# Is the intersection of a set and its set of limit points not empty?

Let $E$ be a set in a metric space $X$ and $E'$ denote the set of all limit points and $\bar{E}$ be the closure of $E$. Then $\bar{E} = E \cup E'$

Prove that $E'$ is closed, and $\bar{E}$ and $E$ have the same limit points

So as sets, they look like?

I know that the $p'$ just solely inside $E'$ is wrong, but is the actual picture even right?

I read a lemma that since $E'$ is the set of all limit points, any point (say) $p'$ must contain another point in $E$ for that point isn't the point itself

-
In $\Bbb R$, what are the limit points of $E=\{1/n: n=1,2,\ldots\}$? – David Mitra Dec 29 '12 at 3:08
Is it just the unit interval? – Hawk Dec 29 '12 at 3:22
It's $0$. But to show that $E'$ is closed you can show that every convergent sequence in $E'$ has its limit in $E'$. – Epsilon Dec 29 '12 at 3:24
Oh sorry I misread his question. I thought he asked what is $E'$ for some reason. I actually have the answers in my hand, and I need to understand the definitions first before I want to continue however – Hawk Dec 29 '12 at 3:32
Alternative hint: to show a set is closed, it is often easier to show its complement is open, – gnometorule Dec 29 '12 at 3:54

To show that $E'$ is closed, we will show $X\setminus E'$ is open. If $x \in X\setminus E'$, then there is $B(x,r) \subseteq X\setminus E$. Then for all $y \in B(x,r)$ since $B(x,r)$ is open there is $B(y,r_2)\subseteq B(x,r)\subseteq X \setminus E$ so $y \in X \setminus E'$ as well. It follows that $x$ is an interior point of $X\setminus E'$, and so $X\setminus E'$ is open, and $E'$ is closed.
Now we show $E$ and $\overline{E}$ have the same limit points. Clearly $E' \subseteq \overline{E}'$ since if every $B(x,r)$ contains an element of $E$ different from $x$, then that element is also an element of $\overline{E}$. Conversely, suppose $B(x,r)$ contains an element $y$ of $\overline{E}$. Since $B(x,r)$ is open, we may choose $B(y,r_2)\subseteq B(x,r)$ and by making $r_2<d(x,y)$ smaller if necessary we may ensure $x\notin B(y,r_2)$. Then since $y \in \overline{E}$ we may choose $z \in E \cap (B(y,r_2)\setminus\{y\})$. Then $z \in E \cap (B(x,r) \setminus \{x\})$ which completes the proof.