Let $M$ denote a connected time-oriented Lorentzian manifold. If $M$ is compact then $M$ must have a closed-curve Problem
Let $M$ denote a connected time-oriented Lorentzian manifold. If $M$ is compact then $M$ must have a closed-curve.
I don't know how to use this information that M is compact to obtain a closed-curve. O'Neill has a demonstration but I didn't get it.
 A: First, consider the set of open sets formed by the chronological future at every point, ie
$$\{ I^{+}(p) | p \in M \}$$
You can show easily enough that this is an open cover of the manifold (just consider the convex normal neighbourhood of every point, you'll be able to show that for any point $q \in M$, there exists a point $p$ such that $q \in I^+(p)$). 
Now, by definition, a manifold is compact if for any open cover, there exists a finite subcover. Therefore, we have a set of points $(p_1, \ldots, p_n)$ such that
$$\bigcup_i I^+(p_n) = M$$
Now consider the problem : for $I^+(p_i)$, to what open set does $p_i$ belong? If $p_i \in I^+(p_i)$, then $p_i$ is on a closed timelike curve. Otherwise, $p_i \in I^{+}(p_j)$, $i \neq j$. But if $p_i \in I^{+}(p_j)$, this means that $p_i$ is to the future of $p_j$, and therefore $I^+(p_i) \subset I^+(p_j)$. The open cover can therefore be reduced to
$$\{ I^+(p_1), \ldots, I^+(p_{i-1}), I^+(p_{i+1}), \ldots, I^+(p_n) \}$$
Using this method $n-1$ times, you can reduce the open cover to $I^+(p_1) = M$, and then there's no avoiding it : $p_1$ is contained in its own future, and therefore there exists a closed timelike curve.
