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

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 2

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

Answer 3

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 $a_{N}$.

Answer 4

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.