In a nutshell, we show that polynomials and nested polytopes are topological, algebraic and algorithmically equivalent. Given two polytops $Asubseteq B$ and a number $k$, the Nested Polytope Problem (NPP) asks, if there exists a polytope $X$ on $k$ vertices such that $Asubseteq X subseteq B$. The polytope $A$ is given by a set of vertices and the polytope $B$ is given by the defining hyperplanes. We show a universality theorem for NPP. Given an instance $I$ of the NPP, we define the solutions set of $I$ as $$ V(I) = {(x_1,ldots,x_k)in mathbb{R}^{kcdot n} : Asubseteq text{conv}(x_1,ldots,x_k) subseteq B}.$$ As there are many symmetries, induced by permutations of the vertices, we will consider the emph{normalized} solution space $V(I)$. Let $F$ be a finite set of polynomials, with bounded solution space. Then there is an instance $I$ of the NPP, which has a rationally-equivalent normalized solution space $V(I)$. Two sets $V$ and $W$ are rationally equivalent if there exists a homeomorphism $f : V rightarrow W$ such that both $f$ and $f^{-1}$ are given by rational functions. A function $f:Vrightarrow W$ is a homeomorphism, if it is continuous, invertible and its inverse is continuous as well. As a corollary, we show that NPP is $exists mathbb{R}$-complete. This implies that unless $exists mathbb{R} =$ NP, the NPP is not contained in the complexity class NP. Note that those results already follow from a recent paper by Shitov. Our proof is geometric and arguably easier.
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