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?)


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.


  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?

  • 1
    $\begingroup$ 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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