I came across the following problem on Cauchy Sequences:
Prove that every compact metric space is complete.
Suppose $X$ is a compact metric space. By definition, every sequence in $X$ has a convergent subsequence. We want to show that every Cauchy sequence in $X$ is convergent in $X$. Let $(x_n)$ be an arbitrary sequence in $X$ and $(x_{n_{k}})$ a subsequence that converges to $a$. Since $(x_{n_{k}}) \to a$ we have the following: $$(\forall \epsilon >0) \ \exists N \ni m,n \geq N \implies |x_{n_{m}}-x_{n_{n}}| < \epsilon$$
Using this, we can conclude that every Cauchy sequence in $X$ is convergent in $X$? Or do we inductively create subsequences and use Cauchy's criterion to show that it converges?
Best Answer
Here's a proof without $\epsilon$'s. Let $X$ be a compact metric space, $\hat{X}$ be its completion, and $i:X \rightarrow \hat{X}$ be the inclusion map. Then $i$ is continuous and hence $i(X)$ is compact. It follows $i(X)$ is closed in $\hat{X}$. But $i(X)$ is dense in $X.$ Therefore $i(X) = \hat{X}.$ We conclude $X$ is complete.