A compact set $K\subseteq\mathbb R^3$ is said to have a smooth boundary, if for all $p\in\partial K$ there is an open neighborhood $U$ of $p$ and a continuously differentiable function $\psi:U\to\mathbb R$ such that $$K\cap U=\left\{x\in U:\psi(x)\le 0\right\}\tag1$$ and $$\psi'(x)\ne0\;\;\;\text{for all }x\in U\tag2.$$ In that case, $\partial K$ is a $2$-dimensional $C^1$-submanifold of $\mathbb R^3$.
Now, if $M$ is a $k$-dimensional $C^1$-submanifold of $\mathbb R^n$, $k\le n$:
- $v\in\mathbb R^n$ is called tangent vector on $M$ at $p\in M$, if there is a $\varepsilon>0$ and a continuously differentiable function $\gamma:(-\varepsilon,\varepsilon)\to M$ such that $\gamma(0)=p$ and $\gamma'(0)=v$. The space of all such $v$ is denoted by $T_p(M)$
- $v\in\mathbb R^n$ is called normal vector on $M$ at $p$, if $v$ is orthogonal to $T_p(M)$. The space of all such $v$ is denoted by $N_p(M)$
We can show that $N_p(M)$ is $(n-k)$-dimensional for all $p\in M$. So, $N_p(\partial K)$ is $1$-dimensional for all $p\in\partial K$. We can select the unique $\nu(p)\in N_p(\partial K)$ of unit length with the following property: There is a $\varepsilon>0$ such that $p+t\nu(p)\not\in K$ for all $t\in(0,\varepsilon)$.
The map $\nu:\partial K\to\mathbb R^3$ is seen to be continuous.
Are we able to find such a unique and continuous normal field $\nu$ even for more general classes of manifolds?
For example, consider the following triangulation of a teapot (just the empty shell):
Since even a single triangle is not a "compact set with smooth boundary", we cannot apply the result described above. But shouldn't we nevertheless be able to select a single "outer" unit normal vector on each point of the teapot?
Since that triangulation is not smooth, there are different normal vectors when you aproach a boundary segment from each of the two triangles of which is boundary. Hence, one cannot define a normal vector field on those segments compatible with the smooth structure of the interiors of the triangles.