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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 $\mathbb R$.

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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
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3 Answers 3

up vote 5 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.

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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.

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I just edited the title of the question. Thanks for pointing that out –  user43901 Nov 5 '12 at 19:45
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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.

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More than slightly different, I’d say. –  Brian M. Scott Nov 5 '12 at 18:40
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