基 (拓撲學)

以下逐條檢驗拓扑的定義：

(1) 等價於「 ${\displaystyle X\in {\mathcal {U}$」的條件

若 ${\displaystyle X\in {\mathcal {U}$ ，則：

${\displaystyle (\exists {\mathcal {A})\left[({\mathcal {A}\subseteq {\mathcal {F})\wedge \left(\bigcup {\mathcal {A}=X\right)\right]}$(a)

考慮到 ${\displaystyle {\mathcal {F}\subseteq {\mathcal {P}(X)}$ ，所以根據有无限并集性質的定理(1)與(2)有

${\displaystyle \bigcup {\mathcal {F}\subseteq X}$

但根據无限并集性質的定理(1)，(a)又等價於：

${\displaystyle (\exists {\mathcal {A})\left[({\mathcal {A}\subseteq {\mathcal {F})\wedge \left(\bigcup {\mathcal {A}=X\right)\wedge \left(\bigcup {\mathcal {A}\subseteq \bigcup {\mathcal {F}\right)\right]}$

所以有：

${\displaystyle X\subseteq \bigcup {\mathcal {F}$

所以從 ${\displaystyle X\in {\mathcal {U}$ 有：

${\displaystyle \bigcup {\mathcal {F}=X}$ (a1)

反之若有 (a1)，因為 ${\displaystyle {\mathcal {F}\subseteq {\mathcal {F}$ ，所以有 ${\displaystyle X\in {\mathcal {U}$ 。故在本定理的前提下，(a1)等價於 ${\displaystyle X\in {\mathcal {U}$ 。

(2) ${\displaystyle \varnothing \in {\mathcal {U}$

首先考慮到 ${\displaystyle \varnothing \subseteq {\mathcal {F}$，然後從无限并集性質的定理(0)有 ${\displaystyle \varnothing =\bigcup \varnothing }$，故 ${\displaystyle \varnothing \in {\mathcal {U}$。

(3) 對任意 ${\displaystyle {\mathfrak {A}\subseteq {\mathcal {U}$ 有 ${\displaystyle \bigcup {\mathfrak {A}\in {\mathcal {U}$

首先，${\displaystyle {\mathfrak {A}\subseteq {\mathcal {U}$ 可等價地展開為

${\displaystyle (\forall A\in {\mathfrak {A})(\exists {\mathcal {B})\left[({\mathcal {B}\subseteq {\mathcal {F})\wedge \left(\bigcup {\mathcal {B}=A\right)\right]}$(b)

上式可直觀地解釋成「 ${\displaystyle A\in {\mathfrak {A}$ 都是 ${\displaystyle {\mathcal {F}$ 內某些集合的并集」，既然如此，取一個蒐集各種不同 ${\displaystyle A}$ 的子集的集族 ${\displaystyle {\mathcal {P}_{\mathfrak {A}$ ：

${\displaystyle {\mathcal {P}_{\mathfrak {A}:=\left\{S\,|\,(\exists A)[(A\in {\mathfrak {A})\wedge (S\subseteq A)]\right\}$

這樣根據有限交集的性質，${\displaystyle x\in \bigcup ({\mathcal {P}_{\mathfrak {A}\cap {\mathcal {F})}$ 等價於

${\displaystyle (\exists S)\left\{(x\in S)\wedge (S\in {\mathcal {F})\wedge (\exists A)\left[(A\in {\mathfrak {A})\wedge (S\subseteq A)\right]\right\}$

考慮到一阶逻辑的定理(Ce)，將${\displaystyle (\exists A)}$ 移至最前，再將${\displaystyle (\exists S)}$移入括弧內 ，上式就依據(Equv)而等價於

${\displaystyle (\exists A)\left\{(A\in {\mathfrak {A})\wedge (\exists S)\left[(x\in S)\wedge (S\in {\mathcal {F})\wedge (S\subseteq A)\right]\right\}$

