is a chain."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "subset_def"
          ],
          "explanation": {
            "zh-Hant": "只宣稱這四個集合成鏈；一般的集合族不必成鏈（τ₂ 中 {a} 與 {b} 不可比較）。",
            "en": "Only these four sets are claimed to form a chain; a general family need not (in τ₂, {a} and {b} are incomparable)."
          }
        },
        "leaf_basis": "subset_def + logic_chain"
      },
      {
        "id": "S05",
        "uses": [
          "S04",
          "L_chain_extremes"
        ],
        "before_tex": "\\varnothing\\neq\\mathcal{S}\\subseteq\\tau_1",
        "after_tex": "\\mathcal{S}\\ \\text{has a largest member}\\ M\\ \\text{and a smallest member}\\ m",
        "relation": "implies",
        "rule": "L_chain_extremes",
        "conditions": [
          "\\mathcal{S}\\neq\\varnothing"
        ],
        "explanation": {
          "zh-Hant": "τ₁ 只有 4 個成員，所以每個子族都有限；子族的成員仍兩兩可比較（S04）。依引理 L_chain_extremes（S05.1 到 S05.3 證明），每個非空子族都有最大與最小成員。",
          "en": "τ₁ has only 4 members, so every subfamily is finite; members of a subfamily are still pairwise comparable (S04). By Lemma L_chain_extremes (proved in S05.1 to S05.3), every nonempty subfamily has a largest and a smallest member."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "L_chain_extremes"
          ],
          "explanation": {
            "zh-Hant": "引理需要有限、非空、可比較三個前提，都已確認。",
            "en": "The lemma needs finiteness, nonemptiness and comparability, all confirmed."
          }
        },
        "leaf_basis": "nat_induction + subset_def + case_analysis"
      },
      {
        "id": "S05.1",
        "uses": [
          "subset_def"
        ],
        "before_tex": "|\\mathcal{S}|=1,\\ \\mathcal{S}=\\{A\\}",
        "after_tex": "M=m=A",
        "relation": "induction_base",
        "rule": "subset_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "只有一個成員 A 時，A⊆A，所以 A 同時是最大與最小成員。",
          "en": "With a single member A, A⊆A, so A is both the largest and the smallest member."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "subset_def"
          ],
          "explanation": {
            "zh-Hant": "只用 ⊆ 的自反性。",
            "en": "Only reflexivity of ⊆ is used."
          }
        },
        "leaf_basis": "subset_def"
      },
      {
        "id": "S05.2",
        "uses": [
          "subset_def",
          "case_analysis"
        ],
        "before_tex": "|\\mathcal{S}|=n+1,\\ \\mathcal{S}=\\mathcal{S}'\\cup\\{A\\},\\ \\mathcal{S}'\\ \\text{has}\\ M',m'",
        "after_tex": "\\mathcal{S}\\ \\text{has a largest and a smallest member}",
        "relation": "induction_step",
        "rule": "case_analysis",
        "conditions": [
          "n\\ge 1"
        ],
        "explanation": {
          "zh-Hant": "取 A∈𝒮，𝒮′=𝒮∖{A} 有 n 個成員且仍成鏈，依歸納假設有最大 M′ 與最小 m′。A 與 M′ 可比較：若 A⊆M′，M′ 是 𝒮 的最大成員；若 M′⊆A，則對 B∈𝒮′ 有 B⊆M′⊆A，A 是最大成員。最小成員同理比較 A 與 m′。",
          "en": "Take A∈𝒮; 𝒮′=𝒮∖{A} has n members and is still a chain, so by the induction hypothesis it has a largest M′ and a smallest m′. A and M′ are comparable: if A⊆M′, then M′ is the largest member of 𝒮; if M′⊆A, then every B∈𝒮′ has B⊆M′⊆A, so A is the largest. The smallest member is found the same way by comparing A with m′."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "subset_def",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "兩種情況由可比較性窮盡；只用到 ⊆ 的遞移性與歸納假設。",
            "en": "Comparability makes the two cases exhaustive; only transitivity of ⊆ and the induction hypothesis are used."
          }
        },
        "leaf_basis": "case_analysis + subset_def + logic_chain"
      },
      {
        "id": "S05.3",
        "uses": [
          "S05.1",
          "S05.2",
          "nat_induction"
        ],
        "before_tex": "P(1)\\ \\land\\ \\forall n\\ge 1\\,(P(n)\\Rightarrow P(n+1))",
        "after_tex": "\\forall n\\ge 1:\\ P(n)",
        "relation": "implies",
        "rule": "nat_induction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "令 P(n) 為「每個 n 個成員的鏈都有最大與最小成員」。由 S05.1、S05.2 與數學歸納原理，P(n) 對每個 n≥1 成立。S05.1 到 S05.3 證得引理 L_chain_extremes。",
          "en": "Let P(n) be 'every chain with n members has a largest and a smallest member'. By S05.1, S05.2 and the induction principle, P(n) holds for every n≥1. S05.1 to S05.3 prove Lemma L_chain_extremes."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05.1",
            "S05.2",
            "nat_induction"
          ],
          "explanation": {
            "zh-Hant": "基礎與歸納步都已證，沒有使用引理本身。",
            "en": "Base and step are both proved, without using the lemma itself."
