# Proof diameter of a set equals diameter of its closure

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.)

• Dec 19, 2018 at 5:27
• I'm asking if a specific proof is valid, the proof in question is not contained/discussed in the thread you linked to. Dec 19, 2018 at 5:35
• 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.
– user507295
Dec 19, 2018 at 5:47
• 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? Dec 19, 2018 at 5:52
• 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.
– user507295
Dec 19, 2018 at 6:29