I am studying geometric theory of foliations from Camacho and Lins and I have stumbled upon an afirmation I can't understand.
There is a theorem which states:
Let F be a leaf of a foliation $\mathcal{F}$ on M and $\Sigma$ a transverse section of $\mathcal{F}$ such that $\Sigma \cap F \neq \varnothing$. We have three possibilities:
- $\Sigma \cap F$ is discrete, and in this case F is an imbedded leaf.
- The closure of $\Sigma \cap F$ in $\Sigma$ contains an open set. This occurs if and only if the closure $\bar{F}$ has non-empty interior and $int (\bar{F}) = \bar{F} - \partial \bar{F}$ is an open set which contains F. In this case we say F is locally dense.
- The closure of $\Sigma \cap F$ is a perfect set (i.e., without isolated points) with empty interior. In this case we say F is an exceptional leaf.)
After this theorem comes the part I don't get. The authours state that the following submanifold $F$ in the picture bellow can't be a leave of any foliation deffined at an open set of $\mathbb{R}^2$
The reason being the point $x$ is an accumulation point of $\Sigma \cap F$ where $\Sigma$ is a transverse segment passing through x, while the same does not happen with the point y.
I really don't get this. Suppose that upper horizontal part of $F$ containing $y$ was not there. In this case I guess I would have the case (2) of the theorem, right? To me, that upper part simply means I would have another transverse segment $\Sigma'$ passing through $y$, (vertical to the figure), giving me case (1) of the theorem.
Since we are talking about 2 distinct transverse segments, I don't see what is the problem.
I would appreciate any help since I just recently started studying the topic (please be kind).
Let $F$ be the space indicated in Figure 2. Note that $F$ is closed, i.e. , $\bar{F}=F$. The boundary $\partial \bar{F}$ is given by \begin{align*} \partial \bar{F} = \bar{F}\cap (\overline{\mathbb{R}^2\setminus F}) = \bar{F}\cap \mathbb{R}^2=\bar{F} \end{align*} Hence we obtain for the interior of $F$ that $\mathrm{int}(\bar{F})=\bar{F}\setminus\bar{F}=\emptyset$ and $2.$ cannot hold.
Let us assume that $\Sigma$ is the vertical line through $x$. Then $F\cap \Sigma$ has isolated points but it is not discrete since $x$ is an accumulation point. Hence neither $1.$ nor $3.$ can hold which implies that $F$ cann't be a leaf of a foliation.