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HELP metric space problem

  1. Feb 16, 2006 #1
    Hi

    I have this here metric space problem which caused me some trouble:

    [tex]S \subseteq \mathbb{R}^n[/tex] then the set

    [tex]\{ \| x - y \| \ | y \in S \} [/tex] has the infimum


    [tex]f(x) = \{ \| x - y \| \ | y \in S \}[/tex]

    where f is defined [tex]f: \mathbb{R}^n \rightarrow \mathbb{R}[/tex]



    I have two problems here which I'm unable to solve:

    (a) show, if S is a closed set and [tex]x \notin S [/tex] then [tex]f(x) > 0[/tex] ????

    (b) show, if S is a closed set, then [tex]S = \{ x \in \mathbb{R}^n | f(x) = 0\}[/tex] ???

    I need to hand this in tomorrow, and I have been strugling this these two problems the last week, therefore I would very much appriciate if anybody could give me an idear on how to solve the two problems above.

    God bless,

    Best Regards,
    Fred
     
    Last edited: Feb 16, 2006
  2. jcsd
  3. Feb 16, 2006 #2

    quasar987

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    [tex]f(x) = \{ \| x - y \| \ | y \in S \}[/tex] ??

    That would mean to each x in R^n, f maps x to ||x-y|| for all y in S. So as soon as card(S)>1, f is not a function.

    Also, what do you mean by "[tex]\{ \| x - y \| \ | y \in S \} [/tex] has the infimum f(x)"?
     
  4. Feb 16, 2006 #3
    Sorry it should have said

    [tex]f(x) = \mathrm{inf} \{ \| x - y \| \ | y \in S \}[/tex]

    Any idears on how to go about this?

    Best Regards

    Fred

    p.s. My problems deals with the distance from [tex]\mathbb{R}^n[/tex] to a point in a subset S of [tex]\mathbb{R}^n[/tex].
     
    Last edited: Feb 16, 2006
  5. Feb 16, 2006 #4

    quasar987

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    a) wouldn't that exeedingly simply argument suffice:

    we know that ||x-y|| = 0 iff x=y. But since x is not in S, x is not equal to y for any y in S. Hence, ||x-y||>0.

    There's probably a problem with this argument as it doesn't even use the closedness of S...
     
  6. Feb 16, 2006 #5
    Hello and thank You for Your answer,

    Then (B) is that the oposite of (A) ??

    Sincerely and God bless

    Fred

     
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