If $X$ is a real normed linear space, then $B(0,r)=r B(0,1)$

Question

If $X$ is a real normed linear space, I want to prove that $$B_{r}(0)=r B_{1}(0)$$

My proof

Let $z\in rB(0,1)$, there exists $x\in B(0,1)$ such that $z=rx$.

$$\|z-0\|=\|rx-0\|=r\|x-0\|<r$$ So, $z\in B(0,r)$ and $$rB(0,1)\subseteq B(0,r).$$ On the other hand, let $z\in B(0,r)$, then $\|z-0\|<r$. This implies that
$$\bigg\|\dfrac{z}{r}-0\bigg\|<1.$$ Thus, $\dfrac{z}{r}\in B(0,1)$ and there exists $x\in B(0,1)$ such that $$\dfrac{z}{r}=x\iff z=rx \in rB(0,1)$$ Hence, $$ B(0,r) \subseteq rB(0,1).$$ Therefore, $$ B(0,r) = rB(0,1).$$

Question

Kindly help check if my proof is correct.

Answer 1

The proof is totally fine. I would have written it thusly:

Suppose $r>0$. Let $x \in B_r(0)$. Then define $y=\frac{1}{r}\cdot x$. Then

$$\|y-0\| = \|y\|=\frac{1}{r}\|x\|= \frac{1}{r}\|x-0\|< \frac{1}{r} \cdot r = 1$$

so that $y \in B_1(0)$ and as $r\cdot y = r \cdot (\frac{1}{r}\cdot x)= (r \frac{1}{r}) \cdot x = 1\cdot x = x$, we have that $x \in rB_1(0)$. So $B_r(0) \subseteq rB_1(0)$.

On the other hand, if $x\in rB_1(0)$ then $x=ry$ for some $y \in B_1(0)$ and then

$$\|x-0\|=\|x\|=\|ry\| = r\|y\| < r \cdot 1= r$$ and so $x \in B_r(0)$. So the other inclusion $rB_1(0) \subseteq B_r(0)$ also holds.

Answer 2

Your proof is fine. Everything is O.K.