Diameter of an open ball in a normed space This is probably a silly question, but I'm reading some class notes that have the following proposition:
In general it's true that $\operatorname{diam}( B(x,r) ) \leq 2r $ but in a normed space $\operatorname{diam}( B(x,r) ) = 2r $. 
The diameter is defined as $ \operatorname{diam} A = \sup\{ d(a,b) : a, b \in A \} $ which is the standard definition.
The proving the inequality is trivial, but I can't seem to solve the equality.
According to the text, we take a vector $v$ so that $\|v\| = r$ and then $x + v \in B(x,r) $ and $x - v \in B(x,r) $. From there we get that $d(x + v, x - v) = 2r$. 
What I can't understand is why would $x + v$ and $x - v$ be inside the open ball if the ball is open. 
I imagine there's an error in the statement and it was meant to say otherwise, but maybe I'm missing something.
Thanks.
 A: If you have a non-zero normed vector space over the field $k = \mathbf{R}$ or $\mathbf{C}$ then you always have a unit vector $u$ (that is, a vector of norm $1$) and then vectors $ru$ and $-ru$ have distance $2r$, so that you have equality indeed. On the zero vector space, the equality is false though, as soon as $r>0$.
But beware, this is false in general. Consider the field $k = \mathbf{Q}_p$ of $p$-adic numbers, for a fixed prime number $p$, and let simply $V = k$ endowed with the $p$-adic absolute value $|\cdot|_p$. Consider then $B_F (0, 1)$ for instance, and let $x,y$ be in it. As $|\cdot|_p$ is non-archimedean, you have $d(x,y) = \|x-y\| = |x-y|_p \leq \max(|x|_p,|y|_p)\leq 1$, and the diameter will never reach $2$.
A: you can take $∥v∥ = r - ϵ$ with $0 <ϵ< r.$
 Then put the question: $x-v ∈ B(x,r)$? well the answer is yes because $ ∥ x-(x-v) ∥= ∥ v ∥=r-ϵ < r.$
the same reason for $x+v$ .
then $ ∥ ( x+v)-(x-v) ∥=2r-2ϵ $ now you can take $ ϵ$ small enought to obtain that :supremum is $2r$
