다음 두 조건을 보이는 것으로 충분하다.
X
∪
f
Y
{\displaystyle X\cup _{f}Y}
는 T1 공간 이다. 즉, 모든 한원소집합 은 닫힌집합 이다.
X
∪
f
Y
{\displaystyle X\cup _{f}Y}
는 단조 정규 공간이다. 즉, 단조 정규성 연산자
G
:
D
X
∪
f
Y
→
T
X
∪
f
Y
{\displaystyle G\colon {\mathcal {D}}_{X\cup _{f}Y}\to {\mathcal {T}}_{X\cup _{f}Y}}
가 존재한다.
T1 . 표준적인 연속 함수
g
:
X
→
X
∪
f
Y
{\displaystyle g\colon X\to X\cup _{f}Y}
h
:
Y
→
X
∪
f
Y
{\displaystyle h\colon Y\to X\cup _{f}Y}
를 생각하자. 그렇다면, 부분 집합
U
⊆
X
∪
f
Y
{\displaystyle U\subseteq X\cup _{f}Y}
이 열린집합 일 필요충분조건 은
g
−
1
(
U
)
⊆
X
{\displaystyle g^{-1}(U)\subseteq X}
와
h
−
1
(
U
)
⊆
Y
{\displaystyle h^{-1}(U)\subseteq Y}
가 둘 다 열린집합 인 것이다. 마찬가지로, 닫힌집합 일 필요충분조건 은
g
−
1
(
U
)
⊆
X
{\displaystyle g^{-1}(U)\subseteq X}
와
h
−
1
(
U
)
⊆
Y
{\displaystyle h^{-1}(U)\subseteq Y}
가 둘 다 닫힌집합 인 것이다. 또한,
g
↾
X
∖
Z
{\displaystyle g\upharpoonright X\setminus Z}
는 열린 위상수학적 매장 이며,
h
{\displaystyle h}
는 닫힌 위상수학적 매장 이며,
g
(
X
∖
Z
)
∪
h
(
Y
)
=
X
∪
f
Y
{\displaystyle g(X\setminus Z)\cup h(Y)=X\cup _{f}Y}
g
(
X
∖
Z
)
∩
h
(
Y
)
=
∅
{\displaystyle g(X\setminus Z)\cap h(Y)=\varnothing }
이다. (그러나
X
∪
f
Y
{\displaystyle X\cup _{f}Y}
는 두 상 의 분리합공간 일 필요가 없다.) 이제, 임의의
x
∈
X
∖
Z
{\displaystyle x\in X\setminus Z}
및
y
∈
Y
{\displaystyle y\in Y}
에 대하여,
g
−
1
(
g
(
x
)
)
=
{
x
}
⊆
X
{\displaystyle g^{-1}(g(x))=\{x\}\subseteq X}
h
−
1
(
g
(
x
)
)
=
∅
⊆
Y
{\displaystyle h^{-1}(g(x))=\varnothing \subseteq Y}
g
−
1
(
h
(
y
)
)
=
f
−
1
(
y
)
⊆
Z
⊆
X
{\displaystyle g^{-1}(h(y))=f^{-1}(y)\subseteq Z\subseteq X}
h
−
1
(
h
(
y
)
)
=
{
y
}
⊆
Y
{\displaystyle h^{-1}(h(y))=\{y\}\subseteq Y}
이다. 따라서,
g
(
x
)
{\displaystyle g(x)}
와
h
(
y
)
{\displaystyle h(y)}
는 닫힌집합 이며,
X
∪
f
Y
{\displaystyle X\cup _{f}Y}
는 T1 공간 이다.
