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Show continuously uniform function

  1. May 11, 2009 #1
    Let F be a non-empty subset of a metric room (X,d) and define the function f: X→R through f(x)= inf_{y∈F} d(x,y)= inf {d(x,y):y ∈ F}. Please show that f is continuously uniform in the entire X.

    Please note that f(x)= inf_{y∈F} d(x,y) means that y∈F is written below the inf...it is hard to show it on one line, so I chose to show as a subscript...Can anyone tell me how I should begin this problem?
     
  2. jcsd
  3. May 11, 2009 #2

    quasar987

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    Strange coincidence, I've been trying to solve the same problem! Well, not exactly the same... I have been trying to show that the map [itex]x\mapsto d(x, F)[/itex] is Lipschitz continuous, meaning that there exists a constant C such that |d(x,F)-d(y,F)|≤Cd(x,y) for all x,y in X. Obviously Lipschitz continuity is stronger than uniform continuity since given ε>0, it suffices to take δ=ε/C. As it turns out, there is an exercise in the book Topology and Geometry of G. Bredon that asks the reader to show that the map [itex]x\mapsto d(x, F)[/itex] is (merely!) continuous. And a hint is provided in suggesting to show that this map is actually Lipschitz of constant C=1. Meaning that, somewhat surprisingly, the easiest way to solve your problem and mine is probably to demonstrate the stronger inequality |d(x,F)-d(y,F)|≤d(x,y) for all x,y in X. So let's try to show this.
     
  4. May 11, 2009 #3

    quasar987

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    The key for me was to consider first the very special case in which F consists of only two points: F={p,q}. In this scenario, the problem is not so overwhelming psychologically and after solving it, I realized that its solutions admits an immediate generalization to the case of arbitrary F.

    The general solution I found is as follows. Tell me what you think.

    First, note that we can assume without loss of generality that d(y,F)≤d(x,F) [if not, interchange the labels of x and y...]. In this case, |d(x,F)-d(y,F)|=d(x,F)-d(y,F) and so we must prove that d(x,F)≤d(x,y)+d(y,F) [i.e., we must prove a kind of triangle inequality in which sets are allowed to occupy one of the variable position].

    Next, consider {e_n}, {f_n} two sequences of points in F such that d(x,F)=lim d(x,e_n) and d(y,F)=lim d(y,f_n) and further assume that d(x,e_n)≤d(x,f_n) for all n in N.

    Then, we have d(x,y)+d(y,f_n) ≥ d(x,f_n) ≥ d(x,e_n) for all n in N and passing to the limit yields the desired conclusion.
     
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