Totally bounded subset in a metric space

[tarheelborn]

Homework Statement

Suppose M is a metric space and A \subseteq M. Then A is totally bounded if and only if, for every \epsilon >0, there is a finite \epsilon-dense subset of A.

Homework Equations

The Attempt at a Solution

I have already done the \Rightarrow but need to verify the other half:
(\Leftarrow ): Now suppose that for \epsilon > 0, A has a finite \epsilon-dense subset. I must prove that A is totally bounded. Since there is an \epsilon-dense set in A, say \{ x_1, x_2, \cdots, x_n \} is \epsilon-dense in A, then B[x_i; \epsilon], \cdots, B[x_n; \epsilon] form a covering of A by sets of diameter < \epsilon. Hence A is totally bounded.

Does this work?

 

[ystael]

This is an essentially correct argument, but for one small technical problem and a few stylistic issues.

The small technical problem is that the diameter of B(x_j; \epsilon) is not necessarily \epsilon; what is it? (Correcting this doesn't change the gist of the argument, but you do need to make adjustments.)

As for the stylistic issues -- Because you're being asked to prove something this simple, it might not hurt to give an explicit argument that the balls \{B(x_j; \epsilon)\} cover all of A -- that is, given any x\in A, why does x lie in one of the balls B(x_j; \epsilon)?

Also, you could be more careful about the quantifiers -- it should run something like: Fix \epsilon > 0; we will exhibit a covering of A by finitely many sets of diameter \leq \epsilon. We know that there exists a finite something-dense set in A, so ... (continue from here). The important point here is that you need to show the covering exists no matter the choice of \epsilon, which isn't clear from what you wrote above.

 

[tarheelborn]

I see. Thank you.