- #1

Buri

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If C subset of X is a compact and A subset of C is closed then A is compact.

Proof: Let U_alpha be an open cover of A.

A subset of X is closed implies that U_0 = X\A is open.

C is a subset of (U_0) U (U_alpha) and covers X.

In particular they cover C (possibly containing U_0).

Therefore, a finite subcover exists U_0, U_1,...,U_k of C and these also cover A.

Therefore, U_1,U_2,...U_k, cover A and hence, A is compact.

I can follow the proof, but don't really see the idea behind it all. I don't see why is it that you just can't get the finite sub cover from C and say it also covers A so its compact. I guess maybe an open cover of A might not be included in the collection of all open covers from C, so it won't be for ALL open covers there is a finite cover.

Furthermore, why do do we kick out the U_0 = X\A at the end? Is there a reason why we have to?

If someone could explain what I'm having problems with, or even go through the entire proof with explanations it would be even better.

Thanks!

EDIT:

Wait, is the reason why I have to get the X\A out is that I'm technically selecting ONE open cover when I add it in? And so I must show that X\A goes for sure, so I only keep the very general U_alpha?