Cartesian product of open sets is a open set

[ELESSAR TELKONT]

Homework Statement

This is not really coursework. Instead, this is some sort of curiosity and proposition formulation rush. Then the initial questions are that if this is a valid result that is worth to be proven.

Let X,Y be metric spaces and X\times Y with another metric the product metric space. A\subseteq X, B\subseteq Y be open sets, then A\times B\subseteq X\times Y is open.

Homework Equations

Definition of open and metric properties.

The Attempt at a Solution

The problem is possible to be reduced to the case where A,B are open balls in X,Y respectively, since one definition of openess is based on the fact that a every point in an open set has some neighborhood contained into the set. Then I think there must be a relationship between metrics on factor spaces and product space one. There is where I'm stuck.

 

[HallsofIvy]

Any point in A x B can be written (x, y) with x in A and y in B. What can you say about balls centered on x in A and on y in B?

 

[ELESSAR TELKONT]

Well, as I have said in my first post this can be reduced to think about balls. I'll go more explicit about this.

As Halls of Ivy said if z\in A\times B, by definition of cartesian product there are some x\in A,\, y\in B such that z=(x,y). Since the factor sets are open in their metric spaces X,\,Y there are some \epsilon_{1}>0,\,\epsilon_{2}>0 such that B_{x,\epsilon_{1}}^{\rho_{1}}\subseteq A,\, B_{y,\epsilon_{2}}^{\rho_{2}}\subseteq B, this is by definition of open set. But then is obvious, from results of cartesian product that B_{x,\epsilon_{1}}^{\rho_{1}}\times B_{y,\epsilon_{2}}^{\rho_{2}}\subseteq A\times B\subseteq X\times Y.

Then, if I prove that B_{x,\epsilon_{1}}^{\rho_{1}}\times B_{y,\epsilon_{2}}^{\rho_{2}} is open and since for every z\in A\times B can construct this then there is some ball B_{z,\epsilon_{3}}^{\rho_{3}}\subseteq B_{x,\epsilon_{1}}^{\rho_{1}}\times B_{y,\epsilon_{2}}^{\rho_{2}}\subseteq A\times B and the cartesian product is open in the product space.

How do I show this with an arbitrary metric \rho_{3} on the product space?

 

[ELESSAR TELKONT]

I tried with contradiction argument on this:

Let B_{x,\epsilon_{1}}^{\rho_{1}}\times B_{y,\epsilon_{2}}^{\rho_{2}} no open in (X\times Y,\rho_{3}). Then there is some z_{0}\in B_{x,\epsilon_{1}}^{\rho_{1}}\times B_{y,\epsilon_{2}}^{\rho_{2}}=C such that for every \delta>0 B_{z_{0},\delta}^{\rho_{3}}\cap C\neq\emptyset\land B_{z_{0},\delta}^{\rho_{3}}\cap((X\times Y)\setminus C)\neq\emptyset.

But that implies the following: For every \delta>0 there are some z_{\delta}\in B_{z_{0},\delta}^{\rho_{4}} such that exists \xi_{\delta}\in X,\, \eta_{\delta}\in Y but \xi_{\delta}\notin B_{x,\epsilon_{1}}^{\rho_{1}}\lor \eta_{\delta}\notin B_{y,\epsilon_{2}}^{\rho_{2}}. Then \rho_{1}(\xi_{\delta},x)\geq\epsilon_{1}\lor \rho_{2}(\eta_{\delta},y)\geq\epsilon_{2} but \rho_{3}(z_{\delta},z)<\delta.

Although I reach this I think the conclusion is not right. The conclusion is: Since this is for every \delta>0 then I have a set of points that approximates to z for decreasing \delta>0 but in the balls on X,Y they go farther or stay at the same distance. This is not possible, then the suposition that the product of balls is not open in (X\times Y,\rho_{3}) leads a contradiction then the product of balls is open in (X\times Y,\rho_{3}) and then the product of any open sets is open in (X\times Y,\rho_{3})

Is it right?