Misner-complete Semi-Riemannian manifold implies inextendible

Question

I need to prove that: Complete Semi-Riemannian manifold implies Misner-Complete and Misner-complete implies inextendible.

This question is from O'Neil, Semi Riemannian Geometry, Q15 in chapter 5.

I know how to prove the first implication, but not sure how prove that Misner-complete implies inextendible.

Answer 1

Suppose $M$ is an extendible semi-Riemannian manifold, i.e., there is an isometric embedding $\iota\colon M \to \tilde{M}$ into some semi-Riemannian manifold $\tilde{M}$ such that $\iota(M)\neq\tilde{M}$ and $\iota(M)$ is an open subset of $\tilde{M}$.

Choose a point $p\in\partial(\iota(M))$ and a point $q\in\iota(M)$ such that $q$ and $p$ can be joined by a finite length timelike or spacelike $\tilde{M}$-geodesic arc $\gamma\colon[0,b]\to\tilde{M}$, i.e., $\gamma(0) = q$, $\gamma(b)=p$, and $0<b<\infty$. (This is possible as $\iota(M)$ is open in $\tilde{M}$.)

As $\iota$ is an isometry $\gamma|_{[0,b)}$ pulls back to a geodesic on $M$: $$\delta\colon[0,b)\to M,\quad s\mapsto \delta(s):=(\iota^{-1}\circ\gamma)(s)$$

But then $\delta([0,b))$ is not contained in a compact subset of $M$ because for any sequence ${s_n}\to b$ in $[0,b)$, $\gamma(s_n)\to p\notin\iota(M)$ and thus $\delta(s_n)$ does not converge in $M$.

This proves extendible $\Rightarrow$ Misner-incomplete.