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.

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    $\begingroup$ It looks correct,if somewhat lengthy. Clever use of the sequential definition of closure. $\endgroup$ – Mathemagician1234 Jun 18 '16 at 21:09
  • $\begingroup$ @Mathemagician1234 Thank you, I can only think of this, do you know if there is a shorter proof?? $\endgroup$ – Khoa ta Jun 18 '16 at 21:14
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    $\begingroup$ Working on it.........lol $\endgroup$ – Mathemagician1234 Jun 18 '16 at 21:15
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    $\begingroup$ 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. $d(p,q)-[d(p,p_n)+d(q+q_n)]>diam(E)$. $\endgroup$ – Surb Jun 18 '16 at 21:17
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    $\begingroup$ @Surb My bad for not spotting the error,but your basic reasoning still looks ok. I'd check it carefully again. $\endgroup$ – Mathemagician1234 Jun 18 '16 at 21:27

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 $$\operatorname{diam}(\operatorname{cl}(E))\leq\operatorname{diam}(E).$$


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

  • $\begingroup$ 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. $\endgroup$ – Khoa ta Jun 18 '16 at 22:45
  • $\begingroup$ @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? $\endgroup$ – Mathemagician1234 Jun 18 '16 at 23:12
  • $\begingroup$ 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. $\endgroup$ – Khoa ta Jun 18 '16 at 23:22
  • $\begingroup$ 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$. $\endgroup$ – Khoa ta Jun 20 '16 at 0:59

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