I need to prove that every compact metric space is complete. I think I need to use the following two facts:
- A set $K$ is compact if and only if every collection $\mathcal{F}$ of closed subsets with finite intersection property has $\bigcap\{F:F\in\mathcal{F}\}\neq\emptyset$.
- 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.
Best Answer
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.