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Is this a valid proof?

Question :

Show that $\mathrm{diam}(A)=\mathrm{diam}(\bar{A})$

proof:

$\forall \, x \in \bar{A},\forall\, \delta\,>0,\, B_\delta(x)\backslash x \, \cap A \neq \varnothing$. This implies $\forall x \in \bar{A},\, \exists \{x_n\}_0^\infty \in A :\lim_{n\to \infty}x_n=x$

Let $\mathrm{diam}(\bar{A}) = \sup\{d(a,b):a,b\in \bar{A}\}$: then $$ \exists \{x_n\}_0^\infty, \{y_n\}_0^\infty : \lim_{n\to\infty}\{x_n\}=a, \quad \lim_{n\to \infty}\{y_n\}=b \text{ and } x_n,y_n\in A $$ and thus, $d(\lim_{n\to \infty}\{x_n\},\lim_{n\to \infty}\{y_n\})=d(a,b)$. Since every $x\in A$ is also in $\bar{A}$, $$ \mathrm{diam}(A)=\mathrm{diam}(\bar{A}). $$

Edit: I want to know if a specific proof is valid, the proof I am asking about is not mentioned or discussed in the 3 year old link referenced as a duplicate by Yadati Kiran (Confused about proof that diameter of a closure of a set is the same as the diameter of the set.)

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  • $\begingroup$ Possible duplicate of Confused about proof that diameter of a closure of a set is the same as the diameter of the set. $\endgroup$
    – Yadati Kiran
    Dec 19, 2018 at 5:27
  • $\begingroup$ I'm asking if a specific proof is valid, the proof in question is not contained/discussed in the thread you linked to. $\endgroup$
    – Frank
    Dec 19, 2018 at 5:35
  • $\begingroup$ You say "let $diam(A) = \sup\{d(a,b):a,b\in \bar{A}\}$." Do you mean $A$ instead of $\bar{A}$? It's also a little weird that you're using the word "let" for a definition that is assumed in the question. $\endgroup$
    – user507295
    Dec 19, 2018 at 5:47
  • $\begingroup$ Thanks for finding that error. Also, in reference to using "let", I did it as an easy way to define $a$ and $b$. What would be a better way? Is there another faster way? $\endgroup$
    – Frank
    Dec 19, 2018 at 5:52
  • $\begingroup$ Oh, I see. That still sounds weird to me, but I'm no professional. More importantly, I feel that "since every $x \in A$ is also in $\bar{A}$" needs to be more justified because it's true that every $x \in \emptyset$ is in $\bar{A}$ but it's not generally true that $diam(\emptyset) = diam(\bar{A})$. Yes, you have elements in $A$ that are arbitrarily close to elements in $\bar{A}$ but I don't feel like you've made that explicit enough to be a proof. $\endgroup$
    – user507295
    Dec 19, 2018 at 6:29

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