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?

  • 6
    $\begingroup$ If $p$ is not a limit point, then it is an element of $E$. In that case, you could just take $p' = p$. $\endgroup$ Jul 25, 2012 at 22:59
  • $\begingroup$ Ah, I see. Thanks! $\endgroup$ Jul 25, 2012 at 23:02
  • $\begingroup$ if $p$ is by itself off somewhere, then it won't be in the closure of $E$. $\endgroup$ Jul 25, 2012 at 23:03

2 Answers 2


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 $\epsilon > 0$.


Basically Rudin needs to write a triangle inequality with two additional arbitrary points $p',q'$.

$d(p,q) \leq d(p,p') + d(p',q) $

$ d(p,p') + d(p',q) \leq d(p,p') + d(p',q') + d(q',q)$

He assumes $E$ is non-empty, so we might pick points $p'$ and $q'$ for this inequality, even if $p = p'$ and $q = q'$.


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