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Cartesian product of open sets is a open set

  1. Mar 7, 2012 #1
    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 [itex]X,Y[/itex] be metric spaces and [itex]X\times Y[/itex] with another metric the product metric space. [itex]A\subseteq X, B\subseteq Y[/itex] be open sets, then [itex]A\times B\subseteq X\times Y[/itex] 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 [itex]A,B[/itex] are open balls in [itex]X,Y[/itex] 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.
     
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  3. Mar 8, 2012 #2

    HallsofIvy

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    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?
     
  4. Mar 8, 2012 #3
    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 [itex]z\in A\times B[/itex], by definition of cartesian product there are some [itex]x\in A,\, y\in B[/itex] such that [itex]z=(x,y)[/itex]. Since the factor sets are open in their metric spaces [itex]X,\,Y[/itex] there are some [itex]\epsilon_{1}>0,\,\epsilon_{2}>0[/itex] such that [itex]B_{x,\epsilon_{1}}^{\rho_{1}}\subseteq A,\, B_{y,\epsilon_{2}}^{\rho_{2}}\subseteq B[/itex], this is by definition of open set. But then is obvious, from results of cartesian product that [itex]B_{x,\epsilon_{1}}^{\rho_{1}}\times B_{y,\epsilon_{2}}^{\rho_{2}}\subseteq A\times B\subseteq X\times Y[/itex].

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

    How do I show this with an arbitrary metric [itex]\rho_{3}[/itex] on the product space?
     
  5. Mar 9, 2012 #4
    I tried with contradiction argument on this:

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

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

    Although I reach this I think the conclusion is not right. The conclusion is: Since this is for every [itex]\delta>0[/itex] then I have a set of points that approximates to [itex]z[/itex] for decreasing [itex]\delta>0[/itex] but in the balls on [itex]X,Y[/itex] 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 [itex](X\times Y,\rho_{3})[/itex] leads a contradiction then the product of balls is open in [itex](X\times Y,\rho_{3})[/itex] and then the product of any open sets is open in [itex](X\times Y,\rho_{3})[/itex]

    Is it right?
     
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