The outward unit normal at a boundary point of a compact solid

Let $E\subseteq {\mathbb{R}}^3$ be compact and let $p\in \mathrm{\partial }E$, the boundary of Interior, closure, boundary, limit point, isolated point and dense subset of a metric space. For a unit vector $\nu \in {\mathbb{R}}^3$, that is one with $\parallel \nu {\parallel }_2=1$ in the norm of The Euclidean inner product $⟨x,y⟩={\sum }_{k<n}{x}_k{y}_k$ on ${\mathbb{R}}^n$, a unit vector $\nu$ is outward at $p$ when there is a real $\epsilon >0$ with $p+t\nu \mathrm{\notin }E$ and $p-t\nu \in E$ for every $t$ with $0<t<\epsilon$.

A plane of unit normals at $p$ is a two-dimensional linear subspace $T\subseteq {\mathbb{R}}^3$; the two unit vectors orthogonal to $T$ are $\pm \nu$ for a single $\nu$, and when one of them is outward at $p$ the other is not, since replacing $\nu$ by $-\nu$ exchanges the two displayed conditions. In that situation the outward one is called the outward unit normal to $T$ at $p$.

• Outwardness alone does not single out one vector. Take $E$ the closed unit ball and $p$ a point of the unit sphere. Every unit vector $\nu$ with $⟨p,\nu ⟩>0$ satisfies the definition, because $\parallel p\pm t\nu {\parallel }_2^2=1\pm 2t⟨p,\nu ⟩+t^2$ is above $1$ for small $t>0$ with the plus sign and below $1$ with the minus sign. So the definition is a condition on a unit vector and not a construction of one; what makes "the outward unit normal" a definite object is the second paragraph, where a plane is supplied and only two candidates remain.

• Existence is not asserted. A boundary point of an arbitrary compact set need admit no outward unit vector: if $E=\{p\}$ is a singleton, then $p-t\nu \mathrm{\notin }E$ for every unit vector $\nu$ and every $t>0$. Nothing below claims outwardness at seams and edges; the claim is made at the interior parameter points of a graph face whose projection lands in the interior of the base.

• Why the condition is one-sided on each side. Requiring only $p+t\nu \mathrm{\notin }E$ would admit a vector tangent to a spike of $E$; requiring only $p-t\nu \in E$ would admit a vector pointing along the surface. Both halves are used where outwardness is proved.