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Since $E \subseteq cl(E) $, then it is immediate that diam $(E) \leq $ diam(cl($E))$.
I only need to show that assuming diam $(E) < $ diam(cl($E))$ will lead to contradiction then I can conclude that diam $(E)= $ diam(cl($E))$.

So suppose that diam(cl($E$)) > diam($E$). Then there exist a $p,q \in $ cl($E$) such that $ d(p,q) > $diam($E$). By def of cl($E$), there exist a sequence of $\{p_n\}, \{ q_n \} \in E$ such that $ p_n , q_n \to p, q ~~ $ respectively as $ n \to \infty $.

Also note that by triangle inequality ,we have

$$ (1) ~~~~~~ d(p,q) - [d(p,p_n) + d(q,q_n)] \leq d(p_n,q_n) \text{ for all } n.$$

Since $ p_n, q_n \to p,q $ , we have $ d(p,p_n) + d(q,q_n) \to 0 $ as $ n \to \infty$.

Since $ d(p,q) > $ diam(cl($E$)), we can choose some $p_n, q_n $ such that $$ d(p,q) - [d(p,p_n) + d(q,q_n)] > \text{diam}(E).$$ Then by (1) above, we have $$ d(p_n,q_n) > \text{diam}(E),$$ a contradiction. Thus the result holds.

Is my proof correct? and is there a shorter way to do this?? thank you very much.

Alex Pozo
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Khoa ta
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    It looks correct,if somewhat lengthy. Clever use of the sequential definition of closure. – Mathemagician1234 Jun 18 '16 at 21:09
  • @Mathemagician1234 Thank you, I can only think of this, do you know if there is a shorter proof?? – Khoa ta Jun 18 '16 at 21:14
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    Working on it.........lol – Mathemagician1234 Jun 18 '16 at 21:15
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    Under the relation $(1)$, I think you wanted to say that $d(p,q)>diam(E)$, no ? Then, I don't really see why you can choose $p_n,q_n$ s.t. $d(p,q)-[d(p,p_n)+d(q+q_n)]>diam(E)$. – Surb Jun 18 '16 at 21:17
  • @Surb thank you for the input, $ d(p,q) > $ diam(E) is the result of my assumption. I think can conclude what you ask since I can choose $p_n,q_n$ so that $ d(p,p_n) + d(q,q_n) < \frac{d(p,q) - diam(E)}{2}$, then it follows from this the result you ask, or at least that's what I think. – Khoa ta Jun 18 '16 at 21:24
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    @Surb My bad for not spotting the error,but your basic reasoning still looks ok. I'd check it carefully again. – Mathemagician1234 Jun 18 '16 at 21:27
  • @Mathemagician1234 it's ok man, but does my explain to "Surb" seem correct to you? It looks correct to me since I thought eventually $d(p,p_n) + d(q,q_n)$ would approach 0 so you can make the expression $ d(p,q) - (d(p,p_n) + d(q,q_n) ) $as close to $d(p,q)$ as you like. – Khoa ta Jun 18 '16 at 21:29
  • @Khoata Looks ok to me too. I posted a shorter proof of the second step below. – Mathemagician1234 Jun 18 '16 at 22:02

3 Answers3

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Suppose that $x,y\in\operatorname{cl}(E)$ and fix $\varepsilon>0$. Then, there exist $a,b\in E$ such that \begin{align*} d(a,x)<&\,\frac{\varepsilon}{2},\\ d(b,y)<&\,\frac{\varepsilon}{2}. \end{align*} The triangle inequality then implies that $$d(x,y)\leq d(x,a)+d(a,b)+d(b,y)\leq d(x,a)+\underbrace{\sup_{\hat a,\hat b\in E}d(\hat a,\hat b)}_{=\operatorname{diam}(E)}+d(b,y)\leq\operatorname{diam}(E)+\varepsilon.$$ Taking supremum over $x,y\in\operatorname{cl}(E)$, one has that $$\operatorname{diam}(\operatorname{cl}(E))=\sup_{x,y\in\operatorname{cl} E}d(x,y)\leq\operatorname{diam}(E)+\varepsilon.$$ Since $\varepsilon>0$ is arbitrary, it follows that $$\operatorname{diam}(\operatorname{cl}(E))\leq\operatorname{diam}(E).$$

