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The proof that I found goes roughly like this : using the negation of the statement of convergence, it is possible to build a subsequence $(a_{n_k})$ that never enters the $\epsilon$-neighbourhood of $V_\epsilon(a)$. Then, since $(a_n)$ is bounded, this implies that $(a_{n_k})$ is also bounded and we can apply the Bolzano-Weierstrass theorem to it to obtain a subsequence of $(a_{n_k})$ which is also a subsequence of $(a_n)$. Since it is a subsequence of $(a_n)$, it is supposed to converge but since it is a subsequence of $(a_{n_k})$, it never enters the $\epsilon$-neighbourhood of $a$ and thus does not converge to $a$. This completes the proof.

My question is why should we use the Bolzano-Weierstrass theorem to produce a subsequence that never enters the $\epsilon$-neighbourhood of $a$ when we already constructed one namely $(a_{n_k})$. Should $(a_{n_k})$ not already be a subsequence of $(a_n)$ that does not converge to $a$ (and hence the proof is completed ?) ? I know there is a logical flaw in the argument (since the fact that $(a_n)$ was bounded was not used) but I cannot find it.

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    $\begingroup$ Your question as stated is trivial: $\{a_n\}$ is a subsequence of $\{a_n\}$, so it converges to $a$. There's nothing to prove. $\endgroup$ – user99914 Jun 15 '15 at 11:53
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    $\begingroup$ @John OP surely meant to say "if $(a_n)$ is a bounded sequence and that if every convergent subsequence of $(a_n)$ converges to $a $, then $(a_n)$ converges to $a$." $\endgroup$ – Gabriel Romon Jun 15 '15 at 11:56
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    $\begingroup$ @LeGrandDODOM Yes this is what I meant. $\endgroup$ – user147834 Jun 15 '15 at 11:57
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    $\begingroup$ @LeGrandDODOM You may be right. Still, John's comment is very appropriate: If OP meant to say what he did, then this is a lesson: if you improperly ask questions, the answers may drastically change. $\endgroup$ – 5xum Jun 15 '15 at 11:57
  • $\begingroup$ Do you know what limsup and liminf are? $\endgroup$ – Matematleta Jun 15 '15 at 12:01
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No you got the rationale of the proof wrong.

Firstly, one assumes that $(a_n)$ does not converge. This yields the existence of some $\epsilon$ such that $\forall N\in \mathbb N$, there is some $n\geq N$ such that $a_n \notin V_{\epsilon}(a)$. This allows you to build a subsequence $a_{n_k}$ such that $\forall k, a_{n_k}\notin V_{\epsilon}(a)$.

Now, since $a_{n_k}$ is bounded, it has a convergence subsequence, say $b_n$, and $b_n$ is therefore a convergent subsequence of $a_{n_k}$, but also a convergent subsequence of $a_n$. Hence $b_n$ converges to $a$. But that is a contradiction, since $b_n$ is a subsequence of $a_{n_k}$.

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  • $\begingroup$ Yes, you are right, now I get it ! Thank you very much for your response. $\endgroup$ – user147834 Jun 15 '15 at 12:10
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In answer to your question, you initially have a subsequence which never enters the $\epsilon$ neighbourhood of $a$, but you do not know that the subsequence converges converges. So the proof uses boundedness and Bolzano-Weierstrass to construct a convergent subsequence which never enters the $\epsilon$ neighbourhood of $a$. Once you have such a convergent subsequence you can show that the limit is not $a$.

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  • $\begingroup$ Yes, I have realised the flaw in my reasoning. I was wrongly assuming that every subsequence was supposed to be convergent. Thank you. $\endgroup$ – user147834 Jun 15 '15 at 12:14
  • $\begingroup$ @P.Munbodh In fact every subsequence of a convergent sequence is convergent (e.g. it satisfies the Cauchy condition, once this is established). It seems so obvious sometimes that it is hard to see where it is being assumed. $\endgroup$ – Mark Bennet Jun 15 '15 at 12:28
  • $\begingroup$ Exactly and here, I assumed it was true while actually trying to prove the convergence of $(a_n)$. I made a circular reasoning. $\endgroup$ – user147834 Jun 15 '15 at 12:32

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