We say that a metric space $M$ is totally bounded if for every $\epsilon>0$, there exist $x_1,\ldots,x_n\in M$ such that $M=B_\epsilon(x_1)\cup\ldots\cup B_\epsilon(x_n)$.
Prove that if $M$ is a totally bounded metric space, then $M$ is bounded. Given an example to show that the converse is false.
The "prove" part is routine. Given points $a,b$, suppose $a$ is in the ball of $x_i$ and $y$ is in the ball of $x_j$. Then $d(a,b)\le d(a,x_i)+d(x_i,x_j)+d(x_j,b)<\epsilon+d(x_i,x_j)+\epsilon$. Since there are only finitely many values of $d(x_i,x_j)$, we are done.
For the "example" part, the space must be bounded, but somehow there exists $\epsilon$ such that the space cannot be covered with finitely many balls of radius $\epsilon$. I can't think of what that space should look like… things like $[a,b]$, open ball, etc. are totally bounded.
Best Answer
Hint: Consider an infinite set in the discrete metric. (Every point is at distance $1$ to every other point.)