The logic of convex sets

This is an interesting question. (Even if there is no finite list of "axioms".)

For example the following is true: Suppose you have a (d+1)-dimensional polytope P. Associate a convex set in $R^d$ to every facet of P, and suppose that every non empty intersection among the facets implies a non empty intersection for the corresponding sets. Then there are some set of facets of the polytope with empty intersection so that the corresponding convex sets have non empty intersection. This is a large collection of statements of the kind you asked that already includes Helly's theorem.

Note that the nerve theorem gives you a lot of information on intersection patterns that cannot be realized. (Including the statement made above.)

A stronger statement for the case described above goes as follows: Suppose that 0 is in the interior of P. Then you can find two faces of F and G of P so that 0 is in their convex hull and the intersection of all sets corresponding to facets containing F and all sets corresponding to facets containing G is nonempty. This can be derived from the Borsuk-Ulam theorem.

Another well known result of the kind you ask about is the colorful Helly theorem (proved by Lovasz and a dual Caratheodory version was proved independently by Barany). It sais that if you have d+1 collections of convex sets in R^d, and if for every choice of one set from each collection there is a point common to all chosen sets then there is a collection all whose members have a point in common.

For n=1 there is a theorem by Lekkerkerker and Boland characterizing interval graphs. The characterization is based on two requirements: (1) there are no induced cycles of length >3. (This leaves us with chordal graphs.) (2) for every 3 vertices, one of the vertices is a neighbor of a vertex in every path between the other two. This gives an infinite set of "axioms" in the sense asked in the question. I suppose there is no finite list.

So overall there are many interesting theorems of the kind you described, and this is a rich area. But a complete description for dimension > 1 does not look realistic.

A more general question is to study the situation where you allow both operation: (1) taking the intersection (2) taking convex hull. This makes the problem even much more complicated and not much is known. An example of a recent result in this more general setting is the following theorem by Novick: Given 7.2k pairwise disjoint convex sets in the plane there is a set in the family that is disjoint to the convex hull of k other sets in the family.

I have a kind of heuristical reason why the case $n=2$ is hard, i. e. for $n=2$ there is no finite set of deduction rules from which we can derive every Horn clause about convex sets (without using any other properties of convex sets). This is not a proof, but it is convincing enough for me to stop considering the $n\geq 2$ case. As for $n=1$, I have no idea.

So here is the argument. For any graph $G$ with (finite) vertex set $V$ and edge set $E$, we can fix a total order on the set $V$, and consider the following Horn clause:

The planarity clause of the graph $G$: LET $P_{u,v}$ be a convex set for any two vertices $u$ and $v$ of $G$ satisfying $u < v$. IF $P_{u,v}\cap P_{t,r}\neq\emptyset$ whenever the sets $\left\lbrace u,v\right\rbrace$ and $\left\lbrace t,r\right\rbrace$ have a common element, THEN we have $P_{u,v}\cap P_{t,r}\neq\emptyset$ for at least one quadruple $\left(u,v,t,r\right)$ of vertices of $G$ such that the sets $\left\lbrace u,v\right\rbrace$ and $\left\lbrace t,r\right\rbrace$ have no common element.

We can easily find that the planarity clause of the graph $G$ is always satisfied if and only if the graph $G$ is not planar. (What we actually are using here is Fáry's theorem.) Now, Kuratowski's theorem shows us that we have non-planar graphs of arbitrarily large size that can only be proven planar by looking at ALL of their vertices together. For example, take $K_5$, and "blow it up" by replacing every edge by a sequence $\cdot - \cdot - \cdot - \cdot - ... - \cdot$ of little edges. Remove any of the edges and you get a planar graph, so if there was a finite family of rules from which all tautologies would follow, at least one of these rules would be talking about a big number of convex sets (namely, at least as many as you have edges in your blown up $K_5$). Contradiction.

As I said, I don't know whether this is really a sound argument, and even if it is, the case $n=1$ still remains open.