# If $p$ is an element of $\overline E$ but not a limit point of $E$, then why is there a $p' \in E$ such that $d(p, p') < \varepsilon$?

I don't understand one of the steps of the proof of Theorem 3.10(a) in Baby Rudin. Here's the theorem and the proof up to where I'm stuck:

Relevant Definitions

The closure of the subset $E$ of some metric space is the union of $E$ with the set of all its limit points.

The diameter of the subset $E$ of some metric space is the supremum of the set of all pairwise distances between its elements.

For the points $x$ and $y$ in some metric space, $d(x, y)$ denotes the distance between $x$ and $y$.

Theorem 3.10(a) If $\overline{E}$ is the closure of a set $E$ in a metric space $X$, then

$$\text{diam} \ \overline{E} = \text{diam} \ E.$$

Proof. Since $E \subseteq \overline{E}$, we have

$$\begin{equation*} \text{diam} \ E \leq \text{diam} \ \overline{E}. \end{equation*}$$

Let $\epsilon > 0$, and pick $p, q \in \overline{E}$. By the definition of $\overline{E}$, there are points $p', q' \in E$ such that $d(p,p') < \epsilon$ and $d(q, q') < \epsilon$...

I see that this works if $p$ and $q$ are limit points of $E$. But how does this work if, say, $p$ isn't a limit point of $E$? What if $E$ is some region in $\mathbb{R}^2$ along with the point $p$ by itself way off somewhere?

-
If $p$ is not a limit point, then it is an element of $E$. In that case, you could just take $p' = p$. –  Dylan Moreland Jul 25 '12 at 22:59
Ah, I see. Thanks! –  Jefferson Huang Jul 25 '12 at 23:02
if $p$ is by itself off somewhere, then it won't be in the closure of $E$. –  Olivier Bégassat Jul 25 '12 at 23:03

## 1 Answer

Dylan answered my question in the question comments above.

In particular, if $p \in \overline{E}$ is not a limit point of $E$, then it has to be in $E$, and so letting $p = p'$, we have $d(p, p') = 0 < \epsilon$ for any $\epsilon > 0$.

-