Q:

Confused about proof that diameter of a closure of a set is the same as the diameter of the set.

Definition Let $E$ be a nonempty subset of a metric space $X$, and let $S$ be the set of all real numbers of the form $d(p,q)$, with $p \in E$ and $q \in E$. The supremum of $S$ is called the diameter of $E$. 
Theorem If $\bar{E}$ is the closure of the set $E$ in a metric space $X$, then $ \text{diam}  \ \bar{E} = \text{diam} \  E.$
Proof
Since $E \subset \bar{E},$ it is clear that $$ \text{diam} \  E \leq \text{diam} \ \bar{E}.$$
Fix $\epsilon > 0,$ and choose $p \in \bar{E}, q \in \bar{E}.$ By the definition of $\bar{E}$, there are points $p', q',$ in $E$ such that $d(p,q') < \epsilon, d(q,q') < \epsilon.$ Hence 
\begin{align*}
d(p,q) &\leq d(p,p') + d(p',q') + d(q',q)\\
&< 2 \epsilon + d(p',q')\\ 
&\leq 2 \epsilon + \text{diam} \ E.
\end{align*}
It follows that $$ \text{diam} \ \bar{E} \leq 2 \epsilon +  \text{diam} \ E,$$
and since $\epsilon$ was arbitrary, (a) was proved. 
The step prior to the last, namely that $ \text{diam} \ \bar{E} \leq 2 \epsilon +  \text{diam} \ E$, was lost to me. We have $d(p,q) \leq \text{diam} \ \bar{E} $, but how do we know $ \text{diam} \ \bar{E}$ is less than or equal to the term on the right in the previous inequality?

A:

Note that the right hand side of the inequality $d(p, q) < 2\epsilon + \operatorname{diam}E$ is a constant independent of $p$ and $q$, so we see that $2\epsilon + \operatorname{diam}E$ is an upper bound for $\{d(p, q) \mid p, q \in \overline{E}\}$. As such, $\operatorname{diam}\overline{E}$, the least upper bound for $\{d(p, q) \mid p, q \in \overline{E}\}$, is less than or equal to this upper bound. That is, $\operatorname{diam}\overline{E} \leq 2\epsilon + \operatorname{diam}E$.
In general, if $S \subset \mathbb{R}$ and $s \leq M$ for all $s \in S$, then $\sup S \leq M$.

A:

Hint: if $S$ is a non-empty (bounded) set of real numbers, to show $\sup S \le b$, it is sufficient to show that for any $x \in S$, $x \le b$.

