# Is an achronal set contained in its own causal future?

+ 2 like - 1 dislike
190 views

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 $S\subset D^+(S)\subset \overline{J^+(S)}$.
This post imported from StackExchange Physics at 2015-02-15 12:13 (UTC), posted by SE-user 0celo7

edited May 2, 2015

+ 3 like - 0 dislike

I do not remember the precise definition in Wald's book, however $J^+(S)$, by definition, always includes $S$. By definition, it is made of $S$ and the points of the spacetime  which do not belong to $S$ and stay along the future directed (so that the trivial curve joining $p$ and $p$ itself is not such a curve) causal curves exiting from $S$.

By the way $J^+(S)$ is not closed in general, though it is true when the spacetime is globally hyperbolic as $J^+(S) = \overline{I^+(S)}$.

answered Feb 19, 2015 by (2,075 points)
+ 1 like - 0 dislike

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

This post imported from StackExchange Physics at 2015-02-15 12:13 (UTC), posted by SE-user ACuriousMind
answered Feb 15, 2015 by (820 points)
+ 0 like - 1 dislike

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.

This post imported from StackExchange Physics at 2015-02-15 12:13 (UTC), posted by SE-user 0celo7
answered Feb 15, 2015 by (50 points)
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.

This post imported from StackExchange Physics at 2015-02-15 12:13 (UTC), posted by SE-user ACuriousMind
@ACuriousMind: Are there any simple counterexamples? I'm not sure how to refine my argument.

This post imported from StackExchange Physics at 2015-02-15 12:13 (UTC), posted by SE-user 0celo7
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.

This post imported from StackExchange Physics at 2015-02-15 12:13 (UTC), posted by SE-user ACuriousMind
@ACuriousMind: What if $S$ is compact? After all, $[0,1)$ is not compact IIRC.

This post imported from StackExchange Physics at 2015-02-15 12:13 (UTC), posted by SE-user 0celo7
@ACuriousMind: A more general question: What is the restriction needed on $\{U_i\}$ such that $\bigcup_iU_i$ is closed?

This post imported from StackExchange Physics at 2015-02-15 12:13 (UTC), posted by SE-user 0celo7
Any compact set in $\mathbb{R}^n$ is bounded and closed, but this does not hold for all metric spaces, in particular, I think it indeed fails on Lorentzian manifolds. That said, I am confused what your actual question here is - you say, $S \subset J^+(S)$ doesn't make sense to you, and yet $J^+$ is the set of points joined by timelike or null curves, and the constant path joining $p$ to $p$ is a null curve, so $p \in J^+(p)$, and hence $S \subset J^+(S)$, right? Generally: Any finite union of closed sets is closed, but nothing else holds in arbitrary topologies.

This post imported from StackExchange Physics at 2015-02-15 12:13 (UTC), posted by SE-user ACuriousMind
@ACuriousMind: Crap. After reading my question again I realize that I don't need $J^+(S)$ to be closed. Post your comment or whatever and I'll accept it as an answer. Thanks. Could you also please answer my more general question (comment above) in your answer? EDIT: Saw your edit.

This post imported from StackExchange Physics at 2015-02-15 12:13 (UTC), posted by SE-user 0celo7