# Every compact metric space is complete, proof verification [duplicate]

This question already has an answer here:

I want to give a proof, that every compact metric space is complete.

Proof:

Let $(X,d)$ be a compact metric space. Let $(x_n)_{n\in\mathbb{N}}$ be a Cauchy sequence. The theorem of Bolzano-Weierstraß yields, that $(x_n)$ has a convergent subsequence $(x_{n_k})_{n_k\in\mathbb{N}}$.

Note $x:=\lim_{n_k\to\infty} x_{n_k}$.

I want to show, that $d(x_n, x)\to 0$.

Let $\epsilon >0$ be arbitrary.

Since $d(x_{n_k}, x)\to 0$ exists for $\epsilon/2 >0$ a $N'\in\mathbb{N}$ such that $d(x_{n_k}, x)<\epsilon/2$ for every $n_k\geq N'$.

Likewise since $(x_n)$ is a Cauchy sequence, we find for $\epsilon/2>0$ a $N''\in\mathbb{N}$ such that for every $n,m\geq N''$ holds, that $d(x_n,x_m)<\epsilon/2$.

Take $N:=\max\{N', N''\}$ and we conclude the proof:

$d(x_n, x)\leq d(x_n, x_{n_k})+d(x_{n_k},x)<\epsilon/2+\epsilon/2=\epsilon$.

Therefor $\lim_{n\to\infty} x_n\to x$ converges and $(X,d)$ is complete.

I would appreciate your thoughts on my proof. Thanks in advance.

## marked as duplicate by uniquesolution, Delta-u, Namaste, user133281, Parcly TaxelJul 20 '18 at 17:01

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• What is Bolzano Weierstrass theorem say in a metric space? – Kavi Rama Murthy Jul 20 '18 at 9:18
• @KaviRamaMurthy: Every sequence in a compact metric space has a convergent subsequence. – Mathematician 42 Jul 20 '18 at 9:19
• @Mathematician42 $B-W$ is a theorem applied on $\mathbb R^n$. Not a general metric space as far as I know. – Vera Jul 20 '18 at 9:19
• In my lecture notes it got called like that. It is the theorem Mathematican 42 stated. – Cornman Jul 20 '18 at 9:20
• It's simply called "sequentially compact" I guess, but you could call the equivalence of compactness and sequentially compactness in metric spaces a Bolzano-Weierstrass type theorem. – Mathematician 42 Jul 20 '18 at 9:20

## 1 Answer

You wrote:

$d(x_n, x)\leq d(x_n, x_{n_k})+d(x_{n_k},x)<\epsilon/2+\epsilon/2=\epsilon$.

You get a correct proof if you wrote: if $n\ge N$ and $n_k \ge N$ then:

$d(x_n, x)\leq d(x_n, x_{n_k})+d(x_{n_k},x)<\epsilon/2+\epsilon/2=\epsilon$.

• The starting point of the proof of OP is wrong. B-W theorem is only for Euclidean spaces. – Kavi Rama Murthy Jul 20 '18 at 9:20
• Since $X$ is compact, every sequence in $X$ has a convergent subsequence. – Fred Jul 20 '18 at 9:23
• But my choice of $N$ is correct, and everything that leads up to it? I just had to mention what you stated? – Cornman Jul 20 '18 at 9:24
• Yes, your choice of $N$ is correct. Just mention what I have stated. – Fred Jul 20 '18 at 9:26