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$X$ is a compact metric space. Let $A \subseteq X$ be a compact set. Let $\{C_n\} \in A$ be a Cauchy sequence. Because it is Cauchy, $\{C_n\}$ must converge somewhere (although it doesn't need to converge in $A$). Let the limit point for $\{C_n\}$ be denoted by $C$. We also have that $A$ is compact $\implies A$ is closed and bounded $\implies$ $C \in A$ (because closed sets include the limit points for all its convergent sequences) $\implies$ A is complete.

lasoon
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  • You need to elaborate on why ${C_n}$ converges in $X$. Just being Cauchy is not sufficient. – Sahiba Arora Jun 03 '17 at 14:46
  • @SahibaArora I wasn't claiming it converged in $X$. I just wanted it to converge somewhere. And then use the fact that $A$ is closed, to say it must contain this limit point. Is this a wrong line of reasoning? Am I allowed to say it converges somewhere without specifying where? – lasoon Jun 03 '17 at 14:53
  • Converging somewhere implies converging in $X$ as $X$ is your metric space. A Cauchy sequence need not converge anywhere. – Sahiba Arora Jun 03 '17 at 14:55
  • @SahibaArora, I don't think I agree that convergence must imply convergence in the metric space. For instance, you could have a sequence in $\mathbb{Q}$ converging to an irrational number that is outside $\mathbb{Q}$. In a similar way, can't you suppose that our Cauchy sequence converges somewhere (possibly outside $X$)? – lasoon Jun 03 '17 at 15:09
  • Another possible proof: let $X'$ be a completion of $X$. Then $X$ is closed in $X'$ since $X$ is compact, and $X$ is dense in $X'$ by definition of completion. Therefore, $X=X'$. – Daniel Schepler Jun 03 '17 at 15:11
  • $A$ is closed in $X$. So for $C \in A$, you need to have convergence in $X$. Also, when you are proving something an arbitrary metric space, you can't go outside the space unless you explicitly specify the space, for example completion of that space as specified by Daniel Schepler. This is to ensure that the set is closed in that space. – Sahiba Arora Jun 03 '17 at 15:13
  • @SahibaArora, my logic was that $A$ is closed $\implies A$ contains all its limit points. So if there exists a Cauchy sequence in $A$ that converges somewhere, it converges in $A$ $\implies$ all Cauchy sequences in $A$ converge in $A$ since all Cauchy sequences converge somewhere. I still don't see the inconsistency between this argument and most of the points you have been making. For instance, you say for $C \in A$ you need convergence in $X$ because $A$ is closed in $X$. But how does this make the argument invalid? – lasoon Jun 03 '17 at 15:25
  • @SahibaArora, I wasn't aware that you couldn't go outside an arbitrarily defined metric space. But this is helpful. Thanks for putting me on the right track. I will think about this some more – lasoon Jun 03 '17 at 15:28
  • Every metric space admits a completion. If a sequence is Cauchy in $X$ it will continue to remain Cauchy in its completion where it converges. I merely asked you to specify this. When you just say "it converges somewhere" without specifying the space it is assumed you mean inside the metric space. – Sahiba Arora Jun 03 '17 at 15:31
  • @SahibaArora, that is very helpful. I wasn't aware of the concept of a completion, but specifying that the sequence converges in the completion seems preferable to saying it converges in an arbitrary "somewhere". Having agreedon that point, is there anything wrong with simply saying it converges somewhere and clarifying that this somewhere could be outside the space? I ask this partly because our class hasn't introduced completion so I am not sure I can use it, and partly because the proof I stated above still seems valid in my mind as long as it is clear that somewhere means in or out of set – lasoon Jun 03 '17 at 15:42
  • Talking about completion guarantees the convergence of that sequence in the completion. Without that concept you can't guarantee that the sequence converges "somewhere". It is intuitively correct. However, when you write down an argument it has to follow logically and not just intuitively. If you can't use completion then you can use an alternative argument. Sam gives a really helpful hint in the answers. – Sahiba Arora Jun 03 '17 at 15:46
  • Thank you. This is very helpful. I understand this much better now – lasoon Jun 03 '17 at 16:02

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Sorry, but if you say that the Cauchy sequence must converge somewhere, you're already using completeness, which you are to prove.

What you know, instead, is that the sequence has a cluster point, because of compactness. But a Cauchy sequence with a cluster point is convergent.

Note that, in general metric spaces, closed and bounded sets need not be compact (the converse is true, but of little use, generally). However you used the converse implication (compact implies closed).

egreg
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