Every compact metric space is complete

Question

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?

Thanks in advance.

Answer 1

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.

Answer 2

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 $m_1<m_2<\cdots<m_k$ 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.

Answer 3

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.

Answer 4

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 $p \in X$.

Answer 5

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

Answer 6

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

Answer 7

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.

Answer 8

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 $(\phi(X),d_{\infty})\rightarrow (X,d)$.

Answer 9

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!

Answer 10

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 $a_{N}$.