Recent content by Ubistvo

Okay, I read this a little bit fast and I want to clarify some stuff: showing $d(x_n,y_n)\to d(x,y)$ follows because $d(x,y)$ is continuous, hence the result follows right? Now since $d(x_n,y_n)\leqslant \text{diam}(A)$ for all $n,$ as $n\to\infty$ we have $d(x,y)\le \text{diam}(A),$ and the...

Ah, so since $A\subset \overline A,$ then obviously $\text{diam}(A)\le\text{diam}(\overline A)$ and the result follows. Thanks Opalg, you're very helpful!

Let $A\subset X$ for $(X,d)$ metric space, then prove that $\text{diam}(A)=\text{diam}(\overline A).$ I know that $\text{diam}(A)=\displaystyle\sup_{x,y\in A}d(x,y),$ but I don't see how to start the proof. The thing I have is to let $\text{diam}(\overline A)=\displaystyle\sup_{x,y\in \overline...