Extremal set theory for the binomial norm

TL;DR

This paper establishes optimal bounds in extremal set theory for families avoiding s pairwise disjoint sets, extending classical results and applying to multiple families.

Contribution

It provides the best possible bounds for these set families, generalizing and unifying several classical results in extremal combinatorics.

Findings

Optimal bounds for families without s disjoint sets

Extension of classical extremal set theory results

Application to multiple families in combinatorics

Abstract

Best possible bounds are established for families without s pairwise disjoint members and the more general problem for several families. The results are shown to apply several classical results.

Figures

Equations

$$|\mathcal{F}|\le \left(\genfrac{}{}{0ex}{}{n}{\lfloor \frac{n}{2}\rfloor }\right).$$

$$\sum_{F\in\mathcal{F}}1\Bigm{/}{n\choose|F|}\leq 1.$$

$$\|\mathcal{F}\|_{n}=\sum_{F\in\mathcal{F}}1\Bigm{/}{n\choose|F|}.$$

$$\mathcal{F}^{(i)}=\{F\in\mathcal{F}:\ |F|=i\}\ \text{ and }\ \varphi(i)=\bigl{|}\mathcal{F}^{(i)}\bigr{|}\Bigm{/}{n\choose i}.$$

$$\Vert \mathcal{F}{\Vert }_n=\sum _{0\le i\le n}\phi (i).$$

$$0\le \Vert \mathcal{F}{\Vert }_n\le n+1\text{ is obvious.}$$

$$0\le {\varrho }_n(\mathcal{F})\le 1.$$

$$\frac{1}{n+1}\cdot \frac{1}{\left(\genfrac{}{}{0ex}{}{n}{\ell }\right)}=\frac{1}{n+2}\cdot \frac{1}{\left(\genfrac{}{}{0ex}{}{n+1}{\ell }\right)}+\frac{1}{n+2}\cdot \frac{1}{\left(\genfrac{}{}{0ex}{}{n+1}{\ell +1}\right)}.$$

$${\varrho }_{n+1}(\mathcal{E}(\mathcal{F}))={\varrho }_n(\mathcal{F}).$$

$$\varrho_{n}(\{F\})=\varrho_{n+1}\bigl{(}\{F,F\cup\{n+1\}\}\bigr{)}.$$

$$\mathcal{G}_{i}=\bigl{\{}G\subset[n]:|G\cap Y|\geq j_{i}\bigr{\}}.$$

$${\varrho }_n({\mathcal{G}}_i)=\frac{k+1-j_i}{k+1},$$

$$\sum _{1\le i\le s}{\varrho }_n({\mathcal{G}}_i)=s-1.$$

$${\varrho }_n({\mathcal{F}}_1)+\cdots +{\varrho }_n({\mathcal{F}}_s)\le s-1.$$

$$\Vert {\mathcal{F}}_1{\Vert }_n+\cdots +\Vert {\mathcal{F}}_s{\Vert }_n\le (n+1)(s-1).$$

$$|\mathcal{F}|\le \frac{1}{2}{2}^n=2^{n-1}.$$

$${\varrho }_n(\mathcal{F})\le \frac{1}{2}.$$

$${\varrho }_n(\mathcal{F})\le \frac{s-1}{s}.$$

$$\sum _{1\le i\le s}{\phi }_i(k_i)\le s-1.$$

$$\sum _{0\le j\le r}\sum _{1\le i\le s}{\phi }_i(k_i+j)\le (r+1)(s-1).$$

$$\mathcal{G}_{i}=\bigl{\{}G\subset[r]:\bigl{(}P_{i}\cup\varrho^{-1}(G)\bigr{)}\in\mathcal{F}_{i}\bigr{\}}.$$

$$\sum _{1\le i\le s}\sum _{0\le j\le r}{\phi }_i(k_i+j)=\sum _{1\le i\le s}\sum _{0\le j\le r}{\gamma }_i(j)\le (s-1)(r+1).$$

$$\sum _{1\le i\le s}\sum _{0\le \ell \le n}{\phi }^{(i)}(k_i+\ell )\le (n+1)(s-1).$$

$$\Vert {\mathcal{F}}_i\Vert =\sum _{0\le \ell \le n}{\phi }^{(i)}(\ell )+{\phi }^{(i)}(n+1)\le 1+\sum _{0\le i\le n}{\phi }^{(i)}(\ell ).$$

$$\Vert {\mathcal{F}}_s\Vert =\sum _{0\le \ell \le n}{\phi }^{(s)}(1+\ell ).$$

$$\sum _{1\le i\le s}\Vert {\mathcal{F}}_i\Vert \le (n+1)s-(d+1).$$

$$\bigl{\|}\mathcal{F}_{1}\square\ldots\square\mathcal{F}_{s}\bigr{\|}\geq\|\mathcal{F}_{1}\|+\dots+\|\mathcal{F}_{s}\|-(s-1)(n+1).$$

$$\Vert {\mathcal{F}}_{s+1}\Vert =(n+1)-\Vert \tilde{\mathcal{F}}\Vert =(n+1)-\Vert {\mathcal{F}}_1\square \dots \square {\mathcal{F}}_s\Vert .$$

$$\Vert {\mathcal{F}}_1\Vert +\cdots +\Vert {\mathcal{F}}_s\Vert -(s-1)(n+1)\le (n+1)-\Vert {\mathcal{F}}_{s+1}\Vert .$$

