以下逐條檢驗拓扑 的定義:
(1) 等價於「 X ∈ U {\displaystyle X\in {\mathcal {U}}} 」的條件
若 X ∈ U {\displaystyle X\in {\mathcal {U}}} ,則:
( ∃ A ) [ ( A ⊆ F ) ∧ ( ⋃ A = X ) ] {\displaystyle (\exists {\mathcal {A}})\left[({\mathcal {A}}\subseteq {\mathcal {F}})\wedge \left(\bigcup {\mathcal {A}}=X\right)\right]} (a)考慮到 F ⊆ P ( X ) {\displaystyle {\mathcal {F}}\subseteq {\mathcal {P}}(X)} ,所以根據有无限并集性質 的定理(1)與(2)有
⋃ F ⊆ X {\displaystyle \bigcup {\mathcal {F}}\subseteq X} 但根據无限并集性質 的定理(1),(a)又等價於:
( ∃ A ) [ ( A ⊆ F ) ∧ ( ⋃ A = X ) ∧ ( ⋃ A ⊆ ⋃ F ) ] {\displaystyle (\exists {\mathcal {A}})\left[({\mathcal {A}}\subseteq {\mathcal {F}})\wedge \left(\bigcup {\mathcal {A}}=X\right)\wedge \left(\bigcup {\mathcal {A}}\subseteq \bigcup {\mathcal {F}}\right)\right]} 所以有:
X ⊆ ⋃ F {\displaystyle X\subseteq \bigcup {\mathcal {F}}} 所以從 X ∈ U {\displaystyle X\in {\mathcal {U}}} 有:
⋃ F = X {\displaystyle \bigcup {\mathcal {F}}=X} (a1)反之若有 (a1),因為 F ⊆ F {\displaystyle {\mathcal {F}}\subseteq {\mathcal {F}}} ,所以有 X ∈ U {\displaystyle X\in {\mathcal {U}}} 。故在本定理的前提下,(a1)等價於 X ∈ U {\displaystyle X\in {\mathcal {U}}} 。
(2) ∅ ∈ U {\displaystyle \varnothing \in {\mathcal {U}}}
首先考慮到 ∅ ⊆ F {\displaystyle \varnothing \subseteq {\mathcal {F}}} ,然後從无限并集性質 的定理(0)有 ∅ = ⋃ ∅ {\displaystyle \varnothing =\bigcup \varnothing } ,故 ∅ ∈ U {\displaystyle \varnothing \in {\mathcal {U}}} 。
(3) 對任意 A ⊆ U {\displaystyle {\mathfrak {A}}\subseteq {\mathcal {U}}} 有 ⋃ A ∈ U {\displaystyle \bigcup {\mathfrak {A}}\in {\mathcal {U}}}
首先, A ⊆ U {\displaystyle {\mathfrak {A}}\subseteq {\mathcal {U}}} 可等價地展開為
( ∀ A ∈ A ) ( ∃ B ) [ ( B ⊆ F ) ∧ ( ⋃ B = A ) ] {\displaystyle (\forall A\in {\mathfrak {A}})(\exists {\mathcal {B}})\left[({\mathcal {B}}\subseteq {\mathcal {F}})\wedge \left(\bigcup {\mathcal {B}}=A\right)\right]} (b)上式可直觀地解釋成「 A ∈ A {\displaystyle A\in {\mathfrak {A}}} 都是 F {\displaystyle {\mathcal {F}}} 內某些集合的并集 」,既然如此,取一個蒐集各種不同 A {\displaystyle A} 的子集 的集族 P A {\displaystyle {\mathcal {P}}_{\mathfrak {A}}} :
P A := { S | ( ∃ A ) [ ( A ∈ A ) ∧ ( S ⊆ A ) ] } {\displaystyle {\mathcal {P}}_{\mathfrak {A}}:=\left\{S\,|\,(\exists A)[(A\in {\mathfrak {A}})\wedge (S\subseteq A)]\right\}} 這樣根據有限交集 的性質, x ∈ ⋃ ( P A ∩ F ) {\displaystyle x\in \bigcup ({\mathcal {P}}_{\mathfrak {A}}\cap {\mathcal {F}})} 等價於
( ∃ S ) { ( x ∈ S ) ∧ ( S ∈ F ) ∧ ( ∃ A ) [ ( A ∈ A ) ∧ ( S ⊆ A ) ] } {\displaystyle (\exists S)\left\{(x\in S)\wedge (S\in {\mathcal {F}})\wedge (\exists A)\left[(A\in {\mathfrak {A}})\wedge (S\subseteq A)\right]\right\}} 考慮到一阶逻辑 的定理(Ce) ,將 ( ∃ A ) {\displaystyle (\exists A)} 移至最前,再將 ( ∃ S ) {\displaystyle (\exists S)} 移入括弧內 ,上式就依據(Equv) 而等價於
