Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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 $S$?

share|cite|improve this answer
How do you know that $S \setminus T$ is open to be able to add it in the open cover for $S$? – Sergey Zykov Mar 21 '15 at 18:06

$T$ is a closed subset of $S$ if and only if $T=C\cap S$ for some $C$ closed in $\mathbb{R}$. But $S$ is closed too, being compact, so $T$ is closed in $\mathbb{R}$ because it is the intersection of two closed sets. This takes care of the remaining part of $(b)$. For $(a)$, $\mathbb{R}\setminus T$ is an open set containing $S$.

share|cite|improve this answer

I will address part (b), since the others address (a) well.

The Heine-Borel theorem says that a subset $V \subset \mathbb{R}$ is compact if and only if it is both closed and bounded.

So suppose that $T \subset S \subset \mathbb{R}$. Since $S$ is compact, it is closed and bounded. What can you now say about the boundedness and closedness of $T$?

share|cite|improve this answer
T is bounded, why it is closed – python3 Mar 26 '14 at 16:36
Well, we know that $S$ is closed, and $T \subset S$ is a closed subset of $S$. So if you can show that a closed subset of a closed subset is closed in $\mathbb{R}$... – Simon Rose Mar 26 '14 at 17:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.