If is any *monotone* graph-edge property, then we can form a union-closed property by taking all graphs such that for every edge . To give a non-trivial example, we could take to be the property “There are disjoint paths of length connecting the two end points of “. Then the union-closed property would be that this is true for every edge , which is clearly not a monotone property but is certainly union closed.

I think it would be an interesting partial result if we could prove FUNC for such properties, or properties of the specific form “ is contained in a subgraph isomorphic to “, or even some weakening of FUNC that is better than is known for general union-closed families. It seems hard to find a counterexample, because as you say one probably needs to look at quite large graphs , but large graphs have many edges and so contribute quite a bit to abundances.

To get round that last problem, it might perhaps help to look at a variant that somehow makes it harder for large graphs. An idea I wondered about was taking a sparse random graph and just looking at subgraphs of . Then perhaps one could have a property like “Every edge is contained in a cycle.” The cycles would have to be quite long.

But actually I don’t really see what would make this work. Basically this set system would consist of unions of cycles in , and I’m not sure why there would be few large ones. (But I still think it would be worth checking, even if not rigorously, that this is not a counterexample.)

]]>Is FUNC true for these union-closed families? For triangles, it looks very likely – nearly all graphs with many edges have every edge in a triangle, and many small graphs don’t. In fact, for any fixed H and a large vertex set, that looks true. So I think to have any chance of finding a counterexample, we would need to look at graphs H which aren’t that much smaller than the vertex set.

]]>More precisely, I claim that there are union-closed with a minimal transversal such that the induced weight function on realizes my worst-case scenario, which is that for all non-singleton , and is independent of and arbitrarily large.

The construction goes like this: fix and take , where the are disjoint copies of a -set. Take to consist of all that satisfy the following implication for every : if , then also for all . The crucial point here is that is a minimal transversal of , and for every there are sets that satisfy . However, the union of any two such sets for different contains all the ‘s. Hence the induced weight function on is if is a singleton, and otherwise. This is the worst-case scenario in the sense that my bound is tight, i.e. it reproduces the exact abundance of the elements of .

So the only further room that I can see for potential strengthening of Knill’s argument is to make a clever choice of . In the above example, take e.g. a minimal transversal that contains one element in each : all elements of such a transversal have abundance close to 1. (Of course, a clever enough choice of will always work if FUNC holds, since one can just choose to contain an element of maximal abundance…)

]]>Ah, I see what’s going on now. So the assumption on the lattice is that it is closed under taking meets.

]]>The goal is to create a finite lattice in which every join-irreducible element is abundant, i.e. the upper set that it generates must contain at least half the lattice elements. In this sense, all the join-irreducibles must be sufficiently far ‘down’ in the lattice; but there should also be sufficiently many join-irreducibles to express every other element as a join of these. (For if one couldn’t, then some other element would be join-irreducible.) This is the challenging tradeoff that one has to achieve.

In order to realize this with the fibre bundle construction, one could try to take to be a power set , and choose the bundle data such that all join-irreducibles of the resulting lattice live over small subsets of . Achieving the required abundance of these join-irreducibles means that for of size , one should strive to make very large. But it also can’t be too large, since every one of its element still needs to be a join of the join-irreducibles further down, and there are only so many possibilities for combining join-irreducibles.

It should help to work out the details of this tradeoff more quantitatively, but I haven’t done so yet.

]]>If instead we put it into conjunctive normal form, we end up saying that for every function that takes vertices other than and to either or , we must satisfy

.

So we get an exponential number of Horn clauses for each edge.

]]>Suppose we have a clause of the form

.

The edges form a graph , and the symmetry of the property we are defining tells us that if is any permutation of the edges that is induced by a permutation of the vertices, then we will also get the clause

.

For example, suppose were the edges , so they made a graph that consisted of a triangle with an edge attached to it. This clause would tell us that if belongs to the graph, then so does at least one of the edges in the attached triangle. But by symmetry that tells us that if *any* edge belongs to the graph, then so does at least one edge from every triangle that overlaps with that edge in precisely one vertex.

That’s a slightly peculiar property, but it is probably already an example of a union-closed property that isn’t monotone.

Actually, I think it *is* monotone. If the graph has at least one edge, then there can be at most one isolated vertex, since otherwise if is an edge and and are isolated vertices, then the triple fails to contain an edge. But then every triple of vertices contains at least one edge, since it must include one non-isolated vertex . If is not already joined to or , then it is joined to some other , from which it follows that is an edge. But that shows that the property above is equivalent to the property of intersecting all triangles, which is a monotone property.

