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.