Towards an optimal hypergraph container lemma
Abstract.
The hypergraph container lemma is a powerful tool in probabilistic combinatorics that has found many applications since it was first proved a decade ago. Roughly speaking, it asserts that the family of independent sets of every uniform hypergraph can be covered by a small number of almost-independent sets, called containers. In this article, we formulate and prove two new versions of the lemma that display the following three attractive features. First, they both admit short and simple proofs that have surprising connections to other well-studied topics in probabilistic combinatorics. Second, they use alternative notions of almost-independence in order to describe the containers. Third, they yield improved dependence of the number of containers on the uniformity of the hypergraph, hitting a natural barrier for second-moment-type approaches.
1. Introduction
The method of hypergraph containers is a powerful and widely-applicable technique in probabilistic combinatorics. The method enables one to control the independent sets of many ‘interesting’ uniform hypergraphs by exploiting the fact that these sets exhibit a certain subtle clustering phenomenon. The survey [2] provides a gentle introduction to the method, illustrated with several example applications.
The heart of the method is a hypergraph container lemma (HCL for short) – a statement that formalises and quantifies the notion of clustering we alluded to above. A generic HCL asserts that every uniform hypergraph admits a collection of containers for the family of independent sets of with the following three properties:
-
(i)
each is contained in some ;
-
(ii)
the family has ‘small’ cardinality; and
-
(iii)
each is ‘almost independent’.
Note that such a statement is vacuously true when is -uniform, as then each is contained in the largest independent set . Although a HCL for -uniform hypergraphs is already implicit in the works of Kleitman and Winston [13] from the early 1980s, an explicit statement was formulated and proved only a decade later by Sapozhenko [22]. General HCLs that are applicable to hypergraphs of an arbitrary uniformity were proved much later, independently and simultaneously, by Balogh, Morris, and Samotij [1] and by Saxton and Thomason [23].
The precise meaning of small in (ii) above depends not only on the notion of almost-independence in (iii), but also on the uniformity of the hypergraph. The dependence on the uniformity in the original HCLs of [1, 23] was rather unfavourable. Roughly speaking, the guaranteed upper bound on for an -uniform hypergraph in [1] and [23] was proportional to and , respectively. To remedy this, Balogh and Samotij [3] proved a more ‘efficient’ HCL that yielded an upper bound on that was proportional to merely a polynomial in . Let us also mention that Morris, Samotij, and Saxton [18] proved an ‘asymmetric’ HCL that was fine-tuned to enumeration of sparse graphs not containing an induced copy of a given subgraph, but has since found additional applications [7, 8, 9, 15].
While the proofs of the four HCLs mentioned in the previous paragraph are elementary, they all are rather lengthy and technical (there are much easier arguments that prove the existence of containers in the -uniform case, see the survey [21]). This motivated several groups of authors to seek HCLs that admit shorter and simpler proofs. Saxton and Thomason [25] found an easy method for building containers for independent sets in simple hypergraphs. In another work [24], the same authors presented a streamlined proof of their original HCL [23] that yields an additional ‘online’ property. A few years later, Bernshteyn, Delcourt, Towsner, and Tserunyan [5] formulated a HCL whose statement adapts notions from nonstandard analysis and supplied a compact (spanning approximately four pages), non-algorithmic proof; unfortunately, their HCL still suffers from unfavourable dependence on the uniformity. Finally, Nenadov [19] recently proved a statement that may be interpreted as a probabilistic HCL; while it does not supply a collection of containers for independent sets, it is sufficiently powerful to replace HCL in most of its typical applications.
1.1. Motivation
We embarked on this project with two independent goals in mind. Our first aim was to further improve the dependence of (the logarithm of) the number of containers on the uniformity of the hypergraph. The second aim was to find a simple(r) construction of containers for general uniform hypergraphs that admits a compact proof. Whereas the first goal has been achieved only partially – each HCL formulated and proved in this paper encounters the same ‘second-moment barrier’ that results in -type dependence of on the uniformity , we do suggest a pathway to improving it further.
