[Math] Closed subset of compact set is compact

analysiscompactnessgeneral-topologyreal-analysis

If S is a compact subset of R and T is a closed subset of S,then T is compact.

(a) Prove this using definition of compactness.

(b) Prove this using the Heine-Borel theorem.

My solution: firstly I should suppose a open cover of T, and I still need to think of the
set S-T. But if S-T is open in R,it can be done because the open cover of T and S-T is a open cover of R. The reality is S-T is not necessarily a open set in R. My question is that How we can find a open cover which covers S-T but misses T! I don't know how to do this thing!

in terms of part (b), I know it is bounded, but How to prove it is closed for T.

Best Answer

Here's how to get started on part a.

Start with an open cover of $T$. You need to show it has a finite subcover. If $U$ is in the open cover, then it's open in $T$, which means that there's an open set $U'$ in $S$ such that $U = T \cap U'$. For every $U$ in your cover, find a corresponding $U'$; now you almost have a cover of $S$. As you observed, if you add in $S - T$, you now have an open cover of $S$. What do you know about every open cover of $S$?

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