# Compactness and sequential compactness in metric spaces

I've got a question: I'm trying to prove that every metric space is compact if and only if the space is sequentially compact. In all the proofs I have found, they used the Bolzano-Weierstrass theorem. Is there a way to prove this fact without using Bolzano-Weierstrass?

Thanks so much!!

• The Bolzano-Weierstraß theorem I know deals only with subsets of $\mathbb{R}^n$. In what form is the Bolzano-Weierstraß theorem used in the proofs you know? Commented Jun 5, 2014 at 21:41
• I need Bolzano-Weierstrass to show that the space is totally bounded. Commented Jun 5, 2014 at 22:05
• As said, the standard Bolzano-Weierstraß theorem is about subsets of $\mathbb{R}^n$. What's the formulation of the version of BW that's used? Commented Jun 5, 2014 at 22:09
• This is the Bolzano-Weierstrass version I have found: Every infinite subset $A$ of a metric space has an accumulation point. Commented Jun 5, 2014 at 22:11
• Is there a "compact" missing? Or something similar? Commented Jun 5, 2014 at 22:14

Theorem Let $$(M,d)$$ be a metric space. The following are equivalent:
(a) $$M$$ is compact;
(b) $$M$$ is sequentially compact;
(c) $$M$$ is complete and totally bounded.

Proof: (a$$\Rightarrow$$b) Suppose $$M$$ is compact, and let $$(x_n)_{n\in\mathbb{N}}$$ be a sequence in $$M$$. Suppose that the sequence $$(x_n)$$ did not have a convergent subsequence, that is, $$(x_n)$$ does not have a cluster point. Then for every $$x\in M$$, there exists some neighbourhood $$U_x$$ of $$x$$ such that $$\left\{n:x_n\in U_x\right\}$$ is finite. Then $$\left\{U_x:x\in M\right\}$$ is an open cover of $$M$$, so by compactness we can find a finite subcover $$U_1,\ldots,U_{k}$$. But notice that $$\mathbb{N}=\left\{n:x_n\in M\right\}=\cup_{i=1}^k\left\{n:x_n\in U_i\right\}$$, and the latter set is finite, a contradiction.

Therefore, $$(x_n)$$ has a cluster point, hence a convergent subsequence.

(b$$\Rightarrow$$c) Suppose $$M$$ is sequentially compact. Since every Cauchy sequence has a convergent subsequence, it follows that $$M$$ is complete. Also, if $$M$$ was not totally bounded, there would exist some $$\varepsilon>0$$ such that no finite collection of open balls of radius $$\varepsilon$$ covers $$M$$. Let $$B_1=B(x_1,\varepsilon)$$ be one such ball. Then, let $$x_2\in M\setminus B_1$$, and let $$B_2=B(x_2,\varepsilon)$$. Then, let $$x_3\in M\setminus (B_1\cup B_2)$$. Continuing this way, we find a sequence $$x_1,x_2,\ldots$$ such that $$x_{n+1}\in M\setminus\cup_{i=1}^nB(x_i,\varepsilon)$$, hence $$d(x_n,x_m)\geq\varepsilon$$ whenever $$n\neq m$$, so $$(x_n)$$ cannot have a convergent subsequence, a contradiction.

Thus, $$M$$ is also totally bounded.

(c$$\Rightarrow$$a) Suppose that $$M$$ is complete and totally bounded. In order to find a contradiction, suppose that $$M$$ was not compact, so there exists some open cover $$\left\{U_i:i\in I\right\}$$ which does not admit a finite subcover.

We can cover $$M$$ by a finite number of sets $$C^1_1,\ldots,C^1_{p_1}$$ of diameter $$\leq 1$$ (since $$M$$ is totally bounded). One of these sets, say $$C^1=C^1_{k_1}$$, cannot be covered by a finite number of sets $$U_i$$ (if all could, we would find a finite subcover for $$M$$). Now, $$C^1$$ can be covered by a finite number of subsets $$C^2_1,\ldots,C^2_{p_2}$$ of diameter $$\leq 1/2$$. Again, one of the sets, say $$C^2=C^2_{k_2}$$, cannot be covered by a finite number of sets $$U_i$$.

Proceeding this way, we find (non-empty) sets $$C^1\supseteq C^2\supseteq C^3\supseteq\cdots$$ such that $$C^k$$ has diameter $$\leq 1/k$$ and $$C^k$$ cannot be covered by a finite number of sets $$U_i$$. Let $$x_k\in C^k$$ be an arbitrary element for every $$k$$. Then $$(x_k)_k$$ is Cauchy (by the condition on the diameters), so it converges to some $$x\in M$$. This $$x$$ belongs to some $$U_i$$, as $$\left\{U_i:i\in I\right\}$$ covers $$M$$, so there exists some $$\delta$$ such that $$B(x,\delta)\subseteq U_i$$. Letting $$N$$ be sufficiently large, so that $$d(x,x_N)<\delta/2$$ and $$1/N<\delta/2$$, we obtain $$C^N\subseteq B(x_N,1/N)\subseteq B(x_N,\delta/2)\subseteq B(x,\delta)\subseteq U_i$$, contradicting the construction of $$C^N$$.

Therefore, $$M$$ is compact.

• One useful way of stating (a) $\iff$ (b) is that a metric space is compact iff it does not have an infinite closed discrete subspace. Commented Jun 13, 2016 at 9:59
• Here is a proof that doesn't go through (c). Commented Jun 10, 2022 at 6:32