          }
        },
        "leaf_basis": "nat_induction"
      },
      {
        "id": "S06",
        "uses": [
          "S05",
          "L_chain_union"
        ],
        "before_tex": "M,\\ m\\ \\text{largest and smallest in}\\ \\mathcal{S}",
        "after_tex": "\\bigcup\\mathcal{S}=M,\\quad \\bigcap\\mathcal{S}=m",
        "relation": "implies",
        "rule": "L_chain_union",
        "conditions": [
          "\\mathcal{S}\\neq\\varnothing"
        ],
        "explanation": {
          "zh-Hant": "依引理 L_chain_union（S06.1、S06.2 證明），非空子族的聯集是它的最大成員，交集是它的最小成員。",
          "en": "By Lemma L_chain_union (proved in S06.1 and S06.2), the union of a nonempty subfamily is its largest member and the intersection is its smallest member."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05",
            "L_chain_union"
          ],
          "explanation": {
            "zh-Hant": "需要 M、m 屬於 𝒮；空族沒有最大成員，所以空族已在 S03 單獨處理。",
            "en": "M and m must belong to 𝒮; the empty family has no largest member, which is why it was handled separately in S03."
          }
        },
        "leaf_basis": "set_extensionality + family_ops + subset_def"
      },
      {
        "id": "S06.1",
        "uses": [
          "family_ops",
          "set_extensionality"
        ],
        "before_tex": "M\\in\\mathcal{S},\\ \\forall A\\in\\mathcal{S}:\\ A\\subseteq M",
        "after_tex": "\\bigcup\\mathcal{S}=M",
        "relation": "implies",
        "rule": "set_extensionality",
        "conditions": [],
        "explanation": {
          "zh-Hant": "M∈𝒮，所以 M⊆⋃𝒮；若 x∈⋃𝒮，則 x∈A 對某個 A∈𝒮，而 A⊆M，故 x∈M，即 ⋃𝒮⊆M。兩向包含得相等。",
          "en": "M∈𝒮, so M⊆⋃𝒮; if x∈⋃𝒮, then x∈A for some A∈𝒮, and A⊆M gives x∈M, so ⋃𝒮⊆M. Inclusion both ways gives equality."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "family_ops",
            "set_extensionality"
          ],
          "explanation": {
            "zh-Hant": "兩個包含各自用到 M∈𝒮 與 M 最大；沒有使用引理本身。",
            "en": "The two inclusions use M∈𝒮 and the fact that M is the largest member, respectively; the lemma itself is not used."
          }
        },
        "leaf_basis": "set_extensionality + family_ops"
      },
      {
        "id": "S06.2",
        "uses": [
          "family_ops",
          "set_extensionality"
        ],
        "before_tex": "m\\in\\mathcal{S},\\ \\forall A\\in\\mathcal{S}:\\ m\\subseteq A,\\ m\\subseteq X",
        "after_tex": "\\bigcap\\mathcal{S}=m",
        "relation": "implies",
        "rule": "set_extensionality",
        "conditions": [],
        "explanation": {
          "zh-Hant": "m∈𝒮，所以 ⋂𝒮⊆m；若 x∈m，則 x∈X 且 x 屬於每個 A∈𝒮（因 m⊆A），故 x∈⋂𝒮。兩向包含得相等。S06.1、S06.2 證得引理 L_chain_union。",
          "en": "m∈𝒮, so ⋂𝒮⊆m; if x∈m, then x∈X and x lies in every A∈𝒮 (since m⊆A), so x∈⋂𝒮. Inclusion both ways gives equality. S06.1 and S06.2 prove Lemma L_chain_union."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "family_ops",
            "set_extensionality"
          ],
          "explanation": {
            "zh-Hant": "用到 m 最小與 m⊆X；沒有使用引理本身。",
            "en": "It uses the fact that m is the smallest member and m⊆X; the lemma itself is not used."
          }
        },
        "leaf_basis": "set_extensionality + family_ops"
      },
      {
        "id": "S07",
        "uses": [
          "S03",
          "S05",
          "S06"
        ],
        "before_tex": "\\mathcal{S}\\subseteq\\tau_1\\ \\text{arbitrary}",
        "after_tex": "\\bigcup\\mathcal{S}\\in\\tau_1",
        "relation": "case_split",
        "rule": "case_analysis",
        "conditions": [],
        "explanation": {
          "zh-Hant": "分兩種情況：𝒮 為空時 ⋃𝒮=∅∈τ₁（S03）；𝒮 非空時 ⋃𝒮=M（S06），而 M∈𝒮⊆τ₁。兩種情況涵蓋 τ₁ 的每個子族，所以 τ₁ 對任意聯集封閉。",
          "en": "Two cases: if 𝒮 is empty, ⋃𝒮=∅∈τ₁ (S03); if 𝒮 is nonempty, ⋃𝒮=M (S06) and M∈𝒮⊆τ₁. The two cases cover every subfamily of τ₁, so τ₁ is closed under arbitrary unions."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "S05",
            "S06",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "「空」與「非空」窮盡所有子族；τ₁ 有限，所以「任意子族」都是有限子族，沒有遺漏無窮子族。",
            "en": "'Empty' and 'nonempty' exhaust all subfamilies; since τ₁ is finite, every subfamily is finite, so no infinite subfamily is missed."
