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A result in topology

  1. Sep 27, 2007 #1

    quasar987

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    I just discovered the following. But since half the things I find in topology turn out to be wrong, I feel I better check with you guys.

    What I convinced myself of this time is that if you have a function f:(X,T)-->(Y,S) btw topological spaces, and S' is a basis for S, then to show f is continuous, is suffices to show that [itex]f^{-1}(S') \subset T[/itex].

    In words, that is because every open set of S can be written as a union of sets of S', and the operations of f^{-1} and union commute. (and that a union of open sets is open)
     
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  3. Sep 27, 2007 #2

    Hurkyl

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    That looks right. In fact, it's rather often used -- e.g. look at the definition of continuity you learned in calc I.
     
  4. Sep 28, 2007 #3
    it's a fact!
     
  5. Sep 28, 2007 #4

    quasar987

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    Thanks.
     
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