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Product of Dense sets

  1. May 9, 2012 #1
    Just a quick question. If Q is a dense set of a metric space X, and P is a dense set of a metric space Y, then is Q x P a dense set of X x Y? I am fairly sure this is the case.

    If this is true, then I want to use this statement to show that the open sets of the product of finite number of separable metric spaces can be written as the union of countably many balls centered around elements in the product of the dense sets.

    Ultimately, this will show that the open sets of the product of finite number of metric spaces is a subset of the product sigma algebra of the separable metric spaces. (Each separable metric space is generated by a countable set of neighborhoods centered around elements in each metric spaces' corresponding dense set)

    Does this argument make sense?

    Thanks!


    EDIT1: Oops, forgot an important part. The metric on the product space is the standard product metric. Specifically, if x and y are elements of the product space, then p(x,y) = max (p1(x1,y1), p2(x2,y2),........, pn(xn,yn)), where x=(x1,x2,x3,x4,....,xn) y=(y1,y2,y3,.....,yn) and pn(x,y) is the metric on the metric space Xn.
     
    Last edited: May 9, 2012
  2. jcsd
  3. May 11, 2012 #2
    Yes. The argument is just a definition chase.

    Let (x,y) be a point in the product X x Y, and fix an r > 0. We need to find a point (p,q) in P x Q so that (p,q) lies in the ball B( (x,y); r). Since P is dense in X, we may find a point p in P so that p is in B(x;r). Similarly, find a q in Q so that q is in B(y;r). Now d((x,y), (p,q)) is by definition the maximum of d(x,p) and d(y,q), which are both less than r by assumption, so (p,q) is in B((x,y); r).

    This shows that any point in the product is in the closure of PxQ, and so PxQ is dense in X x Y.
     
  4. May 12, 2012 #3

    Bacle2

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    Maybe you can try the more general result that the closure of a product is the product of the closures.
     
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