A domain in $\mathbb{R}^n$ with $C^2$-boundary satisfies an "outer spherical condition"

Question

Let $\Omega\subseteq\mathbb{R}^n$ be a domain and $\partial\Omega\in C^2$, i.e.

• $\Omega=\overline{\Omega}^\circ$
• For all $x_0\in\partial\Omega$, there exists a neighbourhood $U\subseteq\mathbb{R}^n$ of $x_0$ and $\phi\in C^2(U)$ such that $$\overline{\Omega}\cap U=\left\{x\in U:\phi(x)\le 0\right\}\tag{1}$$ and $$\nabla\phi(x)\ne 0\;\;\;\text{for all }x\in U\tag{2}$$

I want to show, that for all $x_0\in\partial\Omega$ it holds $$\overline{B}_r(y_0)\cap\overline{\Omega}=\left\{x_0\right\}\;\;\;\text{for some }y_0\in\mathbb{R}^n\setminus\overline{\Omega}\;\text{and }r>0\tag{3}$$

---

This should be trivial, shouldn't it? Let $x_0\in\partial\Omega$. Since $\partial\Omega\in C^2$, there is a $y_0\in\mathbb{R}^n$ and a $r>0$ such that $(1)$ and $(2)$ hold for $U=\overline{B}_r(x_0)$.

Now, I've got two problems

1. I don't see why $\phi\in C^2(U)$ (instead of $\phi\in C^1(U)$) comes into play.
2. It seems like that we need $\phi^{-1}\left((-\infty,0]\right)=\left\{x_0\right\}$, but how can we achieve that?

Answer 1

To see why $C^1$ is not enough, consider

$$f(x) = |x|^{3/2}\cos (1/x^{1/4}),\, x\ne 0,\,\, f(0) = 0.$$

Then $f\in C^1(\mathbb {R}),$ hence $\phi(x,y)= y-f(x) \in C^1(\mathbb {R}^2).$ Define $\Omega = \{\phi < 0\}.$ Then $\Omega $ is the domain below the graph of $f.$ Good to draw a picture here.

Suppose there exists a closed ball $B$ such that $B\cap \overline {\Omega} = \{(0,0)\}.$ Any ball that is a candidate here has to be tangent to the $x$-axis at $(0,0),$ with center on the positive $y$-axis. Then the lower part of $\partial B$ is a circle whose graph, near $0,$ lies under $(x,ax^2)$ for some  $a>0.$ But $\partial \Omega$ contains infinitely many points on the graph of $|x|^{3/2}$ as $x \to 0.$ An infinite number of these will be trapped in $B,$ contradiction.