단조 정규성. 단조 정규성 연산자
G
1
:
S
X
→
T
X
{\displaystyle G_{1}\colon {\mathcal {S}}_{X}\to {\mathcal {T}}_{X}}
G
2
:
S
Y
→
S
Y
{\displaystyle G_{2}\colon {\mathcal {S}}_{Y}\to {\mathcal {S}}_{Y}}
가 주어졌다고 하자. 임의의 서로소 닫힌집합
E
,
F
⊆
X
∪
f
Y
{\displaystyle E,F\subseteq X\cup _{f}Y}
에 대하여,
(
g
−
1
(
E
)
,
g
−
1
(
F
)
)
∈
D
X
⊆
S
X
{\displaystyle (g^{-1}(E),g^{-1}(F))\in {\mathcal {D}}_{X}\subseteq {\mathcal {S}}_{X}}
(
h
−
1
(
E
)
,
h
−
1
(
F
)
)
∈
D
Y
⊆
S
Y
{\displaystyle (h^{-1}(E),h^{-1}(F))\in {\mathcal {D}}_{Y}\subseteq {\mathcal {S}}_{Y}}
이다. 이제,
A
(
E
,
F
)
=
g
−
1
(
E
)
∪
f
−
1
(
G
2
(
h
−
1
(
E
)
,
h
−
1
(
F
)
)
)
⊆
X
{\displaystyle A(E,F)=g^{-1}(E)\cup f^{-1}(G_{2}(h^{-1}(E),h^{-1}(F)))\subseteq X}
B
(
E
,
F
)
=
g
−
1
(
F
)
∪
Z
∖
f
−
1
(
cl
Y
G
2
(
h
−
1
(
E
)
,
h
−
1
(
F
)
)
)
⊆
X
{\displaystyle B(E,F)=g^{-1}(F)\cup Z\setminus f^{-1}(\operatorname {cl} _{Y}G_{2}(h^{-1}(E),h^{-1}(F)))\subseteq X}
라고 하자. 그렇다면,
(
A
,
B
)
∈
S
X
{\displaystyle (A,B)\in {\mathcal {S}}_{X}}
이다. 이는 다음과 같이 보일 수 있다.
cl
X
f
−
1
(
G
2
(
h
−
1
(
E
)
,
h
−
1
(
F
)
)
)
⊆
f
−
1
(
cl
Y
G
2
(
h
−
1
(
E
)
,
h
−
1
(
F
)
)
)
=
Z
∖
(
Z
∖
f
−
1
(
cl
Y
G
2
(
h
−
1
(
E
)
,
h
−
1
(
F
)
)
)
)
{\displaystyle {\begin{aligned}\operatorname {cl} _{X}f^{-1}(G_{2}(h^{-1}(E),h^{-1}(F)))&\subseteq f^{-1}(\operatorname {cl} _{Y}G_{2}(h^{-1}(E),h^{-1}(F)))\\&=Z\setminus (Z\setminus f^{-1}(\operatorname {cl} _{Y}G_{2}(h^{-1}(E),h^{-1}(F))))\end{aligned}}}
cl
X
(
Z
∖
f
−
1
(
cl
Y
G
2
(
h
−
1
(
E
)
,
h
−
1
(
F
)
)
)
)
=
cl
X
(
Z
∖
f
−
1
(
cl
Y
G
2
(
h
−
1
(
E
)
,
h
−
1
(
F
)
)
)
)
∩
Z
=
cl
Z
(
Z
∖
f
−
1
(
cl
Y
G
2
(
h
−
1
(
E
)
,
h
−
1
(
F
)
)
)
)
=
Z
∖
int
Z
f
−
1
(
cl
Y
G
2
(
h
−
1
(
E
)
,
h
−
1
(
F
)
)
)
⊆
X
∖
int
Z
f
−
1
(
G
2
(
h
−
1
(
E
)
,
h
−
1
(
F
)
)
)
=
X
∖
f
−
1
(
G
2
(
h
−
1
(
E
)
,
h
−
1
(
F
)
)
)
