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I posted this question on math.stackexchange.com, but didn't get an answer.

Let $A_1, \ldots, A_k \subseteq \mathbb{R}^n$ be closed convex sets. Is the union $\bigcup_{i=1}^k A_i$ triangulable, that is, homeomorphic to a simplicial complex? If so, why?

This seems plausible, but hard to prove. (Would it be easier if we require the sets to be compact?)

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This is not true even for unions of two compact convex sets.

To construct a suitable counterexample, consider the unit disk $$D=\{z\in\mathbb C:|z|\le 1\}$$in the complex plane $\mathbb C$. By $\partial D:=\{z\in\mathbb C:|Z|=1\}$ we denote the boundary circle of the disk $D$.

Next, choose a decreasing sequence of real numbers $(r_n)_{n\in\omega}$ such that $\lim_{n\to\infty}r_n=\inf_{n\in\omega}r_n=1$ and the points $z_n=r_ne^{i\pi/2^n}$ do not see each other behind the disk, which means that for any distinct numbers $n,m$ the interval $[z_n,z_m]:=\{tz_n+(1-t)z_m:t\in[0,1]\}$ intersects the disk $D$. Let $C$ be the (closed) convex hull of the compact set $D\cup\{z_n\}_{n\in\omega}$. It is easy to see that $C\setminus D$ has infinitely many connected components (homeomorphic to the triangle with a removed side).

Now consider the compact convex sets $A_1=C\times\{0\}$ and $A_2=D\times[-1,1]$ in $\mathbb C\times\mathbb R\cong \mathbb R^3$.

It can be shown that the union $A:=A_1\cup A_2$ is not homeomorphic to a simplicial complex (even to a CW-complex).

Assuming that $A$ is homeomorphic to a CW-complex, we can use the domain invariance theorem to show that the boundary $$\partial A=((C\setminus D)\times\{0\})\cup (\partial D\times [-1,1])\cup (D\times\{-1,1\})$$ of $A$ in $\mathbb C\times\mathbb R$ is contained in the 2-skeleton of the CW-complex. Using the domain invariance theorem once more, it can be shown that $\partial C\cup\partial D$ is contained in the 1-skeleton of the CW-complex and hence it is homeomorphic to a finite graph, which is not true.


Conclusion.

  1. The union of finitely many convex sets in $\mathbb R$ is homeomorphic to a simplicial complex.

  2. For $n\ge 3$ the union of two compact convex set in $\mathbb R^n$ can be non-homeomorphic to a CW-complex.

Problem. What is the situation in dimension 2? Is the union of finitely many compact convex sets in the plane homeomorphic to a simplicial complex?

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    $\begingroup$ To check that I've understood your construction, $A$ is a cylinder with infinitely many teeth protruding from the equator of the cylinder, where the teeth are sufficiently shallow that the intersection of $A$ with the equatorial plane is convex? I'm sufficiently convinced that this works. $\endgroup$ – Adam P. Goucher Feb 28 '18 at 13:25
  • $\begingroup$ @AdamP.Goucher Yes, exactly as you described: cylinder with infinitely many teeth. $\endgroup$ – Taras Banakh Feb 28 '18 at 13:38

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