Let $V$ be a vector space over a field $F$ (real or complex), and let $\Vert\cdot \Vert$ be a norm on $V$ . Further, for $r>0$ , $v\in V$ , let $$ B_r(v) = \{ w\in V: \Vert w-v\Vert < r \}. $$ Then for any non-zero $\lambda\in F$ , we have $$ \lambda B_r(v) = B_{|\lambda| r}(\lambda v). $$

The claim is clear for $\lambda =0$ , so we can assume that $\lambda \neq 0$ . Then \begin{eqnarray*} \lambda B_r(v) &=& \{ z\in V: \Vert w-v\Vert < r\ \mbox{and}\ z=\lambda w \} \\ &=& \{ z\in V: \Vert \frac{z}{\lambda}-v\Vert < r \} \\ &=& \{ z\in V: \Vert z-\lambda v\Vert < |\lambda| r \} \\ &=& B_{|\lambda| r}(\lambda v). \end{eqnarray*}