8
$\begingroup$

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?

$\endgroup$
3
  • 2
    $\begingroup$ Since $p',q' \in E$ then $d(p',q') \le \operatorname{diam} E$. $\endgroup$
    – copper.hat
    May 27, 2015 at 19:28
  • $\begingroup$ I understood that point. I mean that I don't understand how $d(p,q) < 2 \epsilon + \text{diam} \ E$ implies that $\text{diam} \ \bar{E} \leq 2 \epsilon + \text{diam} \ E$ $\endgroup$
    – user110503
    May 27, 2015 at 19:31
  • 2
    $\begingroup$ 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 $\sup.$ $\endgroup$
    – copper.hat
    May 27, 2015 at 19:32

3 Answers 3

9
$\begingroup$

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 $\sup S \leq M$.

$\endgroup$
0
6
$\begingroup$

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 $x \in S$, $x \le b$.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.