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D(x,A) = inf{||x-a|| | a=A}

  1. Jul 4, 2010 #1
    1. The problem statement, all variables and given/known data
    [tex]\underline{x} [/tex] point in [tex] R^{n} A \subset R^{n}[/tex]
    the distanse between [tex] \underline{x}[/tex] to A is definde as [tex] d(\underline{x},A) = \left\{ inf{||\underline{x} = \underline{a}|| | a \in A \right\}[/tex]

    A,B are closed Disjoint sets in R^{n} we define [tex]f(\underline{x}) = \frac{d(\underline{x},B)}{d(\underline{x},A) + d(\underline{x},B)}[/tex]


    to each [tex]\underline{x} \subset R^{n}[/tex] and [tex]\underline{y} \subset R^{n} [/tex] [tex]

    |d(\underline{x},A)-d(\underline{y},A)| \leq||\underline{x}-\underline{y}||[/tex]

    Prove that [tex]f(\underline{x})[/tex] Continuous
    2. Relevant equations
    everything in calculus


    3. The attempt at a solution

    Well I've tried with Cauchy test for limits of function but this does not give me a Continuous function..
    from there I am A bit stuck.
    Thank you.
     
  2. jcsd
  3. Jul 4, 2010 #2
    Show the function g(x) = d(x,E) is continuous for any set E, then the sum of such functions must be continuous, and the quotient as well, where it is defined. And use the disjointness and closedness to show your function is defined everywhere (the denominator is non-zero).
     
  4. Jul 4, 2010 #3
    Thank you
    I know that I need to prove that g(x) = d(x,E) in any set.
    This is why I said I tried with Cauchy test for limits, But I did not managed to do it.
    Any idea of who to prove this?
    Thank you.
     
  5. Jul 4, 2010 #4

    Office_Shredder

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    The last part about x and y is a pretty big hint for how to prove continuity
     
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