Full text

1 Introduction

Let $[n]=\{1,\dots,n\}$ be the standard $n$-element set and $2^{[n]}$ its power set. Subsets of $2^{[n]}$ are called families. For a family $\mathcal{F}\subset 2^{[n]}$, $|\mathcal{F}|$ denotes its size, that is, the number of distinct subsets in $\mathcal{F}$.

The first theorem in extremal set theory was proved by Sperner in 1928. To state it let us say that $\mathcal{F}\subset 2^{[n]}$ is an antichain if there are no distinct sets, $F,G\in\mathcal{F}$ satisfying $F\subset G$.

Sperner’s Theorem** ([S]).**

If $\mathcal{F}\subset 2^{[n]}$ is an antichain, then

[TABLE]

The most important generalisation of (1.1) is the following

LYM Inequality (cf. [Y], [B], [L] and [M]). Suppose that $\mathcal{F}\subset 2^{[n]}$ is an antichain. Then

[TABLE]

To deduce (1.1) from (1.2) just note that ${n\choose|F|}\leq{n\choose\left\lfloor\frac{n}{2}\right\rfloor}$. Sperner proved that equality holds in (1.1) only if $\mathcal{F}={[n]\choose\left\lfloor\frac{n}{2}\right\rfloor}$ or if $n$ is odd and $\mathcal{F}={[n]\choose\frac{(n+1)}{2}}$. Similarly, equality holds in (1.2) only if $\mathcal{F}={[n]\choose i}$ for some $0\leq i\leq n$.

The LYM inequality motivates the following definition.

Definition 1.1**.**

For a family $\mathcal{F}\subset 2^{[n]}$ let us define its binomial norm, $\|\mathcal{F}\|_{n}$ by

[TABLE]

When it causes no confusion we shall omit $n$ and write simply $\|\mathcal{F}\|$.

Often to compute $\|\mathcal{F}\|$, it is convenient to collect terms. For $0\leq i\leq n$ we set

[TABLE]

Obviously, $0\leq\varphi(i)\leq 1$ and $\varphi(i)$ is the probability that (for fixed $i$) if we choose an $i$-subset of $[n]$ according to the uniform distribution on ${[n]\choose i}$, then it belongs to $\mathcal{F}$. Let us note the easy fact

[TABLE]

Both from (1.3) and (1.4)

[TABLE]

For some statements it is more convenient to use the relative measure \varrho_{n}(\mathcal{F}):=\|\mathcal{F}\|_{n}\bigm{/}(n+1). Note that

[TABLE]

Note also the following simple identity valid for $0\leq\ell\leq n$:

[TABLE]

For a family $\mathcal{F}\subset 2^{[n]}$ one can define its natural extension in $2^{[n+1]}$ by \mathcal{E}(\mathcal{F})=\bigl{\{}E\subset[n+1]:E\cap[n]\in\mathcal{F}\bigr{\}}.

Claim 1.2**.**

[TABLE]

Proof.

Simply note that for every $F\in\mathcal{F}$, denoting $|F|$ by $\ell$, (1.7) implies

[TABLE]

Hence (1.8) holds. ∎

The central property for families that we consider in this paper is cross-dependence.

Definition 1.3**.**

The families $\mathcal{F}_{1},\dots,\mathcal{F}_{s}$ are called cross-dependent if there is no choice of $F_{1}\in\mathcal{F}_{1},\dots,F_{s}\in\mathcal{F}_{s}$ such that $F_{1},\dots,F_{s}$ are pairwise disjoint.

Example 1.4**.**

Let $1\leq k\leq n$ and $Y\in{[n]\choose k}$. Let $j_{1},\dots,j_{s}$ be integers, $0\leq j_{i}\leq k$, $j_{1}+\dots+j_{s}=k+1$. Define

[TABLE]

Proposition 1.5**.**

The families $\mathcal{G}_{1},\dots,\mathcal{G}_{s}$ are cross-dependent and

[TABLE]

[TABLE]

Note that (1.9) is an easy consequence of (1.8) while (1.10) follows from (1.9). Our main result states that the above construction is optimal.

