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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 $\mathfrak{F}$.'

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    $\begingroup$ 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). $\endgroup$ – coffeemath Jun 25 at 9:34
  • $\begingroup$ Hey I updated my question with where I'm getting confused $\endgroup$ – jojo Jun 26 at 8:35
  • $\begingroup$ Thanks for including the definition, which makes more sense... $\endgroup$ – coffeemath Jun 26 at 9:04
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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 $U$.

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