1.2. Our results
We formulate two new HCLs that are stronger than the efficient HCL of Balogh and the second author and, at the same time, admit short proofs. The two lemmas use different notions of almost-independence that, unlike all previous works, are not expressed in terms of a ‘balanced supersaturation’ condition; however, both of them are at least as strong as the usual notion (to certify this, we will present short derivations of the efficient HCL from each of our two lemmas). Furthermore, one of our HCLs does not even require the input hypergraph to be uniform.
The statement of our first HCL, A below, is inspired by the recent developments in the study of threshold phenomena in random sets [10, 20]. In order to state it, we need to introduce some notation. First, given a hypergraph , we write
for the up-set generated by . We say that covers a hypergraph if ; in other words, covers is every edge of contains some edge of . Second, for each , define the -weight of a hypergraph to be
which is just the expected number of edges of induced by the -random subset111The -random subset of a finite set is the random set formed by independently retaining each element of with probability . of .
Theorem A.
Let be an -uniform hypergraph with a finite vertex set . For every , there exists a family and functions
such that:
-
(a)
For each , we have .
-
(b)
Each has at most elements.
-
(c)
For every , letting , there exists a hypergraph on with
that covers and satisfies for all .
Let us point out that the above theorem does not assume anything about the input hypergraph, except that it is uniform. The condition that all edges of the cover have size at least two is necessary to prevent the conclusion from being trivial; indeed, every hypergraph with vertex set admits a cover (by singletons) of -weight . An analogous comment applies to the assumption that ; if , then one may simply set for all and let be the empty hypergraph.
Our second HCL, B, has a probabilistic flavour; this stems form the fact that its proof views independent sets as samples from a hard-core model. Perhaps surprisingly, it does not assume anything about the hypergraph, not even uniformity. Given an arbitrary set and a real , we write to denote the -random subset of .
Theorem B.
Let be a hypergraph with a finite vertex set . For all reals and satisfying , there exists a family and functions
such that:
-
(a)
For each , we have .
-
(b)
Each has at most elements.
-
(c)
For every , letting ,
The proofs of both A and B follow the same general strategy that was also used to prove most HCLs to date; it can be traced back to the work of Kleitman and Winston [13]. Namely, we define an algorithm that, given an independent set as input, builds sets and satisfying and, in the proof of A, also a hypergraph that covers . Crucially, the set depends on only through in the following sense: If for some two inputs , the algorithm outputs the same set , it also returns the same set . This property allows one to define the functions and via the algorithm. The main novelty in the algorithm that underlies the proof of A is that it considers queries of the form ‘Is ?’ for sets that are not necessarily singletons. The algorithm used in the proof of B considers only queries of the form ‘Does ?’, but it chooses the vertex based on its occupancy probability in a certain hard-core model.
1.3. Comparing A and B
Even though the descriptions of the container sets in A and B look rather different, they are in fact essentially equivalent. This is a consequence of the following proposition, which quantifies the relation between the smallest -weight of a cover for a uniform hypergraph and the probability that the -random subset of vertices of is independent.
Proposition 1.1.
Suppose that is a hypergraph on a finite vertex set .
-
(i)
For every hypergraph that covers and all ,
-
(ii)
If is -uniform, then, for every , there exists a hypergraph that covers and satisfies
1.4. Relation to earlier HCLs
An attractive feature of A and B is that each of them implies (a slight strengthening of) the ‘efficient’ HCL of Balogh and Samotij [3, Theorem 1.1], stated here as C below, which in turn significantly improves the original HCLs proved in [1, 23]. Given a hypergraph with vertex set and a set , we define
Further, for an integer , we let
The following statement can be easily derived from both A as well as B and Janson’s inequality. We present the two derivations in Section 5.
Theorem C ([3]).
Let be an -uniform hypergraph with a finite vertex set . Suppose that and are such that, for every ,
(1) |
There exists a family and functions
such that:
-
(a)
For each , we have .
