Diameter of a ball in a metric normed space

Question

Could you help me with the following please:

Prove that the diameter of a ball in a normed space it is twice its radius.

My attempt:

$diam(A)=\sup \{d(x,y): x,y\in A\}$

The first inequality is evident $diam(B_\epsilon(a))\leq 2\epsilon$, but for the second I have the following:

Suppose that $diam(B_\epsilon(a))< 2\epsilon$, we have that there exists k such that $diam(B_\epsilon (a))<k<2\epsilon$, we choose $z$ not null and we define $x=a+\dfrac{kz}{2||z||},\ y=a-\dfrac{kz}{2||z||}$, in addition we have that $||x-a||=\dfrac{k}{2}<\epsilon,\ ||y-a||= \dfrac{k}{2}<\epsilon$ and $||x-y||=||k||$

Here it is mentioned that it contradicts the definition of diameter and I would like you to help me understand where the contradiction is, or if you can think of any other demonstration, thank you.

Answer 1

For any $x,y \in B_\epsilon (a)$, $||x - y|| \leq \rm{diam}\, B_\epsilon (a)$. So, $k = ||x - y|| \leq \rm{diam}\, B_\epsilon(a) < k$, a contradiction.

Answer 2

$diam (B_{\epsilon} (a))) \geq \|x-y\|=k $ whenever $k <2\epsilon$. Apply this inequality to $k=2\epsilon-\frac 1 n$ and then let $n \to \infty$ to see that $diam (B_{\epsilon} (a))) \geq 2\epsilon$.