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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?

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4 Answers 4

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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 $\rho(x_i,x_j)\geq \epsilon~\forall i\neq j$). $~_{\square}$

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  • $\begingroup$ I fail to understand the last line. Why need it be that $\forall i \neq j, \rho(x_i, x_j) \geq \epsilon$? I manage to understand that $\forall i \in \mathbb N , \rho(x_i, x_{i+1}) \geq \epsilon $, but how does one further generalize from this? $\endgroup$
    – Melanzio
    Jul 8, 2021 at 8:30
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Let $\epsilon>0$. If $X$ is finite the statement is obvious. If not, take some $x_1\in X$, then take $x_2\in X\setminus B(x_1,\epsilon)$,..., then take $x_{n+1}\in X \setminus\bigcup\limits_{k=1}^n B(x_k,\epsilon)$... (if you eventually run out of points you have proven the statement).

Now use the fact that $\{x_n\}$ contains a subsequence that converges.

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Suppose it is not bounded. Then you can, asuming $X$ is not empty (nor finite) so $x_0 \in X$, you construct this sequence : $x_1 \in B(x_0,1)$, $x_2 \in B(x_0,2)-B(x_0,1)$,... so on. Has this sequence a convergent subsequence?

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Prove first that $X$ sequentially compact implies $X$ compact. With this result, take $\epsilon > 0$ and consider the open cover $\{ B(x,\epsilon) \}_{x \in X}$. By compactness you get: $$X \subset B(x_1, \epsilon)\cup \cdots \cup B(x_n, \epsilon),$$ and the quantity $n$ depends on how small $\epsilon$ is.

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  • $\begingroup$ There is no way to prove it directly? I've already consider that posibility but I have to show the equivalence of 5 statements and I usually do them in order. As an exercise. $\endgroup$ May 23, 2015 at 19:39
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    $\begingroup$ Directly? What comes quickly to mind is something using contradiction. Suppose that $X$ is not totally bounded and try to construct a sequence that doesn't have any convergent subsequence. I usually find these constructions hard.. $\endgroup$
    – Ivo Terek
    May 23, 2015 at 19:41
  • $\begingroup$ Proving that sequential compactness implies compactness is a bit harder than proving the OP's statement (and in fact, many proofs that sequential compactness implies compactness use the OP's statement). $\endgroup$ May 23, 2015 at 19:49

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