Let $(E_1,d_1)$, $(E_2,d_2)$ two metric spaces and $E=E_1\times E_2$ the metric space product. Consider the following metrics in $E$ \begin{eqnarray*} d^{(\infty)}(\mathbf{x},\mathbf{y}) & = & \max(d_1(x_1,y_1),d_2(x_2,y_2)) \\ d^{(1)}(\mathbf{x},\mathbf{y}) & = & d_1(x_1,y_1)+d_2(x_2,y_2) \\ d^{(2)}(\mathbf{x},\mathbf{y}) & = & \sqrt{d_1^2(x_1,y_1)+d_2^2(x_2,y_2)} \\ \end{eqnarray*}I already showed that the topology inducing the three metrics in E are equal, that I did testing that metrics are equivalent.

Now if $\mathcal{O}_1$ is the collection of all open sets of $E_1$ y $\mathcal{O}_2$ is the collection of all open sets of $E_2$.

It is true that the set $$\mathcal{G}=\{A_1\times A_2\subseteq E_1\times E_2:A_1\in \mathcal{O}_1,\,\,A_2\in \mathcal{O}_2\}$$ is a basis for the product topology?

How I can test this? Or how I refute it?

I have a suggestion: Show that the topology induced by any metric $d^{(\infty)}, d^{(1)},d^{(2)}$ is the same that the product topology.

But do not understand why the suggestion test proves my exercise also not know how to test the suggestion

Definition of product topology. The product topology is the collection of sets of $E$ that are unions of products of the form $U_1\times U_2$ with $U_1\in \mathcal{O}_1,\,\,U_2\in \mathcal{O}_2$

  • $\begingroup$ My consfusion is that by definition the set $\mathcal{G}$ is a basis of the product topology then there's nothing to prove, but my analysis teacher asks me to try the suggestion above $\endgroup$ – Roiner Segura Cubero Jun 29 '13 at 19:35
  • $\begingroup$ I think what your teacher wants is to show that these metrics give the product topology as you have defined it. $\endgroup$ – tomasz Jun 29 '13 at 19:56

I agree with @tomasz: I think that you’re supposed to show that each of the three metrics generates the product topology. If you’ve already shown that the metrics are equivalent, it will be enough to show that one of them generates the product topology. I would use $d^{(\infty)}$, since for any $p=\langle x,y\rangle\in E$ and $\epsilon>0$ we have

$$B_{d^{(\infty)}}(p,\epsilon)=\{\langle u,v\rangle\in E:d_1(x,u)<\epsilon\text{ and }d_2(y,v)<\epsilon\}=B_{d_1}(x,\epsilon)\times B_{d_2}(y,\epsilon)\;.$$

Added in response to comments: Suppose that $p=\langle x,y\rangle\in G=U_1\times U_2\in\mathscr{G}$. $U_1$ is open in $E_1$, so there is an $\epsilon_1>0$ such that $B_{d_1}(x,\epsilon_1)\subseteq U_1$. Similarly, there is an $\epsilon_2>0$ such that $B_{d_2}(y,\epsilon_2)\subseteq U_2$. Let $\epsilon=\min\{\epsilon_1,\epsilon_2\}$. What can you say about $B_{d^{(\infty)}}(p,\epsilon)$?

  • $\begingroup$ this idea is used to prove that every element of the topology T is contained in the product topology... $\endgroup$ – Roiner Segura Cubero Jun 29 '13 at 20:29
  • $\begingroup$ @Roiner: Right, and that’s half of what you need. The other half is to show that every member of $\mathscr{G}$ is open in the topology generated by $d^{(\infty)}$, which also uses the fact that I mentioned. $\endgroup$ – Brian M. Scott Jun 29 '13 at 20:30
  • $\begingroup$ This is exactly the part that I can not do.. $\endgroup$ – Roiner Segura Cubero Jun 29 '13 at 20:45
  • $\begingroup$ How are the balls in the product topology? $\endgroup$ – Roiner Segura Cubero Jun 29 '13 at 20:46
  • $\begingroup$ I have tried to do is choose an open set in the product topology and prove that this set is open in the topology d but not how to write $\endgroup$ – Roiner Segura Cubero Jun 29 '13 at 20:48

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