If $X$ is a real normed linear space and $r>0$, then $B_{r}(x_0+y_0)=B_{r}(y_0)+\{x_0\}$ for fixed $x_0,\,y_0\in X.$

Question

Let $X$ be a real normed linear space and for $r>0$, let $$B_{r}(x_0)=\{x\in X:\|x-x_0\|\leq r\}.$$ I want to prove that $B_{r}(x_0+y_0)=B_{r}(y_0)+\{x_0\}$ for fixed $x_0,\,y_0\in X.$

My attempt

Fix $x_0,\,y_0\in X$ and let $x\in B_{r}(x_0+y_0)$. Then, \begin{align}\|x-(x_0+y_0)\|\leq r&\implies \|x-y_0\|-\|x_0\|\leq r\\&\implies \|x-y_0\|\leq r +\|x_0\|\\&\implies x\in B_{r}(y_0)+\{x_0\}\end{align} So, \begin{align}B_{r}(x_0+y_0)\subseteq B_{r}(y_0)+\{x_0\}\end{align} Conversely, let $x\in B_{r}(y_0)+\{x_0\}$, then \begin{align}\|x_0\|+(\|x-y_0\|\leq r)&\implies \|x-(x_0+y_0)\|\leq \|x_0\|+\|x-y_0\|\leq r\\&\implies\|x-(x_0+y_0)\|\leq r\\&\implies x\in B_{r}(x_0+y_0)\end{align} Thus \begin{align} B_{r}(y_0)+\{x_0\}\subseteq B_{r}(x_0+y_0)\end{align} Therefore, $B_{r}(x_0+y_0)=B_{r}(y_0)+\{x_0\}$ for fixed $x_0,\,y_0\in X.$

Question: Can you, please, check if I'm correct? I'm I wrong at some point? Alternative proofs are also welcome.

Answer 1

$z \in \{x_0\} + B_r(y_0)$ means that $z=x_0 + x$ where $\|x-y_0\| < r$. But then $\|z - (x_0 +y_0)\| = \|(x_0 +x) - (x_0+y_0)\|= \|x - y_0\| < r$ so that $z \in B_r(x_0+y_0)$. So $$\{x_0\} + B_r(y_0) \subseteq B_r(x_0+y_0)$$

On the other hand, if $z \in B_r(x_0+y_0)$, then define $x=z-x_0$. Then

$$\|x - y_0\| = \|z-x_0 -y_0\|=\|z - (x_0+y_0)\| < r$$

so $x \in B_r(y_0)$ and as $z=x_0 + x$ by definition, $z \in \{x_0\} + B_r(y_0)$ showing the reverse inclusion.