Set that is bounded but not totally bounded: Reading textbook I've been reading a Real Analysis textbook that my friend loaned to me. I have come across a proposition that says that a totally bounded set is bounded, but a bounded set is not always totally bounded. This makes sense, however I am having trouble thinking of an example of a set that is bounded but not totally bounded. Could anyone shed some light on this? Thanks! 
 A: I know this is an old post, 
but any infinite set M with a discrete metric is bounded by any N>1 but it is not totally bounded for open balls with $\epsilon\leq 1$ because for it to be totally bounded it must have a finite number of points. But such a ball is only a singleton, and you would require infinitely many to cover M. 
A: Take $U=\{e_n|\:n\in\mathbb N\}\subset \ell^\infty (\mathbb R)$.
It is obviously bounded since $\forall x\in U \:\|x\|=1$, but $\forall x,y\in \ell^\infty$ we have $d(x,y)=1$, so obviously for $\epsilon=1$ there is no finite number of open balls with radius $\epsilon$ that cover $U$ - cause each ball would contain at most one member of $U$.
A: Define a new metric $d$ on $\Bbb R$ by $d(x,y)=\min\{|x-y|,1\}$; you can easily check that $d$ generates the usual topology on $\Bbb R$. Every subset of $\Bbb R$ is bounded with respect to $d$, so we need only find a subset that is not totally bounded. $\Bbb N$ will do: if $F\subseteq\Bbb N$, then
$$\Bbb N\cap\bigcup_{x\in F}B_d\left(x,\frac12\right)=F\;,$$
since $\Bbb N\cap B_d\left(x,\frac12\right)=\{x\}$ for each $x\in F$. Thus, no finite family of open $\frac12$-balls centred at points of $F$ can even cover $\Bbb N$.
This idea generalizes. Start with any complete metric space $\langle X,\rho\rangle$ that is not compact. A metric space is compact if and only if it’s complete and totally bounded, so $\langle X,\rho\rangle$ is not totally bounded, and there is some $\epsilon>0$ such that no finite family of open $\epsilon$-balls covers $X$. Without loss of generality we may assume that $\epsilon<1$. Now define a metric $d$ on $X$ by
$$d(x,y)=\min\{\rho(x,y),1\}\;;$$
$d$ generates the same topology as $\rho$, and every subset of $X$ is bounded with respect to $d$. Finally, since $\epsilon<1$, $\epsilon$-balls with respect to $d$ are the same as $\epsilon$-balls with respect to $\rho$, and $X$ is bounded but not totally bounded with respect to $d$.
