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Compact Set in Metric Space

  1. Oct 24, 2010 #1
    Hello all,


    Here is my question while reading a proof.

    For a compact set [tex] K [/tex] in a separable metrizable spce [tex] (E,\rho) [/tex] and a continuous function [tex] t \mapsto f(t) [/tex], if we define

    [tex] D_{K} = \inf \{ t \geq 0 \; : \; f(t) \in K \}[/tex]

    then, [tex] D_{K} \leq t [/tex] if and only if [tex] \inf\{ \rho(f(q),K) : q \in \mathbb{Q} \cap [0,t] \}[/tex] = 0

    May someone shed some light on this? I do not understand it. Thanks very much.


    Wayne
     
  2. jcsd
  3. Oct 25, 2010 #2

    Landau

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    It's not really clear what the domain and codomain of your function f are.
     
  4. Oct 25, 2010 #3
    Domain of [tex] f [/tex] is [tex] \mathbb{R}^{+} [/tex]
    Codomain of [tex] f [/tex] is [tex] \mathbb{R} [/tex]
     
  5. Oct 25, 2010 #4

    Landau

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    Huh? Then I don't understand [itex]f(t)\in K[/itex]...
     
  6. Oct 25, 2010 #5
    Oops...sorry, i misunderstood the term codomain. So codomain here should be [tex] E [/tex] as stated.
     
  7. Oct 25, 2010 #6

    Office_Shredder

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    If [tex] D_k \leq t[/tex] then [tex] f(D_k)\in K[/tex]. We can approximate [tex] D_k[/tex] with rational numbers, and because [tex] D_k \in [0,t][/tex] we can approximate [tex] D_k[/tex] with rational numbers in [tex] \mathbb{Q}\cap [0,t][/tex] If [tex] q_r[/tex] is such a sequence converging to [tex] D_k[/tex], the distance between [tex] f(q_r)[/tex] and [tex] f(D_k)[/tex] goes to zero, which means the distance between [tex]f(q_r)[/tex] and [tex]K[/tex] must go to zero. So the infimum of the distance between [tex] f(q)[/tex] and [tex]K[/tex] for [tex] q\in \mathbb{Q}\cap [0,t][/tex] must be zero because we just found a sequence for which the distance is arbitrarily small.

    This is basically the direction [tex] D_k\leq t[/tex] implies the infimum is zero.
     
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