However, it gives a clue about how to construct non-monotone union-closed properties. Here’s one that seems to work: every edge is contained in a triangle. A triangle has the property, the property is clearly union-closed (after all, a union of unions of triangles is a union of triangles), but a triangle together with one more edge does not have the property.

]]>Here’s a generalization of this construction, which is more conveniently formulated for intersection-closed families. Let be an arbitrary function between finite sets such that every fibre carries the structure of a meet-semilattice. It helps to have a ‘bundle’ picture in mind with as the base space and the total space. Then the idea is to let vary over all ‘partial sections’ of , i.e. functions defined on a subset such that . The set associated to such a section is . The intersection of this with an arising from is .

[I hope that it’s clear how this generalizes your example; if not, let me know.]

I think that all of this should be an instance of the fibre bundle construction from the main post, probably by taking and to be the product of the fibres , but I need to think about this a little bit more.

]]>I wondered whether perhaps one could create a counterexample by taking as the ground set of the set of pairs from some set , so that its elements can be thought of as edges in graphs. Then would be a union-closed set of graphs. When looking for sets of graphs, it is natural to look for *graph properties* — that is, sets that are invariant under permutation of the vertices. Thus, the hope would be to find a union-closed graph property that is for some reason easier for small graphs to satisfy than for large graphs.

There are two classes of properties that come up most often. One is *monotone* properties — that is, ones where if has it and , then has it. Typical examples are “is connected” or “contains a triangle”. For our purposes, these are useless, since every element that occurs at all in a monotone set system occurs with abundance at least 1/2. The other is *hereditary* properties, which are properties such that if is an *induced* subgraph of and has the property, then has the property. A simple example is “contains an induced cycle of length 5”.

The question I was going to ask is whether there are any natural graph properties that are union closed but not monotone. Hereditary properties look promising at first, but at least the example above clearly doesn’t work. We can create a graph with an induced 5-cycle by taking five vertices and putting all edges into except the pairs with mod 5. Taking two such graphs with disjoint sets of five vertices, their union will be the complete graph on vertices, which does not contain an induced 5-cycle.

As I said at the beginning, I don’t expect this question to lead magically to a counterexample to FUNC, but it might give us some insight into what union-closed families can look like.

Maybe we can answer the question by thinking about Horn clauses. The fact that we are talking about a graph property will imply that the Horn-clause description will have a lot of symmetries. Looking at it the other way, perhaps considering sets of Horn clauses with the necessary symmetries will lead to interesting union-closed graph properties.

]]>So instead of taking , I’ll take it to be a disjoint union of copies of a set of size . (This isn’t a significant reformulation of course — maybe redescription would be a better word.)

Given a function , the sets are nested downwards. Conversely, given any nested sequence of sets I can find an that gives rise to them. So the sets in are, in this description, sets of the form where each is a subset of (or more formally of the copy of ) and . I see now why you chose your description — it was to avoid the informality of saying things like that when in reality they are disjoint, or defining an ugly set of bijections.

Be that as it may, with this description it’s obvious that is union closed. Also, since elements of are in a natural one-to-one correspondence with sequences of length that take values in , we get the abundances.

]]>Let . (In the example above, .) Let be a set of size . To construct our family we take as the ground set, and then we take to consist of all sets where ranges over all functions .

The function is injective. Furthermore, (where is defined pointwise). So is a union-closed family of size . It is separating and contains the empty set (corresponding to the function ).

The relative abundance of an element depends only on : it is .

]]>I do also wonder whether more focus on the looking-for-a-counterexample side of things would be sensible. However, it seems very hard to even find a promising direction on that.

But let’s get the ball rolling – here is an attempted counterexample to FUNC.

Let the groundset be , all of equal size. For each , let be a subset of , taken randomly of size . We generate our union-closed family with all sets that contain some , and avoid the corresponding .

These generators are constrained on elements of . Any union of two or more of these generators is constrained on about elements of . So, picking small and then large, almost all the sets in our family are generators. Also, the generators have average size less than – because they exclude slightly more elements than they include.

So, the average size of our sets is small, and the scheme doesn’t favour any particular element of . That’s promising. The problem is that the don’t cover anything like twice – so some (in fact most) elements of are only in a single . It looks very much like those elements are going to be the abundant ones.

]]>For another potential improvement, we could also try to choose the minimal transversal in some clever way, but I don’t have any ideas along these lines.

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