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Convergent Filter Base and Continuous Function

  1. Aug 12, 2009 #1
    Hi All,

    I can't see how the following is proved.

    Given two topological space (X, T), (Y, U) and a function f from X to Y and the following two statements.

    1. f is continuous, i.e. for every open set U in U, the inverse image f-1(U) is in T

    2. For every convergent filter base F -> x, the induced filter base f [[F]] -> f(x)

    it is claimed that statement 1 and statement 2 are equivalent, i.e. 1 if and only if 2

    I can prove 1 -> 2
    but how do i prove 2 -> 1

    thanks a lot !
     
  2. jcsd
  3. Aug 13, 2009 #2
    Suppose the filter-base thingy holds. Let A be an open set in Y. Let B = f^{-1}(A).
    We must show B is open in X. Suppose not. Then there is c in B such that every neighborhood of c meets X minus B. Your filter-base will be all these sets: (neighborhood of c) minus B. This filter-base converges to c. So the image filter-base in Y converges to f(c). But the sets of that image filter-base are outside A and converge to a point of A, contradicting the assumption that A is open.
     
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