Compact Set in Metric Space (1 Viewer)

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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
 

Landau

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It's not really clear what the domain and codomain of your function f are.
 
Domain of [tex] f [/tex] is [tex] \mathbb{R}^{+} [/tex]
Codomain of [tex] f [/tex] is [tex] \mathbb{R} [/tex]
 

Landau

Science Advisor
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Huh? Then I don't understand [itex]f(t)\in K[/itex]...
 
Oops...sorry, i misunderstood the term codomain. So codomain here should be [tex] E [/tex] as stated.
 

Office_Shredder

Staff Emeritus
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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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