# 基 (拓撲學)

## 動機

${\displaystyle {\mathcal {U}}:=\left\{U\in {\mathcal {P}}(X)\,{\bigg |}\,(\exists {\mathcal {A}})\left[({\mathcal {A}}\subseteq {\mathcal {F}})\wedge \left(\bigcup {\mathcal {A}}=U\right)\right]\right\}}$

${\displaystyle X}$ 上的拓扑。事實上有以下的定理：

${\displaystyle \tau :=\left\{U\in {\mathcal {P}}(X)\,{\bigg |}\,(\exists {\mathcal {A}})\left[({\mathcal {A}}\subseteq {\mathcal {F}})\wedge \left(\bigcup {\mathcal {A}}=U\right)\right]\right\}}$

• ${\displaystyle \bigcup {\mathcal {F}}=X}$
• 對所有 ${\displaystyle B_{1},\,B_{2}\in {\mathcal {F}}}$${\displaystyle B_{1}\cap B_{2}\in {\mathcal {U}}}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

「對所有 ${\displaystyle B_{1},\,B_{2}\in {\mathcal {F}}}$${\displaystyle B_{1}\cap B_{2}=\bigcup [{\mathcal {F}}\cap {\mathcal {P}}(B_{1}\cap B_{2})]}$」也就等價於：
「所有的 ${\displaystyle B_{1},\,B_{2}\in {\mathcal {F}}}$ ，對任意 ${\displaystyle x\in B_{1}\cap B_{2}}$ 都存在 ${\displaystyle C\in [{\mathcal {F}}\cap {\mathcal {P}}(B_{1}\cap B_{2})]}$ 使得 ${\displaystyle x\in C}$

## 定義

${\displaystyle {\mathcal {F}}\subseteq {\mathcal {P}}(X)}$集合 ${\displaystyle X}$ 的一個子集族，若滿足：

• ${\displaystyle \bigcup {\mathcal {F}}=X}$　（基的元素覆蓋${\displaystyle X}$
• 所有的 ${\displaystyle B_{1},\,B_{2}\in {\mathcal {F}}}$ ，對任意 ${\displaystyle x\in B_{1}\cap B_{2}}$ 都存在 ${\displaystyle C\in [{\mathcal {F}}\cap {\mathcal {P}}(B_{1}\cap B_{2})]}$ 使得 ${\displaystyle x\in C}$

${\displaystyle \tau :=\left\{U\in {\mathcal {P}}(X)\,{\bigg |}\,(\exists {\mathcal {A}})\left[({\mathcal {A}}\subseteq {\mathcal {F}})\wedge \left(\bigcup {\mathcal {A}}=U\right)\right]\right\}}$

## 重要性質

${\displaystyle {\mathcal {B}}}$集合 ${\displaystyle X}$ 的拓撲基，則 ${\displaystyle {\mathcal {B}}}$ 生成的拓撲是包含 ${\displaystyle {\mathcal {B}}}$最粗拓撲

${\displaystyle {\mathcal {B}}}$ 所生成的拓撲是 ${\displaystyle \tau _{\mathcal {B}}}$ ；另一方面包含 ${\displaystyle {\mathcal {B}}}$最粗拓撲${\displaystyle \tau ({\mathcal {B}})}$

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

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

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

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

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

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

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

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

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

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

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

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

${\displaystyle {\mathcal {B}}_{1}}$${\displaystyle {\mathcal {B}}_{2}}$ 都是集合 ${\displaystyle X}$ 的拓撲基，而 ${\displaystyle \tau _{1}}$${\displaystyle {\mathcal {B}}_{1}}$ 生成的拓撲； ${\displaystyle \tau _{2}}$${\displaystyle {\mathcal {B}}_{2}}$ 生成的拓撲，則以下兩敘述價

• ${\displaystyle \tau _{1}\subseteq \tau _{2}}$
• ${\displaystyle (\forall B_{1}\in {\mathcal {B}}_{1})(\exists B_{2}\in {\mathcal {B}}_{2})(B_{2}\subseteq B_{1})}$

• 如果 B1,B2,...,Bn 是拓撲 T1,T2,...,Tn 的基，則集合積 B1 × B2 × ... × Bn乘積拓撲 T1 × T2 × ... × Tn 的基。在無限乘積的情況下這仍適用，除了出現有限多個基元素之外全部都必須是整個空間之外。
• BX 的基并設 YX子空間。那么如果我們交 B 的每個元素於 Y，結果的集合的搜集是子空間 Y 的基。
• 如果函數 f:XY 映射 X 的所有基元素到 Y 的一個開集，它是一個開映射。類似的，如果 Y 的一個基元素的所有原像在 X 中是開集，則 f連續函數
• X 的子集的搜集是 X 上的拓撲當且僅當它生成自身。
• B 是拓撲空間 X 的基，當且僅當 B 的包含 x 的元素的子搜集形成在 x 上的局部基，對于 X 的任何點 x
• 給定拓撲的一個基，要證明或序列的收斂，在包含假定極限的所有基中的集合中最終證明它就是充分的。

## 閉集基

FX 的閉集基。則

1. F = ∅
2. 對於每個 F1F2F 中，并集 F1F2F 的某個子族的交集(就是說，對于任何不在 F1F2x，存在一個 F3F 包含 F1F2 并不包含 x)。

## 準基

• 實數線上，所有長度為1的開區間便是一個準基。

J.W. 亞歷山大證明了：若每個準基覆盖都有一個有限個元素的子覆蓋，則此空間是緊緻的。

## 參考文獻

• James Munkres (1975) Topology: a First Course. Prentice-Hall.
• Willard, Stephen (1970) General Topology. Addison-Wesley. Reprinted 2004, Dover Publications.