# Let $(x_n)$ be a bounded sequence and $u=\limsup x_n$. Let E be set of limits of convergent subsequences of $(x_n)$. How do I prove $u \in E$?

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?

• 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 $u$. May 24, 2019 at 18:11

Since$$u-1 $$\sup_{k\geqslant n}x_k>u-1$$ and therefore there is some $$n_1\in\mathbb N$$ such that $$u-1. Now, since$$u-\frac12$$\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. Now, you prove by the same approach that there is some $$n_3\in\mathbb N$$ such that $$n_3>n_2$$ and that $$u-\frac13, and so on. So, $$(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$$.