Prove the set of subsequential limits of ($x_t$) is closed A sequence ($x_t$) in metric space ($X,d$). Let $S$ be the set of subsequential limits of ($x_t$) ($S$ could be empty), prove that $S$ is closed.
I need to prove the problem with one of the characterizations of closed: prove that if ($s_t$) is a sequence of subsequential limits with ($s_t \rightarrow s$), then $s$ is a subsequential limit, i.e. $s\in S$.
Could someone help me with this proof please? Thanks.
 A: HINT: This is a ‘follow-your-nose’ proof: try the most straightforward thing, and you find that it works. The most difficult part is keeping the notation straight.
Let $\sigma=\langle x_n:n\in\Bbb N\rangle$ be a sequence in $X$, and let $\langle s_k:k\in\Bbb N\rangle$ be a sequence of subsequential limits of $\sigma$ that converges to $s\in X$. For each $k\in\Bbb N$ there is an infinite $N_k\subseteq\Bbb N$ such that the subsequence $\langle x_n:n\in N_k\rangle$ of $\sigma$ converges to $s_k$. (This is one way to write multiple subsequences without getting lost in a lot of indices: $N_k$ is simply the set of indices appearing in the $k$-th subsequence, and we assume that we take them in increasing order, so that we really do get a subsequence of $\sigma$.)
We want to find an infinite $N\subseteq\Bbb N$ such that the subsequence $\langle x_n:n\in N\rangle$ of $\sigma$ converges to $s$. Once we’ve found $N$, we can index it as $N=\{n_k:k\in\Bbb N\}$, where $n_0<n_1<\ldots\;$, and the idea is to choose the integers $n_k$ one at a time so that $\langle x_{n_k}:k\in\Bbb N\rangle$ really does converge to $s$. Specifically, for each $k\in\Bbb N$ we’ll choose $n_k$ so that 
$$d(x_{n_k},s)<\frac1{2^k}\;.\tag{1}$$
It’s a recursive construction. Imagine that we’ve already chosen $n_k$ for each $k\le\ell$ in such a way that each satisfies $(1)$, and moreover $n_0<\ldots<n_\ell$; we want to choose $n_{\ell+1}>n_\ell$ such that 
$$d(x_{n_{\ell+1}},s)<\frac1{2^{\ell+1}}\;.$$
The sequence $\langle s_k:k\in\Bbb N\rangle$ converges to $s$, so we can find an $m\in\Bbb N$ such that 
$$d(s_m,s)<\frac1{2^{\ell+2}}\;.$$
And $\langle x_k:k\in N_m\rangle$ converges to $s_m$ so we can find an $r\in\Bbb N$ such that 
$$d(x_k,s_m)<\frac1{2^{\ell+2}}$$
whenever $k\ge r$ and $k\in N_m$.


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*Can you finish it off from here by specifying how to choose $n_{\ell+1}$, choosing $n_0$, and rounding out the explanation of why the subsequence $\langle x_n:n\in N\rangle$ of $\sigma$ really does converge to $s$?

