Is this a $C^0$ foliation of $\mathbb{R}^2$?

Let $f(x)=\frac{1}{\sin(\pi x)}$ for $x\in (0, 1)$ and let $\Gamma=\left\{(x,f(x)): x\in (0, 1)\subset \mathbb{R}^2\right\}$ be its graph.

For any set $X\subset \mathbb{R}^2$ and $\lambda>0$ and $\mathbf{v}\in \mathbb{R}^2$, let $\lambda X+\mathbf{v}=\left\{\lambda \mathbf{x}+\mathbf{v}: \mathbf{x}\in X\right\}$.

Let $\mathcal{F}_0$ be the (smooth) foliation of $(1,2)\times \mathbb{R}$ whose leaves are $\Gamma+t \mathbf{e}_2+\mathbf{e}_1$ for $t\in \mathbb{R}$. For any integer $n$ let $\mathcal{F}_n=2^n \mathcal{F}_0$ be the corresponding (smooth) foliation of $(2^n, 2^{n+1})\times\mathbb R$.

We construct a "foliation," $\mathcal{F}$, of $\mathbb{R}^2$ by taking the leaves of $\mathcal{F}$ to be the leaves of $\bigcup_{n=-\infty}^\infty \mathcal{F}_n$ together with the vertical lines at $x=2^n$ for all integers $n$ and the vertical lines $x=t$ for all $t\leq 0$.

My question is whether this is actually a $C^0$ foliation. I'm having a lot of trouble seeing how it could be true for any point on the line $x=0$, but I also don't really know how to show that it isn't (it's obviously not a $C^1$ foliation).

By $C^0$ foliation, I mean that for each point $p\in \mathbb{R}^2$, there is a neighborhood $U_p$ and a homeomorphism $\psi:U_p\to \mathbb{R}_x\times \mathbb{R}_y$ so that $\psi^{-1}(\mathbb{R}\times \{ s\})$ is a connected component of $U_p\cap \sigma$ for $\sigma$ a leaf of the foliation.

No, your partition of $\mathbb R^2$ into curves is not a foliation in the sense you provide. To see this, you only have to notice that any neighborhood $U$ of $(0 , y_0)$ disconnects some curves and does not others. Being given a rectifying chart $(\psi , U)$ and a strict subdomain $U'\subset U$ with smooth boundary, there exists a limiting point $p\notin \{x=0\}$ where two components belonging to the same global curve meet (it is a tangency point between $\partial U'$ and the curve). The image $\psi(U')$ must therefore disconnect some lines $\mathbb R\times\{s\}$ arbitrary close to $\psi(p)$. As a conclusion it is not homeomorphic to another $\mathbb R^2$ through an homeomorphism leaving the vertical lines globally invariant (which is what you want to achieve).

That being said, there does exist a non-constant continuous first integral $F : \mathbb R^2\to\mathbb R$. (This fact underlines the genuine difference with $C^1$ foliations, for which the local inverse theorem applies)

Define $$g : (x,y)\mapsto y-\frac{1}{\sin(\pi x)}.$$ Notice that the function $$F_0 : (x,y)\longmapsto\frac{1}{1+|g(x,y)|}$$ is a continuous first integral of $\mathcal F_0$ on the closed rectangle $[1,2]\times \mathbb R$. Define in the same fashion $F_n$ on $[2^n,2^{n+1}]\times \mathbb R$ by $$F_n(x,y):=2^{-|n|}F_0(2^{-n}x , 2^{-n}y)$$ which is a first integral of $\mathcal F_n$ on the closed rectangle.

These functions patch together, for they vanish along the vertical lines $\{x=2^n\}$, to give a function $F$ continuous on $]0,+\infty[\times \mathbb R$. Simply define $F(x,y):=x$ if $x\leq 0$. This extension is continuous because each $F_n$ is bounded by $2^{-|n|}$.

• Sorry I should have specified. By $C^0$ foliation, I mean that for each point $p\in \mathbb{R}^2$, there is a neighborhood $U_p$ and a homeomorphism $\psi:U_p\to \mathbb{R}_x\times \mathbb{R}_y$ so that $\psi^{-1}(\mathbb{R}\times \{ s\})$ is a connected component of $U_p\cap \sigma$ for $\sigma$ a leaf of the foliation. This seems to be a stronger condition than what you have shown to exist, but I'm not sure... – foliations Feb 18 '15 at 13:08
• @foliations Yes indeed, your requirement is stronger. I will amend my answer. – Loïc Teyssier Feb 18 '15 at 15:17
• Thanks for the update. I'm having a bit of trouble following the logic of the first paragraph, do you mind expanding on it? – foliations Feb 18 '15 at 23:00
• After sleeping on it, I now understand. Thanks so much for your assistance! – foliations Feb 19 '15 at 23:18

Wanted to add a picture of the foliation but could not place the image into a comment (can it be done?) so put it in an answer

(just a sketch of it)