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Cartesian product of separable metric spaces

  1. Jul 26, 2011 #1
    Dear readers,

    Let [itex]X[/itex] be the product space of a countable family [itex]\{X_n:n\in\mathbb{N}\}[/itex] of separable metric spaces.
    If [itex]X[/itex] is endowed with the product topology, we know that it is again separable. Are there other topologies for [itex]X[/itex] such that is separable? Is there a natural metric on [itex]X[/itex] such that [itex]X[/itex] is separable and therefore have a countable base?

    The general question is under what conditions on the product space [itex]X[/itex] the following conclusion holds:
    "For any topological base [itex]\mathcal{B}[/itex] in [itex]X[/itex], the open subsets of [itex]X[/itex] are countable unions of sets in [itex]\mathcal{B}[/itex]"

    Thnx
     
  2. jcsd
  3. Jul 29, 2011 #2
    Let d_n be the metric on Xn. Then d = (sum over n) {(dn/dn+1)/2^n} defines such a metric on X.
    The property, however, owes more to the countability axioms. If each of a finite family of spaces has a countable basis, so must the product space in product topology.
     
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