There exists points in closure of set whose distance is diameter of set

Assume that diameter of set is defined as $$\displaystyle \text{diam}(A) = \sup_{x,y \in A} d(x,y).$$

Prove that if $$\text{diam}(A) < \infty$$ , show that there exists $$x',y' \in A^{-}$$ s.t. $$d(x,y) = \text{diam}(A)$$, where $$A^{-}$$ is closure of $$A$$

Since $$\text{diam}(A)$$ is $$\sup,$$ there exists a sequence $$(x_k,y_k)$$ s.t. $$\displaystyle \lim_{k \rightarrow \infty} d(x_k,y_k) = \text{diam}(A).$$ Now since $$\text{diam}(A) < \infty$$ and $$d(.,.)$$ is continuous we have that $$\displaystyle \lim_{k \rightarrow \infty} d(x_{k}, y_{k}) = d(\lim_{k \rightarrow \infty} x_k , \lim_{k \rightarrow \infty}y_k) = d(x_0,y_0)$$ ,so $$x_0,y_0 \in A^{-}$$ s.t $$d(x_0,y_0) = \text{diam}(A).$$

And we are done.

Is there any thing wrong with above argument?

EDIT : assume we are dealing with euclidean space.

As @GreginGre mentioned, first we need to show that sequences actually converge. since we have $$\sup < \infty$$ we have that $$x_k,y_k$$ are bounded. now use Bolzano Weierstrass theorem to get convergent subsequence. let us rename convergent subsequence to $$x_k,y_k$$ again and it converges to $$x_0,y_0$$, now only part that requires justification is $$\lim_{n \rightarrow \infty} d(x_k,y_k) = diam(A)$$. This is true because subsequence converges to same limit.

• Yes, there is. You didn't prove that the sequences $(x_k)$ and $(y_k)$ are convergent (there are probably not...) Sep 20 '20 at 14:18
• @GreginGre, But can you give example of sequences s.t $d(x_k,y_k)$ (or more generally continuous function) is convergent but $x_k,y_k$ are not ?? Sep 20 '20 at 14:48
• Take $\mathbb{R}$ with classical distance, $A=[-1,1]$ and $x_k=(-1)^k, y_k=-(-1)^k$. Then $d(x_k,y_k)=2=diam(A)$ and $(x_k), (y_k)$ both diverge. Sep 20 '20 at 15:15

You didn't prove that $$(x_k)$$ an $$(y_k)$$ both converge, and in fact you have examples where they do not converge.
For example, take $$\mathbb{R}$$ with classical distance, $$A=[-1,1]$$ and $$x_k=(-1)^k, y_k=-(-1)^k$$. Then $$d(x_k,y_k)=2=diam(A)$$ and $$(x_k), (y_k)$$ both diverge.
Hint. Show the set $$\{(x,y)\in A\times A\mid d(x,y)=diam(A)\}$$ is closed and bounded.
It would be good to have the setting. I'll assume your set is a subset of some metric space. As the commmenter noted, you haven't yet shown $$x_0$$ and $$y_0$$ exist, which is to say, we need to see there are some points $$x_0$$ and $$y_0$$ that your sequence is converging to.