I am working on a problem where I need to prove that, for an open ball $B(x,r) = (x-r, x+r) \subset \mathbb{R}$ has diameter of $2r$ when working with the standard $\mathbb{R}$ metric: $d(x,y)=|x-y|$.
By diameter, I refer to the supremum of the set of all possible values of $d(p,q)$ for $p,q \in B(x,r)$.
I have been proceeding by showing both $\text{diam}(B(x,r)) \leq 2r$ and $\text{diam}(B(x,r)) \geq 2r$.
For $\text{diam}(B(x,r)) \leq 2r$, I was able to take arbitrary $p,q \in B(x,r)$ and apply the triangle inequality to find that $d(p,q) \leq 2r$, thus showing that $2r$ is an upper bound of the set of all distances.
However, I am struggling when working to show $\text{diam}(B(x,r)) \geq 2r$.
I initially thought I could prove by contradiction, and assume $\text{diam}(B(x,r)) <2r$.
I then took $p = c-r+\varepsilon$ and $q=c+r-\varepsilon$ for $\varepsilon >0$,however the triangle inequality ended up just giving me the result $d(p,q) \leq 2r + \varepsilon$ which doesn't help me at all given this is obviously larger than $2r$.
Any tips and hints of how to proceed would be great. 🙂
Best Answer
You have the right idea, but I think you've got yourself muddled up a bit. The definition of supremum is: $s$ is a supremum of the set $A$ if
You showed $2r$ is an upper bound already. Suppose $s'$ is an upper bound on the set of distances. Then, as you say,
$$\forall \varepsilon > 0, \,\, s' > (x+r-\varepsilon) - (x-r+\varepsilon) = 2r-2\varepsilon$$
from which it follows that $s' \ge 2r$.