-
(b)
Each has at most elements.
-
(c)
For every , we have .
In most applications of the hypergraph container method, one recursively applies a basic HCL, such as C, to construct a family of containers that are almost independent in the sense that one can no longer prove a balanced supersaturation result in any of the containers. In order to avoid explicitly performing such a recursive procedure, one may instead invoke one of the several ‘packaged’ HCLs that exist in the literature (see, e.g., [1, Theorem 2.2], [3, Theorem 1.6], or [23, Corollary 3.6]). Both A and B allow one to deduce such a packaged HCL directly, without the need for recursion. For example, the following statement is a straightforward consequence of A.
Theorem D.
Let be an -uniform hypergraph with a finite vertex set . For every , there exists a family and functions
such that:
-
(a)
For each , we have .
-
(b)
Each has at most elements.
-
(c)
For every , there does not exist a hypergraph that satisfies
(2)
We note that condition (c) in the above theorem is the negation of a balanced supersaturation statement that would enable one to construct a small family of containers for independent sets in using a basic HCL such as C.
We present the derivation of D from A in Section 5 and only mention here that the existence of a hypergraph with small -weight that covers precludes the existence of a hypergraph that satisfies the balancedness condition (2). We further note that the derivation of D from A in fact allows one to prove a strengthening of the former, where the assertion (c) is replaced by the stronger assertion that there does not exist a probability distribution on under which the random edge satisfies
Note the striking similarity between this condition and the notion of a -spread measure, introduced by Talagrand [26], which played a key role in the proof [10] of the fractional version of the ‘expectation-threshold’ conjecture of Kahn and Kalai [12], conjectured by Talagrand [26].
Finally, we remark that one may use B to derive an alternative version of D, where assumption (2) is replaced by an analogous sequence of upper bounds on the degrees . The existence of an that satisfies such a modified assumption would imply, via Janson’s inequality, an upper bound on the probability that the -random set is independent in , and thus also in , that is smaller than the upper bound on this probability asserted by B.
1.5. Interpolating between A and B
Before concluding our discussion, we state one more HCL, where the characterisation of almost independence interpolates between those given by A and B.
Theorem E.
Let be a hypergraph with a finite vertex set . For all reals and satisfying , there exists a family and functions
such that:
-
(a)
For each , we have .
-
(b)
Each has at most elements.
-
(c)
For every , letting , there exists a hypergraph with vertex set and for all that covers and satisfies
for all .
1.6. Organisation
In Section 2, we set a few notational conventions, state several auxiliary results, and prove 1.1. In Sections 3 and 4, we prove A and B, respectively. Section 5 is devoted to derivations of standard HCLs (C and D) from A and B. In Section 6, we sketch the proof of E and show that it implies both A and B. In Section 7, we present an attractive conjecture that strengthens 1.1 and discuss some of its implications. Finally, Appendix A contains three proofs of one of our key auxiliary results, 2.2.
2. Preliminaries
2.1. Notation
Given a hypergraph and a positive integer , we define
It will be also convenient to introduce the shorthand notations
Finally, for every , we denote by the link of in , i.e.,
for the sake of brevity, we will write in place of .
2.2. Auxiliary results
While comparing B to other HCLs, we will crucially use the following well-known inequality due to Janson [11].
Theorem 2.1 (Janson’s Inequality).
Suppose that is a hypergraph on a finite set . For every ,
where
and the second sum ranges over all ordered pairs ( and are not necessarily distinct).
Our proof of B, as well as the derivation of this theorem from E, requires the following probabilistic estimate. Since this estimate follows from standard techniques, we postpone its proof to the appendix (where we in fact present three different proofs).
Proposition 2.2.
Suppose that is a finite set and let be a decreasing family of subsets. For every , we have
Finally, we will also employ the following well-known inequality, discovered independently by Bollobás [6], Lubell [16], Meshalkin [17], and Yamamoto [27].
Proposition 2.3 (LYMB inequality).
