The leaf of the foliation has countable basis

Question

This question arises when i'm reading Geometric theory of foliations, Cesar Camacho. In the $\S$2, i need to show that a leaf has sufficient condition to be a manifold, that is Hausdorf and countable basis. Hausdorf is easy to check but i can't find any thing wrong if a leaf has uncountable basis. It need to recommend that we considering a leaf with topology is defined by plaques. So the ony situation to check that when a leaf crosses a chart uncoutably times, what yields contradiction? Thank so much!

Answer 1

Let $(M, \mathscr F)$ be a foliation. The fact the the 2nd countability of the leaves is forced by the 2nd countability of $M$ is a quite subtle point: I'll try to summarise the discussion in Candel & Conlon, Foliations I, §1.2 [BTW, Foliations I and II are good books. If you have access to them, I recommend them whole-heartedly.]

The key technical point is the definition of a regular foliated atlas [Def. I.2.11]:

A foliated atlas $\mathscr U = \{U_\alpha, \phi_\alpha\}_{\alpha \in \mathfrak A}$ of class $C^r$ is said to be regular if

1. for each $\alpha \in \mathfrak A$, $\overline{U_\alpha}$ is a compact subset of a foliated chart $(W_\alpha, \phi_\alpha)$ and $\phi_\alpha = {\psi_\alpha}_{|U_\alpha}$;
2. the cover $\{U_\alpha\}_{\alpha \in\mathfrak A}$ is locally finite;
3. if $(U_\alpha, \phi_\alpha)$ and $(U_\beta, \phi_\beta)$ are elements of $\mathscr U$, then the interior of each closed plaque $P\subset \overline{U_\alpha}$ meets at most one plaque in $\overline{U_\beta}$.

With this kind of atlases, it is quite clear that each leaf is a union of countably many plaques, so it is 2nd countable.

The result you're trying to prove is then a simple corrolary of the following result.

Lemma [1.2.17 in Candel & Conlon]. Every foliated atlas has a coherent refinement that is regular.

["coherent refinement" means that the inital atlas and its refinement both define the same foliation]

I won't copy the proof of this lemma: it's a (quite standard but nontrivial) argument involving Lebesgue's number lemma. If you don't have access to the book and cannot prove the lemma yourself, send me a private message and I'll send you the corresponding pages in the book.