A countable dense subset of the universal development by rational timelike exponential steps from a dense subset of the initial surface

(The set $D$.) Choose a countable dense subset $S_0\subseteq \bar{j}(\mathrm{\Sigma })$ ([L2]) and set $D_0:=S_0$. Recursively, given the countable set $D_n$, choose by Countable Choice a basis $(e_1(x),\dots ,e_4(x))$ of $T_{x}Q$ at every newly added point, retaining all previously chosen bases, form $V_x^{\mathrm{r}\mathrm{a}\mathrm{t}}$ as in the statement, and define $D_{n+1}$ by adding all points ${\exp }_x(v)$ with $x\in D_n$, $v\in V_x^{\mathrm{r}\mathrm{a}\mathrm{t}}$ timelike and ${\exp }_x(v)$ defined. Dependent Choice supplies successive compatible choices of these basis assignments and sets. Each $D_{n+1}$ is countable, being the image of a subset of the countable set $D_n\times {\mathbb{Q}}^4$; hence $D:={\bigcup }_{n\in \mathbb{N}}{D}_n$ is countable by [L6]. Since $D_0\subseteq D$, claim (a) holds.

(Chains of timelike chords, including the past direction.) Fix $p\in Q$ and a member $Q_{\alpha }$ containing it. In that member the complete flow of [L5] joins a point $z_0\in S:=\bar{j}(\mathrm{\Sigma })$ to $p$ with a parameter $t_1\in \mathbb{R}$. If $t_1=0$, then $p\in S$ and is already in the closure of $D_0$. Otherwise finitely many convex normal neighborhoods cover this compact flow segment; a sufficiently fine finite partition of the segment has each subsegment contained in one of those neighborhoods. Write the ordered endpoints along the flow from $S$ to $p$ as $z_0,\dots ,z_k=p$. The inverse velocity $w_i=v(z_{i-1},z_i)$ for each contained joining geodesic is timelike by [L4], using future chords when $t_1>0$ and past chords when $t_1<0$ (reverse the future-directed chord and use the time reversal of the geodesic). The recursion includes both future and past timelike vectors.

(Propagation of closure, without a stage-index assumption.) Work in $Q_{\alpha }$ and prove $z_i\in \overline{D\cap Q_{\alpha }}$ by induction. The case $i=0$ follows from the density of $D_0$ in $S$, since $S$ lies in every member. Suppose the claim holds for $z_{i-1}$, and prescribe any open neighborhood $O$ of $z_i$ within this member. The two-point inverse $v(q,z_i)$ is defined for $q$ close to $z_{i-1}$ in the convex neighborhood of step 1.2 and tends to the timelike vector $w_i$. Hence, for sufficiently close $q$, it is timelike. Choose such a $d\in D\cap Q_{\alpha }$ by the induction hypothesis. There is an open neighborhood of $v(d,z_i)$ in $T_{d}Q$ on which vectors remain timelike, the member geodesic remains defined to time one, and its exponential image lies in $O$, by openness and continuity of the member exponential. Choose a vector $u$ in the fixed rational span at $d$ in this neighborhood. Its quotient exponential equals its member exponential by uniqueness of the geodesic ODE and remains in $O$. As $d\in D_m$ for some $m$, the recursion gives ${\exp }_{d}u\in D_{m+1}\subseteq D$. This proves that every such $O$ meets $D\cap Q_{\alpha }$, completing the induction. The exact inverse identity ${\exp }_d(v(d,z_i))=z_i$ has zero error; only approximation of this vector by $u$ contributes an endpoint error.

(Density.) Step 2.1 gives $p\in \overline{D\cap Q_{\alpha }}$ (and the $t_1=0$ case follows from step 1.2). Since $Q_{\alpha }$ is open in $Q$, every open neighborhood of $p$ in $Q$ therefore meets $D$. Hence $D$ is dense in $Q$.

(Conclusion.) The preceding steps construct a countable set $D\subseteq Q$ with $D\cap \bar{j}(\mathrm{\Sigma })$ dense in $\bar{j}(\mathrm{\Sigma })$ and $D$ dense in $Q$, of the stated explicit recursive form.