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I have seen two proofs for

If $\{x_n\}_{n=1}^\infty$ is a bounded sequence of real numbers and if its every convergent subsequence converges to the same value $L$, then $\{x_n\}$ also converges to the same value.

I have one question about each of the proofs.

First proof

This is the sketch of the first proof.

Suppose that $\{x_n\}$ does not converge to $L$. Then there is an $\epsilon>0$ such that for every positive integer $N$ there is a postive integer $n\geq N$ with $|x_n-L|\geq\epsilon$. We form inductively a subsequence $\{x_{n_k}\}_{k=1}^\infty$ such that $|x_{n_k}-L|\geq\epsilon$. Since $\{x_{n_k}\}$ is bounded, there is a convergent subsequence $\{x_{n_{k_j}}\}_{j=1}^\infty$. This subsequence itself is a subsequence of $\{x_n\}$ and by the hypothesis $\{x_{n_{k_j}}\}_{j=1}^\infty$ is convergent but $|x_{n_{k_j}}-L|\geq\epsilon$ and so $\{x_{n_{k_j}}\}_{j=1}^\infty$ does not converge to $L$. Contradiction.

My question is why did we need to form the subsequence $\{x_{n_{k_j}}\}_{j=1}^\infty$? in other words, what is wrong with this shorter proof

Suppose that $\{x_n\}$ does not converge to $L$. Then there is an $\epsilon>0$ such that for every positive integer $N$ there is a postive integer $n\geq N$ with $|x_n-L|\geq\epsilon$. We form inductively a subsequence $\{x_{n_k}\}_{k=1}^\infty$ such that $|x_{n_k}-L|\geq\epsilon$. Since $\{x_n\}$ is bounded, there is a convergent subsequence $\{x_{n_j}\}$ but $|x_{n_{j}}-L|\geq\epsilon$ and so $\{x_{n_j}\}$ does not converge to $L$. contradiction.

Second Proof

The following proof is taken from this document.

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Is this proof correct? the proof says each mentioned interval contain all but a finite number of elements.

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  • $\begingroup$ I think that in the first line of the first proof "to any $L$" must be replaced by something like: "to the common limit $L$ of the convergent subsequences". Then you find inductively a convergent subsequence that does not converge to $L$ and a contradiction is found. $\endgroup$ – drhab Apr 9 '17 at 11:39
  • $\begingroup$ @drhab: Thanks. I fixed it. $\endgroup$ – user214302 Apr 9 '17 at 11:40
  • $\begingroup$ Sorry, but should focus on a subsequence of $(x_{n_k})_k$ (wich is also a subsequence of $(x_n)_n$ of course). This to make sure that it cannot have $L$ as limit. That's what they did in the longer (and correct) proof. $\endgroup$ – drhab Apr 9 '17 at 11:50
  • $\begingroup$ The statement, as given, "If every convergent subsequence converges to the same limit, L, then the sequence converges to L", is not true! It should be "If every subsequence converges and converges to the same limit, L, then the sequence converge to L". I suspect that was what you meant but your wording, "every convergent subsequence**" leaves the possibility that some subsequences do not convege. $\endgroup$ – user247327 Apr 9 '17 at 11:58
  • $\begingroup$ @user247327 I disagree. It is true and it is not necessary mention the necessary condition that all subsequences converge. $\endgroup$ – drhab Apr 9 '17 at 12:05
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A sequence $(x_{n_k})_k$ is constructed that cannot possibly have any subsequences that converge to $L$.

This sequence has a convergent subsequence (since it is a bounded sequence), and we conclude that this convergent subsequence must have a limit different from $L$. This convergent subsequence is also a subsequence of the original sequence, so a contradiction is found.

In your shorter "proof" you also construct sequence $(x_{n_k})_k$ but you do not really make use of it. It seems that you just leave it aside and step back to the original sequence $(x_n)_n$ and its subsequences.

The second proof works under assumption that the original sequence is convergent and shows that it cannot have a limit that differs from $L$. But what you are after is a proof of this assumption.

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