so we have a total of 2^m sets in total

now assume x belongs to one of the sets then there are 1+2^(m-1)

sets to which x belongs

which is clearly greater than 1/2

this can generalized for overlaping sets by removing all the independent sets from the family

the family is still union closed

in this case the total number of sets become (2^m) – m

but now x belongs 2^(m-1)

which is still greater than 1/2.

clearly, the new family is conformable to any union closed family

]]>Having said that, the second condition doesn’t feel right, since removing elements from some of the can sometimes turn a homomorphism into a non-homomorphism, but it can also sometimes turn a non-homomorphism into a homomorphism. So it doesn’t seem to capture the idea of minimality that I was trying to capture.

Maybe I should have stuck with something simpler like removing all elements with abundance 1.

]]>It’s always a homomorphism if it’s well-defined, but it may not be well-defined. I should have made it clearer that that was the main constraint on .

As for ruling out Tobias’s example, the unique minimal generating set of in his example is , which forces to equal 1, which forces to be a very boring set system — it has to consist of just one set and the empty set, and then by minimality that set has to be a singleton.

]]>I must be missing something – isn’t a map of the form always a homomorphism, whatever we do to the the , so the second condition is never satisfied? Also, presumably this is intended to rule out Tobias’ example with and , but I’m afraid I don’t see how?

]]>Indeed, for union-closed families with transitive symmetry, monotonicity fails for general values of . (We can have the empty set and full set in the family giving .) But I don’t see a reason that this will lead to examples that it fails for which is what Frankl’s conjecture gives. In particular, trying to augment the family with threshold families (as I suggested above) causes violation of the union-closed property. (Maybe, the conjecture extends to all for some , perhaps $p_0=1/2$ but I am not sure it is “truer” for larger p. For the simple example of 0- 2-3- subsets of a 3-element set, monotonicity is violated when .)

]]>I’d like to explain how Thomas’s example can be obtained via the fibre bundle construction developed here: https://gowers.wordpress.com/2016/01/29/func1-strengthenings-variants-potential-counterexamples/#comment-154199

Take minus the singletons, or equivalently the family generated by the 2-subsets of plus the empty set. In specifying the , I’ll restrict myself to only writing down a subset from which the others can be inferred by symmetry.

The maps are just those that take every element to itself.

I need to go now and haven’t checked this in as much detail as I’d like to. But even if some minor detail is off, it should be clear that Thomas’s example is an instance of the general construction.

]]>Ah yes, that’s a good point (and one that I had to think about for a while before understanding why it was correct).

Although I still very much doubt whether any conjecture like this can be true, let me try to refine the conjecture slightly to rule out the two examples so far but to leave something that still implies FUNC.

The refined conjecture says the following. Let be a minimal generating set for a union-closed system (which by convention I will take to include the empty set). Let be some other sets and suppose that the following two statements hold.

(i) The map is a homomorphism.

(ii) If any element is removed from any of the , then the map ceases to be a homomorphism.

Then the maximum abundance of is at most the maximum abundance of .

]]>“let be the union of all the set systems ”

Since and the union includes the case itself, I think that we get , no?

When all the ‘s are injective, then what you describe yields a nice concrete implementation of my abstract fibre bundle construction. But maybe it could also be interesting to consider some non-injective ?

Then you were asking what property characterizes those homomorphisms that arise from the fibre bundle construction, right? That’s an interesting question, and I think that the answer is: it’s precisely those for which for every double inclusion and every , with in there is with . (Think of ‘lifting’ the ‘path’ in the base space to the total space, given lifts of the endpoints.) In particular, this implies that must preserve meets as well.

The condition is clearly necessary for a homomorphism to arise from the fibre bundle construction. Concerning sufficiency, I again only know how to show this in the abstract (without ground sets), as follows.

Given , let . For and , let be given by the meet of all the with in . (Recall that every finite semilattice is a lattice, so that this meet exists.) It’s straightforward to see that this preserves unions (by virtue of these being least upper bounds). It satisfies the required composition law on double inclusions due to the assumption on . Applying the fibre bundle construction to this recovers the original essentially by definition.

(As before, I’ve been abusing set-theoretic notation to talk about join-semilattices.)

I suspect that one can relax the assumed composition law in the fibre bundle construction to (understood pointwise), and then it’s conceivable that every (surjective) homomorphism arises in this way.

]]>That is arguably not a very good counterexample, since all abundances are 0, 1/2 or 1 (again just because there are no coincidences among the unions of the generators). The average abundance becomes large only due to the large number of elements with abundance 1. I want to argue that we shouldn’t really count elements with abundance 1: one reason is that the lattice formulation of FUNC doesn’t even ‘see’ them and would consider your to have a maximal abundance of 1/2. So if one excludes elements of abundance 1 from the strengthened conjecture, then that counterexample doesn’t work.

