# Foliation of $\mathbb{R}^3$.

A foliation of $$\mathbb{R}^3$$ can be given by subspaces of the form $$\{p\} \times \mathbb{R}^2$$ where $$p$$ spans over $$\mathbb{R}$$. I cant see how any open neighbourhood in $$\mathbb{R}^3$$ intersects only a countable number of these subspaces, as is required in the definition of a foliation given in Lee.

Given $$\mathbb{R}$$ is uncountable, every open set of $$\mathbb{R}^3$$ would intersect an uncountable number of such subspaces no?

$$\textbf{Update}$$- I think my confusion is coming the use of the word countable in the following paragraph:

'Let $$M$$ be a smooth $$n$$-manifold, and let $$\mathfrak{F}$$ be any collection of $$k$$-dimensional submanifolds of $$M$$. A smooth chart $$(U, \varphi)$$ for $$M$$ is said to be flat for $$\mathfrak{F}$$ if $$\varphi(U)$$ is a cube in $$\mathbb{R}^n$$, and each submanifold in $$\mathfrak{F}$$ intersects $$U$$ in either the empty set or a countable union of $$k$$-dimensional slices of the form $$x^{k+1}=c^{k+1}, \cdots , x^n=c^n$$. We define a foliation of dimension $$k$$ on $$M$$ to be a collection $$\mathfrak{F}$$ of disjoint, connected, nonempty, immersed $$k$$-dimensional submanifolds of $$M$$ (called the leaves of the foliation), whose union is $$M$$; and such that in a neighborhood of each point $$p\in M$$ there exists a flat chart for $$\mathfrak{F}$$.'

• I'd be interested in an actual quote of Lee's definition where one needs intersection with only countably many such subspaces ("leaves" of foliation). – coffeemath Jun 25 at 9:34
• Hey I updated my question with where I'm getting confused – jojo Jun 26 at 8:35
• Thanks for including the definition, which makes more sense... – coffeemath Jun 26 at 9:04

It's the individual submanifolds and charts that should have "countable" intersections, not the whole $$\mathfrak{F}$$.
The definition says, each submanifold in $$\mathfrak{F}$$ intersects $$U$$ in (something countable). You can easily find charts such that each individual $$p\times\mathbb{R}^2$$ intersects the chart in countable (in this case, even more strongly, empty or single) slices. Of course there are uncountably many submanifolds in $$\mathfrak{F}$$, and every $$U$$ will intersect uncountablu many of them, but this has nothing to do with how it intersects individual submanifolds (members of $$\mathfrak{F}$$) with $$U$$.