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Assume $(a_n)$ is a bounded sequence with the property that every convergent subsequence of $(a_n)$ converges to the same limit $a\in\mathbb{R}$. Show that $(a_n)$ must converge to $a.$

I have tried to prove this by contradiction. That is, assume $(a_n)$ does not converge to $a.$ Then there are two cases:

  1. $(a_n)$ converges to some other value $b.$
  2. $(a_n)$ diverges.

For case1, if $(a_n)$ converges to $b,$ then all the subsequences of $(a_n)$ should converge to $b,$ which contradicts that all convergent subsequences converge to $a.$

For case2, if $(a_n)$ diverges, then I can find counterexamples to show $(a_n)$ does not have the property that all convergent subsequences converge to the same limit. For example $(-1)^n$ has two subsequences with different limits. $(a_n)$

Then I figured out that I did not use the hypothesis that $(a_n)$ is bounded. So there might be something wrong in my proof, but I could not find it out.

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You did not handle case 2) properly.

If $(a_n)$ does not converge to $a$ then there exists $\epsilon >0$ and integers $n_k$ such that $n_1<n_2<...$ and $|a_{n_k}-a| \geq \epsilon$ for all $k$. Now $(a_{n_k})$ is bounded sequence and hence it has a convergent subsequence. But the inequality $|a_{n_k}-a| \geq \epsilon$ holds for this subsequence also. Do you see a contradiction now?

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    $\begingroup$ Yes, I think this is right. But I was just wondering is there any mistakes in my steps above. $\endgroup$
    – Howard
    Mar 21 '21 at 5:31
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    $\begingroup$ Case 2) in your answer does not make sense. $(a_n)$ is a given sequence and an example like $(-1)^{n}$ has no role in this result. @Howard $\endgroup$ Mar 21 '21 at 5:38

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