Concerning the **Lebesgue covering dimension**, absolutely nothing can be said, if you work with manifolds of total space-time dimension 3 or higher.

### Preamble

We will consider spacetimes with closed timelike curves. Let $(M,g)$ be Lorentzian. A point $p\in M$ has a closed timelike curve through it if and only if $p \in I^+(p)$; this is further equivalent to $p\in I(p,p)$ where $I(p,q) = I^+(p)\cap I^-(q)$. Note that if $q\in I(p,p)$, then $q$ both precedes and succeeds $p$, and hence there is a closed timelike curve through $q$. Furthermore, in this case $I^\pm(p) = I^\pm(q)$. And hence we have that

For each $p,q\in M$, either $I(p,p)\cap I(q,q) = \emptyset$ or $I(p,p) = I(q,q)$.

We will let $M_c := \cup_{p\in M} I(p,p)$, the set of all points through which a closed timelike curve passes.

### Trivial case

If $M = M_c$, then the Alexandrov topology is indiscrete, and the covering dimension is 0.

### Finite positive case

#### A basic construction

Let $M = S \times (-1,1) \times \mathbb{S}^1$ be a topological cylinder where $S$ is any Riemannian manifold; parametrize it with $(x,z,t)$. The Riemannian metric on $S$ is denoted by $h$.

Consider the metric on $M$ given by
$$ g = h + ( dt + \sin^2(z) dz) \otimes (-\sin^2(z) dt + dz) + (-\sin^2(z) dt + dz) \otimes (dt + \sin^2(z) dz) \tag{*}$$
or
$$ g = h - 2 \sin^2(z) dt^2 + 2\sin^2(z) dz^2 + (1-\sin^4(z)) (dt \otimes dz + dz\otimes dt) $$
One sees that $g$ is non-degenerate of signature $(-,+,+,\dots)$. The vector field $\partial_t - \partial_z$ is timelike everywhere, and so $g$ is time-orientable. And finally:

When $z \neq 0$, the vector field $\partial_t$ is time-like, but when $z = 0$ the vector field $\partial_t$ is null.

In particular, we have that:

- When $p$ is such that $z \leq 0$, then $I^+(p) = \{z < 0\}$; similarly when $p$ is such that $z \geq 0$, then $I^-(p) = \{z > 0\}$.
- When $p$ is such that $z > 0$, then $I^+(p) = M$; when $p$ is such that $z < 0$, then $I^-(p) = M$.

So the Alexandrov toplogy of $M$ has
$$ \tau = \{ \emptyset, \{z < 0\}, \{z > 0\}, M\} $$

#### The construction refined

Let $n \geq 2$ and $N$ any natural number. Take $N+1$ copies of $\mathbb{S}^n$ and label them $S_0, \ldots, S_N$. Place $S_0$ in the center, and arrange the remaining spheres to surround it. Connect each $S_i$ (when $i > 0$ to $S_0$ with a cylinder that is topologically $\mathbb{S}^{n-1}\times (-1,1)$. So in the case $n = 2$ and $N = 4$ it looks a bit like the chemistry picture below.

Call the resulting manifold (which is still topologically a sphere) $\Sigma$. Let $M_N = \Sigma \times \mathbb{S}^1$. We will place a metric on $M_N$ such that on each of the $S_i$ it is basically just the product metric $h - dt^2$ where $h$ is the round sphere metric. Near the center of the connecting rods the metric will look like (*), with the negative z side near the center and the positive z side near the outside. This can be smoothly glued together.

Let $S'_i$ for $i > 0$ denote the $S_i$ together with the portion of the neck corresponding to $z >0$. (Roughly each of the white regions in the picture above). And let $S'_0$ denote $S_0$ together with all the neck portions corresponding to $z < 0$. (Roughly the magenta portion.) And let $\Gamma_i$ be the boundaries.

To help you think about this: the $\Gamma_i$ (when crossed with $\mathbb{S}^1$) are null hypersurfaces in $M_N$. They function as one-way gates: a future-directed time-like curve can cross from $S_i'$ through $\Gamma_i$ into $S_0'$ in the forward time direction, but cannot go back.

Using a similar analysis as before, given a point $p$, we have that (for this argument, we need that $S_0$ with finitely many disjoint closed discs removed is path connected. This is where we need $n \geq 2$ (and hence spacetime dimension at least 3))

- If $p\in S'_0$, then $I^+(p) = S'_0$ and $I^-(p) = M_N$.
- If $p\in S'_i$, $i > 0$, then $I^+(p) = S'_0 \cup \Gamma_i \cup S'_i$, and $I^-(p) = S'_i$.
- If $p\in \Gamma_i$, then $I^+(p) = S'_0$ and $I^-(p) = S'_i$.

In particular, given $i$, let $p\in S'_i$ and $q\in S'_0$, then $I(p,q) = S'_0 \cup \Gamma_i \cup S'_i$ is open in the Alexandrov topology. In addition, it is the smallest open set in the Alexandrov topology that intersects $\Gamma_i$. Call this set $O_i$.

It is clear now that $\{O_1, \ldots, O_N\}$ is an open cover of $M_N$ in the Alexandrov topology. As $O_i$ are as small as possible, any refinement of the cover will necessarily still contain $\{O_1, \ldots, O_N\}$, and any cover of $M_N$ may be refined to $\{O_1, \ldots, O_N\}$. But now any $p\in S_0$ will be covered $N$ times, this shows that any natural number may arise as the covering dimension of a Lorentzian manifold under the Alexandrov topology.

### The infinite case

I conclude with the case where the covering dimension is infinite. This can be done using a 2 dimensional example, but it can be easily generalized to higher dimensions.

Let $M' = \mathbb{R}\times \mathbb{S}^1 \ni (x,t)$ with the product metric (we identify $\mathbb{S}^1 = \mathbb{R} / \mathbb{Z}$ so has period 1).
Consider $M = M' \setminus (-\infty,0] \times \{0\}$. Then $M_c = \{(x,t) \in \mathbb{R}\times [-1/2,1/2) : x > - |t|\}$.
In particular, if $p\in \partial M_c$, then any open set in the Alexandrov topology that contains $p$ must contain $M_c$.

For $\epsilon > 0$, the set
$$ O_\epsilon = M_c \cup \{(x,t)\in \mathbb{R}\times [-1,0): t < \epsilon \} $$
is open. And altogether they form an open cover of $M$. Key observation is that if $\mathcal{V}$ is a collection of finitely many sets, each of which is a refinement of some $O_\epsilon$, then $\mathcal{V}$ cannot cover $\partial M_c$, never mind $M$. Hence any refinement that covers $M$ needs to contain infinitely many open subsets of the $O_\epsilon$s, each of which intersecting $\partial M_c$; but as every such open set contains $M_c$ we see that the covering dimension must be infinite.

### Remarks

A modification of the procedure in the finite positive case provides a 2 dimensional example with covering dimension 1. I do not know whether in 2 dimensions, other finite covering dimensions are possible.

allof the examples I checked so far). $\endgroup$