          }
        },
        "leaf_basis": "case_analysis + family_ops"
      },
      {
        "id": "S08",
        "uses": [
          "S03",
          "S05",
          "S06"
        ],
        "before_tex": "\\mathcal{S}\\subseteq\\tau_1\\ \\text{finite}",
        "after_tex": "\\bigcap\\mathcal{S}\\in\\tau_1",
        "relation": "case_split",
        "rule": "case_analysis",
        "conditions": [],
        "explanation": {
          "zh-Hant": "𝒮 為空時 ⋂𝒮=X∈τ₁（S03）；𝒮 非空時 ⋂𝒮=m（S06），而 m∈τ₁。所以 τ₁ 對有限交集封閉。",
          "en": "If 𝒮 is empty, ⋂𝒮=X∈τ₁ (S03); if 𝒮 is nonempty, ⋂𝒮=m (S06) with m∈τ₁. So τ₁ is closed under finite intersections."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "S05",
            "S06",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "兩種情況窮盡所有有限子族。",
            "en": "The two cases exhaust all finite subfamilies."
          }
        },
        "leaf_basis": "case_analysis + family_ops"
      },
      {
        "id": "S09",
        "uses": [
          "S02",
          "S03",
          "S07",
          "S08",
          "mko_topology_def"
        ],
        "before_tex": "\\tau_1\\subseteq\\mathcal{P}(X),\\ \\text{unions and finite intersections closed}",
        "after_tex": "(X,\\tau_1)\\ \\text{is a topological space}",
        "relation": "implies",
        "rule": "mko_topology_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "S02、S03、S07、S08 給出定義的全部條件。計數核對：τ₁ 有 4 個成員，共 2⁴=16 個子族；S03 處理空族 1 個，S07、S08 透過 S05、S06 處理其餘 15 個非空子族。這是完整枚舉語義下的證明，不是抽查。",
          "en": "S02, S03, S07 and S08 supply every condition of the definition. Count check: τ₁ has 4 members, hence 2⁴=16 subfamilies; S03 handles the 1 empty subfamily, and S07 and S08 handle the other 15 nonempty ones through S05 and S06. This is a proof covering the complete enumeration, not a sample."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S02",
            "S03",
            "S07",
            "S08",
            "mko_topology_def",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G：τ₁ 是拓撲。足夠條件 H：S02、S03、S07、S08 同時成立；H⇒G 由定義。H 由 G1、G2 正向證成，沒有把 G 當作前提。",
            "en": "Goal G: τ₁ is a topology. Sufficient condition H: S02, S03, S07 and S08 together; H⇒G by the definition. H is proved forwards from G1 and G2; G is never used as a premise."
          }
        },
        "leaf_basis": "mko_topology_def + and_intro_elim"
      },
      {
        "id": "S10",
        "uses": [
          "G1",
          "set_operations_def"
        ],
        "before_tex": "\\{a\\},\\ \\{b\\}\\in\\tau_2",
        "after_tex": "\\{a\\}\\cup\\{b\\}=\\{a,b\\}\\notin\\tau_2",
        "relation": "implies",
        "rule": "set_operations_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "{a}∪{b}={a,b}。逐一比較 τ₂ 的四個成員：{a,b}≠∅（含 a）；≠{a}（含 b 且 b≠a）；≠{b}（含 a 且 a≠b）；≠X（c∉{a,b}，因 c≠a、c≠b）。所以子族 {{a},{b}} 的聯集不在 τ₂ 中，τ₂ 不是拓撲。",
          "en": "{a}∪{b}={a,b}. Compare with the four members of τ₂: {a,b}≠∅ (it contains a); ≠{a} (it contains b, and b≠a); ≠{b} (it contains a, and a≠b); ≠X (c∉{a,b} since c≠a and c≠b). So the union of the subfamily {{a},{b}} is not in τ₂, and τ₂ is not a topology."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "set_operations_def",
            "set_extensionality"
          ],
          "explanation": {
            "zh-Hant": "一個反例子族就足以否定；這用到 a、b、c 兩兩相異（G1）。",
            "en": "One counterexample subfamily suffices to refute; it uses that a, b, c are pairwise distinct (G1)."