For a different family of counterexamples that still work, take to be any union-closed family containing the empty set and with maximal abundance greater than 1/2. Take , and to be given by the empty set mapping to the empty set and everything else mapping to . Then is surjective, but decreases maximal abundance.

]]>Actually, we can add the empty set to and thus the additional term to and then cannot be monotone so it will be decreasing for some . (For this you can take the family with all sets of 0,2,3 elements from .) Now, it looks that the methods described above by replacing each element by a threshold on many elements so that the probability to reach the threshold is p, will transfer the non monotomicity from p to 1/2 as we want.

]]>Why is this “conjecture” even slightly plausible? Well, if we try to disprove it by taking to consist of large sets, then we are likely to find coincidences amongst the unions of sets in that do not correspond to coincidences in .

Actually, writing that sentence has made me realize just how false this “conjecture” is. We can take to be generated by random subsets of for some very large , each of density for some . (The closer is to 1, the larger we need to take .) Then as long as is large enough, all unions of the will be distinct, so the map that takes to will be a homomorphism, and we can make the average abundance in as large as we like.

]]>I think I can convert that failure into a lemma. Let be a union-closed set system. I claim that if contains two disjoint non-empty sets, then it is not possible to find a union-closed set system and a surjective homomorphism such that every has approximately the same number of preimages — except if that number is 1.

The proof is trivial. If and are disjoint elements of and and are preimages, then is a preimage of . Actually, for this argument to be properly trivial I need to assume that and are disjoint, which they don’t have to be, so it’s back to the drawing board yet again — maybe there is some clever way of getting sets to overlap in order to compensate for this effect.

]]>OK, let . I want to estimate its abundance.

First let’s see how many sets in are subsets of a given set but not of any set with a proper subset of . What I need for this construction to work is that I should get roughly the same answer for all .

Unfortunately, I realize as I write this that it doesn’t work. My basic problem is that if and are disjoint sets in , then if I do a construction that creates, in some sense, approximately copies of and another copies of , then when I take their unions I will have to create copies of . I thought I could somehow “couple” the copies of with the copies of , but I can’t.

]]>Let me start with 1. Let and be distinct elements of . (I’m assuming that of course.) If , then choose such that . Then , but since , .

If , then this does not work because we might find that and . That would imply that . But then we can use the set instead. The only thing that can go wrong is if there exist and with . But if is much larger than , then this happens with very low probability. (All I need for this argument is that there exist functions and such that it doesn’t happen.)

The above argument, I forgot to say, was dealing with the case where there is some set that contains both and . If no such exists, then and are already separated.

I’ll deal with abundances in a separate comment.

]]>Here’s a rough proposal for a method of construction that is aimed at showing, roughly speaking, that adding the condition that a set system should separate points doesn’t help, in the sense that it won’t magically make false conjectures true. (Obviously it makes some statements true — e.g. the statement that your family separates points. I haven’t thought about how to characterize the statements that I’m talking about.)

Let be a set system that does not necessarily separate points. Let be the ground set of . Now let be the cyclic group of order and form a new set system that consists of all sets of the form with . I am assuming that is much bigger than .

Now choose two random functions . For each and each , let be the set

and let be the set

.

I now want to let be the set system generated by the sets and .

The two things I want to prove about are as follows.

1. It separates points.

2. The abundance of any element in is approximately the same as the abundance of in .

I will post this comment to make it easier to look at what I’ve just defined and what I want to prove. Of course, this risks that I will find that I’ve got it wrong. (I haven’t written this out in advance.)

]]>Maybe I can get round the small difficulty I was having in the previous comment by insisting, in the converse direction, that each should contain the set itself. Then we can define to be .

]]>I suppose an obvious challenge at this point is to try to find a counterexample that separates points. However, what I would prefer to try to find is a general method for converting set systems that don’t separate points into set systems that do separate points, while keeping the abundances approximately the same (in some sense). I may well be wrong, but it feels to me as though it ought to be possible somehow. A slightly weaker requirement that would still do fine would be to find a way of achieving what can be achieved with duplication but with just a few more sets thrown in so that points are separated.

]]>Yes, that works.

]]>That looks right. So it should convert into a counterexample to the average-overlap-density “conjecture” (the inverted commas because it always seemed a bit unlikely to be true). Let be disjoint sets with a singleton, a set of size and a set of size . Then the counterexample should be

for sufficiently large .

]]>If I’ve understood correctly, I think this is a counterexample with : take the 6-set family . The number of pairs where and is 14, which is less than half of .

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