I posted this answer here which a user pointed out to me is not correct. The question is asking for a proof that a compact metric space is complete.

My answer:

Note that in metric spaces the notions of compactness and sequential compactness coincide. Let $x_n$ be a Cauchy sequence in the metric space $X$. Since $X$ is sequentially compact there is a convergent subsequence $x_{n_k}\to x \in X$. But every subsequence of a convergent sequence converges to the same limit as the original sequence hence $x_n \to x \in X$. Hence $X$ is complete.

I believe this to be correct. By sequential compactness I obtain a convergent subsequence and since every subsequence of a Cauchy sequence converges to the same limit as the original sequence this is an argument that the original sequence converges.

Since I could not understand the comments by the other user I would like to kindly request the assistance of this commnuity to point out my error in different words.

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    $\begingroup$ Your answer is correct. That dude is just picking on you to try to make you prove that if $x_{n_k}\to x$ and $x_n$ is Cauchy, then $x_n\to x$. Which is a triviality $|x_n-x|\leq |x_n-x_{n_k}|+|x_{n_k}-x|$... $\endgroup$ – user144349 Apr 23 '14 at 13:26
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    $\begingroup$ I think the answer is incorrect: the point is in the almost final sentence "But every subsequence of a convergent sequence converges to the same limit as the original sequence hence..." This, of course, is true... if we already know the original sequence converges, which we do not know as this is precisely what we want to prove. $\endgroup$ – DonAntonio Apr 23 '14 at 14:18
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    $\begingroup$ I disagree with user144349:I see no picking on anyone here. What he proposes is precisely what you, @Student, should, or perhaps even meant, to do: to show that if we have a Cauchy sequence s.t. some infinite subsequence converges, then the whole sequence converges and to the same limit as well. $\endgroup$ – DonAntonio Apr 23 '14 at 14:21
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    $\begingroup$ @user144349, that is not a triviality! Indeed, it is the entire point of the problem, and is not what I think Student was basing his/her answer on. See my previous comment here. $\endgroup$ – Santiago Canez Apr 23 '14 at 16:41
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    $\begingroup$ Your new answer is fine, but note that if a student in one of my real analysis courses stated that the fact you used was a "triviality", they would very likely receive no credit unless they justified that "triviality" ;) $\endgroup$ – Santiago Canez Apr 24 '14 at 16:22

You're confused. The fact for Cauchy sequences is: if $(x_n)$ is a Cauchy sequence (no assumptions on anything, like convergence!) and there exists some subsequence $(x_{n_k})$ of $(x_n)$ that converges to some $x \in X$, then $(x_n)$ converges to $x$ as well.

You say "it converges to the same limit as the original sequence", but you (on purpose!) do not assume at all that the original Cauchy sequence converges (this is what is to be shown!), so that statement makes no sense in this context. Correct is: the original sequence also converges to the same (subsequential) limit.

E.g. if we enumerate all rationals in a sequence, for every $x \in \mathbb{R}$ there exists some subsequential limit that converges to $x$. But this says nothing at all about the original sequence (which does not converge at all, and has all reals as subsequential limits). Of course this example sequence is not Cauchy...

So your proof works if we prove this first correct fact about Cauchy sequences and subsequential limits.

  • $\begingroup$ I simply assumed this basic fact about subsequences. But I get it now: my clumsy wording might have lead the user to believe the argument was wrong. I will correct it. $\endgroup$ – Student Apr 23 '14 at 13:36
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    $\begingroup$ @Student, it is not "believing your argument was wrong". Iti s that your argument was wrong, even if we assume you were implicitly assuming the basic proposition in the last sentence in the answer above. $\endgroup$ – DonAntonio Apr 23 '14 at 14:22
  • $\begingroup$ @ Henno: I have edited my answer and included a proof of what I had assumed implicitly. $\endgroup$ – Student Apr 24 '14 at 10:11

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