          }
        },
        "leaf_basis": "set_operations_def + set_extensionality + contradiction"
      },
      {
        "id": "S11",
        "uses": [
          "G1",
          "set_operations_def"
        ],
        "before_tex": "\\tau_2=\\{\\varnothing,\\{a\\},\\{b\\},X\\}",
        "after_tex": "\\varnothing,X\\in\\tau_2,\\ \\text{and}\\ A\\cap B\\in\\tau_2\\ \\text{for all}\\ A,B\\in\\tau_2",
        "relation": "implies",
        "rule": "set_operations_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "τ₂ 含 ∅ 與 X；兩兩交集：{a}∩{b}=∅，任何 A∩X=A，A∩∅=∅，A∩A=A，都在 τ₂ 中（不同成員共 C(4,2)=6 對）。所以只檢查交集或只檢查 ∅、X 會誤判 τ₂ 為拓撲；每一類條件都要檢查。",
          "en": "τ₂ contains ∅ and X; pairwise intersections: {a}∩{b}=∅, and A∩X=A, A∩∅=∅, A∩A=A for every A, all lie in τ₂ (C(4,2)=6 pairs of distinct members). So checking only intersections, or only ∅ and X, would wrongly accept τ₂; every kind of condition must be checked."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "set_operations_def"
          ],
          "explanation": {
            "zh-Hant": "這一步不改變 S10 的結論，只說明 τ₂ 通過了哪些檢查。",
            "en": "This step does not change the conclusion of S10; it only shows which checks τ₂ passes."
          }
        },
        "leaf_basis": "set_operations_def + numeric_evaluation"
      },
      {
        "id": "S12",
        "uses": [
          "S09",
          "subset_def"
        ],
        "before_tex": "\\{\\varnothing,X\\},\\ \\tau_1,\\ \\mathcal{P}(X)",
        "after_tex": "\\{\\varnothing,X\\}\\subsetneq\\tau_1\\subsetneq\\mathcal{P}(X)",
        "relation": "implies",
        "rule": "subset_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "不可分拓撲 {∅,X} 的兩個成員都在 τ₁ 中，而 {a}∈τ₁ 不在其中；τ₁ 的成員都是 X 的子集合，而 {b}∈P(X) 不在 τ₁ 中。P(X) 有 2³=8 個成員，是拓撲：X 的子集合的聯集與交集仍是 X 的子集合。{∅,X} 也是拓撲：其子族的聯集只能是 ∅ 或 X，交集也是。所以同一個 X 上至少有三個不同的拓撲，τ₁ 比不可分拓撲細、比離散拓撲粗。",
          "en": "Both members of the indiscrete topology {∅,X} are in τ₁, while {a}∈τ₁ is not in it; every member of τ₁ is a subset of X, while {b}∈P(X) is not in τ₁. P(X) has 2³=8 members and is a topology: unions and intersections of subsets of X are subsets of X. {∅,X} is a topology too: the union of any of its subfamilies is ∅ or X, and likewise the intersection. So the same X carries at least three different topologies, and τ₁ is finer than the indiscrete and coarser than the discrete topology."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S09",
            "subset_def",
            "set_operations_def"
          ],
          "explanation": {
            "zh-Hant": "粗細比較的是開集族的包含（MKO 段落 2），不是底集合的大小；兩個嚴格包含各由一個見證成員（{a} 與 {b}）確認。",
            "en": "Finer and coarser compare families of open sets by inclusion (MKO paragraph 2), not the size of the underlying set; each strict inclusion is confirmed by a witness member ({a} and {b})."
          }
        },
        "leaf_basis": "subset_def + set_operations_def + family_ops"
      },
      {
        "id": "S13",
        "uses": [
          "G3",
          "finite_sets"
        ],
        "before_tex": "A,B\\in\\tau_{\\mathrm{fin}}",
        "after_tex": "A\\cup B\\in\\tau_{\\mathrm{fin}},\\ A\\cap B\\in\\tau_{\\mathrm{fin}};\\ \\varnothing,\\mathbb{N}\\in\\tau_{\\mathrm{fin}}",
        "relation": "implies",
        "rule": "finite_sets",
        "conditions": [],
        "explanation": {
          "zh-Hant": "∅ 有限，ℕ 直接列入，所以兩者都在 τ_fin。兩兩聯集：A、B 都有限時 A∪B 有限；其一為 ℕ 時聯集為 ℕ。兩兩交集：其一有限時交集是有限集合的子集合，仍有限；兩者都是 ℕ 時交集為 ℕ。所以 τ_fin 對兩兩聯集與兩兩交集封閉。",
          "en": "∅ is finite and ℕ is listed directly, so both are in τ_fin. Pairwise unions: if A and B are finite, A∪B is finite; if one is ℕ, the union is ℕ. Pairwise intersections: if one is finite, the intersection is a subset of a finite set and so finite; if both are ℕ, the intersection is ℕ. So τ_fin is closed under pairwise unions and pairwise intersections."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "finite_sets",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "情況窮盡；只宣稱兩兩（因而有限）封閉，不宣稱任意聯集封閉。",
            "en": "The cases are exhaustive; only pairwise (hence finite) closure is claimed, not closure under arbitrary unions."