Suppose that is a finite set and that is an antichain. Then,
2.3. Proof of 1.1
The first assertion is a straightforward application of Harris’s inequality. Indeed, for every hypergraph that covers , we have
where the final inequality holds as for all .
We now prove the second assertion. Suppose that is -uniform and define, for each , . Let be a maximal subgraph of subject to for all . Define and
Let be the set of minimal elements in the family
and note that by maximality of . Since is an antichain, the LYMB inequality (2.3) yields
Since our upper-bound assumption on yields
for every , we may conclude that
We may now apply Janson’s inequality (Theorem 2.1) to conclude that
as desired.
3. Proof of A
3.1. The algorithm
Assume that is an -uniform hypergraph on a finite set and fix some . The following algorithm takes as input an independent set and returns sets and a hypergraph on .
-
(1)
Let and .
-
(2)
Do the following for :
-
(a)
Let .
-
(b)
If , then set and STOP.
-
(c)
Otherwise, let be the smallest such that for some nonempty and let be an inclusion-maximal set with this property.222It may not be clear at this point that such and exist, but we will prove this shortly.
-
(d)
If , then set and .
-
(e)
Otherwise, if , set and .
-
(f)
Finally, set .
-
(a)
-
(3)
Return , , and .
3.2. Well-definedness
We now show that the algorithm described in the previous section is well defined and that it terminates on every input. Our first lemma takes care of the former and shows that the definition of and in step (2)(2c) is always valid.
Lemma 3.1.
If , then there are and nonempty with .
Proof.
Note first that our assumption implies that
Consequently, there must be some such that
Since the fact that guarantees that , we may take and an index that achieves . ∎
Next, we show that each hypergraph defined in the algorithm is an antichain and that the sequence is strictly increasing. Note that this implies that the algorithm terminates on every input. Indeed, the fact that is an antichain means that it comprises precisely all the minimal elements of and therefore are pairwise distinct.
Lemma 3.2.
The hypergraph is an antichain for every .
Proof.
We prove the assertion of the lemma by induction on . The induction basis holds due to our assumption that is -uniform. For the induction step, assume that is an antichain for some . Observe first that is an antichain as well. Indeed, this is clear when ; otherwise and this is a straightforward consequence of the assumption that is an antichain. Suppose that contained a pair of sets with . Since both and are antichains, then either and , which is clearly impossible, or vice-versa. However, if , then for some . (Indeed, when , we may take ; otherwise, and the existence of such a follows from the fact that .) As a result, if there was with , then for some , contradicting the inductive assumption that is an antichain. ∎
Lemma 3.3.
We have and for every .
Proof.
The definition of implies that . To complete the proof of the lemma, it is therefore enough to show that and . The former statement is clear when ; otherwise, it follows from the assumption that . For the latter statement, since is an antichain (by Lemma 3.2), it is enough to argue that every element of is a proper subset of some element of . When , this is the case since and ; otherwise, if , this follows as . ∎
3.3. Properties
We now turn to establishing some key properties of the algorithm. We will show that, for every input , we have , the set has at most elements, and the hypergraph (which clearly satisfies , see the stopping condition (2)(2b), and for all ) covers . Further, we will prove that the set (and the hypergraph ) depends on only through , which will allow us to define the functions and by and .
Lemma 3.4.
The hypergraph covers .
Proof.
Note first that Lemma 3.3 implies that . Let be an arbitrary edge of . Since covers , there must be some with . However, as , the set cannot be a singleton and thus , as required. ∎
Lemma 3.5.
For each , we have and .
Proof.
It suffices to show that for all ; indeed, implies that , as for every . We prove this by induction on . The induction basis follows as and . For the induction step, assume that for some . Note first that either or and thus . Further, since , it suffices to show that . If , then and thus comprises all subsets of not containing (which includes ). Otherwise, if , we have and thus ; further, contains all sets such that (which again includes ). ∎
Lemma 3.6.
For every , the hypergraph is -uniform for some .
Proof.
