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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?

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    $\begingroup$ Since $p',q' \in E$ then $d(p',q') \le \operatorname{diam} E$. $\endgroup$ – copper.hat May 27 '15 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 '15 at 19:31
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    $\begingroup$ 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 '15 at 19:32
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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 $\sup S \leq M$.

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

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