Show a closed subset of a compact set is also compact

[filter54321]

Homework Statement

Show that if E is a closed subset of a compact set F, then E is also compact.

Homework Equations

I'm pretty sure you refer back to the Heine-Borel theorem to do this.

"A subset of E of R^k is compact iff it is closed and bounded"

The Attempt at a Solution

We are deal with metric spaces here. It should seem that I need to prove the same thing as in the second half of the Heine-Borel theorem. My textbook is proving Heine-Borel in a confusing way without clear statement/reason steps that I can apply to my problem.

-From the given information, E is in F
-From the given information, every open cover of F has a finite subcover of F
-From the given information, S \ E is open

...not sure where this leave me

 

[miqbal]

The Heini-Borel theorem only applies to euclidean spaces. Use the definition of compact.

 

[Dick]

You don't need Heine-Borel for this. Take an open cover of E. If you add the open set S/E to that (S is the whole space, right?), then you have an open cover of F. Since F is compact... Can you continue?

 

[filter54321]

I'm with you so far...you declare O and open cover of E. Then you take the union of O and S \ E and get an open set. This unified open set has to cover all E and all not-E so it has to be an open cover for all of F.

So how does the compactness of F flow back down to E? (Clearly, I have no idea what I'm doing).

 

[Dick]

Since F is compact, there is a finite subcover. So that subcover also covers E. Now you don't need the S\E set to cover E. What's left is a finite subcover of the original cover that covers E.