Proving open set is equivalent across normed spaces Let $A$ be an open set according to $\vert\vert \cdot \vert\vert^{A}$.
Prove that $A$ is open according to a different norm $\vert\vert \cdot \vert\vert^{B}.$
Idea: Since $A$ is open, it follows that for all $a \in A$, $\exists r>0$, such that $B_{r}(a)^{{\vert\vert \cdot \vert\vert}^{A}}\subset A$.
Since all norms are equivalent $\exists c>0$ such that $c\vert\vert x-a \vert\vert^{B}\leq\vert\vert x-a \vert\vert^{A}<r$.
Now I am trying to prove that $B_{\frac{r}{c}}(a)^{\vert\vert \cdot \vert\vert^{B}}\subset A$, but cannot seem to prove it. Any ideas?
 A: All norms are always equivalent in a normed space if it is finite dimensional, so if you assume this property than you can prove that all norms  induce the same Topology.
Assume that $d_1 $ and $d_2$ are distances  on a  topological space X such that exists $C\neq 0$ for which for every $x,y\in X$ than $d_1(x,y)<Cd_2(x,y)$. 
Than $\tau_1\subset\tau_2$ where $\tau_1$ and $\tau_2$ are the respectively Topology induced by themselves.
We can prove it only for open set that are balls because the family of the open balls is a base for the respective topology. 
If $y\in B^{d_1}(x,r)$ than $y\in B^{d_2}(y,s_y)\subset B^{d_1}(x,r)$ where $s_y=\frac{1}{C}(r-d_1(x,y))$ because for every $z\in B^{d_2}(y,s_y)$ you have that $d_1(x,z)<d_1(x,y)+d_1(y,z)<$
$<d_1(x,y)+Cd_2(y,z)<$
$<d_1(x,y)+\frac{1}{C}s_y=$
$=d_1(x,y)+C\frac{1}{C}(r-d_1(x,y))=r$
So $B^{d_1}(x,r)=\bigcup_{y\in B^{d_1}(x,r) } B^{d_2}(y,s_y)\in \tau_2 $
and so $\tau_1\subset \tau_2$.
Now if your normed space X is finite dimensional than the two norms $||\cdot||_1 $ and $||\cdot||_2$ are equivalent and so there exists $k_1,k_2\neq 0$ such that for every $x\in X$
 $k_1||x||_2\leq||x||_1\leq k_2||x||_2$. 
If you consider the distance $d_1$ and $d_2$ induced by themselves than you have that :


*

*there exists $k_2\neq 0$ such that or every $x,y\in X$ than $d_1(x,y)\leq k_2d_2(x,y)$ and  $\tau_1\subset \tau_2$;

*there exists $k_1\neq 0$ such that or every $x,y\in X$ than $d_2(x,y)\leq \frac{1}{k_1}d_1(x,y)$ and  $\tau_2\subset \tau_1$
Than $\tau_1=\tau_2$
