以下逐條檢定拓撲 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)}
,所以根據有無限聯集性質 的定理(1)與(2)有
⋃
F
⊆
X
{\displaystyle \bigcup {\mathcal {F}}\subseteq X}
但根據無限聯集性質 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}}}
,然後從無限聯集性質 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}}}
內某些集合的聯集 」,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\}}
考慮到一階邏輯 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\}}
根據無限聯集性質 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\}}
這樣根據無限聯集性質 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)}
,從無限聯集性質 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\}}
此時考慮到一階邏輯 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 }