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.