# 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?

• You could just show that all the metrics are equivalent, equivalent metrics gives equivalent topologies. – Zelos Malum Feb 6 '16 at 10:55
• You can prove that this induces in each case the product topology. – user42761 Feb 6 '16 at 10:55
• You are absolutely right. Looking closer into that direction, you will notice however that what Zelos suggests is sufficient for this. Recall that metrics $d,d'$ are called equivalent if thee exist constants $c_1,c_2>0$ such that for all $x,y$ we have $c_1d(x,y)\le d'(x,y)\le c_2d(x,y)$. Being equivalent as metrics is the most typical (but not the only) way for two metrics to induce the same topology. – Hagen von Eitzen Feb 6 '16 at 10:59

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.

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 $z \in B_{d_i}(u, \varepsilon_u)$.