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Proving a function is bounded and continuous in a metric space.

  1. May 20, 2013 #1
    1. The problem statement, all variables and given/known data
    Let (X,d) be any metric space. Fix a in X and for each x in X define fx:X→ℝ by:
    fx(z)=d(z,x)-d(z,a) for all z in X.

    Show that fx(z) is bounded and continuous.

    3. The attempt at a solution

    I can't figure out how to tell if it is bounded. Any hints? I'm sure you have to use the triangle inequality but I'm not sure how.

    So to prove f is continuous I'm going to try and use the fact that if U is open fx-1(U) must also be open if f is continuous.

    So suppose U is a open subset of ℝ.
    then fx-1(U) = {z in X: dist(fx(z),ℝ-U)>0}
    Then for all u in fx-1(U) it implies dist(fx(u),ℝ)-U) = r > 0
    so then Bx(z,r/2) is contained in f-1(U)
    So fx-1(U) is open and fx is continuous.
  2. jcsd
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