( ∃ A ) { ( A ∈ A ) ∧ ( ∃ S ) [ ( x ∈ S ) ∧ ( S ∈ F ) ∧ ( S ⊆ A ) ] } {\displaystyle (\exists A)\left\{(A\in {\mathfrak {A}})\wedge (\exists S)\left[(x\in S)\wedge (S\in {\mathcal {F}})\wedge (S\subseteq A)\right]\right\}} 也就等價於
( ∃ A ) { ( A ∈ A ) ∧ ( x ∈ ⋃ [ P ( A ) ∩ F ] ) } {\displaystyle (\exists A)\left\{(A\in {\mathfrak {A}})\wedge \left(x\in \bigcup [{\mathcal {P}}(A)\cap {\mathcal {F}}]\right)\right\}} 根據无限并集性質 的定理(4),從(b)有
( ∀ A ∈ A ) ( ∃ B ) { ( B ⊆ F ) ∧ { A ⊆ ⋃ [ B ∩ P ( A ) ] } } {\displaystyle (\forall A\in {\mathfrak {A}})(\exists {\mathcal {B}})\left\{({\mathcal {B}}\subseteq {\mathcal {F}})\wedge \left\{A\subseteq \bigcup [{\mathcal {B}}\cap {\mathcal {P}}(A)]\right\}\right\}} 這樣根據无限并集性質 的定理(1)又會有
( ∀ A ∈ A ) { A ⊆ ⋃ [ F ∩ P ( A ) ] } {\displaystyle (\forall A\in {\mathfrak {A}})\left\{A\subseteq \bigcup [{\mathcal {F}}\cap {\mathcal {P}}(A)]\right\}} 考慮到 F ∩ P ( A ) ⊆ P ( A ) {\displaystyle {\mathcal {F}}\cap {\mathcal {P}}(A)\subseteq {\mathcal {P}}(A)} ,從无限并集性質 的定理(1)與定理(2)有
⋃ F ∩ P ( A ) ⊆ A {\displaystyle \bigcup {\mathcal {F}}\cap {\mathcal {P}}(A)\subseteq A} 所以最後從(b)有
( ∀ A ∈ A ) { A = ⋃ [ F ∩ P ( A ) ] } {\displaystyle (\forall A\in {\mathfrak {A}})\left\{A=\bigcup [{\mathcal {F}}\cap {\mathcal {P}}(A)]\right\}} 所以 x ∈ ⋃ ( P A ∩ F ) {\displaystyle x\in \bigcup ({\mathcal {P}}_{\mathfrak {A}}\cap {\mathcal {F}})} 最後等價於
( ∃ A ) [ ( A ∈ A ) ∧ ( x ∈ A ) ] {\displaystyle (\exists A)\left[(A\in {\mathfrak {A}})\wedge (x\in A)\right]} 換句話說
x ∈ ⋃ A {\displaystyle x\in \bigcup {\mathfrak {A}}} 這樣考慮到 P A ∩ F ⊆ F {\displaystyle {\mathcal {P}}_{\mathfrak {A}}\cap {\mathcal {F}}\subseteq {\mathcal {F}}} 就有
⋃ A = ⋃ ( P A ∩ F ) ∈ U {\displaystyle \bigcup {\mathfrak {A}}=\bigcup ({\mathcal {P}}_{\mathfrak {A}}\cap {\mathcal {F}})\in {\mathcal {U}}} 所以在本定理的前提下, 對所有 A ⊆ U {\displaystyle {\mathfrak {A}}\subseteq {\mathcal {U}}} 都有 ⋃ A ∈ U {\displaystyle \bigcup {\mathfrak {A}}\in {\mathcal {U}}} 。
(4)等價於「 U , V ∈ U {\displaystyle U,\,V\in {\mathcal {U}}} 則 U ∩ V ∈ U {\displaystyle U\cap V\in {\mathcal {U}}} 」的條件
若
「對所有的 U , V ∈ U {\displaystyle U,\,V\in {\mathcal {U}}} 有 U ∩ V ∈ U {\displaystyle U\cap V\in {\mathcal {U}}} 」(P) 因取任意 B 1 , B 2 ∈ F {\displaystyle B_{1},\,B_{2}\in {\mathcal {F}}} 都有:
B 1 = ⋃ { B 1 } ∈ U {\displaystyle B_{1}=\bigcup \{B_{1}\}\in {\mathcal {U}}} B 2 = ⋃ { B 1 } ∈ U {\displaystyle B_{2}=\bigcup \{B_{1}\}\in {\mathcal {U}}} 故 B 1 ∩ B 2 ∈ U {\displaystyle B_{1}\cap B_{2}\in {\mathcal {U}}} ,換句話說從假設(P)可以推出:
