[Math] If every closed and bounded subset of a metric space $M$ is compact, does it follow that $M$ is complete

real-analysissequences-and-series

I have this question:

If every closed and bounded subset of a metric space $M$ is compact,
does it follow that $M$ is complete?

I don't know how Cauchy sequences interact with compact sets. But by definition,

Since $U$ is a subset of $M$ and $U$ is closed and
bounded and compact, therefore every subsequence $(a_{n_k})$ on $M$ converges to some limit $a$ in
$U$.

Now I need to prove that every Cauchy sequence is convergent. But it seems impossible. What to do?

Best Answer

HINT:

  • Prove (if you’ve not already done so) that a Cauchy sequence in any metric space is bounded.
  • Prove that if a Cauchy sequence in any metric space has a convergent subsequence, then the Cauchy sequence itself converges.
  • Prove that if $S\subseteq M$ is bounded, so is $\operatorname{cl}S$.

Now if $M$ is not complete, it has a Cauchy sequence $\sigma=\langle x_n:n\in\Bbb N\rangle$ that does not converge. Let $S=\{x_n:n\in\Bbb N\}$; $S$ is a bounded subset of $M$, so $\operatorname{cl}S$ is a closed, bounded subset of $M$ and is therefore compact and hence sequentially compact. Therefore ... ?

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