Prove the (open) ball $B_r(a)$ is a subset of $(a-r,a+r)^n$

Question

Let $C_r(a)$=$(a-r,a+r)^n$ be an open cube in $\textbf{R}^n$, I want to

($1$) Show that the (open) ball $B_r(a)\subset C_r(a) \subset B_\sqrt{nr}(a)$

($2$) Define a norm $\left \| \right \|$ in $\textbf{R}^n$ s.t. $C_r(a)= \textbf{B}_r(a)$, where $\textbf{B}_r(a)$ is the ball in the defined norm space.

For ($1$), is it possible to show that the boundary of $B_r(a)$ is contained in $C_r(a)$?

For the ($2$) part, can I define the norm $\left \| a \right \|$ =$a^n$?

Answer 1

Just write the inequalities:

$$x_1^2+\ldots + x_n^2< r^2\implies \max_{1\le i\le n}|x_i|< r$$

But then if the max is less than $r$ all of the $|x_i|< r$, hence $B_r(0)\subseteq (-r,r)^n$. Now shift by $a$ and you get your result. For the second inclusions, just note that $\max |x_i| <r\implies \sum x_i^2< nr^2$ proving the other inclusion. Now show $\displaystyle\lVert x\rVert = \max_{1\le i\le n}|x_i|$ is a norm and your last result follows.