# How to show that the set of accumulation points of $(x_n)$ equals the intersection over all $n$ of the closure of the set of all $x_k, k \geq n$?

I'm trying to solve the following problem:

We call a point $x$ an accumulation point for a sequence $(x_n)$ in a topological space $X$ if every neighbourhood $U$ of $x$ contains infinitely many elements of the sequence. Show that the set of accumulation points equals $\bigcap_n F_n$ where $F_n = \overline{ \{ x_k \vert k \geq n \} }$. Use this to show that the set of accumulation points is closed and nonempty if $X$ is compact.

I'm stuck on the first one:
Show that the set of accumulation points equals $\bigcap_n F_n$ where $F_n = \overline{ \{ x_k \vert k \geq n \} }$.

It seems intuitively obvious, but I don't know how to show it.

Any hints on where to begin greatly appreciated!

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$\newcommand{\cl}{\operatorname{cl}}$I’ll do a large chunk of it and leave the rest for you to try; if you get stuck, just ask.

Let $A$ be the set of accumulation points of the sequence; we want to show that $A=\bigcap_{n\in\Bbb N}F_n$, so let’s try the obvious: show that $A\subseteq\bigcap_{n\in\Bbb N}F_n$ and $\bigcap_{n\in\Bbb N}F_n\subseteq A$.

1. $A\subseteq\bigcap_{n\in\Bbb N}F_n$. One’s first thought is probably to let $x\in A$ and show that $x\in\bigcap_{n\in\Bbb N}F_n$, but here I think that it’s slightly easier to show the contrapositive: if $x\notin\bigcap_{n\in\Bbb N}F_n$, then $x\notin A$. Suppose that $x\notin\bigcap_{n\in\Bbb N}F_n$; then there is some $n\in\Bbb N$ such that $x\notin F_n=\cl\{x_k:k\ge n\}$. Let $U=X\setminus F_n$; then $x\in U$, and $U$ is open, so $U$ is a neighborhood of $x$. If $x_k\in U$, then $k<n$, so $U$ contains only finitely many terms of the sequence. Thus, $x$ has a neighborhood that does not contain infinitely many terms of the sequence, and therefore $x\notin A$.

2. $\bigcap_{n\in\Bbb N}F_n\subseteq A$. Here again I think that it’s slightly easier to show that if $x\notin A$, then $x\notin\bigcap_{n\in\Bbb N}F_n$. Suppose that $x\notin A$; by definition this means that $x$ has a neighborhood $U$ that contains only finitely many terms of the sequence. Can you finish it from here? It’s basically just turning the argument in (1) around.

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Thanks, Brian! Solved it now :) –  Maethor Oct 31 '12 at 22:18
@Maethor: Excellent! You’re welcome. –  Brian M. Scott Oct 31 '12 at 23:54

Well, this does seem intuitively obvious. In such cases, in order to finally convince myself, I sometimes resort to formal symbol manipulations and kind of follow where they lead.

In your case, the definition says that $x$ is an accumulation point for $(x_n)$ iff $$\forall U \in \mathcal{F}_x \,\forall n\,\exists k \colon k \geq n \,\&\,x_k \in U,$$ where $\mathcal{F}_x$ is the set of all neighbourhoods of $x$. Now we just change the order of the first two quantors and mess with the notation a bit: $$\forall n \,\forall U \in \mathcal{F}_x \exists y \in \{x_k|k\geq n\} \colon y \in U,$$ in other words $$\forall n \,\forall U \in \mathcal{F}_x \colon U \cap \{x_k|k\geq n\} \neq \emptyset.$$

Now, the following is just an alternative way to define the closure of a set $A$: $x \in \overline{A}$ if and only if $\forall U \in \mathcal{F}_x \colon U \cap A \neq \emptyset$. So, our last statement can be rewritten: $$\forall n \colon x \in \overline{\{x_k|k\geq n\}},$$ which is in turn equivalent to $$x \in \bigcap_{n} \overline{\{x_k|k\geq n\}},$$ QED.

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