Let $(X,d)$ be a compact metric space. Then $(X,d)$ is both complete and bounded.

Question

Definition: A metric space $(X,d)$ is said to be compact iff every sequence admits a convergent subsequence.

I am having trouble with the following result:

Let $(X,d)$ be a compact metric space. Then $(X,d)$ is both complete and bounded.

So far I have been able to prove that $(X,d)$ is complete as follows.

If $X$ is compact, then every sequence $x_{n}\in X$ admits a convergent subsequence.

In particular, if $x_{n}$ is Cauchy, then it admits a convergence subsequence, which makes it convergent as well.

What concerns me is the second part. How do we prove that it is bounded?

Could someone help me with this?

Answer 1

If it is not bounded there exist $x_n,y_n$ such that $d(x_n,y_n) \to \infty$. There exist subsequences $x_{n_k}$ and $y_{n_k}$ converging to some points $x$ and $y$. Can you show using triangle inequality that $d(x,y)=\lim d(x_{n_k},y_{n_k}$)? That would lead to the contradiction $d(x,y)=\infty$.

Answer 2

Let $x \in X$ and $B_1{x}$ an open ball around $x$ with radius $1$ with respect the metric $d$. Obviously, $X= \cup_{x \in X}B_1(x)$. As $X$ is compact, $X$ is covered by a finite union $B_1(x_1) \cup ... \cup B_1(x_n)$. Each $B_1(x_i)$ is bounded and a finite union of bounded sets is also bounded.