There is a well known fact that a compact spacetime necessarily contains a closed timelike curve (CTC). Proof can be found in several books on GR (e.g. Hawking, Ellis, Proposition 6.4.2), and in essence goes like this:
The spacetime $M$ can be covered by open sets of the form $I^+(p)$, chronological future of the point $p \in M$ (note that a priori $p$ is not an element of the set $I^+(p)$, but such situation can happen in a presence of CTCs). Now, suppose that $M$ is compact. Then there is a finite subcover, say
$$\{ I^+(p_1), \dots, I^+(p_n) \}$$
The point $p_1$ is contained in $I^+(p_{k_1})$ for some $1 \le k_1 \le n$, the point $p_{k_1}$ is contained in $I^+(p_{k_2})$, and so on. Since this subcover is finite, eventually some point $p_{k_r}$ must belong to $I^+(p_{k_s})$, with $s \le r$. Then there is a future directed timelike curve going from $p_{k_r}$ to $p_{k_s}$ (since $s \le r$) and then from $p_{k_s}$ back to $p_{k_r}$ (since $p_{k_r} \in I^+(p_{k_s})$), which gives a closed timelike curve through $p_{k_r}$ (and $p_{k_s}$) in $M$. Q.E.D.
The question that bothers me is: what is implicitly assumed about the spacetime $M$, by saying that the family of sets of the form $I^+(p)$ is indeed a covering of $M$?
Take for example a flat spacetime with compact space part and finite time direction, e.g. $M = [0,1] \times T^3$, where $T^3$ is a (spacelike) 3-torus. This is a compact manifold (since it is a product of two compact manifolds) and there are no CTCs. The loophole in the argument above seems to be in the fact that the "initial points", $\{0\} \times T^3$, are not covered by any set of the form $I^+(p)$.
One can easily modify this example by contracting the initial and final spacelike slices to points ("Big Bang" and "Big Crunch"); the resulting spacetime is still compact and contains no CTCs.
Do these manifolds fail to be "regular spacetimes" for some reason?