Theorem 1.6**.**

Suppose that $\mathcal{F}_{1},\dots,\mathcal{F}_{s}\subset 2^{[n]}$ are cross-dependent, then

[TABLE]

In terms of the binomial norm (1.11) can be restated as

[TABLE]

In the case $\mathcal{F}=\mathcal{F}_{1}=\dots=\mathcal{F}_{s}$ the condition is that $\mathcal{F}$ contains no $s$ pairwise disjoint members. For $s=2$ it is equivalent to $\mathcal{F}$ being intersecting, that is, $F\cap F^{\prime}\neq\emptyset$ for all $F,F^{\prime}\in\mathcal{F}$. Erdős, Ko and Rado [EKR] noted that not both of $F$ and $[n]\setminus F$ can be members of an intersecting family. This implies

[TABLE]

Since ${n\choose\ell}={n\choose n-\ell}$, the same argument implies

[TABLE]

Theorem 1.6 implies the following more general statement.

Corollary 1.7**.**

Suppose that $\mathcal{F}\subset 2^{[n]}$ contains no $s$ pairwise disjoint members. Then

[TABLE]

One can see that (1.15) is best possible for all $n\geq s-1$. This is in great contrast to the problem of estimating $\max|\mathcal{F}|$ for families without $s$ pairwise disjoint members. Even though Kleitman [Kl] determined $\max|\mathcal{F}|$ for $n\equiv 0$ or $-1$ $(\text{\rm mod }s)$ more than 50 years ago, the general case appears to be hopelessly difficult (cf. discussion in later sections). In [FK1] the maximum of $|\mathcal{F}_{1}|+\dots+|\mathcal{F}_{s}|$ was determined if $\mathcal{F}_{1},\dots,\mathcal{F}_{k}\subset 2^{[n]}$ are cross-dependent (cf. Section 5).

For the proof of Theorem 1.6 we are going to need multi-layered inequalities. These are extensions of the following simple inequality proved by Kleitman.

Let us use the notation $\mathcal{F}_{i}^{(\ell)}=\mathcal{F}_{i}\cap{[n]\choose\ell}$ and \varphi_{i}(\ell)=\bigl{|}\mathcal{F}_{i}^{(\ell)}\bigr{|}\bigm{/}{n\choose\ell}, $0\leq\ell\leq n$.

Kleitman Inequality

([Kl]). Suppose that $\mathcal{F}_{1},\dots,\mathcal{F}_{s}\subset 2^{[n]}$ are cross-dependent. Let $k_{1},\dots,k_{s}$ be nonnegative integers satisfying $k_{1}+\dots+k_{s}=n$. Then

[TABLE]

The following is a common generalisation of (1.12) and (1.16).

Multi-layered inequality.

Suppose that $n\geq r\geq 0$ are integers, $\mathcal{F}_{1},\dots,\mathcal{F}_{s}\subset 2^{[n]}$ are cross-dependent. Let $k_{1},\dots,k_{s}$ be nonnegative integers satisfying $k_{1}+\dots+k_{s}+r=n$. Then

[TABLE]

To deduce (1.16) from (1.17), set $r=0$. To deduce (1.12), let $k_{1}=\dots=k_{s}=0$ and $r=n$.

We are going to prove Theorem 1.6 and (1.17) in the next section along with some further generalisations.

In Section 3 we prove the binomial norm version of Harper’s Theorem and the Katona Theorem on $t$-intersecting families. In Section 4 we explore the relation between $|\mathcal{F}|$ and $\|\mathcal{F}\|$. In Section 5 we deduce Kleitman’s bound from (1.15).

2 The proof of the main results

The proof is rather simple. In its core lies a simple probabilistic argument which we present first. Let $k_{1},k_{2},\dots,k_{s}$ and $r$ be nonnegative integers, $n=k_{1}+\dots+k_{s}+r$. Let $\mathcal{F}_{i}\subset 2^{[n]}$ for $1\leq i\leq s$.

Choose a permutation $\pi=(y_{1},\dots,y_{n})$ of $[n]$ uniformly at random. Define P_{1}=\bigl{\{}y_{1},\dots,y_{k_{1}}\bigr{\}},\dots,P_{s}=\bigl{\{}y_{k_{1}+\dots+k_{s-1}+1},\dots,y_{k_{1}+\dots+k_{s}}\bigr{\}} and R=\bigl{\{}y_{n-r+1},\dots,y_{n}\bigr{\}}. Define the map $\varrho:R\to[r]$ by $\varrho(y_{n-r+i})=i$, $1\leq i\leq r$, and $\varrho(G)=\{\varrho(y):y\in G\}$.

