Proof of Rudin's Theorem 3.10

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.

• Definition of closure of $E$: $\overline{E} = E \cup E'$, where $E'$ is the set of limit points of $E$. Commented May 23, 2017 at 0:03
• Doesn't $<$ imply $\leq$?
– avs
Commented May 23, 2017 at 0:03
• 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? Commented May 23, 2017 at 0:05
• 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.
– avs
Commented May 23, 2017 at 0:27

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 $\sup S = 1$.)

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 $$d(p,p') = 0 < \epsilon$$.

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.