How can I prove that $S=\{x_k:k\in \Bbb N\}\cup\{x_0\}$ is closed in $\Bbb R^n$? How can I prove that if $\{x_k\}$ is a convergent sequence in $\Bbb R^n$ with limit $x_0$, then $S=\{x_k:k\in \Bbb N\}\cup\{x_0\}$ is closed in $\Bbb R^n$?
This is my attempt:
It suffices to show that $x_0$ is the only limit point of $S$.
Suppose to the contrary that $\lim_{x\to \infty} x_k=x_0, \lim_{x\to \infty} x_k=x_0' $ and $x_0\neq x_0'$.
Let $\epsilon=\frac1 2 \Vert x_0 -x_0'\Vert >0$ be given.
Then there exists $ k_1, k_2 \in \Bbb N$ such that
$k \ge k_1 \Rightarrow \Vert x_k -x_0\Vert < \frac1 2 \Vert x_0 -x_0'\Vert$
and $k \ge k_2 \Rightarrow \Vert x_k -x_0'\Vert < \frac1 2 \Vert x_0 -x_0'\Vert$
Let $k_0=\max\{k_1,k_2\}$
Then for $k\ge k_0$
$\Vert x_0 -x_0'\Vert \le \Vert x_0 -x_k\Vert + \Vert x_k -x_0'\Vert < \frac1 2 \Vert x_0 -x_0'\Vert+\frac1 2 \Vert x_0 -x_0'\Vert = \Vert x_0 -x_0'\Vert $
(contradiction)
What I proved above is that $x_0$ is the only limit point of $\{x_k\}$.
How can I prove it is the only limit point of S, not just of $\{x_k\}$?

Proceed this way : 
Let $x=\lim_\infty x_k$ and $A=\{x_k\vert k\in\mathbb N\}\cup\{x\}$.
You want to prove that for any sequence $(y_k)_{k\in\mathbb N}\in A^{\mathbb N}$, if $(y_k)$ has a limit $y\in \mathbb R^n$, then $y\in A$.
Now there are two cases :


*

*either $(y_k)$ is stationary, in which case the result is trivial

*Otherwise, you can extract a sequence $(y_{\sigma(k)})$ from $(y_k)$ such that


*

*$\forall k,y_{\sigma(k)}\in \{x_k\vert k\in\mathbb N\}$

*$\sigma$ is strictly increasing



In this case, $(y_{\sigma(k)})$ is also a sub sequence of $(x_k)$, so its limit is $x$, hence $\lim_\infty y_k=\lim_\infty y_{\sigma(k)}=x\in A$
