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A problem from Stephen Abbott's Understanding Analysis:
Exercise 2.5.4.: 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 $a$.

Is the following direct proof valid?

Since all subsequences converge, it means that for any particular $\varepsilon>0$ there exists some number $J\in\mathbb{N}$ in every subsequence, such that, given $j\geq J$ and $n_j\geq n_J$, $\left|a_{n_{j}}-a\right|<\varepsilon$. Let $N$ be equal to the greatest element from the set of specific $n_J$'s from all subsequences and $n\geq N$. Thus, for any $\varepsilon>0$, $\left|a_n-a\right|<\varepsilon$. $\square$

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    $\begingroup$ There are infinitely many subsequences. How could you guarantee that the set of $n_J$s is bounded? $\endgroup$ – Daniel Fischer Jan 5 '17 at 15:31
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Your proof has two problems: first, the problem does not state that every subsequence converges, but rather that if a given subsequence has a limit then this limit must be $a$.

Second, since there are infinitely many subsequences of $\{a_n\}$, there are infinitely many $n_J$, and so the $N$ in your proof is not necessarily defined.

Instead of your approach, I suggest trying to prove that $\limsup_{n\to\infty}a_n=a=\liminf_{n\to\infty}a_n$, using the fact that there are subsequences of $\{a_n\}$ converging to the limsup and to the liminf.

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