# Boundedness and total boundedness

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.

Hint: Consider an infinite set in the discrete metric. (Every point is at distance $1$ to every other point.)
Consider the set of real numbers $$\mathbb{R}$$ with the discrete metric defined as
$$d(x,y)= \begin{cases} 1,&\quad x\ne y \\ 0,&\quad x=y \end{cases}.$$
$$B_{\mathbb{R}}(x,r)= \{y\,|\,y\in\mathbb{R}\,\land d(x,y)1 \end{cases}.$$ Accordingly, every subset of $$\mathbb{R}$$ is closed, open and bounded with the discrete metric, including the $$\mathbb{R}$$ itself! But clearly $$\mathbb{R}$$ is not totally bounded for if we choose $$\epsilon = \frac{1}{2}$$ then there won't be any finite covering for it!