# Homework Help: Subspace topology and Closed Sets

1. Apr 3, 2013

### Shaggydog4242

1. The problem statement, all variables and given/known data

Hi,

This is my first post. I had a question regarding open/closed sets and subspace topology.

Let A be a subset of a topological space X and give A the subspace topology. Prove that if a set C is closed then C= A intersect K for some closed subset K of X.

2. Relevant equations

3. The attempt at a solution

I know that if C is closed then its compliment is open. I started by letting V be an open set in X and then (A/C) = A intersect V . From there I think it’s fair to assume X is an element of C which implies that x is an element of A intersect K.

I just can’t seem to prove that C = A intersect K from the information given.

I also can show that x is an element of A intersect K then X is an element of C but I don't know if that would help.

2. Apr 3, 2013

### CompuChip

I'm not sure I understand this question.

Is C supposed to be a subset of A?
If it is, then the statement is trivial - take K = C.
If not, then the statement makes no sense - take X = R, A = [-1, 1], C = [0, 2]. Clearly C is closed in X (its complement is open), but $C \not\subseteq A$ so the intersection of any K with A can never yield C.

Or am I missing something?

3. Apr 3, 2013

### Shaggydog4242

In this case C would be in A. I'm not sure why I would be allowed to set K=C if K can be any closed subset in X while C has to be closed in A.

4. Apr 4, 2013

### Fredrik

Staff Emeritus
You're supposed to show that if $C\subset A$ is closed with respect to the subspace topology, then there exists a closed $K\subset X$ (closed with respect to the topology of X) such that $C=A\cap K$. So one thing that really stands out when I read your ideas about how to approach the problem is that you're not saying anything about how to define K. Another thing that looks really strange is the statement that X is an element of C. Maybe you meant to type x, but you haven't said what x is.

This looks like a good way to start: By assumption C is closed with respect to the subspace topology. This implies that A-C is open with respect to the subspace topology. By definition of "subspace topology", this implies that...

5. Apr 4, 2013

### Shaggydog4242

Thanks again for the help!

Yeah, i meant x $\in$X. I know that if C is open in A then automatically C = A$\bigcap$K for some closed set K of X but I don't see how I can go from assuming A-C is open in A to C being open in A.

6. Apr 4, 2013

### Shaggydog4242

I think I got it right after I posted that.

C is closed in A implies that A-C is open in X. Then by subspace topology A-C=A$\bigcap$K compliment for some closed K in X. A$\bigcap$K compliment implies C = A - (A-C) = A-K compliment = A$\bigcap$K! unless I messed up somewhere...

7. Apr 5, 2013

### Fredrik

Staff Emeritus
Don't you mean open in A? (Open with respect to the topology of A).

Yes, this works. You may want to include some more details though, just to make your proof easier to follow. I would say that since A-C is open with respect to the topology of A, there's an open set E such that $A-C=A\cap E$. Then I would define $K=E^c$, and finally prove that $C=A\cap K$.