Suppose first that . In this case, is clearly -uniform and as and . Suppose now that . In this case, is clearly -uniform and as is nonempty. ∎
Lemma 3.7.
For every and all and with ,
Proof.
We prove the lemma by induction on . The base case is trivial, as is -uniform and thus the hypergraphs are all empty. For the induction step, assume that the assertion holds for some (and all and ). Since , we have
(3) |
for all and all . Further, since the hypergraph is -uniform for some (by Lemma 3.6), we have unless . We may therefore assume from now on that , as otherwise the desired inequality (for all with ) follows from (3) and the inductive assumption. The minimality of and the fact that imply that . It thus suffices to show that for every with . Fix some such set . If , then the maximality of implies that . Otherwise, if , we have
as desired. ∎
Corollary 3.8.
For every and all ,
Lemma 3.9.
For every ,
Proof.
Since the hypergraph is -uniform for some (by Lemma 3.6), it follows from the definition of that
Further, since (by Lemma 3.3), Corollary 3.8 implies that for each and thus . The assertion of the lemma follows as and when . ∎
Lemma 3.10.
We have .
Proof.
Let denote the number of rounds of the algorithm in which . Since is -uniform, it follows from Lemma 3.9 that
On the other hand, since and , we have
We conclude that . Since when and otherwise, the assertion of the lemma follows. ∎
Lemma 3.11.
Suppose that the algorithm with inputs outputs and , respectively. If , then .
Proof.
Observe first that the execution of the algorithm on a given input (an independent set of ) depends solely on how the set and the hypergraph are defined in each round (via step (2)(2d) or step (2)(2e)). Thus, if the outputs of the algorithm applied to and are different, then there must be some for which but , or vice-versa. Let be the earliest such round and note that and thus also . We may clearly assume that and . However, Lemma 3.5 implies that , a contradiction. ∎
We finish our discussion with a formal proof of A.
Proof of A.
For each independent set , set and , where is the output of the algorithm with input . (Recall that Lemmas 3.2 and 3.3 imply that the algorithm terminates on every input.) Further, set . Lemma 3.11 guarantees that the function is well-defined. By Lemma 3.5, we have for every whereas Lemma 3.10 shows that for every . Finally, writing , Lemma 3.4 guarantees that the hypergraph is a cover for with ; the property that for all is straightforward from the definition of . This concludes the proof of the theorem. ∎
4. Proof of B
4.1. The algorithm
Assume that is a hypergraph on a finite set and fix some . The following algorithm takes as input an independent set and returns sets .
-
(1)
Let and .
-
(2)
Do the following for :
-
(a)
If there exists a vertex such that and
let be some such vertex; otherwise, set and STOP.
-
(b)
If , set and .
-
(c)
Otherwise, if , set and .
-
(a)
-
(3)
Return and .
4.2. Properties
We now establish some key properties of the algorithm. We first claim that the algorithm always terminates with . Indeed, for each , the vertex is either added to or becomes an edge of ; in both cases, is never considered again in step (2)(2a). Further, we claim that the set can be reconstructed with the knowledge of only. Indeed, for all , the hypergraph depends on only. Therefore, running the algorithm with input instead of results in the same sets and . This allows us to define the functions and by and . It remains to show that, for every input , we have , the set has at most elements, and that with probability at least .
Lemma 4.1.
For each , we have and .
Proof.
We prove the assertion of the lemma by induction on . The induction basis follows as and . For the induction step, assume that the assertion holds for some . Suppose first that . In this case, and , which means that . Further
Suppose now that . In this case, and , which means that . Further, , as . ∎
The above lemma clearly implies that . Further, , as no vertex satisfying can belong to an independent set of . This establishes .
Lemma 4.2.
We have .
Proof.
We claim that the sequence of probabilities satisfies
for all , which gives
(4) |
Indeed, if , then the inequality follows as , and thus . On the other hand, if , then , and thus . Consequently, by the choice of ,
which gives the desired inequality. We claim that (4) implies the desired inequality . Indeed, if , then , where the second inequality follows as and the map is decreasing on . ∎
Lemma 4.3.
