3
$\begingroup$

Definition. Let $E$ be a nonempty subset of $X$, and let $S$ be the set of all real numbers of the form $d(p, q)$, with $p,q\in E$. The sup of $S$ is called the diameter of $E$.

Theorem 3.10. If $\overline{E}$ is the closure of a set $E$ in a metric space $X$, then $$\text{diam }\overline{E} = \text{diam }E.$$

Proof: Fix $\varepsilon>0$, and choose $p, q \in \overline{E}$. By the definition of $\overline{E}$, there are points $p',q' \in E$ such that $d(p,p') < \varepsilon$ and $d(q,q') < \varepsilon$. Hence $$d(p, q) \le d(p,p') + d(p', q') + d(q', q) < 2\varepsilon + d(p', q') \le 2\varepsilon + \text{diam }E.$$

Ok until here. But then they use the inequality above to come up with $$\text{diam }\overline{E} \le 2\varepsilon + \text{diam }E$$

Where I can only see the strict inequality because of the strict inequality relation made in the inequalities above.

$\endgroup$
4
  • $\begingroup$ Definition of closure of $E$: $\overline{E} = E \cup E' $, where $E'$ is the set of limit points of $E$. $\endgroup$ Commented May 23, 2017 at 0:03
  • 1
    $\begingroup$ Doesn't $<$ imply $\leq$? $\endgroup$
    – avs
    Commented May 23, 2017 at 0:03
  • $\begingroup$ Yes, but I think if you have $A \ge B$ (they does it in the beginning of the proof) and $A < B$ you have a contradition? $\endgroup$ Commented May 23, 2017 at 0:05
  • 1
    $\begingroup$ Ok, you seem comfortable getting to this: for every $\epsilon > 0$ and for all $p, q \in \overline{E}$, one has $$ d(p, q) < \text{diam }E + 2 \epsilon $$ Doesn't that imply that $$ \sup_{p, q \in \overline{E}} d(p, q) \leq \text{diam }E + 2 \epsilon? $$ The strict inequality becomes a sharp one upon taking the $\sup$ on the left-hand side. $\endgroup$
    – avs
    Commented May 23, 2017 at 0:27

3 Answers 3

12
$\begingroup$

You're missing an important point. If $x<c$ for all $x\in S$, then $\sup S\le c$. (For example, take $S = (0,1)$. For every $x\in S$, we have $x<1$, but $\sup S = 1$.)

$\endgroup$
5
$\begingroup$

Another subtlety to this proof, just in case anyone missed it. The reason that we know that there are $p, q \in E$ such that $d(p, p') < \epsilon$, $d(q, q') < \epsilon$ is the following:

Every point in $\overline{E}$ is either

(1.) a limit point of $E$

(2.) a point in $E$

If $p$ is a limit point of $E$, then we can find a point $p' \in N_{\epsilon}(p)$ such that $p' \not = p.$ (By the definition of a limit point). And since $p'$ is in that neighborhood with radius $\epsilon$, $d(p, p') < \epsilon$.

If $p \in E$, then we set $p' = p$ and we have that $d(p,p') = 0 < \epsilon$.

$\endgroup$
0
$\begingroup$

The logical flow in a more generic setting makes it clearer.

Given $\epsilon >0$, and a non-empty $S\subseteq \mathbb{R}$.

  1. If for all $s \in S$ we have $s < 2\epsilon+b$, then $2\epsilon + b$ is an upper bound on $S$. (Defn of upper bound).
  2. Then $\sup S \leq 2 \epsilon + b$, as the supremum is always less than or equal to all other upper bounds (Defn of supremum).
  3. Then $\sup S \leq b$. Proof: Consider not, then $\sup S > b$. Then choosing $\epsilon \in (0, \frac{\sup S - b}{2} ) $ implies $2 \epsilon + b < \sup S + b - b = \sup S$, a contradiction to 1.

Identifying $S = \{ d(p, q) \mid p, q \in \bar{E} \}$, and $ b =$ diam $ \bar{E} $ proves the result.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .