Existence of a convex covering on a Semi-Riemannian Manifold

Question

A convex covering $R$ of a Semi-Riemannian manifold $M$ is a covering of $M$ by open geodesically convex sets, such that if elements $a$ and $b$ of $R$ meet, then the intersection is also convex.

In the Book of O'Neill on Semi-Riemannian manifolds on p.131 it is shown in Lemma 10, that for every open covering $C$ of $M$, there exists a convex refinement $R$ of $C$. Let $C^*$ be the open covering of $M$ consisting of all convex open sets contained in any element of $C$.

What I don't understand, is the following step in this proof: "Since $M$ is second countable, hence paracompact, there is an open covering $B$ such that if two elements meet then their union is contained in some element of $C^*$"

Would be glad, if someone could give me a hint...

Answer 1

At first I thought that the following "proof" would work. But this is clearly fake since $A^*\cap B^*$ is not convex. The correct proof uses what the topologists call barycentric covering, it can be seen on Engelking's book and is a kind of characterization of paracompact spaces (with the additional hypothesis of $T1$.)

Let $\cal B$ be a refinement of $\cal C^*$. Then, take $A,B \in \cal B$ and suppose that $A\cap B \neq \emptyset.$ Note that since $\cal B$ is a refinement, there are $A^*, B^* \in \cal C^*$ such that $A\subset A^*$ and $B\subset B^*$. Then, once $A\cap B \neq \emptyset,$ necessarily $A^* \cap B^* \neq \emptyset$. Furthermore, $A^*\cap B^* \subset A^*$ that is convex. Therefore, $A^*\cap B^*$ is convex.

Now, one observes that $A^*\cup B^*$ is convex. In fact, take two points $p,q \in A^*\cup B^*.$ Note that since $A^*, B^*$ are convex its only remain to consider $p \in A^*$ and $q \in B^*.$ Take $o \in A^*\cap B^*$ and the unique minimizing geodesics $\gamma_1$ and $\gamma_2$ that connects $p$ to $o$ and $o$ to $q$. Note that $\gamma_1\ast\gamma_2$ is the unique minimizing geodesic connecting $p$ to $q$, hence, $A^*\cup B^*$ is convex, as desired. Once $A\cup B$ is contained on $A^*\cup B^*$, the claim follows.

Answer 2

Since every smooth manifold is metrizable, the following suffices:

Claim: Let $X$ be a metric space, and $C$ an open cover of $X$. Then there exists an open cover $D$ of $X$, such that if two elements of $D$ meet, their union is contained in an element of $C$.

Proof: For each $p$ in $X$, let $\varepsilon_p > 0$ be such that $B_{\varepsilon_p}(p)$ is contained in an element of $C$. Let $\varepsilon_p' = \varepsilon_p/4$. We claim that the set of $B_{\varepsilon_p'}$ satisfies the result. Let $V_p = B_{\varepsilon_p'}$. If $V_p$ and $V_q$ meet, then there exists a point in their intersection. Suppose, without loss of generality, that $\varepsilon_q \geq \varepsilon_p.$ Then $d(p,q) < \varepsilon_p' + \varepsilon_q' \leq 2 \varepsilon_q'$. Suppose $r$ is in the union of $V_p$ and $V_q$. Then either $r$ is in $V_q$ or $V_p$. In the latter case, $d(r,q) \le d(r,p) + d(p,q) < 3 \varepsilon_q' < \varepsilon_q.$ Thus in either case, $r$ is contained in $B_{\varepsilon_q}(q)$. But $r$ was an arbitrary element of the union, so the union of $V_p$ and $V_q$ is contained in $B_{\varepsilon_q}(q)$, which is contained in an element of $C$. Done.