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

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!

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 $$x_j$$.