# 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 $$p,q \in E$$. 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 $$p,q \in \bar{E}.$$ By the definition of $$\bar{E}$$, there are points $$p', q',$$ in $$E$$ such that $$d(p,q') < \epsilon$$ and $$d(q,q') < \epsilon.$$ 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?

• Since $p',q' \in E$ then $d(p',q') \le \operatorname{diam} E$. May 27, 2015 at 19:28
• I understood that point. I mean that I don't understand how $d(p,q) < 2 \epsilon + \text{diam} \ E$ implies that $\text{diam} \ \bar{E} \leq 2 \epsilon + \text{diam} \ E$ May 27, 2015 at 19:31
• I see what you are asking. If you have $x \le L$ for all $x \in S$, then you have $\sup S \le L$. Here we have $d(p,q) \le ...$, where the right hand side is a fixed quantity, hence it is true for the $\sup.$ May 27, 2015 at 19:32

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 $\sup S \leq M$.
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 $x \in S$, $x \le b$.
It's true for any $$\epsilon$$, so $$\epsilon$$ can be infinitesimally small. If your definition of "less than" has an error bound of $$n$$, choose $$\epsilon = \frac{n}4$$, which is certainly less than $$\frac{\epsilon}2$$, which gives $$2\epsilon$$ is less than or equal to your error bound.