Diameter of Open Ball in the Reals 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. :)
 A: If $0<\varepsilon<2r$, then $x+r-\frac\varepsilon2,x-r+\frac\varepsilon2\in B(x,r)$. Besides,$$\left|\left(x+r-\frac\varepsilon2\right)-\left(x-r+\frac\varepsilon2\right)\right|=2r-\varepsilon.$$So,$$\operatorname{diam}\bigl(B(x,r)\bigr)\geqslant2r-\varepsilon,$$and therefore, since this takes place for each $\varepsilon\in\left(0,2r\right)$,$$\operatorname{diam}\bigl(B(x,r)\bigr)\geqslant\lim_{\varepsilon\to0}2r-\varepsilon=2r.$$
A: 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

*

*$s$ is an upper bound of $A$.

*If $s'$ is any upper bound of $A$, then $s' \ge s$.

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 $s' \ge 2r$.
A: Since for each $\ n\in\mathbb{N},\ x-r+\frac{1}{n}\in B(x,r)\ $ and $\ x+r-\frac{1}{n}\in B(x,r),\ $ it follows that
$\text{diam}(B(x,r)) \geq \sup\left\{\ d\left(\ x-r+\frac{1}{n},\ x+r-\frac{1}{n}\right):\ n\in\mathbb{N}\right\} = 2r.$
