I'd like to know if this demonstration is correct.

Let $X$ be a metric space and $K \subseteq X$. Show that if $K$ is compact, then $K$ is sequentially compact.

$K$ is compact, therefore every open cover has a finite subcover. Consider then a sequence $\{x_n\}_{n \in \mathbb{N}} \subset K$ and suppose (to find a contradiction) that it has no covergent subsequence, i.e., no element of $K$ is accumulation point for $\{x_n\}_{n \in \mathbb{N}}$.

This means that, for every $x \in K$ exists a $\varepsilon_x$ such that $B_{\varepsilon_x}(x)\cap \{x_n, n \in \mathbb{N}\}$ is finite, where $B_r(x)$ denotes the open ball with radius $r$ centered in $x$.

Note that every set $B_{\varepsilon_x}(x)$ is open and the union over all $x \in K$ obviously covers $K$.

Now, as $K$ is compact by hypothesis, there exists a finite set $K_0 \subset K$ such that $$K = \bigcup_{x \in K_0}B_{\varepsilon_x}(x).$$

Now, observe that $$\{x_n,n \in \mathbb{N}\} = \{x_n,n \in \mathbb{N}\}\cap K = \{x_n,n \in \mathbb{N}\}\cap \left[\bigcup_{x \in K_0}B_{\varepsilon_x}(x)\right]$$

$$=\bigcup_{x \in K_0}\left[\{x_n,n\in \mathbb{N}\}\cap B_{\varepsilon_x}(x)\right].$$

But this last set is finite, as it is a finite union of finite sets. This is absurd as $\{x_n, n\in \mathbb{N}\}$ is infinite, therefore $\{x_n\}_{n \in \mathbb{N}}$ must have an accumulation point.

This shows that $K$ compact implies $K$ sequentially compact.

  • 3
    $\begingroup$ It's fine. Just in case the range set of $\{x_n\}$ is finite, then there is still a convergent subsequence since at least one value is repeated infinitely often and thus the conclusion still holds. $\endgroup$ Commented Jul 9, 2020 at 19:49
  • $\begingroup$ @user710290: That is true, but it should be pointed out explicitly in the proof. $\endgroup$ Commented Jul 9, 2020 at 19:56

1 Answer 1


It’s basically correct, but I would tighten up one point. The fact that the sequence has no accumulation point actually means that for each $x\in K$ there is an $\epsilon_x>0$ such that $\{n\in\Bbb N:x_n\in B_{\epsilon_x}(x)\}$ is finite; this is strictly stronger than the statement that $B_{\epsilon_x}(x)\cap\{x_n:n\in\Bbb N\}$ is finite, since $\{x_n:n\in\Bbb N\}$ itself might be finite. However, the fact that $\{n\in\Bbb N:x_n\in B_{\epsilon_{x_k}}(x_k)\}$ is finite for each $k\in\Bbb N$ actually ensures that $\{x_n:n\in\Bbb N\}$ is infinite. You really should show this in your argument, however, since you need it at the end, when you get your contradiction by noting that $\{x_n:n\in\Bbb N\}$ is infinite.

  • $\begingroup$ Can you explain why exactly it follows that $\{x_n : n \in N\}$ is infinite? Do you mean that the $\cup_{x_k} \{ n \in ℕ : x_n \in 𝐵_{e_x}(x_k)\}$ would then be a finite set and thus would imply that $(x_n)$ is not an infinite sequence, which would lead to absurd? I'm confused because when you say "for each $k \in N$" you already imply that the sequence $(x_n)$ is infinite. $\endgroup$
    – Aelx
    Commented Dec 19, 2020 at 19:11
  • 1
    $\begingroup$ @СССР: Yes, that’s right: you get an immediate contradiction. $\endgroup$ Commented Dec 19, 2020 at 19:17

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