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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 $\{x_k\}$?

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    $\begingroup$ Be careful that naming the limit $x_0$ can be confusing since it can also be the first term of your sequence. $\endgroup$ May 9, 2016 at 14:42
  • $\begingroup$ @Mandrathax Then you mean the proof above is done? No need to proceed further? $\endgroup$
    – Chloe
    May 9, 2016 at 15:23
  • $\begingroup$ No this is just a notation issue, you should name the limit $x$ or $l$ so that you don't get confused with $x_n$ for $n=0$ $\endgroup$ May 9, 2016 at 15:29

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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
    1. $\forall k,y_{\sigma(k)}\in \{x_k\vert k\in\mathbb N\}$
    2. $\sigma$ is strictly increasing

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

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