MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is a lattice of convex sets in $R^2$ distributive?

share|cite|improve this question
Take three distinct collinear points. The middle one intersects the join of the two extremes, but the meet of any two is empty. – François G. Dorais Jan 25 '10 at 0:12
Did you mean to ask, «is the lattice of convex sets in $R^2$ distributive?» – Mariano Suárez-Alvarez Jan 25 '10 at 0:57
Why this is tagged category-theory? – sdcvvc Jan 25 '10 at 13:43
I retagged this. – David Speyer Jan 25 '10 at 15:57

In the paper "On non-modular n-distributive lattices: Lattices of convex sets" (Acta Sci. Math. (Szeged) 52 (1987), 35-45) A. Hunh proves that the lattice of convex sets in $\mathbb R^n$, $Co(\mathbb R ^n)$, is $n+1$-distributive but not $n$-distributive. This means that the lattice of convex sets in $\mathbb R^n $ is only distributive for $n=0$, as the example by François G. Dorais shows.

Here we define a lattice $L$ to be $n$-distributive if for any $x,y_0,\dots,y_n$ the following identity holds: $$x\wedge \bigvee_{i=0}^n y_i =\bigvee_{i=0}^n (x \wedge \bigvee_{i\neq j} y_i).$$ This nice result shows that the dimension of a Euclidean space has a lattice theoretical characterization. The proof relies on Caratheodory's theorem. The same holds for the dual of $Co(\mathbb{R}^n)$, it is $n+1$-distributive but not $n$-distributive, and the proof uses Helly's theorem.

share|cite|improve this answer

I googled "lattice of convex sets"+distributive and the second link led me to the abstract for the paper: "Geometric Condition for Local Finiteness of a Lattice of Convex Sets" in Mathematica Moravica, Vol. 1 (1997), 35–40 by Matt Insall, which contains the following sentence:

We give elementary examples which establish the following facts: the lattice of convex subsets of a Hilbert space is not locally finite, it is not modular (hence not distributive), and locally finite lattices of closed convex sets in any Hilbert space have very restrictive geometric arrangements of their members.

This means that the answer to your question is no, since $\mathbb{R}^2$ is certainly a Hilbert space.

Unfortunately, the journal did not have the paper online, but I was able to find Insall's webpage and from there, the PDF for the paper.

The paper is relatively easy to read and has illustrations for the examples.

share|cite|improve this answer
Let's see....{0} is also a Hilbert space, so.... – Joel David Hamkins Jan 25 '10 at 3:58
Yeah, having glanced at the paper, it appears that what they actually prove is explicitly in the plane, but apparently they were more concerned with the result for general nonzero complex Hilbert spaces. However, as François G. Dorais's comment shows, this is overkill for this particular question. – Jonas Meyer Jan 25 '10 at 6:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.