也就等價於

${\displaystyle (\exists A)\left\{(A\in {\mathfrak {A})\wedge \left(x\in \bigcup [{\mathcal {P}(A)\cap {\mathcal {F}]\right)\right\}$

根據无限并集性質的定理(4)，從(b)有

${\displaystyle (\forall A\in {\mathfrak {A})(\exists {\mathcal {B})\left\{({\mathcal {B}\subseteq {\mathcal {F})\wedge \left\{A\subseteq \bigcup [{\mathcal {B}\cap {\mathcal {P}(A)]\right\}\right\}$

這樣根據无限并集性質的定理(1)又會有

${\displaystyle (\forall A\in {\mathfrak {A})\left\{A\subseteq \bigcup [{\mathcal {F}\cap {\mathcal {P}(A)]\right\}$

考慮到 ${\displaystyle {\mathcal {F}\cap {\mathcal {P}(A)\subseteq {\mathcal {P}(A)}$ ，從无限并集性質的定理(1)與定理(2)有

${\displaystyle \bigcup {\mathcal {F}\cap {\mathcal {P}(A)\subseteq A}$

所以最後從(b)有

${\displaystyle (\forall A\in {\mathfrak {A})\left\{A=\bigcup [{\mathcal {F}\cap {\mathcal {P}(A)]\right\}$

所以 ${\displaystyle x\in \bigcup ({\mathcal {P}_{\mathfrak {A}\cap {\mathcal {F})}$ 最後等價於

${\displaystyle (\exists A)\left[(A\in {\mathfrak {A})\wedge (x\in A)\right]}$

換句話說

${\displaystyle x\in \bigcup {\mathfrak {A}$

這樣考慮到 ${\displaystyle {\mathcal {P}_{\mathfrak {A}\cap {\mathcal {F}\subseteq {\mathcal {F}$ 就有

${\displaystyle \bigcup {\mathfrak {A}=\bigcup ({\mathcal {P}_{\mathfrak {A}\cap {\mathcal {F})\in {\mathcal {U}$

所以在本定理的前提下， 對所有 ${\displaystyle {\mathfrak {A}\subseteq {\mathcal {U}$ 都有 ${\displaystyle \bigcup {\mathfrak {A}\in {\mathcal {U}$ 。

(4)等價於「 ${\displaystyle U,\,V\in {\mathcal {U}$ 則 ${\displaystyle U\cap V\in {\mathcal {U}$」的條件

若

「對所有的 ${\displaystyle U,\,V\in {\mathcal {U}$ 有 ${\displaystyle U\cap V\in {\mathcal {U}$」(P)

因取任意 ${\displaystyle B_{1},\,B_{2}\in {\mathcal {F}$ 都有：

${\displaystyle B_{1}=\bigcup \{B_{1}\}\in {\mathcal {U}$

${\displaystyle B_{2}=\bigcup \{B_{1}\}\in {\mathcal {U}$

故 ${\displaystyle B_{1}\cap B_{2}\in {\mathcal {U}$ ，換句話說從假設(P)可以推出：

「對所有 ${\displaystyle B_{1},\,B_{2}\in {\mathcal {F}$ ，${\displaystyle B_{1}\cap B_{2}\in {\mathcal {U}$」(P')

另一方面， ${\displaystyle U\cap V\in {\mathcal {U}$ 可等價地展開為：

${\displaystyle (\exists {\mathcal {E})\left\{({\mathcal {E}\subseteq {\mathcal {F})\wedge \left(U\cap V=\bigcup {\mathcal {E}\right)\right\}$

因為 ${\displaystyle U,\,V\in {\mathcal {U}$ 可等價地展開為：

${\displaystyle (\exists {\mathcal {A})\left[\left(U=\bigcup {\mathcal {A}\right)\wedge ({\mathcal {A}\subseteq {\mathcal {F})\right]}$

