Infinite Sperner's theorem

One of the most classical results in extremal set theory is Sperner's theorem, which says that the largest antichain in the Boolean lattice $2^{[n]}$ has size $\Theta\big(\frac{2^n}{\sqrt{n}}\big)$. Motivated by an old problem of Erd\H{o}s on the growth of infinite Sidon sequences, in this note we study the growth rate of maximum infinite antichains. Using the well known Kraft's inequality for prefix codes, it is not difficult to show that infinite antichains should be"thinner"than the corresponding finite ones. More precisely, if $\mathcal{F}\subset 2^{\mathbb{N}}$ is an antichain, then $$\liminf_{n\rightarrow \infty}\big|\mathcal{F} \cap 2^{[n]}\big|\left(\frac{2^n}{n\log n}\right)^{-1}=0.$$ Our main result shows that this bound is essentially tight, that is, we construct an antichain $\mathcal{F}$ such that $$\liminf_{n\rightarrow \infty}\big|\mathcal{F} \cap 2^{[n]}\big|\left(\frac{2^n}{n\log^{C} n}\right)^{-1}>0$$ holds for some absolute constant $C>0$.


Introduction
A typical question in extremal combinatorics asks to determine or estimate the maximum or minimum possible size of a combinatorial structure which satisfies certain requirements.Once such a result is established one can also ask about the growth rate of the infinite structure that satisfies the same requirements and compare its behavior with the one in the finite setting.Probably the first problem of this kind was posed by Erdős in the 1940s.A Sidon set is a set of natural numbers not containing any non-trivial solutions to the equation a + b = c + d.Denote by [n] the set {1, . . ., n}.It is known (see [5]) that for all large n there exist Sidon sets S ⊂ [n] of size at least |S| ≥ (1− o(1)) √ n.On the other hand, it was already observed in 1941 by Erdős and Turán [6] that no infinite Sidon set S ⊂ N can achieve S ∩ [n] = Θ( √ n) simultaneously for all n.This was further refined by Erdős who showed (see [7], Chapter II, 3) that if S is a Sidon set, then In particular, this shows that infinite Sidon sequences are thinner than the densest finite ones.The proof of this result appeared in a 1953 letter from Erdős to Stöhr.This letter is cited in [14], which studies a number of other problems in additive combinatorics in the same spirit of "finite versus infinite behavior".A remarkable construction of Ruzsa [11] shows that there are Sidon sets S with S ∩ [n] ≥ n for all n.Closing this gap is a fascinating open problem.For further generalizations of this problem, we refer the reader to [9,8].
It is only natural to study this phenomenon of "finite versus infinite" for other extremal problems as well.
It is well known that Sidon sets and C 4 -free graphs are intimately related.Hence, the corresponding question in the graph theoretic setting is that given an infinite C 4 -free graph G on vertex set N, how large can the minimum degree δ n of G n be, where G n is the restriction of G to the set [n]? Conlon-Fox-Sudakov [2] proved the graph-theoretic analogue of Erdős' result, that is, They also extended this result to the more general setting of K s,t -free graphs.
In this short note we consider a similar problem for maximal antichains.A family F ⊂ 2 N is an antichain if for all distinct A, B ∈ F we have A ⊂ B. One of the most classical results of extremal combinatorics is Sperner's theorem [13], which states that if F ⊂ 2 [n] is an antichain then Following Erdős, one can naturally ask what happens with Sperner's problem in an infinite setting?Using the well known Kraft's inequality [10] from 1949, it is not difficult to show that there is a polynomial drop in density in this case.
Theorem 1.1.Let F ⊂ 2 N be an antichain.Then While in the case of Sidon sets there is a wide gap between the lower and upper bound, perhaps surprisingly, in our problem we can provide bounds that match up to a polylogarithmic term.
Theorem 1.2.There exists an antichain F ⊂ 2 N such that We give the proof of the upper bound, Theorem 1.1, in Section 2. The construction that shows the lower bound in Theorem 1.2 is presented in Section 3. Some open questions and future research directions are discussed in Section 4.

Infinite antichains cannot be too large
Consider a family S of {0, 1} sequences of finite length.Say that S is a prefix code if no element of S is a prefix of another.Denoting by |s| the length of a {0, 1} sequence, the well known Krafts's inequality [10] tells us that if S is a prefix code, then We show that we can use this inequality to prove the following proposition, which then implies Theorem 1.1.
Proof.It is enough to show that for every positive integer N , we have For a set F ∈ F, let max(F ) denote the maximum element of F , and for 1 ≤ n ≤ N , let A n = {F ∈ F : max(F ) = n}.We can identify each set A ∈ A n with a {0, 1} sequence s A of length n such that the i-th element of s A is 0 if i ∈ A, and 1 otherwise.Then the fact that F is an antichain implies that the family } is a prefix code.Therefore, by the above mentioned Kraft's inequality, we get This can be rewritten as which implies the desired inequality.
Proof of Theorem 1.1.
. Then by Proposition 2.1, we have As the sum ∞ n=1 1 n log n diverges, we conclude that lim inf n→∞ f n = 0.