          }
        },
        "leaf_basis": "finite_sets + case_analysis + set_operations_def"
      },
      {
        "id": "S14",
        "uses": [
          "G3",
          "family_ops"
        ],
        "before_tex": "\\mathcal{S}=\\{\\{2k\\}:k\\in\\mathbb{N}\\}\\subseteq\\tau_{\\mathrm{fin}}",
        "after_tex": "\\bigcup\\mathcal{S}=E\\notin\\tau_{\\mathrm{fin}}",
        "relation": "implies",
        "rule": "family_ops",
        "conditions": [],
        "explanation": {
          "zh-Hant": "每個單點集 {2k} 有限，屬於 τ_fin。依聯集定義，⋃𝒮=E 是全部偶數。E 不在 τ_fin 中：S14.1 證 E≠ℕ，S14.2 證 E 不是有限集合。所以 τ_fin 對這個無窮子族的聯集不封閉，不是拓撲，雖然 S13 的兩兩檢查都通過。",
          "en": "Each singleton {2k} is finite and belongs to τ_fin. By the definition of union, ⋃𝒮=E is the set of all even numbers. E is not in τ_fin: S14.1 shows E≠ℕ and S14.2 shows E is not finite. So τ_fin is not closed under the union of this infinite subfamily and is not a topology, although every pairwise check of S13 passes."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "family_ops"
          ],
          "explanation": {
            "zh-Hant": "反例需要一個無窮子族；有限子族的聯集不會出現這個問題（S13）。結論依賴 S14.1、S14.2。",
            "en": "The counterexample needs an infinite subfamily; unions of finite subfamilies cause no problem (S13). The conclusion depends on S14.1 and S14.2."
          }
        },
        "leaf_basis": "family_ops + contradiction"
      },
      {
        "id": "S14.1",
        "uses": [
          "G3"
        ],
        "before_tex": "1\\in E\\ ?",
        "after_tex": "1\\notin E,\\ \\text{so}\\ E\\neq\\mathbb{N}",
        "relation": "implies",
        "rule": "set_extensionality",
        "conditions": [],
        "explanation": {
          "zh-Hant": "若 1=2k，k=0 給 0，k≥1 給至少 2，都不是 1；所以 1∉E，而 1∈ℕ，故 E≠ℕ。",
          "en": "If 1=2k, then k=0 gives 0 and k≥1 gives at least 2, neither of which is 1; so 1∉E, while 1∈ℕ, hence E≠ℕ."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "set_extensionality",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "一個不屬於 E 的自然數就足以說明兩集合不等（N5）。",
            "en": "One natural number outside E suffices to show the sets differ (N5)."
          }
        },
        "leaf_basis": "set_extensionality + order_basics + case_analysis"
      },
      {
        "id": "S14.2",
        "uses": [
          "G3",
          "finite_sets"
        ],
        "before_tex": "E\\ \\text{finite}\\ ?",
        "after_tex": "E\\ \\text{is infinite}",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "假設 E 有限。0∈E，所以 E 非空，依 finite_sets 有最大元素 m=2j。但 m+2=2(j+1)∈E 且 m+2>m，與 m 最大矛盾。所以 E 是無窮集合。",
          "en": "Suppose E were finite. 0∈E, so E is nonempty and by finite_sets has a largest element m=2j. But m+2=2(j+1)∈E and m+2>m, contradicting maximality. So E is infinite."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "finite_sets",
            "contradiction"
          ],
          "explanation": {
            "zh-Hant": "矛盾依賴「有限非空自然數集合有最大元素」與 2j+2=2(j+1)（N6）。",
            "en": "The contradiction relies on 'a finite nonempty set of naturals has a largest element' and 2j+2=2(j+1) (N6)."
