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). 


**Added**


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. 



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


