# Diameter

#### Ubistvo

##### New member
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 $\text{diam}(\overline A)=\displaystyle\sup_{x,y\in \overline A}d(x,y),$ so for $x,y\in \overline A,$ there exists sequences $x_n$ and $y_n$ such that $x_n\to x$ and $y_n\to y,$ but I don't know if I'm on the right track.

#### Opalg

##### MHB Oldtimer
Staff member
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 $\text{diam}(\overline A)=\displaystyle\sup_{x,y\in \overline A}d(x,y),$ so for $x,y\in \overline A,$ there exists sequences $x_n$ and $y_n$ such that $x_n\to x$ and $y_n\to y,$ but I don't know if I'm on the right track.
Yes, that is the way to start. You then need to show that $d(x_n,y_n)\to d(x,y)$. But $d(x_n,y_n)\leqslant \text{diam}(A)$ for all $n$, hence $d(x,y)\leqslant \text{diam}(A)$. That holds for all $x,y\in \overline A$, so $\text{diam}(\overline A)\leqslant \text{diam}(A)$. The reverse inequality is obvious.

#### Ubistvo

##### New member
Ah, so since $A\subset \overline A,$ then obviously $\text{diam}(A)\le\text{diam}(\overline A)$ and the result follows.

You then need to show that $d(x_n,y_n)\to d(x,y)$. But $d(x_n,y_n)\leqslant \text{diam}(A)$ for all $n$, hence $d(x,y)\leqslant \text{diam}(A)$.
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 rest follows.