Prove this partition of the plane is not a foliation Let us define a partition of the plane as follows: for the points $(x_0,y_0)$ with $y_0\leq0$ we have leaves that are straight lines ($y=y_0$) and for $y_0>0$ we have leaves $e^{x+\operatorname{ln}y_0-x_0}$. This is indeed a partition. But I am being asked whether it is a foliation.
A rank $k$ foliation is a collection $\{L_\alpha\}_{\alpha\in A}$ of leaves of a connected immersed submanifold such that it forms a partition of that manifold and for every point of the manifold there exists a chart $(U,\phi=(x_1,\ldots,x_n)$ such that $U\cap L_\alpha$ is a countable union of slices $\{x_{k+1}=\text{ constant},\ldots,x_n=\text{ constant}\}$ or empty.
Now I do not believe the given partition is a foliation and that the problem lies on the $y=0$ line. When we take a chart around $(x_0,0)$ we have leaves which are lines and leaves which are exponential curves, which will contradict the continuity of $\phi$ the homeomorphism of the chart, I think. But I do not know how to rigorously prove this. 
edit: The Frobenius theorem tells us that there exists a bijection between foliations and involutive distributions which maps a foliation to a distribution $D$, where $D_p := T_pL_\alpha$ for all $p\in \mathbb{R}^2$, where $L_\alpha$ is the leaf through $p$. 
A distribution on $\mathbb{R}^2$ is a subbundle of $T\mathbb{R}^2$. Lemma 10.32 of Lee gives us a good criterium for subbundles. It says that in our case, we have a subbundle if and only if for every $p\in \mathbb{R}^2$ there exists a neighborhood $U$ on which there exists sections $\sigma_1,\ldots,\sigma_m: U\rightarrow T\mathbb{R}^2$ with the property that $\sigma_1(q),\ldots,\sigma_m(q)$  form a basis for $D_q$ at each $q\in U$.
The tangent space to a leaf at a point $(x,y)$ will be a straight line with slope $y$ if $y>0$ and slope $0$ if $y\leq 0$. They are linear spaces of dimension $1$. What we wanted to show is that for a neighborhood around $(0,0)$ (or some other point on $y=0$), there is no smooth section such that the above property holds. I believe a continuous section always exists, but I am not sure how to show whether it is smooth or not.
 A: Suppose it is a foliation. By the global Frobenius theorem, it will correspond to an involutive distribution $D$. Where $D_p$ is given by $T_pL_\alpha$, where $L_\alpha$ is the leaf through $p$. So $D$ is a subbundle of the tangent bundle which means in particular that for a neighborhood $U$ around $(0,0)$ we have a smooth section $\sigma: U\rightarrow T\mathbb{R}^2$ such that $\{\sigma(q)\}$ forms a basis for $D_q$ for $q\in U$. 
Now if $q= (q_1,q_2)$ and $q_2>0$, $D_q$ will be the straight line with slope $q_2$ since $\frac{d}{dx}e^{x+\operatorname{ln}q_2-q_1}= q_2$. If $q_2\leq 0$, $D_q$ will be the straight line with slope $0$. But this means clearly that $\sigma$ can never be smooth since its first derivative will have a discontinuity. Indeed, $\sigma$ will be of the form $q\mapsto (\lambda_q, \lambda_qq_2)$ where $\lambda_q$ is a smooth function from $q$ to an element of $\mathbb{R}$. Deriving the second component, we see that there is a discontinuity for $q_2 = 0$. Hence, $\sigma$ is not smooth.
