Why is the set of singular points of starlike boundary $\Gamma$ closed?

Question

I'm reading Geometric Inequalities of Yu. D. Burago and V. A. Zalgaller. I don't understand why $E_1$ is closed in the proof of the following lemma.

Several definition.

Suppose $ \Omega $ is a compact $ m $-dimensional submanifold with non-empty boundary $ \Gamma $ in an $ m $-dimensional ($m \geqslant 2$) Riemannian manifold $ M $, where $ \Gamma $ is a $ C^{1} $-smooth $ (m-1) $-dimensional submanifold. We shall say that $ \Omega $ has a boundary $ \Gamma $ starlike with respect to the point $ q \in \Omega $ if in the metric $ \rho $ induced in $ \Omega $ by the inclusion $ \Omega \subset M $, each point $ x \in \Gamma $ can be joined to $ q $ by at least one shortest arc $ q x $, and each such shortest arc lies in $ \{x\} \cup {\rm int} ~\Omega $ and is transversal to $ \Gamma $ at the point $ x $. The point $ x \in \Gamma $ is called non-singular if the shortest arc $ q x $ is unique and on it $ q $ and $ x $ are not conjugate. The other points of $ \Gamma $ are considered singular.

Lemma. The set of singular points of a smooth starlike boundary $ \Gamma $ of a domain $ \Omega $ is closed and possesses a zero $ (m-1) $-dimensional measure on $ \Gamma $.

Partial proof. Suppose $ E_{1} $ is the set of those singular points of $ \Gamma $ for which $ \Omega $ possesses at least one shortest arc $ q x $ along which $ x $ is conjugate to $ q $; $E_{2} $ is the set of all the other singular points.

Suppose $ Q_{0} $ is the set of unit vectors in $ T_{q} M $ tangent to those lines $ q x $ for which $ x \in E_{1} $ and $ x $ is conjugate to $ q $. Since $ x q $ is transversal to $ \Gamma $, according to the implicit function theorem, we can find open sets $ Q \supset Q_{0}$, $Q \subset S^{m-1}(0,1) \subset T_{q} M $ and a smooth function $ l: Q \rightarrow \mathbb{R}_{+}^{1} $ such that for $ y \in Q $, the geodesic line $ \exp _{q} t y $ intersects $ \Gamma $ at the point $ \exp _{q} l(y) y $ and if $ y \in Q_{0} $, then $ \exp _{q} t y $ is the shortest line joining $ q $ to $ x=\exp _{q} l(y) y \in E_{1} $. Since $ E_{1} $ is contained in the image of the critical points of the map $ f: Q \rightarrow \Gamma $ defined by the relation $ f(y)=\exp _{q} l(y) y $, it follows from Sard's theorem that $ \operatorname{mes}_{m-1}\left(E_{1}\right)=0 $. Since all the $ q x $ are transversal to $ \Gamma $, the set $ E_{1} $ is closed.

The last sentence states that "Since all the $ q x $ are transversal to $ \Gamma $, the set $ E_{1} $ is closed." This makes me very confused. I've been thinking for a long time, but I don't know why. Any help would be appreciated!

Answer 1

I agree with you, the sentence looks a bit strange. In fact, the conclusion that $E_1$ is closed is true without the assumption of transversality. Here is how I would prove it.

Let $(x_j \mid j \geq 1)$ be a sequence of points in $E_1$, converging to some point $x \in \Gamma$. Write $\gamma_j$ for the minimizing geodesic from $q$ to $x_j$, each of which has a non-zero Jacobi field by assumption.

We can extract a subsequence in order to guarantee that $\gamma_j \to \gamma$, a minimizing geodesic from $q$ to $x$. Let $v \in T_q M$ be so that $\gamma = t \in [0,1] \mapsto \exp_q(tv)$, and likewise define $v_j \in T_q M$ corresponding to $\gamma_j$. Then $v_j \to v$.

Then we can argue by contradiction: if $x$ were not conjugate to $q$ along $\gamma$, then $(\mathrm{d} \exp_q)(v): T_v( T_q M) \to T_x M$ would be invertible. But then $(\mathrm{d} \exp_q)(v_j)$ would also be invertible, once $j$ is large enough that $v_j$ lies close to $v$, and this would contradict our assumption about the $x_j$.