Convex set that cannot be approximated by a polygon

Question

In $\mathbb{R}^2$, show that there exist no polygon containing the set $C = \{ (x,y) \in \mathbb{R}^2 | y \geq x^2\}$ and included in $C + B(0,1)$ where $B(0,1)$ is the open unit ball.

Intuitively, we could say that we can't find a finite number of affine inequalities containing the parabola $C$ and included in a parabola that includes $C + B(0,1)$. However I struggle to formalize it. Thanks.

Answer 1

The answer to this is going to depend on what counts as a "polygon".

• If a "polygon" has its most traditional meaning of a closed piecewise-linear curve, then all polygons are bounded and cannot contain an unbounded such as $C$.
• If a "polygon" is any simple piecewise-linear curve with only finitely many vertices, so it can be open ending in two rays extending to $\infty$ (usually this is called a "polygonal curve", not a "polygon"). Then it is a matter of showing that the two end-rays must either intersect $y = x^2$ or eventually deviate from it by a distance of more than $1$. This is a matter of showing that $y = x^2$ deviates from linear by more than that, no matter how far out you go.
• If a "polygon" is a piecewise-linear curve that can have infinitely many vertices, then the statement is false. One can easily approximate $y=x^2$ to any accuracy with infinitely many line segments.

I have to assume that you must mean the second case, since the first is trivial and the third is false. If you can clarify, then I or someone else can share more details on it.