My Question: Let $A_1, \ldots, A_k \subseteq \mathbb{R}^n$ be closed convex sets.
Is the union $\bigcup_{i=1}^k A_i$ triangulable$^1$? If so, why?
Background:
I'm trying to better understand the Nerve Theorem from Topology. In his Book Computational Topology (p.71) Edelsbrunner presents the following Nerve theorem without giving a proof:
Let F be a finite collection of closed, convex sets in Euclidean Space. Then the nerve$^2$ of $F$ is homotopy equivalent to $\bigcup F$.
He also mentions the following classical nerve theorem, which one can find in Topological methods (p.1850), and which is supposed to be more general:
Let $X$ be a triangulable space and let
$\mathcal{A} = \{A_1, \ldots, A_k\}$ be a finite closed cover$^3$ of $X$ such that
every non-empty intersection of the ${A_i}'s$ is contractible. Then the nerve of $\mathcal{A}$ is homotopy equivalent to $X$.
So does the former follow from the later?
$^1$ A space is called triangulable if it is homeomorphic to some simplicial complex.
$^2$ The nerve of a collection $F$ of sets is the abstract simplicial complex $\{Y \subseteq F : \bigcap Y \neq \emptyset\}$.
$^3$ A closed cover is a covering by closed subset of a topological space.
Best Answer
It seems like this is a few years too late, but yes, the former follows for the latter. To see this, we need the following two things:
From here, the former statement follows from the latter, since a collection of closed convex sets in Euclidean space is a closed cover of their union such that each non-empty intersection of the sets is contractible.