Given a metric space $X$ which is sequentially compact (i.e every sequence has a converging subsequence), show that $X$ is complete and totally bounded.
I've already shown that $X$ is complete, since given $(x_n)$ a Cauchy sequence in $X$, there exist a converging subsequence $x_{n_{k}} \to x \in X$ which implies $x_n \to x$ (I have already proven this result previously). However when trying to prove that it is totally bounded, Im finding it quite difficult, since I don't know what strategy to take to write in some way, my space $X$ in terms of sequences to therefore use my hypothesis. Any hint?
Best Answer
Just to cover both results:
Assume that $(x_n)_{n\in\mathbb{N}}$ is a Cauchy sequence in a sequentially compact space $X$. Introduce a convergent subsequence $(x_{n_k})_{k\in\mathbb{N}}$ of $(x_n)$ with $x_{n_k}\to x.$ Let $\epsilon>0$ be given. Choose $N$ such that $\rho(x_i,x_j)<\epsilon/2,~i,j\geq N$. Choose $n_k>N$ such that $\rho(x_{n_k},x)<\epsilon/2$. Then we have \begin{equation*} \rho(x,x_N)\leq\rho(x,x_{n_k})+\rho(x_{n_k},x_N)<\epsilon/2+\epsilon/2=\epsilon \end{equation*} proving completeness.
Assume by contradiction that $X$ is not totally bounded. Take $\epsilon>0$ such that $X$ cannot be covered by a collection of finitely many $\epsilon -$balls. Choose $x_1\in X,~x_2\in X\setminus B_{\epsilon}(x_1),~x_3\in X\setminus B_{\epsilon}(x_1)\setminus B_{\epsilon}(x_2),...~$ .Then we have a sequence $(x_n)$ which cannot contain a convergent subsequence (since $\rho(x_i,x_j)\geq \epsilon~\forall i\neq j$). $~_{\square}$