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Show f is continuous

  1. Feb 16, 2010 #1
    A problem on the final exam is to show for a metric space (X,d) and a compact subset C in X prove that the function [tex]f(x) = min_{y \in C} d(x,y) [/tex] is continuous.

    Now, there are two approches you can take. One is to go to the episolon delta definition of continuous, and the other is to use open sets.

    Seeing as how C is compact, I think the better approach is to use open sets. That is, to show that for a point y = f(x), make a neighborhood around it, call it U. Then [tex]f^{-1}(U)[/tex] must be shown to be open somehow.

    Taking the second approach, I can see that for any point p in [tex]f^{-1}(U)[/tex] we can construct a neighborhood V around it, so that [tex]p \subset V \subset f^{-1}(U)[/tex]. Um.. let me think... I know I can cover [tex]f^{-1}(U)[/tex] with finitely many open sets, due to the compactness of C, but I really am stuck. And the thing is, I have no idea where to begin to use the definition of f, [tex]f(x) = min_{y \in C} d(x,y) [/tex]. I'm pretty sure I'm appraoching this totally wrong but I can't think of anything else to do. Any help is greatly appreciated.
  2. jcsd
  3. Feb 17, 2010 #2
    I think the epsilon-delta approach may be better. In fact, for this function show you can use delta=epsilon. That is, prove [tex]|f(x)-f(x')| \le d(x,x')[/tex]
  4. Feb 18, 2010 #3
    That question is somewhat odd - by using "min" rather than infinum they must be trying to give a hint. Here's how to make use of that:

    Supposing [tex]f^{-1}(U)[/tex] is not empty for some open set [tex]U[/tex] (otherwise we are done), fix some point [tex]x_0 \in f^{-1}(U)[/tex].

    As [tex]U[/tex] open, there exists [tex]r > 0[/tex] such that the interval [tex] ( f(x_0) - r, f(x_0) + r ) \subset U[/tex].

    As [tex]C[/tex] is compact, there exists a [tex]y_0 \in C[/tex] such that [tex]d(x_0, y_0) = f(x_0)[/tex].

    Let [tex]x \in B(x_0, r/2)[/tex].

    Then, [tex]f(x) \leq d(x, y_0) \leq d(x, x_0) + d(x_0, y_0) \leq r/2 + f(x_0)[/tex].
    Similarily, [tex]f(x) \geq f(x_0) - r/2[/tex], and so [tex]x \in f^{-1}(U)[/tex].

    Therefore, [tex]B(x_0, r/2) \subset f^{-1}(U)[/tex], hence [tex]f^{-1}(U)[/tex] is open. It follows that [tex]f[/tex] is continuous.
    Last edited: Feb 18, 2010
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