자세한 정의는 https://arca.live/b/math/20969485 참조.

연습 문제) S ⊆ P(X)=(X의 멱집합) 이라고 하자.  

그리고 S가 다음의 3가지 조건을 만족한다고 하자. 

1.  U A  (A∈S) =X     (S는 X의 cover이다.)

2. ∀A∈S, A is closed in (X, T).   (S는 closed set들의 집합이다.)

3. S is locally finite in (X, T).    (모든 X의 원소 x에 대해, x를 원소로 갖는 열린 집합 G_x가 존재하여, S의 원소들 중에 G_x와 서로소가 아닌 원소들은 유한 개이다.)


그리고 B⊆X라고 하고, ∀A∈S, (A, T_A)를 (X, T)의 subspace라고 하고, 

그러면, 다음의 2가지를 보여라. 

(B is open in (X, T) ⇔ ∀A∈S, B∩A is open in (A, T_A) )

(B is closed in (X, T) ⇔ ∀A∈S, B∩A is closed in (A, T_A)


여기서 


1) claim : B is open in (X, T) ⇒ ∀A∈S, B∩A is open in (A, T_A)

pf of 1) Let B be open in (X, T). Let A∈S. Note that T_A={G∩A | G∈T}. 

So, B∩A ∈T_A. So, B∩A is open in (A, T_A). 

Since A is arbitrary element of S, ∀A∈S, B∩A is open in (A, T_A).


2) claim : B is closed in (X, T) ⇒ ∀A∈S, B∩A is closed in (A, T_A) 

pf of 2)

Let B be closed in (X, T). Let A∈S. Note that T_A={G∩A | G∈T}. 

Then, X-B is open in (X, T). So, by 1), (X-B)∩A = (A-B) is open in (A, T_A). 

So, A-(A-B) = A∩B is closed in (A, T_A). 

Since A is arbitrary element of S, ∀A∈S, B∩A is closed in (A, T_A).



3) claim : if ∀D⊆X, (D is open in (X, T) ⇔ ∀A∈S, D∩A is open in (A, T_A)) , then (∀A∈S, B∩A is closed in (A, T_A)⇒  B is closed in (X, T)).

pf of 3) Let (∀D⊆X, (D is open in (X, T) ⇔ ∀A∈S, D∩A is open in (A, T_A))) be true.

And let ∀A∈S, B∩A be closed in (A, T_A). 

Then, ∀A∈S, A-(B∩A)=A-B=A∩(X-B) is open in (A, T_A). And X-B ⊆X.

So, by the hypothesis,  X-B is open in (X, T). Therefore, X-(X-B)=B is closed in (X, T).


4) claim :∀A∈S, B∩A is open in (A, T_A)⇒  B is open in (X, T)

 pf of 4) This is left as exercise