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I am studying the global causality of the spacetime. Here, I come across a problem.

Suppose a point $r\in \partial I^+(p)$. $I^+(p)$ is the chronological future of a different point $p$ in spacetime. Then, it is claimed that $I^+(r)\subset I^+(p)$. But why?

Let me first try to prove this conclusion. I notice there is a theorem:

Let a subset $S\subset M$ ($M$ the spacetime manifold) and set $B=\partial I^+[S]$. Then, if $x\in B-\bar S$, there exists a null geodesic $\eta\subset B$ with future endpoint $x$ and which is either past-endless or has a past endpoint on $\bar S$.

So we can set $S=\{p\}=\bar S$. Since $r\in\partial I^+(p)$ and $r\ne p$, $r\in B-\{p\}$ with $B=\partial I^+(p)$. Therefore, there is a null geodesics $\eta$ lying on $B$ and passing through $p$. Is this correct?

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  • $\begingroup$ Is this homework? What definition are you using for the chronological future? Depending on the definition, the transitivity of the relation might just be trivial. $\endgroup$
    – user4552
    Aug 15, 2014 at 2:38
  • $\begingroup$ No, it's not homework. I am learning it by myself. I actually read Penrose's Techniques of Differential Topology in Relativity. In this book, he defined the chronological future of a point $p$ as the set of points which are connected to $p$ via trips. A trip is a curve which is piecewise a funture-oriented timelike geodesic. We can also use timelike curves to define the chronological future. $\endgroup$ Aug 15, 2014 at 2:54
  • $\begingroup$ @DrakeMarquis: I haven't read the book, but by that definition, $r$ isn't (or at least isn't necessarily) in the chronological future of $p$. Are you sure the segments aren't allowed to be null? If they are, then I think your proof can work: you show that $r\in I^+(p)$ by exhibiting a trip (though you need to show that it's not past-endless), and then the result follows from the transitivity of "is in the chronological future of", which is trivial (glue the trips together). If they aren't, then I don't see how to use the null geodesic you found, or how else to prove this result. $\endgroup$
    – benrg
    Aug 15, 2014 at 3:52
  • $\begingroup$ @benrg The segments shouldn't be null. I should show that the trip should end at $p$, which is difficult for me... $\endgroup$ Aug 15, 2014 at 4:00
  • $\begingroup$ @benrg I realized just now it might be a better idea to use the property of boundary. $r\in\partial I^+(p)$, so $r$ is a limiting point which implies that any open set $O$ containing $r$ intersects $I^+(p)$. I only have to show that $O\cap I^+(r)$ belongs to $I^+(p)$... but, I am not sure how to do so... By the way, $r\notin I^+(p)$ bc $r$ is on the boundary. $\endgroup$ Aug 15, 2014 at 4:09

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@benrg I got it. Follow your first half suggestion. Pick $s\in I^+(r)$, so $r\in I^-(s)$. $I^-(s)\cap I^+(p)\ne\emptyset$. Choose any point $q\in I^-(s)\cap I^+(p)$. There is a trip from $q$ to $s$, and at the same time, there is trip from $p$ to $q$, so glue the 2 trips to get a 3rd one from $p$ to $s$. Therefore, $s\in I^+(p)$. $s$ is an arbitrary point in $I^+(r)$, so $I^+(r)\subset I^+(p)$. Thank you!

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