Homework Help: Cartesian product of open sets is a open set

1. Mar 7, 2012

ELESSAR TELKONT

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

Definition of open and metric properties.

3. 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.

2. Mar 8, 2012

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?

3. Mar 8, 2012

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?

4. Mar 9, 2012

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?