Intersecting a chronological past boundary with a spacelike section

The source's results apply componentwise to $E$: a connected component cannot be compact, since $v$ would attain a maximum there with $dv=0$, contradicting its timelike gradient. A future timelike AC-curve is causal AC and has strictly positive Lorentzian length; the imported continuous-curve characterization and positive-length result give a piecewise smooth future timelike curve with the same endpoints. Conversely a piecewise smooth future timelike curve is AC after a finite piecewise affine parametrization. Thus the two chronological relations coincide. Similarly the source's causal relation agrees with AC causal reachability by the imported curve characterization. Consequently $P$ is open, as a union of open chronological pasts, and push-up is valid for the curves used below, in every stated dimension.

Set $W={\mathrm{grad}}_{g}v$ and $V=W\mathrm{/}g(W,W)$. In coordinates the inverse matrix of the smooth nondegenerate metric is smooth, so $W$ and $V$ are smooth. The negative denominator makes $V$ future timelike and $dv(V)=1$. Moreover $dv(U)>0$ for every nonzero future causal vector $U$: in a Lorentz-orthonormal frame write $W=(-\alpha ,w)$ and $U=(\beta ,u)$ with $\alpha >\mathrm{\mid }w\mathrm{\mid }$ and $\beta \ge \mathrm{\mid }u\mathrm{\mid }>0$ for a null $U$, or $\beta >\mathrm{\mid }u\mathrm{\mid }\ge 0$ for a timelike $U$; then $g(W,U)=\alpha \beta +w\cdot u\ge (\alpha -\mathrm{\mid }w\mathrm{\mid })\beta >0$. In particular $v\circ \gamma$ is strictly increasing after $a$.

Let $y_j=\gamma (a+(b-a)\mathrm{/}(j+1))$ for integers $j\ge 1$. Then $y_j\to p$, $v(y_j)>0$ and $v(y_j)\to 0$. By F2, for all large $j$ the local integral curve $s\mapsto \mathrm{\Phi }(s,y_j)$ of $V$ is defined for $-v(y_j)\le s\le 0$. Since $dv(V)=1$, its initial point $z_j=\mathrm{\Phi }(-v(y_j),y_j)$ satisfies $v(z_j)=0$, and smooth dependence gives $z_j\to p$. This curve is future timelike from $z_j$ to $y_j$, while the remaining nondegenerate restriction of $\gamma$ is future causal from $y_j$ to $\gamma (b)\in Q$. Push-up gives $z_j\in P\cap {\mathrm{\Sigma }}^{\circ }$. Openness of $P$ implies $p\mathrm{\notin }P$. Hence every section neighborhood of $p$ meets $P\cap {\mathrm{\Sigma }}^{\circ }$ and its complement, proving the first assertion.

Put $A=P\cap {\mathrm{\Sigma }}^{\circ }$ and $L=K\cap {\mathrm{\Sigma }}^{\circ }=B^{\circ }\setminus A$. The set $A$ is section-open, $B^{\circ }$ is section-closed, and therefore $L$ is section-closed. Take $x\in {\mathrm{\partial }}_{{\mathrm{\Sigma }}^{\circ }}L$. Then $x\in L\subset B^{\circ }$. It cannot be in $O$, since the assumed section neighborhood there is contained in $A$ and is disjoint from $L$. Hence $x\in {\mathrm{int}}_{{\mathrm{\Sigma }}^{\circ }}{B}^{\circ }$. In that interior $L$ is locally the complement of $A$, so every section neighborhood of $x$ meets $A$, while $x\mathrm{\notin }P$. For every ambient neighborhood $U$ of $x$, its trace $U\cap {\mathrm{\Sigma }}^{\circ }$ is a section neighborhood and therefore meets $P$; it also contains $x\mathrm{\notin }P$. Thus $x\in B^{\circ }\cap {\mathrm{\partial }}_{E}P$.

Conversely, take $p\in B^{\circ }\cap {\mathrm{\partial }}_{E}P$. Since $P$ is open, $p\mathrm{\notin }P$ and hence $p\in L$. The assumed null segment lets step 2.1 apply, so every section neighborhood of $p$ meets $A$, which is disjoint from $L$. Each such neighborhood also meets $L$ at $p$ itself. Thus $p\in {\mathrm{\partial }}_{{\mathrm{\Sigma }}^{\circ }}L$. It cannot belong to $O\subset A$, so it lies in ${\mathrm{int}}_{{\mathrm{\Sigma }}^{\circ }}{B}^{\circ }$. Together with step 2.2 this proves the equality and its stated inside interpretation.