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?

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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.

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  • $\begingroup$ What do you mean, inductively define a subsequence $a_{n_{k}}$? $\endgroup$ – ALannister Sep 22 '15 at 17:18
  • $\begingroup$ could you please show me what inductively defining the subsequence $a_{n_{k}}$ would look like in terms of the $c$ subscripts? Also, I'm a little confused: in order to show that $l$, which we're assuming is in $E^{\prime}$ is also contained in $E$ (see how I defined those in my original question), we're showing that the sequence $a_{n}$ itself converges? I'd like to accept this answer, but I need to understand it fully before doing so. $\endgroup$ – ALannister Sep 23 '15 at 12:35
  • $\begingroup$ Assuming that $\infty$ is a limit of a sequence of cluster points, I produced a subsequence $a_{n_k}$ converging to it, thereby proving that $\infty$ is itself a cluster point. $\endgroup$ – Sonner Sep 23 '15 at 13:04
  • $\begingroup$ Because if a sequence converges to m, every subsequence must also converge to m? $\endgroup$ – ALannister Sep 23 '15 at 13:08
  • $\begingroup$ I like what you did. Wondering if this was equivalent: suppose $\{a_{j_{k}}\}$ is a subsequence of $a_{j}$. Then, $\{j_{k}\}$ is an increasing sequence in $\mathbb{N}$, so $j_{k}\geq k$. Since $a_{j} \to \infty$, $a_{j}>\epsilon$, $\forall j > N \in \mathbb{N}$. Then, $k>N$ implies that $j_{k}>N$, since $j_{k}\geq k$. So, $a_{j_{k}}>\epsilon$, $\forall k>N$. So, $l \in E^{\prime}$ is also $\in E$, which implies that $E$ is closed? $\endgroup$ – ALannister Sep 23 '15 at 13:31

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.

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