Prove that all three metrics induces the same topology on $X_1\times X_2$ Prove that if $(X_1,d_1)$ and $(X_2,d_2)$ are metric spaces on $X_1\times X_2$ and metric $d:(X_1\times X_2)\times (X_1\times X_2)\rightarrow R$  is defined in following way:
(i)$d((x_1,x_2),(y_1,y_2))=((d_1(x_1,y_1)^2 +(d_2(x_2,y_2)^2)^{1/2}$ 
(ii)$d((x_1,x_2),(y_1,y_2))=d_1(x_1,y_1) +d_2(x_2,y_2)$ 
(iii)$d((x_1,x_2),(y_1,y_2))=max (d_1(x_1,y_1), d_2(x_2,y_2))$ 
induced that same topology on $X_1\times X_2$
I am beginner in Topology, perhaps we have to take a point in $X_1\times X_2$, and find an open ball containing this point and then have to prove that there exist an open ball in other metric space containing this open ball and vice-versa.
Am I right? 
 A: Equivalent metrics gives the same topology, so we can show that the metrics are equivalent, I'll replace $d(x_1,y_1)=x$ and $d(x_2,y_2)=y$ and show that they are equivalent. Remember 2 metrics are equivalent if $c d_2(x,y)\leq d_1(x,y)\leq C d_1(x,y)$ for some $c,C\in\mathbb{R}$ always holds for all $x,y\in M$. We will show that i and iii are eqivalent and ii and iii are equivalent.
i $=$ iii
we have
$$x^2+y^2 \leq 2\max(x,y)^2$$
which gives us $C=\sqrt{2}$, we also have
$$\max(x,y)^2\leq x^2+y^2$$
so $c=1$ and we're done there
ii $=$ iii
Again we have
$$x+y \leq 2\max(x,y)$$
so $C=2$ and
$$\max(x,y) \leq x+y$$
so $c=1$ again. This shows that all 3 are equivalent and ergo gives the same topology.
A: Yes you are right. It will be enough to show that if a set is open with respect to one metric then it is open with respect to another. 
Symbolically one way to show this would be to show
$$U \,\,\text{Open w.r.t. } d_i \implies U\,\, \text{Open w.r.t. }d_{ii}\implies U \,\,\text{Open w.r.t. }d_{iii}\implies U \,\,\text{Open w.r.t. }d_i$$
For example we can show
$$U \,\,\text{Open w.r.t. } d_i \implies U\,\, \text{Open w.r.t. }d_{ii}$$
in the following way.
As $U$ is open w.r.t. $d_i$ we know for all $u \in U$  there exists an open ball centred at $u$ and contained contained in $U$:
$$B_{d_i}(u, \varepsilon_u) \subseteq U.$$
If we can find a ball $B_{d_{ii}}(u, \delta_u) \subseteq U$ then $U$ will also be open w.r.t. $d_{ii}$. Now
\begin{eqnarray}
(d_1(x_1,x_2)^{2} + d_2(y_1,y_2)^2)^{1/2} \le (d_1(x_1,x_2) + d_2(y_1,y_2)) &\iff&\\
d_1(x_1,x_2)^{2} + d_2(y_1,y_2)^2 \le \Big( d_1(x_1,x_2)^{2} + d_2(y_1,y_2)^2 + 2*d_1(x_1,x_2)d_2(y_1,y_2)\Big)&\iff&\\
-2*d_1(x_1,x_2)d_2(y_1,y_2) \le 0
\end{eqnarray}
The last line is true and we can work our way back up the equivalences and conclude
$$d_i \le d_{ii}.$$
Thus we have 
$$B_{d_{ii}}(u,\varepsilon_u) \subseteq B_{d_i}(u, \varepsilon_u)$$
as if $z \in B_{d_{ii}}(u,\varepsilon_u)$ then 
$$d_{ii}(z,u) <\varepsilon_u$$
and we know
$$d_i(z,u) \le d_{ii}(z,u) < \varepsilon_u \implies d_{i}(z,u) < \varepsilon_u$$
so that $z \in B_{d_i}(u, \varepsilon_u)$.