We have .
Proof.
Define . Since clearly , we have
In view of 2.2, the assertion of the lemma will follow if we prove that
(5) |
To this end, fix an arbitrary . Since , as we have already observed above, and since , by Lemma 4.1, we have
Finally, since , the stopping condition (2)(2a) implies that the above probability is at least . Summing this inequality over all yields (5). ∎
We conclude our discussion with a short derivation of B.
Proof of B.
For each independent set , set and , where is the output of the algorithm with input . Further, set . Since the algorithm will perform the exact same operations when we replace the input set with , the function is well-defined. Now, Lemma 4.1 implies assertion (a), Lemma 4.2 proves assertion (b), and Lemma 4.3 establishes assertion (c). ∎
5. Derivations of the standard HCLs
Derivation of C from A.
We apply A with to obtain a family , functions and , and a hypergraph for every . Assertions (a) and (b) follow immediately from the respective assertions in A, so we only need to argue that (c) holds as well.
To this end, fix some , let , and let be the hypergraph covering that satisfies and whose all edges have size at least two. By the assumption of the theorem, for every ,
where the second inequality follows as and . The fact that covers implies that
On the other hand,
which yields or, equivalently, . ∎
Remark.
A careful examination of the proof allows one to improve the bounds in assumption (1) in C somewhat further. First, note that we only needed the bounds
Further, we claim that changing step (2)(2b) in the algorithm from the proof of A to
-
(2b’)
If or , then set and STOP
allows us to derive a stronger version of C, with the upper bound on in (1) increased by a factor of roughly . To see this, notice that the only modifications needed in the proof of A occur in Lemmas 3.1 and 3.10. The conclusion of Lemma 3.1 still holds since
and, consequently, there must be some such that
which implies that for some . On the other hand, the assertion of Lemma 3.10 may now be strengthened. Indeed, since now
we may deduce the stronger bound , which allows the conclusion of C to be reached under the weaker assumption
Derivation of C from B.
We apply B with and to obtain a family and functions and . Assertions (a) and (b) follow immediately from the respective assertions in B, so we only need to argue that (c) holds as well.
Suppose that and let . We may assume that , as otherwise follows from the assumption that , as in the previous derivation. Preparing to apply Janson’s inequality (Theorem 2.1), we define and
By the assumption of the theorem,
and consequently, by Janson’s inequality,
where the last inequality holds thanks to our assumption that . On the other hand, the assertion of B gives
a contradiction. ∎
Derivation of D from A.
We apply A to obtain a family , functions and , and a hypergraph for every . Assertions (a) and (b) follow immediately from the respective assertions in A, so we only need to argue that (c) holds as well.
To this end, fix some , let , and let be the hypergraph covering that satisfies and whose all edges have size at least two. Assume toward contradiction that there is a hypergraph that satisfies (2). Since covers , and thus also , we have
a contradiction. ∎
6. Interpolating between A and B
In this section, we sketch the proof of E, which is a fairly straightforward adaptation of the proof of B (given in Section 4), and present derivations of A and B from E.
6.1. The algorithm
Assume that is a hypergraph on a finite set and fix some . The following algorithm takes as input an independent set and returns sets and a hypergraph .
-
(1)
Let and .
-
(2)
Do the following for :
-
(a)
If there exists a set such that and
let be a minimal such set; otherwise, set and STOP.
-
(b)
If , set and .
-
(c)
Otherwise, if , set and .
-
(a)
-
(3)
Return , and .
6.2. Properties
The above algorithm always terminates because no set is ever considered more than once. Indeed, either , and thus for all , or , and thus for all . Moreover, the set and the hypergraph can be reconstructed with the knowledge of only, as running the algorithm with input instead of results in the same sets and and hypergraph . This allows us to define and . It remains to show that, for every input , we have , the set S has at most elements, and that is a cover of with the desired property.
