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Let $S \subset \mathcal{M}$ (where $\mathcal{M}$ is our manifold), the future Cauchy development of $S$ is defined as

$$D^+ (S) := \{p \in \mathcal{M} |\ \text{every past inextendible causal curve through} \ p \ \text{intersects} \ S \}$$

analogously, the same name is used for the following set

$$\tilde{D}^+ (S) := \{p \in \mathcal{M} |\ \text{every past inextendible timelike curve through} \ p \ \text{intersects} \ S \}$$

Both Minguzzi (2019) and Hawking & Ellis (1973) (in Remark 3.2 and Proposition 6.5.1 respectively) state that $D^+ (S) \subset \tilde{D}^+ (S)$. Why is this true? Both sources state it as a direct consequence from the former definitions with no further explanation, but I don't see it. A causal curve is either a null curve or a timelike curve. This means that for $p \in D^+ (S)$, there are more curves that pass through $p$ that intersect $S$ compared to the points in $\tilde{D}^+ (S)$, shouldn't this mean that $\tilde{D}^+ (S) \subset D^+ (S)$? After all, the timelike curves mentioned in the definition of $\tilde{D}^+ (S)$ are included in the definition of $D^+ (S)$.

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    $\begingroup$ It all causal curves etc passing through $p$ intersect $S$, then, in particular, all timelike curves etc passing through $p$ intersect $S$. In other words, if $p\in D^+(S)$, then $p$ also belongs to the other set. That is just the thesis. $\endgroup$ Commented Sep 1, 2021 at 15:17
  • $\begingroup$ But doesn't it work the other way around too? If all timelike curves passing through $p$ intersect $S$ (thus $p \in \tilde{D}^+ (S)$), said curves are also causal by defintion, then $p$ belongs in $D^+ (S)$. I don't see where I'm wrong here, but I must be, otherwise both sets would be the same. $\endgroup$
    – R. M.
    Commented Sep 1, 2021 at 15:22
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    $\begingroup$ Consider an open spacelike disk $S$ in Minkowski spacetime. Then $\tilde{D}^+(S)$ includes all points in the future conical surface whose basis is $S$. Think of the tip of the cone in particular. These points do not belong to ${D}^+(S)$, so that the inclusion is strict in general. $\endgroup$ Commented Sep 1, 2021 at 16:07

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Suppose there exists a point $p\in\mathcal{M}$ for which every past inextendible timelike curve through $p$ intersects $S$, but there exist an (inextendible) null curve through $p$ that does not intersect $S$. Then $p \in \tilde{D}^{+}(S)$, but $p \notin D^{+}(S)$.

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  • $\begingroup$ I see. However, let's put it the other way around. Let $p \in \tilde{D}^+ (S) \Rightarrow$ every timelike curve that passes through $p$ intersects $S$. In particular, since these curves are timelike, they're also causal, doesn't this mean that $p \in D^+ (S)$? $\endgroup$
    – R. M.
    Commented Sep 1, 2021 at 15:26
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    $\begingroup$ No, because of the potential case above. There could be causal curves through $p$ that do not intersect $S$. $\endgroup$
    – TimRias
    Commented Sep 1, 2021 at 15:28
  • $\begingroup$ Indeed. I was ignoring the requirement that $\underline{\text{every}}$ causal curve through $p$ must intersect $S$ in the definition of $D^+ (S)$. Thank you for your explanation. $\endgroup$
    – R. M.
    Commented Sep 1, 2021 at 15:37
  • $\begingroup$ I explicitly described a counterexample in my comment above. $\endgroup$ Commented Sep 1, 2021 at 16:13

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