I am taking a second pass at Lee's book, dotting the i's in the proofs and setting myself the objective of doing all the exercises in the body of the text. My question has to do with a step in the proof of the claim:
let $D$ be an involutive distribution on a smooth manifold $M.$ The collection of all maximal connected integral manifolds of $D$ forms a foliation of $M.$
Start by taking an involutive distribution $D\subseteq TM$, and let $\{N_{\alpha}\}_{\alpha}$ be any collection of connected integral manifolds of $D$ with a point in common, and consider $N=\bigcup N_{\alpha}.$ Give $N$ the final topology induced by ${\displaystyle i_{\alpha }:N_{\alpha }\to N}.$ This works because $N_{\alpha}\cap N_{\beta}$ is open in $N_{\alpha}$ and in $N_{\beta}.$
The idea is to show that $N$ is an integral manifold of $D.$ Then, by previously proved results, the claim follows.
My question has to do with the step in the proof of second-countablility:
We can cover $M$ with countably many flat charts for $D$, say ${W_i}.$ It suffices to show that $N\cap W_i$ is contained in a countable union of slices for each $i$ , because any open subset of a single slice is second-countable and thus $N$ can be expressed as a union of countably many subsets, each of which is second-countable and open in $N.$
I do not quite understand this statement. Isn't it in fact true that if a slice $S$ intersects $N\cap W_i$, it must be contained in $N?$
The following picture, taken from Prof. Lee's book got me to think about this approach:
If $H$ is some $N_{\beta}$ and $S$ is a slice, and therefore, an integral manifold of $D$, then $S\cup N_{\beta}$ is one of the $N_{\alpha}.$
And if a slice $S$ of $W_i$ meets $N$ then it meets some $N_{\beta}.$ Now, $S\cap N_{\beta}$ is open in $S$ (again, by a previous result) and second-countable. I believe $S\cap N_{\beta}$ is also open in $N.$ Assuming that this is so, then all that remains is to show that $N$ contains an at most countable number of slices, which is not hard to prove. It's mostly a copy of Lee's proof: fix $p\in \bigcap _{\alpha}N_{\alpha}.$ Of course, $p$ is contained in some slice $S_p.$ Now, consider the collection $\mathscr C$ all finite sequences of slices $S_{i_j}\subseteq W_{j}$, such that successive elements of a given sequence have non-empty intersection, and such that $S_p$ is the first element of any sequence. If $S$ is an arbitrary slice contained in some $W_i\cap N,$ then we may choose a $q\in S\cap N_{\alpha}\cap W_i\subseteq W_i\cap N$ and produce a path in $N_{\alpha}$ from $p$ to $q.$ And now a compactness argument shows that there is a sequence in $\mathscr C$ from $S_p$ to $S.$ In this way, we get a function from $\mathscr C$ onto the set of slices of flat charts of the $N_{\alpha}.$ Now, $\{W_i\}_i$ is countable and since at each step in the sequence, when it comes to choosing the next one, there are only countably many slices to choose from, because there are only countably many slices in any $W_i.$ So, in fact, $\mathscr C$ is countable. This concludes the proof.
I'd like to know if this idea will work, and if possible, a clarification of the second annotated remark cited above.