The assertion of Lemma 4.1, that is, and for all , remains true, with essentially the same proof. Consequently, and thus , as no vertex in can belong to an independent set of . Observe additionally that, for all , we have and thus . The desired upper bound on can be established by adapting the proof of Lemma 4.2. Indeed, note first that it is enough to show that the sequence satisfies the inequality for all . The inequality follows as . Further, if , then
where the equality holds as .
Last but not least, we argue that possesses all the claimed properties. First, covers since . Second, each satisfies as otherwise we would have , by the definition of . We also remark that, for all with , we have
by the terminating condition in the algorithm and the fact that .
Finally, we show that for every . Since and , it is enough to prove the stronger statement that for all and all . We show this by induction on . The induction basis is vacuously true as . Assume that . If , then and the only edge in is , which is disjoint from . Else, if , then and we replace each , which is disjoint from , with the edge , which is disjoint from .
6.3. Derivations of A and B
Derivation of B from E.
Since the assumptions of the two theorems are identical, and their assertions differ only in the characterisation of the sets , we just need to show that assertion (c) in E implies that, for every , we have
where . Let and note that the fact that is a cover of implies that
Further, the fact that for all implies that, for every
It follows that
and consequently, by 2.2, that . ∎
The derivation of A from E is decidedly more complicated. We will show that that the cover of given by E has the property . We will argue by contradiction, showing that, if , some maximal set among those whose degree in is large satisfies ; the proof of this inequality will employ a second-moment argument.
Derivation of A from E.
Suppose that is an -uniform hypergraph with vertex set and let be arbitrary. We apply E with and density parameter to obtain a family of sets satisfying for each and functions and such that for all . It suffices to show that, for every , there is a hypergraph on with all edges of size at least two that covers and satisfies , where . To this end, fix some and let be the set of all inclusion-minimal elements in the cover given by (c) in E. As covers and satisfies for all , we only need to show that .
Assume by contradiction that . We claim that there exists a set and an such that
(6) |
To this end, note that
which means that (6) must hold with for some and ; note that for every since all edges of have at least two vertices. Let be an inclusion-maximal set that is independent in and satisfies (6) for some and let . Such choice of guarantees that, on the one hand,
and, on the other hand, for every ,
where the last inequality follows as every set that has fewer than elements and satisfies is independent in (otherwise, would be a proper subset of some edge of , contradicting the fact that is an antichain).
Since , assertion (c) of E yields
whereas, by Harris’s inequality, writing for conditioned on the event ,
Let denote the number of edges that induces in . Using the Paley–Zygmund inequality, we may bound
which means that .
7. Concluding remarks
Comparing the two notions of almost-independence that are used to describe the containers in A and B led us to proving 1.1. While part (ii) of this proposition is sufficient for our purposes, we hope that a much stronger statement is true. We will say that a hypergraph is -bounded, for some positive integer , if for all .
Question 7.1.
What is the smallest function such that any -bounded hypergraph on a finite vertex set admits a cover that satisfies
1.1 (ii) shows that when one assumes that is -uniform rather than -bounded. However, a fairly straightforward adaptation of the proof of this proposition extends its validity to all -bounded hypergraphs. On the other hand, it is not hard to see that . (Indeed, consider the hypergraph with vertex set whose edges are all perfect matchings and for some small constant .) Originally, we conjectured that does not have to depend on , at the cost of multiplying the above upper bound on the probability by a factor polynomial in . Soon after the original version of this work was made public, Quentin Dubroff found a counterexample to this stronger conjecture. (His counterexample is the hypergraph with vertex set whose edges are all -factors in , for some , and slightly larger than .) We later realised that a universal upper bound on the probability that that has the form , for some that depends only on , implies the corresponding stronger upper bound of . To see this, consider the hypergraph that comprises vertex-disjoint copies of . It is easy to verify that and that, for every , the smallest -weight of a cover of is times the smallest -weight of a cover for . This observation leads to many further counterexamples to our (very naive) conjecture.