Constructing dense infinite antichains
There are many easy constructions that match the upper bound of Kraft's inequality.Note that we cannot use these for our problem, because being an antichain is a much stronger requirement than being prefix-free, e.g. the family {{1, 2, 3}, {1, 3}} is prefix-free but not an antichain.
Before we give the construction that achieves the bound in Theorem 1.2, it will be helpful to first look at a slightly worse construction.While this only achieves a density of 2 n /n 3/2 , it will illustrate some of the ideas of the proof of Theorem 1.2.Consider the family and let F = n F n .Observe that F is an antichain, and that there is a bijection between elements of F n and lattice paths from (0, 0) to (n, n) where every step is to the right or up, and the path is fully under the line y = x except at the endpoints.The enumeration of such paths is given by the Catalan numbers, and hence One can then verify that We will generalize the previous construction as follows.Fix an arbitrary monotone increasing function f : N → N, and define the family Again one can show that F = ∞ n=1 F n is an antichain.Indeed, suppose to the contrary that there is A ∈ F n and B ∈ F m for some n, m such that A ⊂ B. As each F n is a uniform family, we must have n = m.Moreover as f is a monotone increasing function, we must have n < m.Then A contains n + f (n) elements from [2n], and since A ⊂ B, B must contain at least n + f (n) elements from [2n] as well.This however contradicts the definition of F m .
We show that if we pick f (n) = Θ( √ n log log n), then F n has the desired size.The main difficulty of our proof comes from estimating the size of F n .
Proof of Theorem 1.2.Let f (x) := ⌊3 x log log(x + 3)⌋ + 100 and Set F := ∞ n=1 F n .Then F is an antichain, so it only remains to show that It will be helpful for us to identify sets with lattice paths that start at (0, 0) and take steps in directions (1, 1) and (1, −1).By identifying the terms "contains element i" and "the i-th step is in direction (1, 1)", we find a bijection between F n and the family G n , defined as the collection of paths starting at (0, 0), ending at (2n, 2f (n)), taking steps in directions (1, 1) and (1, −1), and staying strictly below the curve y = 2f (x/2) for all even x satisfying 0 < x < 2n.See Figure 1 for an illustration.
The following statement is a version of the Law of the Iterated Logarithm (see e.g.[1]).For the sake of the reader we prove here a specific form which we need for our construction.
Proof.Let C > 0 be a positive constant.For all i, let A i be the event that i is the smallest integer with S i > C i log log(i + 3) + 100, and observe that for i ≤ 100 the event A i is empty.Fix an arbitrary integer n, and in the first step of the proof we will show an upper bound for the probability that A i occurs for some i with ⌈n/2⌉ ≤ i ≤ n.Let U j be the event that the number of 1's amongst X j+1 , X j+2 , . . ., X n is at least (n − j)/2, thus clearly P(U j ) ≥ 1/2 for all j.From definitions, one can easily check that for all j ≥ n/2 ≥ 100, we have Therefore, As A j and U j are independent, we can write Putting equations ( 4) and ( 5) together, and using that the A j -s, and hence the A j ∩ U j -s, are disjoint events, we get We can bound the last quantity using the standard Chernoff bound, to get Setting C = 3 then gives P N j=100 A j ≤ 1 2 .As this bound holds uniformly for all N , the claim follows.
For k ∈ Z, denote by P (k) the collection of paths starting at (0, 0), ending at (n, k), taking steps in directions (1,1) and (1, −1) (such paths exist precisely if n + k is even), and staying strictly below the curve y = 2f (x/2) for all even x satisfying 0 < x ≤ n.Similarly, let Q(k) denote the collection of paths starting at (n, k), ending at (2n, 2f (n)), taking steps in directions (1, 1) and (1, −1), and staying strictly below the curve y = 2f (x/2) for all even x satisfying x < 2n.Observe that hence, it is enough to find suitable lower bounds for Proof.It is a straightforward calculation to verify that 2f (n/2) ≥ A ≥ 3 n log log(n + 3) + 100.
Hence, by Lemma 3.1, we have The more difficult part is to derive a lower bound for |Q(k)|.
