Is an achronal set contained in its own causal future? I use Wald's notation: $I^+$ is the chronological future and $J^+$ is the causal future.
My confusion arises from the following passage in Wald (1984):

Now, let $S$ be a closed, achronal set (possibly with edge). We define the future domain of dependence of $S$, denoted $D^+(S)$, by
  $$D^+(S)=\{p\in M|\, \text{Every past inextendible causal curve through $p$ intersects $S$}\}$$
  Note that we always have $S\subset D^+(S)\subset J^+(S)$.

I have to disagree with the last statement. We know that $S$ is achronal, i.e. $I^+(S)\cap S=\emptyset$. The relation $S\subset D^+(S)\subset J^+(S)$ implies $S\subset J^+(S)$, i.e. $J^+(S)\cap S\ne\emptyset.$ But I cannot see how a set can be both achronal and contained in its causal future. Hence the title of my question. 
I think Wald meant to write $S\subset D^+(S)\subset \overline{J^+(S)}$. [EDIT: Disregard this statement.]
 A: As the causal future of $p$ is the set of points joined to $p$ by timelike or null curves, and the constant path $\gamma(t) = p$ joining $p$ to $p$ itself has vanishing tangent vector and hence is a null curve (though a rather silly one), $p \in J^+(p)$, and so, $S \subset J^+(S)$.
A: The causal relation $J$ is defined so as to be reflexive.
For instance, the causal future is defined as follows
$$
J^+(S)=\cup_{p\in S} J^+(p)
$$
where
$$
J^+(p)=\{q: q=p \textrm{ or  there is  a causal curve from } p \textrm{ to } q \}
$$
(this is really the commonly accepted definition). So it becomes clear that
$$
S=\cup_{p\in S}\{p\}\subset  \cup_{p\in S} J^+(p) \subset J^+(S).
$$
Usually by causal curve one understands a piecewise regular curve, so a constant curve is usually not seen as a causal curve, hence the definition of $J^+(p)$ give above.
A: I'm fairly sure I got it.
The causal future $J^+(p)$ of a point $p$ is defined as the set of all points $q$ connected by a future pointing timelike or null curve to $p$. I think the secret lies in that this is a closed set in Minkowski spacetime. To see this, we see that the curves connecting the points in $J^+(p)$ are the timelike curves (negative length) plus the null curves (zero length). Included in the set of null curves is the trivial curve connecting $p$ to $p$, which has zero length. Thus $J^+(p)$ is closed. Since $J^+(S)$ is just the union of all $J^+(p)$, $p\in M$, it is also closed. This means it contains $S$, because $S\cap\partial J^+(S)\ne\emptyset$ and $J^+(S)$ is closed. I think this generalizes nicely to a general spacetime, because even though $J^+(S)$ need no longer be closed, $S\cap\partial J^+(S)\ne\emptyset$ should still hold.