          }
        },
        "leaf_basis": "contradiction + finite_sets + distributive"
      },
      {
        "id": "S15",
        "uses": [
          "S09",
          "S10",
          "S12",
          "S14"
        ],
        "before_tex": "\\text{goals (1) to (4)}",
        "after_tex": "\\tau_1\\ \\text{topology};\\ \\tau_2,\\ \\tau_{\\mathrm{fin}}\\ \\text{not topologies};\\ \\{\\varnothing,X\\}\\subsetneq\\tau_1\\subsetneq\\mathcal{P}(X)",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "合併：S09 給 (1)；S10 給 (2)（S11 說明它通過的檢查）；S12 給 (3)；S13、S14 給 (4)。",
          "en": "Combine: S09 gives (1); S10 gives (2) (with S11 showing which checks it passes); S12 gives (3); S13 and S14 give (4)."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S09",
            "S10",
            "S12",
            "S14",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G：四個小題。足夠條件 H：S09、S10、S12、S14 的結論；H⇒G 只是合取，H 由已知正向證成。",
            "en": "Goal G: the four parts. Sufficient condition H: the conclusions of S09, S10, S12 and S14; H⇒G is conjunction, and H is proved forwards from the givens."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "τ₁={∅,{a},{a,b},X} 是 X={a,b,c} 上的拓撲：它是一條鏈，非空子族的聯集與交集分別是最大與最小成員，空族給出 ∅ 與 X，16 個子族全部涵蓋。τ₂={∅,{a},{b},X} 不是拓撲，因 {a}∪{b}={a,b}∉τ₂，雖然它含 ∅、X 且對交集封閉。{∅,X}⊊τ₁⊊P(X)。ℕ 上的 τ_fin 對兩兩聯集與有限交集封閉，但偶數集合是其單點集的聯集且不在 τ_fin 中，所以不是拓撲。",
      "en": "τ₁={∅,{a},{a,b},X} is a topology on X={a,b,c}: it is a chain, the union and intersection of a nonempty subfamily are its largest and smallest members, the empty family gives ∅ and X, and all 16 subfamilies are covered. τ₂={∅,{a},{b},X} is not a topology, since {a}∪{b}={a,b}∉τ₂, although it contains ∅ and X and is closed under intersections. {∅,X}⊊τ₁⊊P(X). On ℕ, τ_fin is closed under pairwise unions and finite intersections, but the set of even numbers is a union of its singletons and is not in τ_fin, so τ_fin is not a topology."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "τ₂={∅,{a},{b},X} 含 ∅、X 且對交集封閉，所以是拓撲",
          "en": "τ₂={∅,{a},{b},X} contains ∅ and X and is closed under intersections, so it is a topology"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "{a}∪{b}={a,b}∉τ₂，聯集條件失敗。",
          "en": "{a}∪{b}={a,b}∉τ₂, so the union condition fails."
        }
      },
      {
        "claim": {
          "zh-Hant": "τ_fin 對兩兩聯集封閉，所以對任意聯集封閉，是 ℕ 上的拓撲",
          "en": "τ_fin is closed under pairwise unions, hence under arbitrary unions, so it is a topology on ℕ"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "單點集 {2k} 的聯集是偶數集合，既非有限也不等於 ℕ。",
          "en": "The union of the singletons {2k} is the set of even numbers, which is neither finite nor equal to ℕ."
        }
      },
      {
        "claim": {
          "zh-Hant": "{{a},{a,b},X} 是 X 上的拓撲",
          "en": "{{a},{a,b},X} is a topology on X"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "缺少 ∅，也就是空索引族的聯集不在族中；僅缺少 ∅ 這一點，就足以否定它是拓撲。",
          "en": "It lacks ∅, that is, the union of the empty family is not in it; this alone refutes it."
        }
      },
      {
        "claim": {
          "zh-Hant": "{∅,{a},{a,b}} 是 X 上的拓撲",
          "en": "{∅,{a},{a,b}} is a topology on X"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "缺少 X，也就是空族的交集（按 X 解讀）不在族中。",
          "en": "It lacks X, that is, the intersection of the empty family (read as X) is not in it."
        }
      },
      {
        "claim": {
          "zh-Hant": "τ₁ 比 P(X) 細",
          "en": "τ₁ is finer than P(X)"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "τ₁⊊P(X)，所以 P(X) 較細；例如 {b}∈P(X) 但 {b}∉τ₁。",
          "en": "τ₁⊊P(X), so P(X) is finer; for example {b}∈P(X) but {b}∉τ₁."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "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": "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": "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": "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": "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": "logic_chain",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "推論的傳遞：由 P⇒Q 與 Q⇒R 得 P⇒R；⇔ 同理。",
          "en": "Chaining: from P⇒Q and Q⇒R infer P⇒R; likewise for ⇔."
        }
      },
      {
        "id": "case_analysis",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分情況論證：若 P∨Q 成立，且 P⇒R、Q⇒R 都已證，則 R 成立。所有情況必須涵蓋全部可能。",
          "en": "Proof by cases: if P∨Q holds and both P⇒R and Q⇒R are proved, then R holds. The cases must cover every possibility."
        }
      },
      {
        "id": "and_intro_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "合取：由 P 與 Q 得 P∧Q；由 P∧Q 可分別得 P 與 Q。",
          "en": "Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q."
        }
      },
      {
        "id": "contradiction",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "反證法：若假設 ¬P 推出矛盾（某命題與其否定同時成立），則 P 成立。",
          "en": "Proof by contradiction: if assuming ¬P leads to a contradiction (a statement and its negation both hold), then P holds."
        }
      },
      {
        "id": "order_basics",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數的序：任兩數恰有 u<v、u=v、u>v 之一；u<v 且 v<w 推出 u<w；u<v 推出 u+w<v+w；u<v 且 w>0 推出 uw<vw；u<v 且 w<0 推出 uw>vw。",
          "en": "Order of the reals: exactly one of u<v, u=v, u>v holds; u<v and v<w give u<w; u<v gives u+w<v+w; u<v and w>0 give uw<vw; u<v and w<0 give uw>vw."
