Question with regards to proof about diameter of a closed ball.

Question

I have a question with regards to my understanding: Consider the following: If a,z $\in$ X and r,s$\in$ $\mathbb{R}^+$ then diam(B[a,r]) $\leq$ 2r. Where B[a,r] is the closed ball and X is a metric space, why is it okay to start the proof with the following: "Let x,y $\in$ X and d(a,x) $\leq$ r and d(a,y) $\leq$ r.."? what if X or the ball has only one element? Is it because im taking arbitrary elements that means i'm not excluding the possibility that x=y? and so it won't matter if I consider even more than 2 as well? Could someone please explain further, in terms of logic (if possible?, please?)

Answer 1

You start by picking two arbitrary elements of $B[a,r]$, because we are proving the statement $\forall x,y \in B[a,r]: d(x,y) \le 2r$ which will imply the diameter fact immediately. We make no assumptions on $x$ or $y$ beyond what is given by the definition of $B[a,r]$, namely $d(x,a) \le r$. They could indeed by equal. Logically speaking, $B[a,r]$ could even be empty, as a statement of the form $\forall x,y \in A: \phi(x)$ is always true. But of course it's not, in this particular case (but we don't need it), as $a \in B[a,r]$ when $r \ge 0$.

But $\operatorname{diam}(A)$ is only well-defined for $A \neq \emptyset$ as all elements of $\mathbb{R}$ are upper bounds for $\emptyset=\{d(x,y): x,y \in \emptyset\}$ and so the least upper bound does not exist (or is defined to be $-\infty$), so it's comforting that that $B[a,r]$ is non-empty, as guaranteed by $r\ge 0$ (if $r <0$ the metric space axioms imply that it is empty).