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Characteristic function

  1. Nov 12, 2008 #1
    1. The problem statement, all variables and given/known data

    Let (X,d) be a metric space, A subset of X, x_A: X->R the characteristic
    function of A. (R is the set of all real numbers)

    Let V_d(x) denote the set of neighbourhoods of x with respect the metric d.
    Prove that x_A is continuous in x (x in X) if and only if there
    exists an element V in V_d(x) such that V is a subset of A and V is a
    subset of X\A.




    3. The attempt at a solution

    OK, so assume x_A is continuous then for each closed subset of R
    the preimage of this closed subset under x_A must be closed.
    Take V= {0} then V is closed so (x_A)^(-1) = { y in X: x_A(y) = 0} = X\A.

    But I don't see how this helps..I don't know hwo to find such V in V_d(x).
     
  2. jcsd
  3. Nov 12, 2008 #2
    My work:
    So assume it is continuous at a point c.
    Then put e=1/2 then there exists delta > 0 such that if d(x,c)< delta
    then |x_A(x) - x_A(c)|< 1/2.
    So we have 2 cases: x and c in A or x,c in X\A.
    Then I don't know what to take as V..
     
  4. Nov 14, 2008 #3

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    Am I reading this right? V is a subset of both A and X\A? This is only possible if V is empty, and the empty set is certainly not a neighborhood of anything.
     
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