Proving $\text{diam}(A)=\text{diam}(\overline A)$ for Metric Spaces

[Ubistvo]

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]

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]

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!

 

[Ubistvo]

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.

Is that how you did it?

 

[blue_raver22]

I would approach this problem by first defining the concept of diameter for a metric space. In a metric space, the diameter of a set $A$ is defined as the supremum of the distance between any two points in $A$. This means that $\text{diam}(A)=\displaystyle\sup_{x,y\in A}d(x,y)$.

Next, I would define the closure of a set $A$ in a metric space as the set of all limit points of $A$. In other words, $\overline A$ is the set of all points that can be approached arbitrarily closely by points in $A$.

Now, to prove that $\text{diam}(A)=\text{diam}(\overline A)$, we need to show that for any two points $x,y\in A$, the distance between them is the same as the distance between their limit points in $\overline A$. In other words, we need to show that $\displaystyle\sup_{x,y\in A}d(x,y)=\displaystyle\sup_{x,y\in \overline A}d(x,y)$.

To do this, we can use the fact that for any two points $x,y\in A$, there exists sequences $x_n$ and $y_n$ in $A$ such that $x_n\to x$ and $y_n\to y$. This means that the distance between $x$ and $y$ can be approximated by the distance between $x_n$ and $y_n$ for large enough $n$. Since $x_n$ and $y_n$ are in $A$, we have $\displaystyle\sup_{x,y\in A}d(x,y)\geq\displaystyle\sup_{x,y\in \overline A}d(x,y)$.

On the other hand, since $x_n$ and $y_n$ are in $\overline A$, we also have $\displaystyle\sup_{x,y\in \overline A}d(x,y)\geq\displaystyle\sup_{x,y\in A}d(x,y)$. This is because the limit points in $\overline A$ are closer to each other than the points in $A$.

Combining these two inequalities, we get $\displaystyle\sup_{x,y\in A}d(x,y)=\