# Compact Set in Metric Space

1. Oct 24, 2010

### wayneckm

Hello all,

Here is my question while reading a proof.

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

$$D_{K} = \inf \{ t \geq 0 \; : \; f(t) \in K \}$$

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

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

Wayne

2. Oct 25, 2010

### Landau

It's not really clear what the domain and codomain of your function f are.

3. Oct 25, 2010

### wayneckm

Domain of $$f$$ is $$\mathbb{R}^{+}$$
Codomain of $$f$$ is $$\mathbb{R}$$

4. Oct 25, 2010

### Landau

Huh? Then I don't understand $f(t)\in K$...

5. Oct 25, 2010

### wayneckm

Oops...sorry, i misunderstood the term codomain. So codomain here should be $$E$$ as stated.

6. Oct 25, 2010

### Office_Shredder

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

This is basically the direction $$D_k\leq t$$ implies the infimum is zero.