{\displaystyle {\begin{aligned}\operatorname {cl} _{X}(Z\setminus f^{-1}(\operatorname {cl} _{Y}G_{2}(h^{-1}(E),h^{-1}(F))))&=\operatorname {cl} _{X}(Z\setminus f^{-1}(\operatorname {cl} _{Y}G_{2}(h^{-1}(E),h^{-1}(F))))\cap Z\\&=\operatorname {cl} _{Z}(Z\setminus f^{-1}(\operatorname {cl} _{Y}G_{2}(h^{-1}(E),h^{-1}(F))))\\&=Z\setminus \operatorname {int} _{Z}f^{-1}(\operatorname {cl} _{Y}G_{2}(h^{-1}(E),h^{-1}(F)))\\&\subseteq X\setminus \operatorname {int} _{Z}f^{-1}(G_{2}(h^{-1}(E),h^{-1}(F)))\\&=X\setminus f^{-1}(G_{2}(h^{-1}(E),h^{-1}(F)))\end{aligned}}}
cl
X
g
−
1
(
E
)
⊆
g
−
1
(
cl
X
∪
f
Y
E
)
=
g
−
1
(
E
)
⊆
(
X
∖
Z
)
∪
(
g
−
1
(
E
)
∩
Z
)
=
(
X
∖
Z
)
∪
f
−
1
(
h
−
1
(
E
)
)
⊆
(
X
∖
Z
)
∪
f
−
1
(
G
2
(
h
−
1
(
E
)
,
h
−
1
(
F
)
)
)
⊆
(
X
∖
cl
X
(
Z
∖
f
−
1
(
cl
Y
G
2
(
h
−
1
(
E
)
,
h
−
1
(
F
)
)
)
)
)
∪
(
X
∖
cl
X
(
Z
∖
f
−
1
(
cl
Y
G
2
(
h
−
1
(
E
)
,
h
−
1
(
F
)
)
)
)
)
=
X
∖
cl
X
(
Z
∖
f
−
1
(
cl
Y
G
2
(
h
−
1
(
E
)
,
h
−
1
(
F
)
)
)
)
{\displaystyle {\begin{aligned}\operatorname {cl} _{X}g^{-1}(E)&\subseteq g^{-1}(\operatorname {cl} _{X\cup _{f}Y}E)\\&=g^{-1}(E)\\&\subseteq (X\setminus Z)\cup (g^{-1}(E)\cap Z)\\&=(X\setminus Z)\cup f^{-1}(h^{-1}(E))\\&\subseteq (X\setminus Z)\cup f^{-1}(G_{2}(h^{-1}(E),h^{-1}(F)))\\&\subseteq (X\setminus \operatorname {cl} _{X}(Z\setminus f^{-1}(\operatorname {cl} _{Y}G_{2}(h^{-1}(E),h^{-1}(F)))))\cup (X\setminus \operatorname {cl} _{X}(Z\setminus f^{-1}(\operatorname {cl} _{Y}G_{2}(h^{-1}(E),h^{-1}(F)))))\\&=X\setminus \operatorname {cl} _{X}(Z\setminus f^{-1}(\operatorname {cl} _{Y}G_{2}(h^{-1}(E),h^{-1}(F))))\end{aligned}}}
cl
X
g
−
1
(
F
)
⊆
g
−
1
(
cl
X
∪
f
Y
F
)
=
g
−
1
(
F
)
⊆
(
X
∖
Z
)
∪
(
g
−
1
(
F
)
∩
Z
)
=
(
X
∖
Z
)
∪
f
−
1
(
h
−
1
(
F
)
)
⊆
(
X
∖
Z
)
∪
f
−
1
(
Y
∖
cl
Y
G
2
(
h
−
1
(
E
)
,
h
−
1
(
F
)
)
)
⊆
(
X
∖
cl
X
f
−
1
(
G
2
(
h
−
1
(
E
)
,
h
−
1
(
F
)
)
)
)
∪
(
X
∖
cl
X
f
−
1
(
G
2
(
h
−
1
(
E
)
,
h
−
1
(
F
)
)
)
)
=
X
∖
cl
X
f
−
1
(
G
2
(
h
−
1
(
E
)
,
h
−
1
(
F
)
)
)
{\displaystyle {\begin{aligned}\operatorname {cl} _{X}g^{-1}(F)&\subseteq g^{-1}(\operatorname {cl} _{X\cup _{f}Y}F)\\&=g^{-1}(F)\\&\subseteq (X\setminus Z)\cup (g^{-1}(F)\cap Z)\\&=(X\setminus Z)\cup f^{-1}(h^{-1}(F))\\&\subseteq (X\setminus Z)\cup f^{-1}(Y\setminus \operatorname {cl} _{Y}G_{2}(h^{-1}(E),h^{-1}(F)))\\&\subseteq (X\setminus \operatorname {cl} _{X}f^{-1}(G_{2}(h^{-1}(E),h^{-1}(F))))\cup (X\setminus \operatorname {cl} _{X}f^{-1}(G_{2}(h^{-1}(E),h^{-1}(F))))\\&=X\setminus \operatorname {cl} _{X}f^{-1}(G_{2}(h^{-1}(E),h^{-1}(F)))\end{aligned}}}