${\displaystyle (\exists {\mathcal {B})\left[\left(V=\bigcup {\mathcal {B}\right)\wedge ({\mathcal {B}\subseteq {\mathcal {F})\right]}$

所以在 ${\displaystyle U,\,V\in {\mathcal {U}$ 的前提下 ${\displaystyle U\cap V\in {\mathcal {U}$ 又可更進一步等價地展開為：

${\displaystyle (\exists {\mathcal {A})(\exists {\mathcal {B})(\exists {\mathcal {E})\left\{({\mathcal {A},\,{\mathcal {B},\,{\mathcal {E}\subseteq {\mathcal {F})\wedge \left(U=\bigcup {\mathcal {A}\right)\wedge \left(V=\bigcup {\mathcal {B}\right)\wedge \left[\left(\bigcup {\mathcal {A}\right)\cap \left(\bigcup {\mathcal {B}\right)=\bigcup {\mathcal {E}\right]\right\}$

此時考慮到一阶逻辑的定理(Ce)，連續使用兩次會有：

${\displaystyle \left[x\in \left(\bigcup {\mathcal {A}\right)\cap \left(\bigcup {\mathcal {B}\right)\right]\Leftrightarrow (\exists A)(\exists B)[(A\in {\mathcal {A})\wedge (B\in {\mathcal {B})\wedge (x\in A\cap B)]}$

這樣的話，若取一個包含所有 ${\displaystyle A\cap B}$ 的集族：

${\displaystyle {\mathcal {C}:=\left\{S\in {\mathcal {P}(X)\,{\big |}\,(\exists A)(\exists B)[(A\in {\mathcal {A})\wedge (B\in {\mathcal {B})\wedge (S=A\cap B)]\right\}$

這樣就有：

${\displaystyle \bigcup {\mathcal {C}=\left(\bigcup {\mathcal {A}\right)\cap \left(\bigcup {\mathcal {B}\right)}$

而且考慮到 ${\displaystyle {\mathcal {A}\subseteq {\mathcal {F}$ 和 ${\displaystyle {\mathcal {B}\subseteq {\mathcal {F}$ ，所以在(P')的前提下，所有的 ${\displaystyle A\cap B}$ 都在 ${\displaystyle {\mathcal {U}$ 裡，換句話說，${\displaystyle {\mathcal {C}\subseteq {\mathcal {U}$ ，故從上小結的結果有：

${\displaystyle U\cap V\in {\mathcal {U}$

所以，(P')跟(P)等價。

綜合上面的(a1)、(a2)、和(P')，本定理得證。${\displaystyle \Box }$

${\displaystyle {\mathcal {B}$

${\displaystyle \tau _{\mathcal {B}$

${\displaystyle {\mathcal {B}$

${\displaystyle \tau ({\mathcal {B})}$

根據最粗拓撲的定義有：

${\displaystyle \vdash (\forall S){\big \{}[S\in \tau ({\mathcal {B})]\Leftrightarrow (\forall {\mathfrak {T})\{[({\mathfrak {T}{\text{ is a topology of }X)\wedge ({\mathcal {B}\subseteq {\mathfrak {T})]\Rightarrow (S\in {\mathfrak {T})\}{\big \}$(a)

那以量词公理(A4) 將 ${\displaystyle \forall S}$ 去掉會有：

${\displaystyle \vdash [S\in \tau ({\mathcal {B})]\Leftrightarrow (\forall {\mathfrak {T})\{[({\mathfrak {T}{\text{ is a topology of }X)\wedge ({\mathcal {B}\subseteq {\mathfrak {T})]\Rightarrow (S\in {\mathfrak {T})\}$

那再使用量词公理(A4)，配合(D1)會有：

${\displaystyle \vdash [S\in \tau ({\mathcal {B})]\Rightarrow \{[(\tau _{B}{\text{ is a topology of }X)\wedge ({\mathcal {B}\subseteq \tau _{B})]\Rightarrow (S\in \tau _{B})\}$