        }
      },
      {
        "id": "distributive",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分配律：u(v+w)=uv+uw；配合乘法交換律也有 (v+w)u=vu+wu。",
          "en": "Distributivity: u(v+w)=uv+uw; with commutativity also (v+w)u=vu+wu."
        }
      },
      {
        "id": "mko_topology_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-topological-space",
          "sha256": "95f78b4f323b3139dd05068842f7209ac33b39b3d86feb5b73e0eb8849cdb43d",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "給定集合 X，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；(X,τ) 稱為拓撲空間。",
          "en": "For a set X, a topology on X is a family τ⊆P(X) such that ∅∈τ, X∈τ, the union of any subfamily of τ belongs to τ, and the intersection of any finite subfamily of τ belongs to τ. The pair (X,τ) is a topological space."
        }
      },
      {
        "id": "family_ops",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "對 X 的子集合族 𝒮：x∈⋃𝒮 ⇔ 存在 A∈𝒮 使 x∈A；x∈⋂𝒮 ⇔ x∈X 且對每個 A∈𝒮 都有 x∈A。因此 ⋃∅=∅、⋂∅=X。",
          "en": "For a family 𝒮 of subsets of X: x∈⋃𝒮 ⇔ x∈A for some A∈𝒮; x∈⋂𝒮 ⇔ x∈X and x∈A for every A∈𝒮. Hence ⋃∅=∅ and ⋂∅=X."
        }
      },
      {
        "id": "finite_sets",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "兩個有限集合的聯集有限；有限集合的子集合有限；ℕ 的每個有限非空子集都有最大元素。",
          "en": "The union of two finite sets is finite; a subset of a finite set is finite; every finite nonempty subset of ℕ has a largest element."
        }
      },
      {
        "id": "L_chain_extremes",
        "kind": "lemma",
        "proved_in": [
          "S05.1",
          "S05.2",
          "S05.3"
        ],
        "statement": {
          "zh-Hant": "若有限非空集合族 𝒮 中任兩成員都可用 ⊆ 比較，則 𝒮 有最大成員 M（每個 A∈𝒮 有 A⊆M）與最小成員 m（每個 A∈𝒮 有 m⊆A）。",
          "en": "If any two members of a finite nonempty family 𝒮 are comparable under ⊆, then 𝒮 has a largest member M (A⊆M for every A∈𝒮) and a smallest member m (m⊆A for every A∈𝒮)."
        }
      },
      {
        "id": "L_chain_union",
        "kind": "lemma",
        "proved_in": [
          "S06.1",
          "S06.2"
        ],
        "statement": {
          "zh-Hant": "若 M∈𝒮 是 𝒮 的最大成員，則 ⋃𝒮=M；若 m∈𝒮 是最小成員，則 ⋂𝒮=m（𝒮 的成員都是 X 的子集）。",
          "en": "If M∈𝒮 is the largest member of 𝒮 then ⋃𝒮=M; if m∈𝒮 is the smallest member then ⋂𝒮=m (members of 𝒮 being subsets of X)."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S09",
        "lhs": "2**4",
        "rhs": "16"
      },
      {
        "id": "N2",
        "step": "S09",
        "lhs": "2**4 - 1",
        "rhs": "15"
      },
      {
        "id": "N3",
        "step": "S12",
        "lhs": "2**3",
        "rhs": "8"
      },
      {
        "id": "N4",
        "step": "S11",
        "lhs": "comb(4,2)",
        "rhs": "6"
      },
      {
        "id": "N5",
        "step": "S14.1",
        "lhs": "1 % 2",
        "relation": "!=",
        "rhs": "0"
      },
      {
        "id": "N6",
        "step": "S14.2",
        "vars": [
          "j"
        ],
        "degree": 1,
        "lhs": "2*j + 2",
        "rhs": "2*(j+1)"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "依 MKO 的定義（含空族約定 a2）逐條驗證。有限例以「鏈」引理一次處理所有非空子族，再計數確認 16 個子族全部涵蓋，不是抽查。無窮例只用來說明「兩兩封閉」與「任意封閉」的差別。不處理連續映射、基或度量拓撲；不改動 canonical MKO 或英文 edition。",
      "en": "Verifies each condition from the MKO's definition (with the empty-family convention a2). In the finite example a chain lemma handles all nonempty subfamilies at once, and a count confirms that all 16 subfamilies are covered, not sampled. The infinite example serves only to separate 'pairwise closure' from 'arbitrary closure'. Continuous maps, bases and metric topologies are not treated; the canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "X=\\{a,b,c\\}",
        "底集合，a、b、c 相異",
        "the underlying set, with a, b, c distinct"
      ],
      [
        "\\tau_1,\\ \\tau_2",
        "兩個候選子集合族",
