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

Is there a way to prove this using sequential compactness instead of open cover definitions?

My first gut reaction was that the fact was obvious since we can show that the closed subset $[a,b]$ is compact in $\mathbb R$.

share|cite|improve this question
Your guess is correct. It is true. – Mhenni Benghorbal Nov 5 '12 at 18:31
I would echo Sanchez' remarks below. – copper.hat Nov 5 '12 at 18:37
up vote 6 down vote accepted

Although they coincide in metric spaces, in general sequential compactness and compactness are very different properties: neither implies the other, so you can’t expect to use sequential compactness to prove something about compactness. Thus, you’re really talking about two different theorems.

Theorem 1. If $\langle X,\tau\rangle$ is a compact space, and $K$ is a closed subset of $X$, then $K$ is compact.

The proof using open covers is trivial.

Theorem 2. If $\langle X,\tau\rangle$ is a sequentially compact space, and $K$ is a closed subset of $X$, then $K$ is sequentially compact.

Proof. Let $\sigma=\langle x_k:k\in\Bbb N\rangle$ be a sequence in $K$. Since $\sigma$ is also a sequence in $X$, it has a subsequence $\langle x_{n_k}:k\in\Bbb N\rangle$ that converges to some $x\in X$. If $U$ is an open neighborhood of $x$, there is a $k\in\Bbb N$ such that $x_{n_k}\in U$, so $U\cap K\ne\varnothing$. Thus, $x\in\operatorname{cl}K=K$, and the subsequence $\langle x_{n_k}:k\in\Bbb N\rangle$ is convergent in $K$ as well as in $X$. $\dashv$

As you can see, it’s not hard to prove that sequential compactness is inherited by closed sets, but the proof is at least a little more involved than the trivial proof for compactness.

share|cite|improve this answer

Since we're dealing with subsets of a metric space, sequential compactness and compactness are the same thing, so we can instead show that a closed subset of a sequentially compact set is also sequentially compact. To see this, suppose $X$ is sequentially compact and $Y$ a closed subset of $X$. Take any sequence $\{x_n\}$ of points of $Y$. This is a sequence of points of $X$, so what can we say about it based on the sequential compactness of $X$? Since $Y$ is closed (contains all its limit points), what can we then conclude?

In general, "sequentially compact" and "compact" are not the same thing. (Neither of them even need be a consequence of the other!) In such settings, you're going to need to show that open covers have finite subcovers, or show that non-empty collections of closed sets with the finite intersection property have non-empty intersections, or something like that.

share|cite|improve this answer
I just edited the title of the question. Thanks for pointing that out – user43901 Nov 5 '12 at 19:45

Yes. Closed subset of (sequentially) compact set is (sequentially) compact. However, sequential compactness is a slightly different thing from compactness, so I don't see how you can evade open covers.

share|cite|improve this answer
More than slightly different, I’d say. – Brian M. Scott Nov 5 '12 at 18:40

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.