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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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Student
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  • How do you know every subsequence of $(x_n)$ has a subsequence converging to the same $x$. This needs justification. I think this is what you're trying to say in your second to last sentence. In any case, this sentence is, at best, vague. (It would be a bit easier to show that a Cauchy sequence with a convergent subsequence is itself convergent.) – David Mitra Apr 23 '14 at 13:23
  • The point is to identify a potential limit of the sequence, this is what you get from your subsequence. After you obtain the convergent subsequence you may make use of the fact that the original sequence is Cauchy. – Thomas Apr 23 '14 at 13:26
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    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 $|x_n-x|\leq |x_n-x_{n_k}|+|x_{n_k}-x|$... –  Apr 23 '14 at 13:26
  • Now that I read them, your answer it actually better than most answers to that question. Certainly better than the selected answer, the answer that only quotes a theorem, and also the answer in the comments to the question. –  Apr 23 '14 at 13:33
  • @user144349 Thank you. I only posted it because I thought I could contribute something to that thread. I now also see that it can still be improved, in particular, that the wording is so clumsy that the reader might well think that it is wrong. – Student Apr 23 '14 at 13:37
  • I thank you all for your help. I will rewrite and improve the answer. – Student Apr 23 '14 at 13:38
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    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. – DonAntonio Apr 23 '14 at 14:18
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    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. – DonAntonio Apr 23 '14 at 14:21
  • @Student, There originally was a comment to the effect that: "a Cauchy sequence always converges, just maybe to something outside the set." What does "outside the set" mean for a general metric space $X$? This is certainly true for Cauchy sequences in $\mathbb Q$ since you know beforehand that $\mathbb Q$ sits inside of $\mathbb R$, but why can you view a general metric space as always sitting inside of another complete one? This is the point which my comments were meant to clarify. I appreciate you taking the time to improve your answer. – Santiago Canez Apr 23 '14 at 16:39
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    @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. – Santiago Canez Apr 23 '14 at 16:41
  • @SantiagoCanez Yes, of course a general metric space does not automatically sit in another metric space except perhaps its completion. However, the candidate limit $x$ is already in $X$ since $x_{n_k}\to x$. But the key to the misunderstanding here is that I do think it is a triviality that the sequence $x_n$ converges to $x$ if $x_{n_k}\to x$. – Student Apr 24 '14 at 10:02
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    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" ;) – Santiago Canez Apr 24 '14 at 16:22
  • Nope, it is not correct. You assumed the Cauchy sequence converges to prove that Cauchy sequence converges. – Geralt of Rivia Oct 19 '17 at 16:42
  • The essential fact is as follows: if a Cauchy sequence has a convergent subssequence then the original sequence was also convergent (necessarily to the same limit). – Henno Brandsma Oct 21 '17 at 13:43

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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.

Henno Brandsma
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  • 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. – Student Apr 23 '14 at 13:36
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    @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. – DonAntonio Apr 23 '14 at 14:22
  • @ Henno: I have edited my answer and included a proof of what I had assumed implicitly. – Student Apr 24 '14 at 10:11