# 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 Tamaroff Jan 5 '14 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 Tamaroff Jan 5 '14 at 5:30

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.

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 $|x_{n_k}-x|<{\varepsilon\over 2}$. Let $N_2$ be such that $n,m\ge N_2$ implies $|x_n-x_m|<{\varepsilon \over 2}$.

Then $n>N=\max(N_1,N_2)$ implies $$|x_n-x|\le |x_n-x_N|+|x_N-x|<\varepsilon$$

Hence $X$ is complete.

• This is not yet correct, since you haven't yet shown that the original sequence converges. The result which you need to establish first is that if a Cauchy sequence has a convergent subsequence, then the original sequence converges. – Santiago Canez Apr 23 '14 at 13:00
• @SantiagoCanez I believe this is a misunderstanding due to my clumsy writing. I will rewrite and improve the answer. Thank you for your comments. – Student Apr 23 '14 at 13:39
• @SantiagoCanez I now included the proof of what I assumed and consider a triviality. – Student Apr 24 '14 at 10:10
• @SantiagoCanez Do you find the edited answer more palatable? – Student Apr 24 '14 at 15:46
• Yes, this looks great, thanks. – Santiago Canez Apr 24 '14 at 16:21

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.

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 – DanielWainfleet Jul 10 at 11:24

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 Tamaroff Jan 5 '14 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. – Student Apr 23 '14 at 12:10

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. – DanielWainfleet Jul 10 at 10:21

I do not have the sufficient score to comment to answers, so could I ask my question w.r.t. one of the answers here?

It is about @Student's answer which has received the most votes. I understand everything there except for such a thing: My understanding is that, in the last step, $$|x_N-x|<\epsilon/2$$ can only be true when $$x_N$$ is an element of the convergent subsequence $$\{x_{n_k}\}$$, but here we need $$x_N$$ to be in $$\{x_n\}$$, the general sequence. So I think in the original proof the $$|x_N-x|<\epsilon/2$$ is not warranted.

I am an amateur (engineering background, not mathematician). So my above argument must be wrong somewhere. Please help me in understanding! Thanks in advance!

• I think you are right that there is a mistake. It should say that if $n>N$ then there exists $k$ such that $n_k>N$. Then $|x_n-x_{n_k}|<\epsilon/2$ (because both $n$ and $n_k$ are greater than $N_2$) and $|x_{n_k}-x|<\epsilon$ (because $n_k>N_1$)... BTW. If a space X is metrizable (that is, if its topology can be generated by a metric) but X is not compact then its topology can be generated by an incomplete metric. This is very hard. – DanielWainfleet Jul 10 at 11:16
• Thank you very much for your kind help @DanielWainfleet ! – camio Jul 10 at 11:49

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 $a_{N}$.