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Show a closed subset of a compact set is also compact

  • #1

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 Rk 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
 

Answers and Replies

  • #2
28
0
The Heini-Borel theorem only applies to euclidean spaces. Use the definition of compact.
 
  • #3
Dick
Science Advisor
Homework Helper
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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?
 
  • #4
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).
 
  • #5
Dick
Science Advisor
Homework Helper
26,258
618
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.
 

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