이제,
f
−
1
(
G
2
(
h
−
1
(
E
)
,
h
−
1
(
F
)
)
)
=
V
(
E
,
F
)
∩
Z
{\displaystyle f^{-1}(G_{2}(h^{-1}(E),h^{-1}(F)))=V(E,F)\cap Z}
인 열린집합
V
(
E
,
F
)
⊆
X
{\displaystyle V(E,F)\subseteq X}
을 잡고,
U
(
E
,
F
)
=
G
1
(
A
(
E
,
F
)
,
B
(
E
,
F
)
)
∩
(
X
∖
Z
∪
V
(
E
,
F
)
)
⊆
X
{\displaystyle U(E,F)=G_{1}(A(E,F),B(E,F))\cap (X\setminus Z\cup V(E,F))\subseteq X}
G
(
E
,
F
)
=
g
(
U
(
E
,
F
)
)
∪
h
(
G
2
(
h
−
1
(
E
)
,
h
−
1
(
F
)
)
)
⊆
X
∪
f
Y
{\displaystyle G(E,F)=g(U(E,F))\cup h(G_{2}(h^{-1}(E),h^{-1}(F)))\subseteq X\cup _{f}Y}
라고 하자.
g
−
1
(
G
(
E
,
F
)
)
=
U
(
E
,
F
)
∪
f
−
1
(
G
2
(
h
−
1
(
E
)
,
h
−
1
(
F
)
)
)
=
U
(
E
,
F
)
∪
(
f
−
1
(
G
2
(
h
−
1
(
E
)
,
h
−
1
(
F
)
)
)
∩
G
1
(
A
(
E
,
F
)
,
B
(
E
,
F
)
)
)
∩
=
U
(
E
,
F
)
∪
(
V
(
E
,
F
)
∩
Z
∩
G
1
(
A
(
E
,
F
)
,
B
(
E
,
F
)
)
)
=
U
(
E
,
F
)
∪
(
U
(
E
,
F
)
∩
Z
)
=
U
(
E
,
F
)
⊆
X
{\displaystyle {\begin{aligned}g^{-1}(G(E,F))&=U(E,F)\cup f^{-1}(G_{2}(h^{-1}(E),h^{-1}(F)))\\&=U(E,F)\cup (f^{-1}(G_{2}(h^{-1}(E),h^{-1}(F)))\cap G_{1}(A(E,F),B(E,F)))\cap \\&=U(E,F)\cup (V(E,F)\cap Z\cap G_{1}(A(E,F),B(E,F)))\\&=U(E,F)\cup (U(E,F)\cap Z)\\&=U(E,F)\subseteq X\end{aligned}}}
h
−
1
(
G
(
E
,
F
)
)
=
h
−
1
(
g
(
U
(
E
,
F
)
∩
Z
)
)
∪
h
−
1
(
h
(
G
2
(
h
−
1
(
E
)
,
h
−
1
(
F
)
)
)
)
=
h
−
1
(
g
(
f
−
1
(
G
2
(
h
−
1
(
E
)
,
h
−
1
(
F
)
)
)
)
∪
G
2
(
h
−
1
(
E
)
,
h
−
1
(
F
)
)
=
G
2
(
h
−
1
(
E
)
,
h
−
1
(
F
)
)
⊆
Y
{\displaystyle {\begin{aligned}h^{-1}(G(E,F))&=h^{-1}(g(U(E,F)\cap Z))\cup h^{-1}(h(G_{2}(h^{-1}(E),h^{-1}(F))))\\&=h^{-1}(g(f^{-1}(G_{2}(h^{-1}(E),h^{-1}(F))))\cup G_{2}(h^{-1}(E),h^{-1}(F))\\&=G_{2}(h^{-1}(E),h^{-1}(F))\subseteq Y\end{aligned}}}
는 모두 열린집합 이므로,
G
(
E
,
F
)
⊆
X
∪
f
Y
{\displaystyle G(E,F)\subseteq X\cup _{f}Y}
는 열린집합 이다. 또한,
E
=
(
E
∩
g
(
X
∖
Z
)
