$X$ is a metric space such that every closed and bounded subset of $X$ is compact . Then $X$ is complete.

Let $(x_{n})_{n}$ be a Cauchy sequence of $X$. Now every Cauchy sequence in a metric space is bounded. If $(x_{n})$ is convergent with limit, say, $x_{0}$, then $\{x_{n} : n \}\cup \{x_{0}\}$ is closed, otherwise the set $\{x_{n} : n \}$ is a closed set. If the sequence was convergent there was nothing to do. So when the sequence's convergent is not known, being closed and bounded, $\{x_{n} : n \}$ is a compact set. In metric spaces, compactness is equivalent to sequential compactness. Hence, the sequence $(x_{n})_n$ has a convergence subsequence $(x_{n_{k}})_k$ converging to, say, $x_{0}$. For being Cauchy, $(x_{n}$ also converges to $x_{0}$.

Thus we see that every Cauchy sequence in $X$ is convergent . Hence $X$ is complete.

Is my proof correct? For I have doubts regarding the step where I assumed that the range of the sequence, i.e. $\{x_{n} : n \}$ is closed. This would be true if I were in $\mathbb R$ but can this be assured for an arbitrary metric space? Or does that step require modification in some other respect?

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    $\begingroup$ Why the set $\{x_n\}$ is closed? $\endgroup$ – Quang Hoang Sep 22 '15 at 16:07
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    $\begingroup$ I'm pretty sure the sequence $x_n = 1/n$ does not form a closed subset of $\Bbb R$. $\endgroup$ – user98602 Sep 22 '15 at 16:08
  • $\begingroup$ @MikeMiller : I will edit my question . I had in mind if the sequence were not assumed to be convergent . But $\{{{1}\over {n}}\}\cup \{0\}$ is closed in $\mathbb R$ right $?$ $\endgroup$ – user118494 Sep 22 '15 at 16:13
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    $\begingroup$ Ah, that makes more sense. Important detail; with it, your proof is valid. To show this you need to exploit the fact that it's a Cauchy sequence to show that the only possible accumulation point would be the limit of the sequence, and then the fact that it has no limit can be used to show the sequence forms a closed subset. E: Dominik's answer is much clearer and less painful than mine. $\endgroup$ – user98602 Sep 22 '15 at 16:15
  • $\begingroup$ I'm struggling here. Why must the closure $\overline{\{x_n\}}$ be in X? As far as I can see $X = (0,1)$ and $x_n = 1/n$ provides a counterexample. Every closed subset of this is bounded and compact, yet X is not complete? $\endgroup$ – George Moore Nov 25 '15 at 9:28

The set $\{x_n\}$ is in general not closed, but its closure $\overline{\{x_n\}}$ is. You just need to prove that the closure of a bounded set is again bounded to correct your proof.


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