        "two candidate families of subsets of X"
      ],
      [
        "\\mathcal{S}\\subseteq\\tau",
        "τ 的子族",
        "a subfamily of τ"
      ],
      [
        "\\bigcup\\mathcal{S},\\ \\bigcap\\mathcal{S}",
        "子族的聯集與交集（交集相對於 X）",
        "union and intersection of a subfamily (intersection relative to X)"
      ],
      [
        "\\mathcal{P}(X)",
        "冪集，即離散拓撲",
        "the power set, i.e. the discrete topology"
      ],
      [
        "\\tau_{\\mathrm{fin}}",
        "ℕ 上的候選族",
        "the candidate family on ℕ"
      ],
      [
        "E",
        "ℕ 中偶數的集合",
        "the set of even numbers in ℕ"
      ]
    ],
    "md_error_controls": [
      {
        "title": "不是任何子集合族都是拓撲 / Not every family of subsets is a topology",
        "zh": "MKO 指出的誤解：把任何子集合族當成拓撲。τ₂ 含 ∅ 與 X，也對交集封閉，卻因 {a}∪{b}={a,b}∉τ₂ 而不是拓撲（S10、S11）。必須逐類檢查全部條件。",
        "en": "The misconception the MKO names: treating any family of subsets as a topology. τ₂ contains ∅ and X and is closed under intersections, yet it is not a topology because {a}∪{b}={a,b}∉τ₂ (S10, S11). Every kind of condition must be checked."
      },
      {
        "title": "兩兩封閉不等於任意封閉 / Pairwise closure is not arbitrary closure",
        "zh": "MKO 指出的另一個誤解：把只對兩兩聯集封閉當成對任意聯集封閉。τ_fin 對兩兩聯集封閉（S13），但無窮多個單點集 {0}、{2}、{4} 等的聯集是偶數集合，不在 τ_fin 中（S14）。在含 ∅ 的有限族上，每個子族都有限，二元聯集封閉再加空聯集條件即可涵蓋任意聯集；若未含 ∅，空子族仍可能使它失敗；本稿對 τ₁ 改用鏈引理一次處理全部子族。",
        "en": "The other misconception the MKO names: taking closure under pairwise unions for closure under arbitrary unions. τ_fin is closed under pairwise unions (S13), but the union of the infinitely many singletons {0}, {2}, {4} and so on is the set of even numbers, which is not in τ_fin (S14). For a finite family containing ∅, every subfamily is finite and binary-union closure together with the empty-union condition suffices; without ∅ the empty subfamily can still fail; for τ₁ this guide uses the chain lemma to handle all subfamilies at once."
      },
      {
        "title": "別漏掉空族 / Do not forget the empty family",
        "zh": "公理 ∅∈τ、X∈τ 正是空族聯集與空族交集的情形（S03）。漏掉 ∅ 的 {{a},{a,b},X} 或漏掉 X 的 {∅,{a},{a,b}} 都不是拓撲。",
        "en": "The axioms ∅∈τ and X∈τ are exactly the cases of the empty union and the empty intersection (S03). {{a},{a,b},X}, which lacks ∅, and {∅,{a},{a,b}}, which lacks X, are not topologies."
      },
      {
        "title": "粗細比較的是開集族 / Finer and coarser compare families of open sets",
        "zh": "τ₁⊆P(X) 表示 P(X) 較細、τ₁ 較粗；這是開集族的包含關係，與底集合 X 的大小無關（MKO 段落 2）。",
        "en": "τ₁⊆P(X) means P(X) is finer and τ₁ coarser; this is inclusion of families of open sets and has nothing to do with the size of X (MKO paragraph 2)."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "finite_sets（兩有限集合之聯集有限、有限集之子集有限、有限非空自然數集有最大元素）作為前置知識使用，未在本稿證明。",
        "en": "finite_sets (a union of two finite sets is finite, a subset of a finite set is finite, a finite nonempty set of naturals has a largest element) is used as prerequisite knowledge and not proved here."
      },
      {
        "zh-Hant": "一般敘述須補上空族條件：有限集合族若含 ∅ 且對二元聯集封閉，則對其任意子族的聯集封閉；∅ 的成員條件用來處理空子族。本稿沒有另證或使用這個一般命題，τ₁ 的完整證明仍採鏈引理並單獨處理空族。",
        "en": "The general remark needs its empty-family qualification: a finite family of sets that contains ∅ and is closed under binary unions is closed under unions of arbitrary subfamilies. Membership of ∅ handles the empty subfamily. This guide does not separately prove or use that general result; the complete proof for τ₁ still uses the chain lemma and treats the empty family separately."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "4da430d185bcefe3f83821030d9b54994119c5304ce3b95397b135d2151f8de7",
    "original_source_sha256": "9fc206f52faa8c93d175434ba7de7802035b74beb7ba12cd2a3c31cc736a538d",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [
      "BR-M030"
    ],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}