Diameter closure set

Question

Let $(X,d)$ be a metric space. Let $A$ be a subset of X.

My question is when $diam(A)=diam(\bar{A})$?

Such that: $$diam(A)= sup\{d(x,y): x, y \in A\}$$

Answer 1

If you correct your definition ($\sup$ instead of $\inf$) then the equality always holds. Indeed, $diam(\bar{A}) \ge diam(A)$ and for every $x$, $y$ in $\bar{A}$, let $(x_n)$ and $(y_n)$ two sequences of $A$ converge to $x$ and $y$. You have $d(x_n, y_n) \le diam(A)$ so $d(x,y) \le diam(A)$ and this implies $diam(\bar A) \le diam(A)$

Answer 2

It happens always. Suppose otherwise, that is, suppose that, for some set $A$, $\operatorname{diam}(A)<\operatorname{diam}\left(\overline A\right)$. Then there are $x,y\in\overline A$ such that $\operatorname{diam}(A)<d(x,y)$. Let $(x_n)_{n\in\mathbb N}$ and $(y_n)_{n\in\mathbb N}$ be sequences of elements of $A$ such that $\lim_{n\to\infty}x_n=x$ and that $\lim_{n\to\infty}y_n=y$. Then$$(\forall n\in\mathbb{N}):d(x_n,y_n)\leqslant\operatorname{diam}(A)<\operatorname{diam}\left(\overline A\right).$$But$$\lim_{n\to\infty}d(x_n,y_n)=d(x,y)>\operatorname{diam}(A).$$