)
∪
(
E
∩
h
(
Y
)
)
⊆
g
(
g
−
1
(
E
)
∖
Z
)
∪
h
(
h
−
1
(
E
)
)
⊆
g
(
A
(
E
,
F
)
∖
Z
)
∪
h
(
h
−
1
(
E
)
)
⊆
g
(
G
1
(
A
(
E
,
F
)
,
B
(
E
,
F
)
)
∖
Z
)
∪
h
(
G
2
(
h
−
1
(
E
)
,
h
−
1
(
F
)
)
)
⊆
g
(
U
(
E
,
F
)
)
∪
h
(
G
2
(
h
−
1
(
E
)
,
h
−
1
(
F
)
)
)
=
G
(
E
,
F
)
{\displaystyle {\begin{aligned}E&=(E\cap g(X\setminus Z))\cup (E\cap h(Y))\\&\subseteq g(g^{-1}(E)\setminus Z)\cup h(h^{-1}(E))\\&\subseteq g(A(E,F)\setminus Z)\cup h(h^{-1}(E))\\&\subseteq g(G_{1}(A(E,F),B(E,F))\setminus Z)\cup h(G_{2}(h^{-1}(E),h^{-1}(F)))\\&\subseteq g(U(E,F))\cup h(G_{2}(h^{-1}(E),h^{-1}(F)))\\&=G(E,F)\end{aligned}}}
F
=
(
F
∩
g
(
X
∖
Z
)
)
∪
(
F
∩
h
(
Y
)
)
⊆
g
(
g
−
1
(
F
)
∖
Z
)
∪
h
(
h
−
1
(
F
)
)
⊆
g
(
B
(
E
,
F
)
∖
Z
)
∪
h
(
h
−
1
(
F
)
)
⊆
g
(
X
∖
cl
X
G
1
(
A
(
E
,
F
)
,
B
(
E
,
F
)
)
∖
Z
)
∪
(
X
∪
f
Y
∖
h
(
cl
Y
G
2
(
h
−
1
(
E
)
,
h
−
1
(
F
)
)
)
)
⊆
X
∪
f
Y
∖
cl
X
∪
f
Y
G
(
E
,
F
)
{\displaystyle {\begin{aligned}F&=(F\cap g(X\setminus Z))\cup (F\cap h(Y))\\&\subseteq g(g^{-1}(F)\setminus Z)\cup h(h^{-1}(F))\\&\subseteq g(B(E,F)\setminus Z)\cup h(h^{-1}(F))\\&\subseteq g(X\setminus \operatorname {cl} _{X}G_{1}(A(E,F),B(E,F))\setminus Z)\cup (X\cup _{f}Y\setminus h(\operatorname {cl} _{Y}G_{2}(h^{-1}(E),h^{-1}(F))))\\&\subseteq X\cup _{f}Y\setminus \operatorname {cl} _{X\cup _{f}Y}G(E,F)\end{aligned}}}
이며 (
g
↾
X
∖
Z
{\displaystyle g\upharpoonright X\setminus Z}
가 열린 매장 이며
h
{\displaystyle h}
가 닫힌 매장 이라는 사실을 사용하였다), 만약
(
E
,
F
)
,
(
E
′
,
F
′
)
∈
D
X
∪
f
Y
{\displaystyle (E,F),(E',F')\in {\mathcal {D}}_{X\cup _{f}Y}}
E
⊆
E
′
{\displaystyle E\subseteq E'}
F
⊇
F
′
{\displaystyle F\supseteq F'}
라면
U
(
E
,
F
)
∖
Z
⊆
U
(
E
′
,
F
′
)
∖
Z
{\displaystyle U(E,F)\setminus Z\subseteq U(E',F')\setminus Z}
U
(
E
,
F
)
∩
Z
⊆
U
(
E
′
,
F
′
)
∩
Z
{\displaystyle U(E,F)\cap Z\subseteq U(E',F')\cap Z}
이므로
G
(
E
,
F
)
⊆
G
(
E
′
,
F
′
)
{\displaystyle G(E,F)\subseteq G(E',F')}
이다. 따라서,
G
:
D
X
∪
f
Y
→
T
X
∪
f
Y
{\displaystyle G\colon {\mathcal {D}}_{X\cup _{f}Y}\to {\mathcal {T}}_{X\cup _{f}Y}}
는 단조 정규성 연산자이다.