Equivalent definition of "totally boundedness"

Question

A metric space (X, d) is said to be totally bounded (or precompact) if, for every $\epsilon > 0,$ the space X can be covered by a finite family of open balls of radius $\epsilon $.

Another way of saying this is: A metric space (X, d) is totally bounded if, for every $\epsilon > 0,$ there exists a finite subset A ⊆ X such that d(x, A) < $\epsilon $ for all x ∈ X.

So my question is : Can we say that a metric space is totally bounded if and only if it has a finite dense subset? (I guess we cannot because I have never heard of it. But according to the meaning of dense subset, I cannot find where the problem is.)

Thanks a lot!

Answer 1

"But according to the meaning of dense subset, I cannot find where the problem is."

The problem is that a dense subset is "good" for all choices of $\epsilon$, whereas in your case, for every $\epsilon>0$ there is a certain finite subset that will do, but this particular finite subset may not do for all smaller choices of $\epsilon$, and in fact, will never do, because it is finite.

Answer 2

No, we cannot. If $F$ is a finite subset of a metric space $X$, then $\overline F=F$ and therefore $F$ is dense if and only if $F=X$.

If $X$ is totally bounded, then there is a finite set $F$ such that $X=\bigcup_{x\in F}B_1(x)$. And there is another set $F^\star$ such that $X=\bigcup_{x\in F^\star}B_{1/2}(x)$. It doesn't have to be the same set. And it usually isn't.