Subspace topology and Closed Sets

[Shaggydog4242]

Homework Statement

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.

Homework Equations

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.

 

[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 [itex]C \not\subseteq A[/itex] so the intersection of any K with A can never yield C.

Or am I missing something?

 

[Shaggydog4242]

Thanks for replying!

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.

 

[Fredrik]

Homework Statement

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.

Homework Equations

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.

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...

 

[Shaggydog4242]

Thanks again for the help!

Yeah, i meant x [itex]\in[/itex]X. I know that if C is open in A then automatically C = A[itex]\bigcap[/itex]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.

 

[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[itex]\bigcap[/itex]K compliment for some closed K in X. A[itex]\bigcap[/itex]K compliment implies C = A - (A-C) = A-K compliment = A[itex]\bigcap[/itex]K! unless I messed up somewhere...

 

[Fredrik]

C is closed in A implies that A-C is open in X.

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

Then by subspace topology A-C=A[itex]\bigcap[/itex]K compliment for some closed K in X. A[itex]\bigcap[/itex]K compliment implies C = A - (A-C) = A-K compliment = A[itex]\bigcap[/itex]K! unless I messed up somewhere...

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##.