因為 ${\displaystyle \tau _{\mathcal {B}$ 是 ${\displaystyle {\mathcal {B}$ 所生成的拓撲，配合(D2)有：（${\displaystyle {\mathcal {P}$ 為 「${\displaystyle {\mathcal {B}$ 是 ${\displaystyle X}$ 的拓撲基」的正式敘述）

${\displaystyle {\mathcal {P}\vdash [S\in \tau ({\mathcal {B})]\Rightarrow (S\in \tau _{B})}$

另一方面，根據拓撲基的定義有：

${\displaystyle {\mathcal {P},\,S\in \tau _{B}\vdash (\exists {\mathcal {C})\left[\left(S=\bigcup {\mathcal {C}\right)\wedge ({\mathcal {C}\subseteq {\mathcal {B})\right]}$

而根據拓扑的定義(關於聯集的部分)與演繹定理會有：

${\displaystyle {\mathcal {P},\,[({\mathfrak {T}{\text{ is a topology of }X)\wedge ({\mathcal {B}\subseteq {\mathfrak {T})],\,\left[\left(S=\bigcup {\mathcal {C}\right)\wedge ({\mathcal {C}\subseteq {\mathcal {B})\right]\vdash (S\in {\mathfrak {T})}$

這樣根據(GEN)與演繹定理就有：

${\displaystyle {\mathcal {P},\,[({\mathfrak {T}{\text{ is a topology of }X)\wedge ({\mathcal {B}\subseteq {\mathfrak {T})]\vdash (\exists C)\left[\left(S=\bigcup {\mathcal {C}\right)\wedge ({\mathcal {C}\subseteq {\mathcal {B})\right]\Rightarrow (S\in {\mathfrak {T})}$

換句話說，從演繹定理與(D1)有：

${\displaystyle {\mathcal {P},\,S\in \tau _{B}\vdash [({\mathfrak {T}{\text{ is a topology of }X)\wedge ({\mathcal {B}\subseteq {\mathfrak {T})]\Rightarrow (S\in {\mathfrak {T})}$

那從普遍化元定理就有：

${\displaystyle {\mathcal {P},\,S\in \tau _{B}\vdash (\forall {\mathfrak {T})\{[({\mathfrak {T}{\text{ is a topology of }X)\wedge ({\mathcal {B}\subseteq {\mathfrak {T})]\Rightarrow (S\in {\mathfrak {T})\}$

這樣從(a)，配合(AND)與(D1)就有：

${\displaystyle {\mathcal {P},\,S\in \tau _{B}\vdash S\in \tau ({\mathcal {B})}$

這樣從(AND)和演繹定理就有：

${\displaystyle {\mathcal {P}\vdash (S\in \tau _{B})\Leftrightarrow [S\in \tau ({\mathcal {B})]}$

套用(GEN)將 ${\displaystyle \forall S}$ 重新加入就會有：

${\displaystyle {\mathcal {P}\vdash \tau _{B}=\tau ({\mathcal {B})}$

故本定理得証。${\displaystyle \Box }$

${\displaystyle O\in \tau _{Y}$

${\displaystyle {\mathcal {F}\subseteq {\mathfrak {B}$

${\displaystyle O=\bigcup {\mathcal {F}$

這樣的話，若取：

${\displaystyle f^{-1}({\mathcal {F})=\{A\,|(\exists B\in {\mathcal {F})(A=f^{-1}(B))\,\}$

則有：

${\displaystyle f^{-1}(O)=\bigcup f^{-1}({\mathcal {F})}$

${\displaystyle f^{-1}({\mathcal {F})\subseteq \tau _{X}$

這樣根據拓撲空間的定義就有：

${\displaystyle f^{-1}(O)=\bigcup f^{-1}({\mathcal {F})\in \tau _{X}$

故${\displaystyle f}$是${\displaystyle \tau _{X}$-${\displaystyle \tau _{Y}$连续。${\displaystyle \Box }$