Is every subset of a totally bounded set also totally bounded?

[Fredrik]

Not really homework, but a textbook-style question...

Homework Statement

Is every subset of a totally bounded set (of a metric space) totally bounded?

Homework Equations

F is said to be totally bounded if, for every $\epsilon>0$, there's a finite subset $F_0\subset F$ such that $$F\subset\bigcup_{x\in F_0}B(x,\epsilon)$$, where $B(x,\epsilon)$ is the open ball of radius $\epsilon$ around x.

The Attempt at a Solution

Suppose that $E\subset F$, and that F is totally bounded. Let $\epsilon>0$ be arbitrary. We know that there exists a finite set $F_0\subset F$ such that $$E\subset F\subset\bigcup_{x\in F_0}B(x,\epsilon)$$, but this doesn't seem to help, since $F_0$ doesn't have to be a subset of E. We might even have $F_0\cap E=\emptyset$. So now I'm starting to think that maybe E doesn't have to be totally bounded at all. For example, if F is some open ball in $\mathbb R^2$ and E is some kind of fractal or something.

 

[Dick]

No, a totally bounded subset of a totally bounded set is totally bounded. Most definitions are a bit looser than yours. But you can still prove it. Pick a finite set F0 to cover with balls of radius e/2. Can you use that to construct a finite set E0 that covers E with radius e?

 

[Fredrik]

Ah, I get it now. I'll just pick one point from each non-empty $B(x,\epsilon/2)\cap E$ with $x\in F_0$, and take those points to be my E₀. Then I consider open balls around those points, and I need to take these balls to have twice the radius to ensure that they cover E (by covering the old balls that had non-empty intersection with E). Thanks.