What is the relationship between induced topologies $ \tau _{d_{1}} $, $ \tau_{d_{2}} $ and $ \tau_{d_{3}} $? In $ R^{2} $ consider the following norms $ \Vert (x_{1}, x_{2}) \Vert_{1} = \vert x_{1} \vert + \vert x_{2} \vert$, $ \Vert(x_{1}, x_{2}) \Vert _{2} = \sqrt{x_{1}^{2} + x_{2}^{2}}$  and $ \Vert (x_{1}, x_{2}) \Vert_{3} = max\lbrace  \vert  x_{1} \vert, \vert  x_{2} \vert\rbrace$. Let $ d_{1} $ be the metric induced by $ \Vert  \cdot \Vert_{1} $, $ d_{2} $ the metric induced by $ \Vert  \cdot \Vert_{2} $ and $ d_{3} $ the metric induced by $ \Vert  \cdot \Vert_{3} $. What is the relationship between $ \tau _{d_{1}} $, $ \tau_{d_{2}} $ and $ \tau_{d_{3}} $?
$ \tau _{d_{1}} = \tau_{d_{2}} = \tau_{d_{3}} $ because the metrics are equivalent. This is true?
 A: All norms in finite dimensional vector spaces are equivalent and so the topologies that they induce coincide:
Claim: Let $\Vert \cdot \Vert_1$, $\Vert \cdot \Vert_2$ be two norms on $X$ and let $d_1$ and $d_2$ be the induced metrics. If $\Vert \cdot \Vert_1$ and $\Vert \cdot \Vert_2$ are equivalent, that is, there exist $A, C>0$ such that 
$$ A\Vert \cdot \Vert_1 \leq \Vert \cdot \Vert_2 \leq C\Vert \cdot \Vert_1,$$
then the topologies $\tau_1$ and $\tau_2$ induced by $d_1$ and $d_2$ coincide.
Proof: Notice that the equivalence of norms translates into the following relation between balls with respect to the different metrics: for all $x\in X$ and $r>0$
$$ B_{d_1}(x,\frac{1}{C}r)\subset B_{d_2}(x, r),$$
$$ B_{d_2}(x, Ar)\subset B_{d_1}(x,r).$$
Indeed, let $y\in B_{d_1}(x,\frac{1}{C}r)$ then $d_1(x,y)\leq \frac{1}{C}r$ or equivalently $Cd_1(x,y)\leq r$. This implies that $d_2(x,y)\leq r$ and thus $y\in B_{d_2}(x,r)$. The second containment is proved analogously.
Now let's prove that $\tau_1=\tau_2$.
$\tau_1\subset t_2$: Let $U$ in $\tau_1$. We need to show that $U\in \tau_2$. So let $x\in U$; since $U\in \tau_1$ there exists $r>0$ such that 
$$ x\in B_{d_1}(x,r)\subset U.$$
By the second containment above we have
$$ B_{d_2}(x, Ar)\subset B_{d_1}(x,r)\subset U.$$
Since $x\in U$ was arbitrary , $U$ is open with respect to $d_2$, or equivalently , $U\in \tau_2$.
$\tau_2\subset \tau_1$: This is proved in the exact same way.
