Open–closed continuation for the set of individually admissible parameters

By [F1] and [F3], the actual original closed exterior section ${\mathrm{\Sigma }}_1=C\cup S$ is connected and smooth with boundary $S$, spacelike and maximal for $g$. Since $g_1=g$ on the exterior and near $S$, these properties hold for $g_1$ up to the boundary. The product in [F3] identifies its interior with the zero stationary graph $u_1=0$. The normalized height in [F1] is therefore identically zero, so all its end derivatives through order four vanish. This verifies every admissibility condition in [F2] and gives $1\in \mathcal{A}$.

Fix $0<\epsilon <1$, set $I=(\epsilon ,1]$ and $B=\mathcal{A}\cap I$, and equip $I$ with $d_I(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$. The three metric axioms in [F6] restrict from $\mathbb{R}$ to $I$. Its radius-$r$ balls are $I\cap (x-r,x+r)$. Thus every trace of an ambient open set satisfies the metric ball condition on $I$; conversely, if $U\subseteq I$ satisfies that condition, choose for each $x\in U$ a radius $r_x>0$ with $I\cap (x-r_x,x+r_x)\subseteq U$. The union of these ambient intervals has trace exactly $U$. Hence this metric topology is the relative topology of [F6].

For any $q_{\ast }\in B$, choose one individually admissible section there. Its hypotheses for [F4] are precisely [F1]–[F2], and $0<\epsilon <q_{\ast }\le 1$. Apply [F4] with $\delta =\epsilon$. The nearby smooth fixed-boundary flow graphs are diffeomorphic to the connected reference section, hence connected; they retain all the graph, maximality, boundary and end conditions in [F2]. Every sufficiently close $q\in I$ therefore belongs to $B$. Thus $B$ is relatively open in $I$, and it is nonempty by step 1.1.

Let $(q_j)$ be any sequence in $B$ converging in $(I,d_I)$ to $q_{\ast }\in I$. By [F7], choose one admissible section ${\mathrm{\Sigma }}_j$ for each term; these are choices only at parameters already in $\mathcal{A}$. All $q_j$ lie in $[\epsilon ,1]$ and $q_j\to q_{\ast }$ there. Thus [F5] applies on that fixed positive interval and supplies an individually admissible section at $q_{\ast }$. It follows that $q_{\ast }\in B$. By the metric sequential-closedness theorem in [F7], $B$ is closed in $(I,d_I)$, hence relatively closed by step 1.2.

The interval $I$ is order-convex, hence connected by [F8]. To apply its separated-set formulation explicitly, suppose $C=I\setminus B$ were nonempty. Steps 2.1 and 2.2 make both $B$ and $C$ relatively open. For each point of either set, step 1.2 gives an ambient interval whose trace on $I$ avoids the other set. Since both sets lie in $I$, that ambient interval itself avoids the other set. By [F8], no point of $B$ lies in the ambient closure of $C$ and no point of $C$ lies in the ambient closure of $B$. Thus the nonempty sets $B,C$ would disconnect $I$, a contradiction. Consequently $B=I$. Given any $q\in (0,1]$, take $\epsilon =q\mathrm{/}2$ to obtain $q\in \mathcal{A}$. This proves the full positive-parameter conclusion without assuming an interval-wide family.