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?

  • 1
    $\begingroup$ You could just show that all the metrics are equivalent, equivalent metrics gives equivalent topologies. $\endgroup$ – Zelos Malum Feb 6 '16 at 10:55
  • $\begingroup$ You can prove that this induces in each case the product topology. $\endgroup$ – user42761 Feb 6 '16 at 10:55
  • $\begingroup$ 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. $\endgroup$ – 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 $z \in B_{d_i}(u, \varepsilon_u)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.