Let $M$ be a connected smooth manifold without boundary. Are the following equivalent?
I recall wondering about this as a student. Some of the implications are well know but I never figured out the others. This question reminded of this.
That 1 implies 3 is standard and appears in most textbooks on manifolds that include vector field flows, it follows by a standard existence and uniqueness argument.
3 implies 1 is a little more work but it's a fairly standard argument as well. The idea is that if the manifold is non-compact you can take a proper Morse function on the manifold, $f : M \to [0,\infty)$. If you re-scale the gradient like $\frac{f^2}{1+|\nabla f|^2}\nabla f$, its flow lines can't be complete since it hits infinity in a finite time. And technically you don't need any Morse theory to make this conclusion. It's enough to have a proper function $f : M \to [0,\infty)$ then show you can embed an arc $p : [0,\infty) \to M$ in $M$ so that $f(p(x)) = x$ for all $x$. You can define an incomplete vector field whose flow line is the image of $p$, then extend it to $M$.
To relate 1 and 2 notice that in a manifold, a subspace is compact, connected and open means its a path-component. If you take a manifold and collapse each path-component to a point (the path-component space) you get a discrete space. So this gives you 1 implies 2. The converse 2 implies 1 follows from the argument in my comment to Mariano above -- if it's non-compact you can take a properly embedded arc, cut along it and self-embed. In reverse this embeds your manifold in a bigger manifold.
