General Topology – Is a Closed Subset of a Compact Set Compact?

Question

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 $\mathbb R$.

Answer 1

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.