Showing $B_{r}(0)=rB_{1}(0)$, scaling the open ball

Question

Let $X$ be a normed space, and $r > 0$. Show that $B_{r}(0)=rB_{1}(0)$

My idea:

The first inclusion is easy for me: "$\supseteq $" let $y \in rB_{1}(0)$, then there exists $x\in B_{1}(0)$ so that $y=rx$. Hence, $\vert \vert y\vert \vert=r \vert\vert x\vert\vert<r$ and hence$ y \in B_{r}(0)$

I am having more difficulty with the second.

"$\subseteq $" let $x \in B_{r}(0)$, then by definition $\vert \vert x \vert \vert < r$. We also know $x=\frac{x}{\vert\vert x \vert \vert}\vert\vert x \vert \vert$.

And it is clear that: $\vert \vert \frac{x}{\vert\vert x \vert \vert}\vert\vert x \vert \vert\vert\vert<r$ but how do I show that $x \in rB_{1}(0)$?

Answer 1

If $x\in B_r(0)$, then $x=r\dfrac xr$ and $\dfrac xr\in B_1(0)$, since $\displaystyle\left\lVert\frac xr\right\rVert=\frac{\lVert x\rVert}r<1$ .