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A family of sets for some is a sunflower if there is a *core set* contained in each of the such that the *petal sets* are disjoint. If , let denote the smallest natural number with the property that any family of distinct sets of cardinality at most contains distinct elements that form a sunflower. The celebrated Erdös-Rado theorem asserts that is finite; in fact Erdös and Rado gave the bounds

*sunflower conjecture*asserts in fact that the upper bound can be improved to . This remains open at present despite much effort (including a Polymath project); after a long series of improvements to the upper bound, the best general bound known currently is for all , established in 2019 by Rao (building upon a recent breakthrough a month previously of Alweiss, Lovett, Wu, and Zhang). Here we remove the easy cases or in order to make the logarithmic factor a little cleaner.

Rao’s argument used the Shannon noiseless coding theorem. It turns out that the argument can be arranged in the very slightly different language of Shannon entropy, and I would like to present it here. The argument proceeds by locating the core and petals of the sunflower separately (this strategy is also followed in Alweiss-Lovett-Wu-Zhang). In both cases the following definition will be key. In this post all random variables, such as random sets, will be understood to be discrete random variables taking values in a finite range. We always use boldface symbols to denote random variables, and non-boldface for deterministic quantities.

Definition 1 (Spread set)Let . A random set is said to be -spread if one has for all sets . A family of sets is said to be -spread if is non-empty and the random variable is -spread, where is drawn uniformly from .

The core can then be selected greedily in such a way that the remainder of a family becomes spread:

Lemma 2 (Locating the core)Let be a family of subsets of a finite set , each of cardinality at most , and let . Then there exists a “core” set of cardinality at most such that the set has cardinality at least , and such that the family is -spread. Furthermore, if and the are distinct, then .

*Proof:* We may assume is non-empty, as the claim is trivial otherwise. For any , define the quantity

Let be the set (3). Since , is non-empty. It remains to check that the family is -spread. But for any and drawn uniformly at random from one has

Since and , we obtain the claimIn view of the above lemma, the bound (2) will then follow from

Proposition 3 (Locating the petals)Let be natural numbers, and suppose that for a sufficiently large constant . Let be a finite family of subsets of a finite set , each of cardinality at most which is -spread. Then there exist such that is disjoint.

Indeed, to prove (2), we assume that is a family of sets of cardinality greater than for some ; by discarding redundant elements and sets we may assume that is finite and that all the are contained in a common finite set . Apply Lemma 2 to find a set of cardinality such that the family is -spread. By Proposition 3 we can find such that are disjoint; since these sets have cardinality , this implies that the are distinct. Hence form a sunflower as required.

Remark 4Proposition 3 is easy to prove if we strengthen the condition on to . In this case, we have for every , hence by the union bound we see that for any with there exists such that is disjoint from the set , which has cardinality at most . Iterating this, we obtain the conclusion of Proposition 3 in this case. This recovers a bound of the form , and by pursuing this idea a little further one can recover the original upper bound (1) of Erdös and Rado.

It remains to prove Proposition 3. In fact we can locate the petals one at a time, placing each petal inside a random set.

Proposition 5 (Locating a single petal)Let the notation and hypotheses be as in Proposition 3. Let be a random subset of , such that each lies in with an independent probability of . Then with probability greater than , contains one of the .

To see that Proposition 5 implies Proposition 3, we randomly partition into by placing each into one of the , chosen uniformly and independently at random. By Proposition 5 and the union bound, we see that with positive probability, it is simultaneously true for all that each contains one of the . Selecting one such for each , we obtain the required disjoint petals.

We will prove Proposition 5 by gradually increasing the density of the random set and arranging the sets to get quickly absorbed by this random set. The key iteration step is

Proposition 6 (Refinement inequality)Let and . Let be a random subset of a finite set which is -spread, and let be a random subset of independent of , such that each lies in with an independent probability of . Then there exists another random subset of with the same distribution as , such that and

Note that a direct application of the first moment method gives only the bound

but the point is that by switching from to an equivalent we can replace the factor by a quantity significantly smaller than .One can iterate the above proposition, repeatedly replacing with (noting that this preserves the -spread nature ) to conclude

Corollary 7 (Iterated refinement inequality)Let , , and . Let be a random subset of a finite set which is -spread, and let be a random subset of independent of , such that each lies in with an independent probability of . Then there exists another random subset of with the same distribution as , such that

Now we can prove Proposition 5. Let be chosen shortly. Applying Corollary 7 with drawn uniformly at random from the , and setting , or equivalently , we have

In particular, if we set , so that , then by choice of we have , hence In particular with probability at least , there must exist such that , giving the proposition.It remains to establish Proposition 6. This is the difficult step, and requires a clever way to find the variant of that has better containment properties in than does. The main trick is to make a conditional copy of that is conditionally independent of subject to the constraint . The point here is that this constrant implies the inclusions

and Because of the -spread hypothesis, it is hard for to contain any fixed large set. If we could apply this observation in the contrapositive to we could hope to get a good upper bound on the size of and hence on thanks to (4). One can also hope to improve such an upper bound by also employing (5), since it is also hard for the random set to contain a fixed large set. There are however difficulties with implementing this approach due to the fact that the random sets are coupled with in a moderately complicated fashion. In Rao’s argument a somewhat complicated encoding scheme was created to give information-theoretic control on these random variables; below thefold we accomplish a similar effect by using Shannon entropy inequalities in place of explicit encoding. A certain amount of information-theoretic sleight of hand is required to decouple certain random variables to the extent that the Shannon inequalities can be effectively applied. The argument bears some resemblance to the “entropy compression method” discussed in this previous blog post; there may be a way to more explicitly express the argument below in terms of that method. (There is also some kinship with the method of dependent random choice, which is used for instance to establish the Balog-Szemerédi-Gowers lemma, and was also translated into information theoretic language in these unpublished notes of Van Vu and myself.)
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