The average (over ) of the number of sets containing a random element is, but there are three sets. Unfortunately, it is very easy to see that that is false: consider, for example, the three sets, , and. A proof idea that may not be completely hopelessДЄn immediate reaction of any probabilistic combinatorialist is likely to be to wonder whether in order to prove that there exists a point in at least half the sets it might be easier to show that in fact an average point belongs to half the sets. IвЂ™ll present these just in case one or other of them stimulates a useful idea in somebody else. The first is a natural idea for generalizing the problem to make it easier that completely fails, at least initially, but can perhaps be rescued, and the second is a failed attempt to produce a counterexample. It includes some quite interesting reformulations (interesting because you donвЂ™t just look at them and see that they are trivially equivalent to the original problem).ДЇor the remainder of this post, I want to discuss a couple of failures. It is very nicely written and seems to be a pretty comprehensive account of the current state of knowledge about the problem. If you are potentially interested, then one good thing to do by way of preparation is look at this survey article by Henning Bruhn and Oliver Schaudt. Must there be an element of that belongs to at least of the sets? This seems like the sort of question that ought to have an easy answer one way or the other, but it has turned out to be surprisingly difficult. Suppose also that it is union closed, meaning that if, then as well.

Suppose you have a family that consists of distinct subsets of a set. The problem, for anyone who doesnвЂ™t know, is this. This post, if the project were to go ahead, could be FUNC0 in general I like the idea that we would be engaged in a funky line of research. That general approach runs into difficulties with FranklвЂ™s union-closed conjecture, so I suggest FUNC. For the density Hales-Jewett problem we went for DHJ, and for the ErdЕ‘s discrepancy problem we used EDP. A notable feature of the polymath experiment has been that it throws up surprises, so while I wouldnвЂ™t expect a polymath project to solve FranklвЂ™s union-closed conjecture, I also know that I need to be rather cautious about my expectations - which in this case is an argument in favour of giving it a try.ДЄ less serious problem is what acronym one would use for the project. But, as one of the commenters said, that is not necessarily an argument against trying it.

A potentially worrying aspect of the idea is that the problem is extremely elementary to state, does not seem to yield to any standard techniques, and is rather notorious. Although it was from only a couple of people, I had an enthusiastic response to a very tentative suggestion that it might be rewarding to see whether a polymath project could say anything useful about FranklвЂ™s union-closed conjecture.

Polymath program freezes