# In a metric space, compactness implies completness

Proposition Let $(M,d)$ be a metric space. If $K\subset M$ is compact, then $K$ is complete.

Proof Let $\{x_n\}_{n=1}^\infty \subset K$ be a Cauchy sequence, then $$\forall \varepsilon > 0 \quad \exists n_0 \in \mathbb N \quad \forall m,n\geq n_0 \quad d(x_n,x_m) < \varepsilon.$$ $(\text{Step } 1)$ Since $K$ is compact, $\{x_n\}_{n=1}^\infty$ has an accumulation point $x^*\in K$ (already proved for every metric space). Let's see that $\lim_{n\to\infty} x_n = x^*$.

$(\text{Step } 2)$ Since $x^*$ is an accumulation point $$\forall k\in \mathbb N \quad \exists x_{n_k} \quad\text{s.t.}\quad d(x_{n_k}, x^*) <\frac{1}{k}$$

$(\text{Step } 3)$ Let $\varepsilon > 0$, then $\exists k_0 \quad n_{k_0} \geq n_0 \quad d(x_{n_{k_0}} x^*) < \varepsilon$. Now if $n \geq n_0$, then $$d(x_n,x*) \leq d(x_n, x_{n_{k_0}}) + d(x_{n_{k_0}}, x*) \leq \varepsilon + \varepsilon = 2\varepsilon$$ so every Cauchy sequence converges and $K$ is complete.

Question Is this proof correct? I can't understand steps $(2)$ and $(3)$ (why uses $\frac{1}{k}$?).

Edit Is this proof correct?

Let $\{x_n\}_{n=1}^\infty \subset K$ be a Cauchy sequence, by definition $\forall \varepsilon >0 \;\, \exists n_0\in \mathbb N \;\, \forall m,n \geq n_0 \;\, d(x_n, x_m) < \varepsilon$. Since $K$ is compact, the sequence has an accumulation point $x^*\in K$ (already proved). Let's see that $\lim_{n\to\infty}x_n = x^*$: since $x^*$ is an accumulation point, then exists a subsequence $\{x_{n_k}\}_{k=1}^{\infty}\subset \{x_{n}\}_{n=1}^{\infty}$ that converges to $x^*$, in other words $\forall \varepsilon > 0 \;\, \exists k_0\in\mathbb N \;\, \forall k \geq k_0 \;\, d(x_{n_{k}}, x^*) < \varepsilon$. Because of the triangular property $d(x_n,x^*) \leq d(x_n, x_{n_{k}}) + d(x_{n_{k}}, x^*) \leq 2\varepsilon$, so every Cauchy sequence converges and $K$ is complete.

• What's your definition of compactness for metric spaces? There are 4 "competing" ones out there: (1) complete and totally bounded, (2) every sequence has a convergent subsequence, (3) every infinite subset has a limit point, and (4) every open cover has a finite subcover. Dec 14 '12 at 22:13
• Here, definition 3 Dec 14 '12 at 22:14

It contains some errors, though the basic idea is right. Here’s how I would prove that $\lim\limits_{n\to\infty}x_n=x^*$.
Let $\epsilon>0$. Since the sequence is Cauchy, there is an $n_0\in\Bbb N$ such that $d(x_m,x_n)<\frac{\epsilon}2$ for all $m,n\ge n_0$. Since $x^*$ is an accumulation point of the sequence, it is also an accumulation point of the tail sequence $\langle x_n:n\ge n_0\rangle$. (Why?) Thus, there is an $n_1\ge n_0$ such that $d(x_{n_1},x^*)<\frac{\epsilon}2$. The triangle inequality then implies that $d(x_n,x^*)\le d(x_n,x_{n_1})+d(x_{n_1},x^*)<\frac{\epsilon}2+\frac{\epsilon}2=\epsilon$ for all $n\ge n_0$. Since $\epsilon>0$ was arbitrary, $\langle x_n:n\in\Bbb N\rangle\to x^*$.
In (Step 2) you are essentially constructing a subsequence of $\{ x_n \}_{n=1}^\infty$ which converges to the accumulation point $x^*$ of $\{ x_n : n \in \mathbb{N} \}$. I say essentially because you are not ensuring that $n_{k+1} > n_k$, but the basic idea is there. The use of $\frac 1k$ is not overly important, but you do want some sequence of positive real numbers converging to $0$.
• @user50554: I'm not quite certain if I understand your question. The main use of the $\frac 1k$ is in (Step 3) where you take some $k_0$ such that $\frac{1}{k_0} < \epsilon$ (this is implicit with $d ( x_{n_{k_0}} , x^* ) < \epsilon$). Replacing $\frac 1k$ everywhere with, say, $\frac{1}{2^k}$ would not really change the proof. Dec 14 '12 at 22:37