Let $M = \mathbb R^{D+1}$ be Minkowski space. Recall that the causal complement of a set $A \subseteq M$ is the set $A^\perp \subseteq M$ where $p \in A^\perp$ there is no timelike path between $p$ and any point of $A$. Recall that $A$ is causally closed if $A = A^{\perp\perp}$. Any set $A \subseteq M$ is contained in a unique minimal causally closed set, its causal closure $A^{\perp\perp}$. It is known [1] that the causally closed subsets of $M$ form an orthomodular lattice with orthocomplement $(-)^\perp$.
For instance,
The causal complement of a singleton $\{p\}^\perp$ is neither open nor closed -- it is the complement of the union of the future and past timelike cones of $p$, minus $p$ itself (which sort of "tunnels" between the two cones).
The causal closure of a singleton is itself.
The causal closure of two-point set $\{p,q\}$ is $\{p,q\}$ if $p,q$ are lightlike or spacelike separated, and shaped like the space of revolution of a "diamond" shape if $p,q$ are timelike separated, with $p,q$ at the opposite tips of the diamond.
Let $C \subset M$ be a Cauchy surface, and let $A \subset C$ be both dense and codense in $C$. Then $A$ is causally closed (in $M$).
On account of examples like (3), it is not possible to fully "classify" causally closed sets. But I wonder if it's possible to obtain a classification "up to" such issues. For instance,
Question 1: If $A \subseteq M$ is causally closed, then it's not hard to show that $A \cup A^\perp$ is topologically closed. Is it possible to "classify" which closed subsets $C \subseteq M$ are of the form $C = A \cup A^\perp$? Does such a set $C$ always contain a Cauchy surface? Is it the union of a Cauchy surface with a set with nonempty interior, about which more might be said?
If $A \subseteq M$ is causally closed, let $\partial_{spacelike}(A) = \overline{A} \cap \overline{A^\perp}$, where $\overline{(-)}$ denotes the topological closure.
Question 2: Which sets $B \subseteq M$ are of the form $B = \partial_{spacelike} A$?
Given a set $C \subseteq M$ of the form $C = A \cup A^\perp$ as in Question 1, I think we can identify $B = \partial_{spacelike} A \subseteq C$ by the condition that $B$ comprises those points $p \in C$ such that any timelike path through $p$ intersects $C$ only at $p$. I believe that every way of choosing $A$ such that $C = A \cup A^\perp$ corresponds to choosing some connected components of $C \setminus B$ to lie in $A$, and letting the remaining ones lie in $A^\perp$. But I'm a bit confused about what the rules are about divvying up $B$ into $A$ and $A^\perp$.
Question 3: Given a set $C \subseteq M$ answering to Question 1, and given its subset $B \subseteq C$ answering to Question 2, what exactly are the rules governing the possible choices of $A \subseteq C$ such that $C = A \cup A^\perp$?
I'm kind of hoping that the requirement boils down to some density / codensity conditions, so that we get a "classification modulo the wildness of choice of some dense-codense subset".
[1] See Cegła and Jadczyk, and also Casini for a generalization to arbitrary Lorentzian manifolds.
