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.]

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, $S \subset J^+(S)$.

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.

• Careful, an infinite union of closed sets is not necessarily closed - just as an infinite intersection of open sets is not necessarily open, so you need another argument than just "$J^+(p)$ is closed" to conclude that $J^+(S)$ is closed. – ACuriousMind Feb 15 '15 at 2:54
• @ACuriousMind: Are there any simple counterexamples? I'm not sure how to refine my argument. – Ryan Unger Feb 15 '15 at 2:56
• Take the intervals $[0,1 - \frac{1}{n}]$. They are closed, but the union over all $n$ is $[0,1)$, which is neither open nor closed. I am not a home with causal sets, so I cannot tell you which way this argument is supposed to go. – ACuriousMind Feb 15 '15 at 2:58
• @ACuriousMind: What if $S$ is compact? After all, $[0,1)$ is not compact IIRC. – Ryan Unger Feb 15 '15 at 3:02
• @ACuriousMind: A more general question: What is the restriction needed on $\{U_i\}$ such that $\bigcup_iU_i$ is closed? – Ryan Unger Feb 15 '15 at 3:08

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 $$J^+(p)=\{q: q=p \textrm{ or there is a causal curve from } p \textrm{ to } q \}$$ (I am not familiar with Wald's book, but 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.