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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.

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Hint: Consider an infinite set in the discrete metric. (Every point is at distance $1$ to every other point.)

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Consider the set of real numbers $\mathbb{R}$ with the discrete metric defined as

$$ d(x,y)= \begin{cases} 1,&\quad x\ne y \\ 0,&\quad x=y \end{cases}. $$

Then every open ball with this metric is

$$B_{\mathbb{R}}(x,r)= \{y\,|\,y\in\mathbb{R}\,\land d(x,y)<r\}= \begin{cases} \{x\},& \quad r\leq 1 \\ \mathbb{R},& \quad r>1 \end{cases}. $$ 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!

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