Let $(M,g)$ be a finite dimensional Riemannian manifold. Suppose it is bounded as a metric space, i.e. $$\sup \{ d(x,y) | x,y \in M \} < \infty$$ where $d$ is the distance induced by $g$. Is $M$ also totally bounded?
I know bounded spaces are not totally bounded in general, but the only examples I know are in some infinite dimensional Banach space. I think the answer should be yes for manifolds, since by Nash Embedding Theorem they are (isometric to) subspaces of some Euclidean space, and bounded subspaces of $\mathbb{R}^n$ are also totally bounded. If this is true (which assumes I correctly understand the statement of Nash Theorem), is there a more concrete proof?
This is false. Take universal cover $M$ of the open 2 dimensional Euclidean unit disk $D^2$ with center removed, equipped with pull back flat metric. I let $d$ denote the distance function on $M$ induced by the pull-back Riemannian metric. I claim that the diameter of $M$ is $\le 2$ (actually, it is exactly 2 but we do not need this). To see this it is convenient to identify $M$ with the complex half-plane $\{z: Re(z)<0\}$ and the universal cover $M\to D^2\setminus 0$ with the complex exponential function. Note also that for every $y, t\in {\mathbb R}$, $$ \lim_{x\to-\infty} d(x+ iy, x+it)=0. $$ Furthermore, the length of every path $(x+iy), -\infty<x<\infty$ equals $1$. Therefore, to connect any two points $p, q\in M$ by a path of length $<2$, first connect $p$ and $q$ by horizontal lines to some points $p', q'$ with $Re(p')=Re(q')\ll 0$ and then connect $p'$ to $q'$ by a vertical interval.