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Hausdorffness of the product topology

  1. Feb 29, 2008 #1

    quasar987

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    [SOLVED] Hausdorffness of the product topology

    Is it me, or is the product of an infinite number of Hausdorff spaces never Hausdorff???

    Recall that the product topology on

    [tex]\Pi_{i\in I}X_i[/tex]

    has for a basis the products of open sets

    [tex]\Pi_{i\in I}O_i[/tex]

    where all but finitely many of those O_i are not the whole X_i.

    ---

    say I is countable for simplicity and consider x=(x1,...) and y=(y1,...) two distinct points in the product space. I don't see how we can find two ngbh of x and y that do not intersect!
     
  2. jcsd
  3. Feb 29, 2008 #2
    Hausdorff, given either the product topology or box topology. This has been proven here before. The converse is true as well (though mathwonk remarked that the product must not be empty).
     
  4. Feb 29, 2008 #3

    quasar987

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    I found the thread thx.
     
    Last edited: Feb 29, 2008
  5. Feb 29, 2008 #4

    quasar987

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    But it seems to deal only with the converse.

    Given any 2 open set of the basis containing x and y resp., since infinitely many O_i are X_i in both ngbh, they will always have a huge intersection. Is this not the case?
     
  6. Feb 29, 2008 #5

    morphism

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    Ask yourself this question: Is there anything in [itex]\emptyset \times X \times X \times \cdots[/itex]?
     
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