Every compact metric space is complete

I need to prove that every compact metric space is complete. I think I need to use the following two facts:

1. A set $K$ is compact if and only if every collection $\mathcal{F}$ of closed subsets with finite intersection property has $\bigcap\{F:F\in\mathcal{F}\}\neq\emptyset$.
2. A metric space $(X,d)$ is complete if and only if for any sequence $\{F_n\}$ of non-empty closed sets with $F_1\supset F_2\supset\cdots$ and $\text{diam}~F_n\rightarrow0$, $\bigcap_{n=1}^{\infty}F_n$ contains a single point.

I do not know how to arrive at my result that every compact metric space is complete. Any help?

• Closed subsets of compact spaces are compact. If $F=\{K_\alpha\}$ is a centered family of compact sets, $\bigcap F\neq \varnothing$. If, moreover, ${\rm diam}K_n\to 0$, it is easily seen $\bigcap F$ must consist of only one point. Nested sets are centered. Hence the result.
– Pedro
Commented Jan 5, 2014 at 5:26
• As a side note, a metric space is compact iff it is complete and totally bounded. You have a part of the implication, so maybe you're interested in finishing things off. =)
– Pedro
Commented Jan 5, 2014 at 5:30

If you do not wish to use the Heine-Borel theorem for metric spaces (as suggested in the answer by Igor Rivin) then here is another way of proving that a compact metric space is complete:

Note that in metric spaces the notions of compactness and sequential compactness coincide. Let $$x_n$$ be a Cauchy sequence in the metric space $$X$$. Since $$X$$ is sequentially compact there is a convergent subsequence $$x_{n_k}\to x \in X$$.

All that now remains to be shown is that $$x_n \to x$$. Since $$x_{n_k}\to x$$ there is $$N_1$$ with $$n_k \ge N_1$$ implies $$d(x_{n_k},x)<{\varepsilon\over 2}$$. Let $$N_2$$ be such that $$n,m\ge N_2$$ implies $$d(x_n,x_m)<{\varepsilon \over 2}$$.

Let $$n\geq N=\max(N_1,N_2)$$ and pick some $$n_k\geq N$$. Then $$d(x_n,x)\le d(x_n,x_{n_k})+d(x_{n_k},x)<\varepsilon.$$

Hence $$X$$ is complete.

Let $$\langle F_n\rangle_{n\in\Bbb{N}}$$ be a descending sequence of nonempty closed sets satisfy that $$\operatorname{diam} F_n\to 0$$ as $$n\to\infty$$. You can easily check that if $$m_1 then $$F_{m_1}\cap F_{m_2}\cap\cdots\cap F_{m_k} =F_{m_k}\neq \varnothing$$ so $$\langle F_n\rangle_{n\in\Bbb{N}}$$ satisfies finite intersection property. Since $$(X,d)$$ is a compact metric space, $$\bigcap_{n\in\Bbb{N}} F_n$$ is not empty. Since $$\operatorname{diam} \bigcap_{n\in\Bbb{N}} F_n \le \operatorname{diam} F_m\to 0\qquad \text{as }\> m\to\infty$$ so $$\bigcap_{n\in\Bbb{N}} F_n$$ contains at most one point. So $$\bigcap_{n\in\Bbb{N}} F_n$$ is singleton.

Let $X$ be a compact metric space and let $\{p_n\}$ be a Cauchy sequence in $X$. Then define $E_N$ as $\{p_N, p_{N+1}, p_{N+2}, \ldots\}$. Let $\overline{E_N}$ be the closure of $E_N$. Since it is a closed subset of compact metric space, it is compact as well.

By definition of Cauchy sequence, we have $\lim_{N\to\infty} \text{diam } E_N = \lim_{N\to\infty} \text{diam } \overline{E_N} = 0$. Let $E = \cap_{n=1}^\infty \overline{E_n}$. Because $E_N \supset E_{N+1}$ and $\overline{E_N} \supset \overline{E_{N+1}}$ for all $N$, we have that $E$ is not empty. $E$ cannot have more than $1$ point because otherwise, $\lim_{N\to\infty} \text{diam } \overline{E_N} > 0$, which is a contradiction. Therefore $E$ contains exactly one point $p \in \overline{E_N}$ for all $N$. Therefore $p \in X$.

For all $\epsilon > 0$, there exists an $N$ such that $\text{diam } \overline{E_n} < \epsilon$ for all $n > N$. Thus, $d(p,q) < \epsilon$ for all $q \in \overline{E_n}$. Since $E_n \subset \overline{E_n}$, we have $d(p,q) < \epsilon$ for all $q \in E_n$. Therefore, $\{p_n\}$ converges to $p \in X$.

• This A directly uses the material in the Q.... +1 for that Commented Jul 10, 2019 at 11:24

Here's a not-so-fancy way:

Let $\{a_n\}$ be a Cauchy sequence.

If the set of values in the (image of the) sequence is finite, then use the Cauchy criterion to show that the sequence is eventually constant (and so converges).

If the set of values of the sequence is infinite, then use compactness to finite a limit point of this set. Use this limit point to construct a convergent subsequence of the original sequence. Then use the Cauchy criterion to show the original sequence converges to the same limit as the subsequence.

Here is a simple proof:

Note that a compact metric space is sequentially compact. So that every sequence, including Cauchy sequences, have convergent subsequences. Since all Cauchy sequences have convergent subsequences, we must find that all Cauchy sequences converge. Meaning that the metric space is complete.

This is true because a Cauchy sequence with a convergent subsequence is convergent.

For the separable case, here's a cute proof.

If $$(X,d)$$ is non-empty compact and separable, then Banach's embedding theorem implies that there exists an isometric embedding $$\phi$$ of $$(X,d)$$ into $$(C[0,1],d_{\infty})$$ where $$d_{\infty}$$ is the sup norm.

Since any isometric embedding is a homeomorphism onto its image, then $$\phi(X)$$ is compact in $$(C[0,1],d_{\infty})$$. Since the latter is complete and any non-empty compact subset of a complete metric space is complete then $$(\phi(X),d_{\infty})$$ is complete. Since $$\phi$$ is an isometry, then it takes convergent (resp. Cauchy) sequences to convergent (resp. Cauchy) sequences. The result follows by noting that since $$\phi^{-1}$$ is also an isometry from $$(\phi(X),d_{\infty})\rightarrow (X,d)$$.

• All compact metric spaces are separable. Commented Jul 10, 2019 at 10:21

This follows from Heine-Borel (see the wiki page for the relevant proofs).

• How does this follow from Heine-Borel's theorem? Were you trying to answer another question?
– Pedro
Commented Jan 5, 2014 at 5:29
• @PedroTamaroff The Heine-Borel for metric spaces says that a set is compact iff it is complete and totally bounded. So there is nothing to prove as the desired claim is an immediate consequence. Commented Apr 23, 2014 at 12:10
• But the OP is asking for a general metric space. If the space is not Heine-Borel, I don't see why this follows from Heine-Borel.
– Bach
Commented Dec 9, 2020 at 18:25
• Isn't Heine-Borel only applicable to subspaces R^n? Commented Jun 22, 2022 at 11:06

Hint: If you have a sequence wherein $d(a_{N},a_{k})<r$ for all $k\geq N$, then the limit of that sequence, if exist, must be in the closure of the open ball radius $r$ around $a_{N}$.