# In a metric space, compact implies sequentially compact

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.

• 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. Commented Jul 9, 2020 at 19:49
• @user710290: That is true, but it should be pointed out explicitly in the proof. Commented Jul 9, 2020 at 19:56

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.
• 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.