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I've been trying to attempt this problem for a long time now. At fist I tried to show that the sequence $(u_n)$, where $u_n = \sup_{i \geq n} x_i$, is a subsequence of $(x_n)$. But this is not true for all sequences. Any suggestions?

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    $\begingroup$ Given $u_k$, take $x_{n_k}$ such that $u_k - 1/k < x_{n_k} \le u_k$. This $x_{n_k}$ converges to $u$. $\endgroup$ May 24, 2019 at 18:11

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Since$$u-1<u=\limsup_nx_n=\lim_{n\to\infty}\sup_{k\geqslant n}x_k,$$ $\sup_{k\geqslant n}x_k>u-1$ and therefore there is some $n_1\in\mathbb N$ such that $u-1<x_{n_1}\leqslant u$. Now, since$$u-\frac12<u=\limsup_nx_n=\lim_{n\to\infty}\sup_{k\geqslant n}x_k,$$$\sup_{k\geqslant n}x_k>u-\frac12$ and therefore there is some $n_2\in\mathbb N$ such that $n_2>n_1$ and that $u-\frac12<x_{n_2}\leqslant u$. Now, you prove by the same approach that there is some $n_3\in\mathbb N$ such that $n_3>n_2$ and that $u-\frac13<x_{n_3}\leqslant u$, and so on. So, $(x_{n_k})_{k\in\mathbb N}$ is a subsequence of $(x_n)_{n\in\mathbb N}$ and $\lim_{k\to\infty}x_{n_k}=u=\limsup_nx_n$.

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