Show this is not a manifold with boundary Consider a curve $\alpha: \mathbb R \to \mathbb R^2$ defined by $t \mapsto (e^t \cos(t), e^t \sin(t))$. Show the closure of $\alpha(\mathbb R )$ is not a manifold with boundary. 
Denote $\alpha(\mathbb R )$ by M. The closure of M is the spiral plus the limit point, the origin. I want to show there is no boundary coordinate chart for the origin. Intuitively,for every open neighbourhood $B$ of $(0,0)$, for every given direction $(a,b)$, there is a point $(x,y)$ in $\overline M  \cap B$ s.t. $(x,y)= \lambda (a,b)$. Then $\overline M  \cap B$ is not the graph of a smooth function $g$ because if $g'(0)=b/a$, then there are points in $\overline M  \cap B$ that are not in that direction. But how can I write this rigourously?

Suppose $M$ is a manifold with boundary. Then there is an open set $U$ in $\mathbb{R}^2$ and a diffeomorphism $F:U\to\mathbb{R}^2$ such that $F(M\cap U)$ is either 


*

*the line $\{(x,0):x\in\mathbb{R}\}$, or 

*the half-line $\{(x,0):x\ge 0\}$ 


Consider the $y$ component of $F$, call it $g$. It vanishes on $M$. Consequently, $g_x$ vanishes at every point where the tangent to $M$ is horizontal. As such points accumulate toward the origin, continuity implies $g_x(0,0)=0$. Similarly $g_y(0,0)=0$. But then $DF(0,0)$ is degenerate, contradicting it being a diffeomorphism.
