Showing set of all cluster points of sequence in extended $\mathbb R $ is closed. An extended real number ($c \in \mathbb{R}, c = \infty, c=-\infty$) is said to be a cluster point of a sequence $\{a_{n}\}$ if a subsequence $\{a_{n_{k}}\}$ converges to $c$. 
I need to show that the set of all cluster points is a closed set.
So far, I have the following:
Let $E=\{$set of all real cluster points of $\{a_{n}\}\}$.
Let $E^{\prime}=\{\text{set of cluster points of}\,E\}$.
Suppose $p$ is a cluster point of $E^{\prime}$. Then, $\forall \frac{\epsilon}{2} >0 \exists q \in N_{\frac{\epsilon}{2}}(p)$, where $q \neq p$ And $q \in E^{\prime}$.
We need to show that $p \in E^{\prime}$:
Since $q \in E^{\prime}$, it is a cluster point of $E$, so $\forall \frac{\epsilon}{2}>0$, $\exists x \in N_{\frac{\epsilon}{2}}(q)$, where $x\neq q$, $x \in E$.
Now, $|p-x|\leq |p-q|+|q-x|<\frac{\epsilon}{2} + \frac{\epsilon}{2} =2$.
So, $p$ is a cluster point of $E$, thus $E$ is closed.
Problem is, I'm not sure how to handle the cases where $c=\pm \infty$. I know that $\{a_{n}\}$ is said to converge to $\infty$ if $\forall \epsilon >0, \exists N \in \mathbb{N}$ such that $\forall n>N$, $a_{n}>\epsilon$ (Same for $-\infty$, but then, $a_{n}<\epsilon$. But, how do I get around the absence of a nice epsilon neighborhood in these cases?
 A: Call $C$ the set of cluster points. Take $(c_n)\subseteq C$ and suppose $c_n\to l$. If $l$ is real you proved that $l\in C$, so let's assume that $l=\infty$. 
Fix $N\in\mathbb{N}$. Since $c_n\to\infty$, there exists a certain $c_k>N+1$ and since such $c_k$ is a cluster point there is some $a_h$ such that $|a_h-c_k|<1$, and this implies $a_h>N$. So I proved that for any $N\in\mathbb{N}$ there exists an $h$ such that $a_h>N$. From here one can easily define inductively a subsequence $a_{n_k}\to\infty$.

Now fix $a_{n_0}>0$. There are infinitely many $m$ such that $a_{m}>\max(a_{n_0},1)$, so choose $n_1>n_0$ such that $a_{n_1}>\max(a_{n_0},1)$, next choose $n_2>n_1$ such that $a_{n_2}>\max(a_{n_1},2)$ and so on.
A: Denoting the extended reals by $\overline{\Bbb R},$ consider the function $f:\overline{\Bbb R}\to[-1,1]$ given by $$f(x)=\begin{cases}1 & x=\infty\\-1 & x=-\infty\\\frac2\pi\arctan x & x\in\Bbb R.\end{cases}$$ You should be able to show that this is a bijection, and is convergence-preserving in the sense that for a sequence of points $x_n\in\overline{\Bbb R},$ we have that $x_n\to c$ if and only if $f(x_n)\to f(c).$ Furthermore, it is a continuous, closed map, meaning in particular that for $C\subseteq\overline{\Bbb R},$ we have that $C$ is closed in $\overline{\Bbb R}$ if and only if $f(C)$ is closed in $[-1,1].$
Thus, it suffices to prove the result for $[-1,1],$ which you should be able to do easily.
Alternately, we can define a metric on $\overline{\Bbb R}$ by $d(x,y)=\bigl|f(x)-f(y)\bigr|,$ and show that this metric yields exactly the same open sets in $\overline{\Bbb R}$ as usual. This will allow us to use the same sort of argument that you made already.