「對所有 B 1 , B 2 ∈ F {\displaystyle B_{1},\,B_{2}\in {\mathcal {F}}} , B 1 ∩ B 2 ∈ U {\displaystyle B_{1}\cap B_{2}\in {\mathcal {U}}} 」(P') 另一方面, U ∩ V ∈ U {\displaystyle U\cap V\in {\mathcal {U}}} 可等價地展開為:
( ∃ E ) { ( E ⊆ F ) ∧ ( U ∩ V = ⋃ E ) } {\displaystyle (\exists {\mathcal {E}})\left\{({\mathcal {E}}\subseteq {\mathcal {F}})\wedge \left(U\cap V=\bigcup {\mathcal {E}}\right)\right\}} 因為 U , V ∈ U {\displaystyle U,\,V\in {\mathcal {U}}} 可等價地展開為:
( ∃ A ) [ ( U = ⋃ A ) ∧ ( A ⊆ F ) ] {\displaystyle (\exists {\mathcal {A}})\left[\left(U=\bigcup {\mathcal {A}}\right)\wedge ({\mathcal {A}}\subseteq {\mathcal {F}})\right]} ( ∃ B ) [ ( V = ⋃ B ) ∧ ( B ⊆ F ) ] {\displaystyle (\exists {\mathcal {B}})\left[\left(V=\bigcup {\mathcal {B}}\right)\wedge ({\mathcal {B}}\subseteq {\mathcal {F}})\right]} 所以在 U , V ∈ U {\displaystyle U,\,V\in {\mathcal {U}}} 的前提下 U ∩ V ∈ U {\displaystyle U\cap V\in {\mathcal {U}}} 又可更進一步等價地展開為:
( ∃ A ) ( ∃ B ) ( ∃ E ) { ( A , B , E ⊆ F ) ∧ ( U = ⋃ A ) ∧ ( V = ⋃ B ) ∧ [ ( ⋃ A ) ∩ ( ⋃ B ) = ⋃ E ] } {\displaystyle (\exists {\mathcal {A}})(\exists {\mathcal {B}})(\exists {\mathcal {E}})\left\{({\mathcal {A}},\,{\mathcal {B}},\,{\mathcal {E}}\subseteq {\mathcal {F}})\wedge \left(U=\bigcup {\mathcal {A}}\right)\wedge \left(V=\bigcup {\mathcal {B}}\right)\wedge \left[\left(\bigcup {\mathcal {A}}\right)\cap \left(\bigcup {\mathcal {B}}\right)=\bigcup {\mathcal {E}}\right]\right\}} 此時考慮到一阶逻辑 的定理(Ce) ,連續使用兩次會有:
[ x ∈ ( ⋃ A ) ∩ ( ⋃ B ) ] ⇔ ( ∃ A ) ( ∃ B ) [ ( A ∈ A ) ∧ ( B ∈ B ) ∧ ( x ∈ A ∩ B ) ] {\displaystyle \left[x\in \left(\bigcup {\mathcal {A}}\right)\cap \left(\bigcup {\mathcal {B}}\right)\right]\Leftrightarrow (\exists A)(\exists B)[(A\in {\mathcal {A}})\wedge (B\in {\mathcal {B}})\wedge (x\in A\cap B)]} 這樣的話,若取一個包含所有 A ∩ B {\displaystyle A\cap B} 的集族:
C := { S ∈ P ( X ) | ( ∃ A ) ( ∃ B ) [ ( A ∈ A ) ∧ ( B ∈ B ) ∧ ( S = A ∩ B ) ] } {\displaystyle {\mathcal {C}}:=\left\{S\in {\mathcal {P}}(X)\,{\big |}\,(\exists A)(\exists B)[(A\in {\mathcal {A}})\wedge (B\in {\mathcal {B}})\wedge (S=A\cap B)]\right\}} 這樣就有:
⋃ C = ( ⋃ A ) ∩ ( ⋃ B ) {\displaystyle \bigcup {\mathcal {C}}=\left(\bigcup {\mathcal {A}}\right)\cap \left(\bigcup {\mathcal {B}}\right)} 而且考慮到 A ⊆ F {\displaystyle {\mathcal {A}}\subseteq {\mathcal {F}}} 和 B ⊆ F {\displaystyle {\mathcal {B}}\subseteq {\mathcal {F}}} ,所以在(P')的前提下,所有的 A ∩ B {\displaystyle A\cap B} 都在 U {\displaystyle {\mathcal {U}}} 裡,換句話說, C ⊆ U {\displaystyle {\mathcal {C}}\subseteq {\mathcal {U}}} ,故從上小結的結果有:
U ∩ V ∈ U {\displaystyle U\cap V\in {\mathcal {U}}} 所以,(P')跟(P)等價 。
綜合上面的(a1)、(a2)、和(P'),本定理得證。 ◻ {\displaystyle \Box }