Proof.Say that a path is good if it starts at (0, 0), ends at (n, 2f (n) − k), and with the exception of the second endpoint, stays strictly below the straight line connecting (0, 2f (n/2) − k) and (n, 2f (n) − k).This line is given by the formula Then the number of good paths is a lower bound for |Q(k)|, as every good path shifted by the vector (n, k) gives rise to a path in Q(k).
Note that the number of good paths is equal to the number of paths that start at (0, 0), end at (n, 2f (n)− k), and with the exception of the first endpoints, stay strictly above the line y = µx.Indeed, given a good path with steps p 1 , . . ., p n where p i ∈ {(1, 1), (1, −1)} for all i, the reverse path p n , . . ., p 1 is of the second type, and vice versa.We will count the number of these paths as follows.Take any path from the origin to (n, 2f (n) − k) using steps (1, 1) and (1, −1).Such a path contains p = 1 2 (n + 2f (n) − k) steps in the (1, 1) direction and q = 1 2 (n − 2f (n) + k) steps in the (1, −1) direction.This path can be represented by a sequence x 1 , . . ., x n of +1's and -1's, where x i = 1 precisely if the i'th step was in the (1, 1) direction and x i = −1 otherwise.Say that the sequence x 1 , . . ., x n , where the number of +1's is p and the number of −1's is q, is good if j i=1 x i > µj for all 1 ≤ j ≤ n.Then the number of good sequences is equal to the number of good paths.
Given a sequence x 1 , . . ., x n of +1's and −1's, arrange them in clockwise order around a circle, looping around only once, so that x 1 comes after x n in the clockwise direction.Say that two {−1, +1} sequences are equivalent if we can get one from the other by clockwise rotation, that is, x 1 , . . ., x n is equivalent to y 1 , . . ., y n if there exists r ∈ [n] such that x i = y i+r for every i ∈ [n], where indices are meant modulo n.We show that each equivalence class contains at least p − ⌊ 1+µ 1−µ q⌋ good sequences.In order to prove this, we need a variant of the well known "cycle lemma" (see e.g.[3]).Although this lemma appeared in many forms in different papers, the variant which we need is not so common and can be found in [4] (corollary of Theorem 3).
Given a sequence x 1 , . . ., x n of +1's and −1's in circular order and the number µ, say that an index r is a head if r+j−1 i=r x i > µj for all j ∈ [n].Equivalently, the number of +1's on every nonempty arc beginning with x r and proceeding in the positive direction is greater than 1+µ 1−µ times the number of −1's on the same arc.Clearly, if there are p +1's and q −1's in the sequence, then each head is the starting point of a good sequence in the circular order.Here we give a different proof from that in [4] which has an additional benefit of explicitly pointing to the locations of heads.Lemma 3.4.Let x 1 , x 2 , . . ., x p+q be a sequence with values in {−1, +1} such that the number of +1's is p and the number of −1's is q = 0. Arrange this sequence around a circle as above, and let µ ∈ R satisfy 0 < µ ≤ p−q p+q .Then this sequence has at least p − ⌊ 1+µ 1−µ q⌋ distinct heads.
Proof.For all i, let S i = (x 1 + . . .+ x i ) − µi.Let t := min 0≤i≤n S i , and note that since we set S 0 = 0 we have t ≤ 0. Observe that Note that for i in this range, d i ≤ S n ; moreover the γ i -s are well-defined and distinct because S j+1 − S j ≤ 1 − µ for all j.We show that γ i + 1 is a head for 1 ≤ i ≤ p − ⌊ 1+µ 1−µ q⌋.
Fix an i with 0 ≤ i ≤ p − ⌊ 1+µ 1−µ q⌋ and for j ∈ [n], let T j = (x γ i +1 + x γ i +1 + . . .+ x γ i +j ) − µj, where the subscripts are interpreted modulo n.We claim that T j is strictly positive.If j is such that γ i + j ≤ n then T j > 0 because γ i was the largest index with S i < d i .If j is larger, then we get Thus T j > 0 for all j ∈ [n], and hence γ i + 1 is a head as claimed.
For the rest of the proof we will assume that n is large.Recall that A natural question, raised in [2] whether it is true for any infinite C 6 -free graph G that lim inf n→∞ δ(G n ) n 1/3 = 0.
Finally, it would be interesting to obtain further results in this spirit of "finite versus infinite" for some other problems in extremal combinatorics.

Figure 1 :
Figure 1: An element of the family G n

2 9 C 2 ≤ e − 2 9 C 2
log log n .Now we are ready to bound the probability that an A i occurs for some i ≤ N , where N is an arbitrary large integer of the form N = 100 • 2 k .log log N + e − 2 9 C 2 log log(N/2) + . . .+ e − 2 9 C 2 log log 200 = k i=1 (log 100 + i log 2) − 2 9 C 2