I have the following problem:
Let $s$ be a real number and $(s_n)$ be a sequence of real numbers. Suppose that for any subsequence $(s_{n_{k}})$ of $(s_n)$, $(s_{n_{k}})$ has a subsequence $(s_{n_{k_{l}}}$) satisfying $$ \lim_{l \to \infty} s_{n_{k_{l}}}=s. \qquad (1)$$ Please show that $\lim\sup s_n = \lim \inf s_n = s$
I don't know if my proof is right. I have the following:
Let $S$ be the set of subsequential limits of $(s_n)$. Then, $S$ contains $s$ since it is the limit of some subsequence of $(s_n)$, namely $(s_{n_{k_{l}}})$, which is a subsequence of $(s_n)$. Now, I claim $S$ only contains $s$. Suppose, as a contradiction, that $S$ contains some other element $t$. That implies, that there exists a subsequence $(s_{n_{t}})$ with limit $t$. This implies that any subsequence of $(s_{n_{t}})$, namely $(s_{n_{t_{p}}})$ converges to $t$. But, that is a contradiction to the supposition that any subsequence of $(s_n)$ has a subsequence that converges to $s$. Hence, $S$ has only the element $s$. Therefore, $\lim\sup s_n=\sup S =s $ and $\lim\inf s_n=\inf S=s$, so $\lim\sup s_n = \lim \inf s_n = s$.
Would this be a complete valid proof?