If topology induced by two norms are the same, then the norms are equivalent Suppose $\| \cdot \|_{\alpha}$ and $\| \cdot \|_{\beta}$ are two norms on a vector space $X.$
    Show that the two norms are equivalent if and only if the normed spaces $(X,\|\cdot\|_{\alpha})$ and $(X,\|\cdot\|_{\beta})$ have the same open sets.
My attempt:
Denote $\tau_{\alpha}$ and $\tau_{\beta}$ as topology induced by $\| \cdot \|_{\alpha}$ and $\| \cdot \|_{\beta}$ respectively.
$(\Rightarrow):$
Suppose that there exists $c_1>0$ and $c_2>0$ such that $c_1 \| x \|_{\alpha} \leq \|x\|_{\beta} \leq c_2 \| x \|_{\alpha}.$
We wish to show that $\tau_{\alpha} = \tau_{\beta}.$
Let $U_{\alpha} \in \tau_{\alpha}$ and $x \in U_{\alpha}.$
Since $U_{\alpha}$ is open, there exists $\delta>0$ such that $B(x,\delta) \subseteq U_{\alpha}.$
By assumption, we have $B_{\beta}(x,\delta c_1) \subseteq B_{\alpha}(x,\delta) \subseteq U_{\alpha}.$
Hence, $U_{\alpha} \subseteq \tau_{\beta}.$
Let $U_{\beta} \in \tau_{\beta}$ and $x \in U_{\beta}.$
Since $U_{\beta}$ is open, there exists $\delta>0$ such that $B_{\beta}(x,\delta) \subseteq U_{\beta}.$
By assumption, we have $B_{\alpha}(x,\delta/c_2) \subseteq B_{\beta}(x,\delta)\subseteq U_{\beta}.$
Hence, $U_{\beta} \in \tau_{\alpha}.$
Therefore, $\tau_{\alpha} = \tau_{\beta}.$
Is my proof above correct? I do not know how to prove the other direction. 
Any hint would be appreciated. 
 A: Yes, it is correct.
In order to prove the reverse implication, suppose that both spaces have the same open sets. Consider the ball $B_\beta(0,1)$. Since it is an open set and a neighborhood of $0$, there is a $c_1>0$ such $B_\alpha(0,c)\subset B_\beta(0,1)$. In other words, $\|x\|_\alpha<c_1\Longrightarrow\|x\|_\beta<1$. For the same reason, there is a $c_2>0$ such that $\|x\|_\beta<c_2\Longrightarrow\|x\|_\alpha<1$. Therefore, the norms are equivalent. In fact, asserting that $\|x\|_\alpha<c_1\Longrightarrow\|x\|_\beta<1$ is equivalent to asserting that $\|x\|_\alpha\leqslant c_1\Longrightarrow\|x\|_\beta\leqslant1$, which is equivalent to $\|x\|_\alpha\leqslant1\Longrightarrow\|x\|_\beta\leqslant\frac1{c_1}$. So, if $x\neq0$, then $\left\|\frac x{\|x\|_\alpha}\right\|_\alpha=1$ and so $\left\|\frac x{\|x\|_\alpha}\right\|_\beta\leqslant\frac1{c_1}$. This proves that $\|x\|_\beta\leqslant\frac1{c_1}\|x\|_\alpha$, which is still true if $x=0$. The other inequality has a similar proof.
A: In this answer, I discuss in general the conditions for two metrics (so more general than norms) to be equivalent in the topology sense. 
We can apply this to the norm-based metrics $d_1(x,y) = ||x-y||_\alpha$ and $d_2(x,y) = ||x-y||_\beta$, at the point $0$ and $\epsilon =1$ say, to get that if
the topologies induced by $d_1$ and $d_2$ are the same there exists $\delta_1 > 0$ such that $B_{\delta_1}(0, \delta_1) \subseteq B_{d_2}(0, 1)$ and $\delta_2 > 0$ such that $B_{\delta_2}(0, \delta_1) \subseteq B_{d_1}(0, 1)$.
Now use norm properties like $||tx|| = t\cdot||x||$ for $t>0$ to use these inclusions to find your $c_1$ and $c_2$: (scaling trickery)
Suppose that we have some $x \neq 0$, then $||x||_\alpha > 0$. Then the vector $tx$ with $t = \frac{\delta_1}{2||x||_\alpha}$ has the property that $$d_\alpha(0, tx) = ||tx||_\alpha = t\cdot ||x||_\alpha = \frac{\delta_1}{2} < \delta_1$$ which says that $tx \in B_{d_1}(0,\delta_1) \subseteq B_{d_2}(0,1)$, and this again says that $||tx||_\beta < 1$. So pulling $t$ out the $||.||_\beta$ now:
$$\frac{\delta_1}{2||x||_{\alpha}}||x||_\beta < 1 \rightarrow \frac{\delta_1}{2}||x||_\beta \le ||x||_\alpha \leftrightarrow ||x||_\beta \le \frac{2}{\delta_1}||x||_\alpha $$ where the last inequality also holds for $x=0$ (hence the $\le$). So $c_2= \frac{2}{\delta_1}$ works. 
Try to find $c_1$ yourself from the other ball inclusion with $\delta_2$ etc.
It's a nice exercise. 
