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Math Help - Borel sigma Algebra

  1. #1
    C.E
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    Borel sigma Algebra

    Could someone please help me with the following question?

    Show that the Borel sigma algebra of the product of two seperable metric spaces is generated by products of Borel sets.

    I am not sure how to start. I think I need to show that any open set in the product space is a countable union of products of Borel sets but I don't know how to do this. Any ideas?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by C.E View Post
    Could someone please help me with the following question?

    Show that the Borel sigma algebra of the product of two seperable metric spaces is generated by products of Borel sets.

    I am not sure how to start. I think I need to show that any open set in the product space is a countable union of products of Borel sets but I don't know how to do this. Any ideas?
    It depends on how you're topologizing the product space--but I'll assume you mean you're giving it the product topology. Then, let B be Borel in X\times Y. Then, by definition there exists countably many open sets \left\{O_n\right\}_{n\in\mathbb{N}} such that, say for this case, \displaystyle B=\bigcap_{n\in\mathbb{N}}O_n. But, by definition of the product topology each O_n may be written as \displaystyle \bigcup_{\alpha\in\mathcal{A}}U_{\alpha\in\mathcal  {A}}\times V_{\alpha,n} where U_{\alpha,n},V_{\alpha,n} are open in X,Y respectively. But, for any second countable topological space (and thus in particular any separable metric space--since second countability and separability are equivalent for metric spaces) every open cover of a subset has a countable subcover (this is Lindelof's theorem--a proof can be found here at problem 26). Thus, we may assume that \#\left(\mathcal{A}\right)\leqslant\aleph_0. The result then just falls out since


    \displaystyle \begin{aligned}B &=\bigcap_{n\in\mathbb{N}}O_n\\ &=\bigcap_{n\in\mathbb{N}}\bigcup_{\alpha\in\mathc  al{A}}\left(U_{\alpha,n}\times V_{\alpha,n}\right)\\ &=\bigcup_{\alpha\in\mathcal{A}}\bigcap_{n\in\math  bb{N}}\left(U_{\alpha,n}\times V_{\alpha,n}\right)\\ &= \bigcup_{\alpha\in\mathcal{A}}\underbrace{\left(\b  igcup_{n\in\mathbb{N}}U_{\alpha,n}\right)}_{\text{  Borel}}\times\underbrace{\left(\bigcup_{n\in\mathb  b{N}}V_{\alpha,n}\right)}_{\text{Borel}}\end{align  ed}
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