triple_sec
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Suppose $diam(\overline{E})>diam(E)$. Then $diam(\overline{E}-E)> 0$. Therefore, for every $p,q\in \overline{E} - E$ , $\exists d\in \mathbb R\geq 0$ such that $d(p,q)< d$. Now consider the following: By the definition of closure, $\overline{E}=E\cup E'$ where $E' = \{p| p \text{ is an accumulation point of E}\}$. Therefore, $p,q \in \overline{E}- E $ are accumulation points of $E$ where $p,q\notin E$. Clearly, there exists an open ball where $q\in B_d(p)\subset \overline{E} -E$. Let $N_p(q)$ be a neihborhood of $p$ containing $q$ in $\overline{E}-E$. Then since $p$ is an accumulation point of $E$, there exists $z\in E$ such that $z\in B_l(p)\subset N_p(q)$. where $B_l(p)\subset N_p(q)$ where $l\leq d\in \mathbb R \geq 0 $.

Let $r= diam(E)$. Then:

$$d(p,q) \leq d(p,z) + d(z,q) = l + d = r. $$

But this means $diam(\overline{E}) \leq diam(E)$ and we have a contradiction! Q.E.D.

Alex Pozo
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Mathemagician1234
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  • Thank you for your answer, but I have a question about this, by your claim that "there exist an open ball $B_d(p) \subseteq $ cl($E$) - $E$ , since $p$ is an accumulation point of $E$, then $ B_d(p)$ contain a point $x$ in $E$ but then this will violate the fact that $B_d(p) \subseteq $ cl($E$) - $E$ ?? I'm sorry if I don't understand your idea. – Khoa ta Jun 18 '16 at 22:45
  • @Khoata cl(E) - E is defined as the set of all points in the closure of E which does not contain any points of E i.e. cl(E) $\cap E^c$. Therefore if the open ball contains a point of E, then the ball is not a subset of cl(E)-E and we have a contradiction. Clear? – Mathemagician1234 Jun 18 '16 at 23:12
  • thank you for the reply, but how do you know cl($E$) - $E$ is an open set so that you can conclude for any point in the set there is an open ball contain in that set?? I know closure of a set is closed but I don't see how that leads to that fact. – Khoa ta Jun 18 '16 at 23:22
  • I kept thinking about what I asked you yesterday but I still don't get how you can conclude there is an open ball in cl($E$) $- E$, but here is my counterexample, suppose $E$ is the interval $ (0,1) \in \mathbb{R} $, then you can see that cl($E$) $ - E$= $ {0,1 } $ and by def, diam(cl($E$)) $= 1 - 0 = 1 > 0$. But as you can see there is no open ball contain in cl($E$) $ - E$. – Khoa ta Jun 20 '16 at 00:59
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Suppose there exists $x, y \in \bar{E}$ such that $r = \rho(x, y) - \text{diam}(E) > 0$. Then there exists $u \in E \cap B(x, \frac{r}{2})$ and $v \in E \cap B(y, \frac{r}{2})$ (since $x, y \in \bar{E}$). Therefore \begin{align*} && \rho(x, y) - \rho(u, v) \leq \rho(x, u) + \rho(y, v) &< r = \rho(x, y) - \text{diam}(E) && \\ \Longrightarrow && \text{diam}(E) &< \rho(u, v), && \end{align*} a clear contradiction. Thus, $\rho(x, y) \leq \text{diam}(E)$ for all $x, y \in \bar{E}$ and so $\text{diam}(\bar{E}) \leq \text{diam}(E)$.

Regular space
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