An interior compact causal trajectory has a finite-time endpoint

The smooth ADM data make $N+\mathrm{\mid }X{\mathrm{\mid }}_h$ continuous on $K$. Since $x(v_0)\in K$, this compact set is nonempty. The extreme-value theorem therefore bounds $N+\mathrm{\mid }X{\mathrm{\mid }}_h$ on $K$, say by $B$. Choose times $v_j\uparrow v_{\ast }$; compactness gives a subsequence for which $x(v_j)\to p\in K$. Take an interior coordinate ball centered at $p$, with coordinate radius $r>0$ and compact closure in its chart. Metric comparison on its closure gives $h(Y,Y)\ge c\mathrm{\mid }Y{\mathrm{\mid }}_{\mathrm{e}}^2$ for $c>0$. Thus while the curve remains in the ball, $\mathrm{\mid }x\mathrm{\prime }{\mathrm{\mid }}_{\mathrm{e}}\le B\mathrm{/}\sqrt{c}=:D$ almost everywhere.

Choose a subsequence time $v_j$ so late that its coordinate distance from the center is less than $r\mathrm{/}4$ and $D(v_{\ast }-v_j)<r\mathrm{/}4$; for $D=0$ only the first condition is needed. Were there a subsequent exit from the coordinate ball, continuity would give a first exit, but integration of the coordinate speed up to that exit would give total displacement less than $r\mathrm{/}4$, contradicting the required displacement of more than $3r\mathrm{/}4$. Hence the whole remaining tail stays in that chart. Its coordinates are $D$-Lipschitz and are Cauchy as $v\uparrow v_{\ast }$. They converge to the coordinate of $p$, since the chosen subsequence does. Thus $(v,x(v))\to (v_{\ast },p)\in E$. The argument with reversed parameter direction treats a finite lower endpoint. Such a limit contradicts precisely the endpoint definition of inextendibility.