Suppose I have a non-empty open set $A\in\mathbb{R}^N$ with a piecewise smooth boundary $\partial A$. For any point $p\in \partial A$ at which $\partial A$ is locally smooth, let $\vec{N_p}$ be the vector which is normal to $\partial A$ and points into $A$.
Let $p^*$ be a point on the boundary $\partial A$ at which two or more smooth components of $A$ meet. Define the set of normal vectors at $p^*$ as:
$$\tilde{N}(p^*) = \{\vec{v}\in\mathbb{R}^N:\vec{v}=\lim_{n\rightarrow\infty}\vec{N}(p_n) \text{ for some sequence }p_n\rightarrow p^* \text{ at which }\vec{N}(p_n) \text{ is well defined. }\}$$
This is probably a really elementary question... but how can I rigourously show that you can always find a $N-1$ dimensional hyperplane $P$, such that all vectors in $\tilde{N}(p^*)$ lie on or to one side of $P$?
I've drawn a picture to help.
This is not true without some more hypotheses on $A$.
If $A = \{ x \in \mathbb R^2 |x_1 x_2 >0 \}$, then $\partial A = \{x\in \mathbb R^2 | x_1 = 0 \text{ or }x_2 = 0\}$ and $\tilde N(0) = \{ \pm e_1 , \pm e_2\}$, which is not contained in any half-space.
I don't know which additional property of $A$ would make this work. Connectedness is not enough : the previous $A$ is not connected, but $A\cup \{ x \in \mathbb R^2 | x_1 - x_2 >1\}$ works just the same.