Scope of smooth field equations at an exterior horizon boundary

Fix $q$. The residuals are $d{F}_q$, $d({\star }_{g_q}{F}_q)$, and the covariant two-tensor with components $R_{ab}=\mathrm{Ric}(g_q{)}_{ab}-2((F_q{)}_{ac}(F_q{)}_b{}^c-\frac{1}{4}(g_q{)}_{ab}(F_q{)}_{cd}(F_q{)}^{cd})$. Their components are continuous in every smooth chart in $U$. Indeed the inverse of the smooth nondegenerate matrix $g_q$ is smooth, its oriented volume density and Hodge map are smooth, and the Christoffel and Ricci formulas use at most two derivatives of $g_q$. All contractions and exterior derivatives therefore produce smooth tensor components.

To establish non-implication, use $M={\mathbb{R}}^4$ with coordinates $(t,x,y,z)$ and orientation $dt\wedge dx\wedge dy\wedge dz$, take $E=\{x<0\}$ and $O=M$, and choose the constant family $g_q=-dt\otimes dt+dx\otimes dx+dy\otimes dy+dz\otimes dz$, $N_q={\mathrm{\partial }}_t$, $F_q=\beta (x)dx\wedge dy$. By [F4] the two-forms are smooth across $x=0$, and the metrics are smooth, nondegenerate and Lorentzian everywhere. Their coefficients are constant, so their Christoffel symbols and Ricci tensors vanish. On $E$, $F_q=0$ and all the electrovacuum equations hold.

Let $p\in O\cap \bar{E}$. If a residual were nonzero at $p$, some component $r$ in a chart in $U$ containing $p$ would satisfy $r(p)\mathrm{\ne }0$. Continuity gives an open neighborhood $W$ of $p$ in that chart with $\mathrm{\mid }r\mathrm{\mid }>\mathrm{\mid }r(p)\mathrm{\mid }\mathrm{/}2$. Since $U$ is open in $M$, $W$ is open in $M$. The equations on $E$ imply $W\cap E=\mathrm{\varnothing }$, so the closed set $M\setminus W$ contains $E$. By [F3] it contains $\bar{E}$, contradicting $p\in W\cap \bar{E}$. Thus every residual vanishes at $p$. This proof uses the defining closed-set property of closure directly.

In this example $\bar{E}=\{x\le 0\}$: this closed set contains $E$, and every neighborhood of a point with $x=0$ intersects $E$, since its coordinate image is open. At a point with $x>0$ put $b=\beta (x)>0$. The only independent nonzero field component is $(F_q{)}_{xy}=b$, and raising spatial indices leaves it unchanged. Hence $(F_q{)}_{cd}(F_q{)}^{cd}=2b^2$ and $(F_q{)}_{tc}=0$, so the prescribed right-hand side of the $tt$ Ricci equation is $2(0-\frac{1}{4}(-1)2b^2)=b^2>0$, whereas $\mathrm{Ric}(g_q{)}_{tt}=0$. Thus the Einstein equation fails throughout the nonempty open complement $O\setminus \bar{E}=\{x>0\}$ despite smooth extension across the boundary and validity on $E$. This supplies the asserted non-implication without weakening the boundary conclusion.