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Let $(M,g)$ be a spacetime. Let's say that $F$ is a future set if $F = I^+(S)$ for some set $S$. I'm trying to check the equivalence "$F$ is a future set if and only if $I^+(F) \subseteq F$".

If $F$ is a future set, then $F = I^+(S)$ implies $I^+(F) = I^+(I^+(S))\subseteq I^+(S) = F$, as wanted, since the last inclusion is easy to check and holds for arbitrary sets.

I can check the other implication if we assume from the start that $F$ is open. Then $S = F$ would work, since the inclusion $I^+(F) \subseteq F$ is given, and the inclusion $F \subseteq I^+(F)$ is verified by taking $x \in F$ and a small geodesic ball centered in $x$ contained in $F$ (this is possible since we assume $F$ open); then we use the exponential map to send a timelike geodesic to the past of $x$, and any point on that curve will testify that $x \in I^+(F)$, as wanted.

I do not know how to proceed if we don't assume that $F$ is open from the start. Help?

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The equivalence you propose is false if $F$ is not open. Actually, your second statement is the true definition: $F\subset M$ is a future set if $I^+(F)\subset F$. Your first statement clearly can only be true if $F$ is open, for $I^+(x)$ is open for all $x\in M$, and only in such a case can it be equivalent to your second statement. However, there are subsets $F\subset M$ that are not open and nonetheless satisfy $I^+(F)\subset F$ - take e.g. $$F=\{x=(x^0,\ldots,x^{n-1})\in\mathbb{R}^{1,n-1}=(\mathbb{R}^n,g=\text{diag}(-+\cdots+))\ |\ x^0\geq 0\}\ ,$$ which is clearly closed and indeed satisfies $I^+(F)\subset F$. Of course, the inclusion is strict if $F$ is not open.

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  • $\begingroup$ Imho you should consider upvoting the question if you deem it good enough to spend time writing an answer. $\endgroup$ – joshphysics Jul 25 '17 at 3:48
  • $\begingroup$ Fair enough... Sorry for that. $\endgroup$ – Pedro Lauridsen Ribeiro Jul 25 '17 at 3:48

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