Is $\overline{u^{-1}((-\infty,a))}$ a $C^k$-manifold with boundary for a $C^k$ function $u$?

Question

Let $u\in C^k(\mathbf{R}^d)$, $M:=\overline{u^{-1}((-\infty,a))}=\overline{\{x\in\mathbf{R}^d \mid u(x)<a\}}$, where $a$ is a constant such that $M\neq\emptyset$. Is $M$ a $C^k$-manifold with boundary?

Here are my thoughts. If $a$ is not the critical value of $u$, namely, $\nabla u(x)\neq\mathbf 0$ for any $x\in u^{-1}(a)$, then the Implicit Functionn Theorem can be used to construct the $C^k$-diffeomorphism of $M$ to the upper half space $\mathbf H^n_+=\{x=(x^1,x^2,...,x^n)\mid x^n\ge0\}$.

But how to deal with the case that $a$ is exactly the critical value of $u$? Is the following function a counter-example? \begin{equation} w(x)= \begin{cases} \exp(-\frac{1}{x^2})\sin\frac{1}{x}, &x>0; \\ 0, &x\le0. \end{cases} \end{equation} In the other words, is $N=\overline{w^{-1}((-\infty,0))}$ a smooth manifold with boundary?

Answer 1

You don't have to go to wild functions in order to get a counter example. Just take $w(x,y)=xy$ on $\mathbb R^2$ and $a=0$. Then what you get is the union of two closed quadrants in $\mathbb R^2$, which certainly is not a smooth manifold with boundary around $(0,0)$.