We remark that showing that would imply the Kahn–Kalai Conjecture / Park–Pham Theorem [20, Theorem 1.1]; it would also be consistent with the strengthening of the Park–Pham Theorem due to Bell [4]. Indeed, let be a finite set and let be an arbitrary increasing property. Recall that the expectation threshold of is the number defined as follows:
Let be the set of all minimal elements of and let , so that is an -bounded hypergraph on and if and only if . Let , suppose that , and let be the cover of from the statement of 7.1. Observe that
Consequently, we may conclude that
Finally, showing that would imply an improvement of the upper bound on the number of containers in A, as the following theorem shows.
Theorem F.
Let be an -bounded hypergraph with a finite vertex set . For all reals and that satisfy , there exists a family and functions
such that:
-
(a)
For each , we have .
-
(b)
Each has at most elements.
-
(c)
For every , letting , there exists a hypergraph with
that covers .
Proof.
Let , , and be the collection and the functions given by B with in place of and in place of . Fix some , let , and let be a cover of with smallest -weight. The assertion of B and the definition of yield the following two inequalities:
Taking logarithms of both sides yields the inequality
as required. ∎
8. Acknowledgements
First and foremost, we are indebted to Quentin Dubroff for supplying a counterexample to a conjecture stated in a previous version of this article. Further, we would like to thank Jinyoung Park for helpful discussions of this conjecture and the counterexample. Last but not least, we would like to thank Rob Morris, Peter Keevash, and Huy Pham for helpful comments and suggestions.
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Appendix A Three proofs of 2.2
One can see the assertion of the proposition is best-possible by picking an arbitrary subset and considering the family . Indeed, in this case and
We provide three proofs of the proposition. The first proof uses the chain rule for relative entropy, the second proof employs a compression argument combined with the Kruskal–Katona theorem, and the third proof uses a generalisation of the edge-isoperimetric inequality for the hypercube due to Kahn and Kalai [12].
First proof.
Without loss of generality, we may assume that . Let be the indicator function of conditioned on and note that
where denotes the Kullback–Leibler divergence between two probability distributions. For any , write in place of . With this notational convention, . The first key property of that we will use is that it obeys the familiar chain rule (see, e.g., [14, Section 4]):
where we wrote for the conditional relative entropy and
Since the function is convex, we have, for every random variable ,
Finally, since conditioned on , the vector is the indicator function of conditioned on belonging to a decreasing family of sets, we have almost surely. This implies that
as claimed. ∎
Second proof.
We are going to prove the following equivalent inequality:
(7) |
Let . We first argue that it is enough to establish (7) in the case where is an integer. To see this, suppose that . We claim that the (continuous) function cannot be constant on any open interval containing and thus it takes rational values for arbitrarily close to . Suppose that this were not true and for all . We would then have, for all ,
contradicting the fact that is a polynomial in . Since both sides of (7) are continuous in , it is enough to prove this inequality when is rational.
Assume now that and let be a positive integer such that . Consider the family defined by
and observe that is decreasing and that . Invoking (7) with replaced by , we conclude that
Assume , let , and let . Our first key observation is that
Write , so that . For each , let be the unique number in such that . The assumption that is decreasing implies that for all and, consequently, for each . Therefore, and
Our second key observation is that
Since is decreasing, the Kruskal–Katona theorem implies that . In particular, letting
we have
Let be a vector that achieves the above maximum. We claim that . Indeed, suppose that for some and let be the unique number satisfying and
Let be the vector obtained from by replacing the th and the th coordinates by and note that
and, similarly,
contradicting the maximality of . Let be the number such that . Since
we have and thus
as claimed. ∎
Third proof.
Let and let be the function defined by if and only if . Note that is increasing and that . Recall that the total influence of with respect to is the quantity
Observe now that, for every ,
and, similarly,
Consequently, letting , we have
The desired inequality follows from the edge-isoperimetric inequality for , which states that
see [12, Section 4]. ∎