Let $\mathcal{F}_{1},\dots,\mathcal{F}_{s}\subset 2^{[n]}$. For $1\leq i\leq r$ let us define the family $\mathcal{G}_{i}=\mathcal{G}_{i}(\pi)\subset 2^{[r]}$ by

[TABLE]

Transfer Lemma**.**

• (i)

If $\mathcal{F}_{1},\dots,\mathcal{F}_{s}$ are cross-dependent, then $\mathcal{G}_{1},\dots,\mathcal{G}_{s}$ are cross-dependent as well.

• (ii)

The expected value $\gamma_{i}(j)$ of \bigl{|}\mathcal{G}_{i}^{(j)}\bigr{|}\bigm{/}{r\choose j} satisfies $\gamma_{i}(j)=\varphi_{i}(k_{i}+j)$.

Proof.

(i) Let $G_{i}\in\mathcal{G}_{i}$, $i=1,\dots,s$. Then \bigl{(}P_{i}\cup\varrho^{-1}(G_{i})\bigr{)}\in\mathcal{F}_{i}. Noting that $P_{1},\dots,P_{s}$ are pairwise disjoint and disjoint to $R$ as well, the cross-dependence of $\mathcal{F}_{1},\dots,\mathcal{F}_{s}$ implies the existence of $i\neq i^{\prime}$ with $\varrho^{-1}(G_{i})\cap\varrho^{-1}(G_{i^{\prime}})\neq\emptyset$. This in turn implies $G_{i}\cap G_{i^{\prime}}\neq\emptyset$ as desired.

(ii) For any fixed $0\leq j\leq r$ and $G\in{[r]\choose j}$ the $(k_{i}+j)$-element set $P_{i}\cup\varrho^{-1}(G)$ is a uniformly random subset of $[n]$. Thus the probability of $P_{i}\cup\varrho^{-1}(G)$ being in $\mathcal{F}_{i}$ is $\varphi_{i}(k_{i}+j)$. This proves (ii). ∎

Let us turn to the proof of (1.12) and (1.17). First note that for $r=0$ inequality (1.17) is the same as (1.11). Let us apply induction on $n$. For $n=0$ the statement amounts to saying that $\emptyset$ can be a member of at most $s-1$ of the $s$ families.

Assuming that (1.12) holds for $n$ replaced by $r$ let us prove (1.17) for all pairs $(r,n)$, $r\leq n$.

Let $\mathcal{F}_{1},\dots,\mathcal{F}_{s}\subset 2^{[n]}$ be cross-dependent and define the random permutation $\pi$ and the sets $P_{1},\dots,P_{s}$, $R$ and the families $\mathcal{G}_{i}\subset 2^{[r]}$ as above. Applying the induction hypothesis on (1.12) to $\mathcal{G}_{1},\dots,\mathcal{G}_{s}$ in view of the Transfer Lemma we obtain

[TABLE]

This proves (1.17).

Next we assume that (1.17) holds for $r=n$ and prove (1.12) for $n+1$. Assume that $\mathcal{F}_{i}\subset 2^{[n]}$, $i=1,\dots,s$, are cross-dependent. By symmetry we may assume that $\emptyset\notin\mathcal{F}_{s}$. Applying (1.17) with $k_{1}=\dots=k_{s-1}=0$, $k_{s}=1$, $r=n$ yields

[TABLE]

For $1\leq i\leq s-1$ we have

[TABLE]

For $i=s$ we know that $\varphi^{(s)}(0)=0$, i.e.,

[TABLE]

Comparing these with (2.1) yields

[TABLE]

For an integer $0\leq d\leq n$ and families $\mathcal{F}_{i}\subset 2^{[n]}$, $1\leq i\leq s$, we say that $\mathcal{F}_{1},\dots,\mathcal{F}_{s}$ are $d$-cross-dependent if there is no choice of pairwise disjoint $F_{1}\in\mathcal{F}_{1},\dots,F_{s}\in\mathcal{F}_{s}$ satisfying $|F_{1}\sqcup\ldots\sqcup F_{s}|\leq d$.

Theorem 2.2**.**

If $\mathcal{F}_{1},\dots,\mathcal{F}_{s}\subset 2^{[n]}$ are $d$-cross-dependent, then

[TABLE]

Note that for $d=n$ we get back the inequality (1.12).

Proof.

Define \mathcal{F}_{s+1}=\bigl{\{}F\subset[n]:|F|\geq n-d\bigr{\}}. Then $\|\mathcal{F}_{s+1}\|=d+1$.

Claim 2.3**.**

$\mathcal{F}_{1},\dots,\mathcal{F}_{s+1}$* are cross-dependent.*

Proof of the claim.

The opposite would mean the existence of pairwise disjoint sets $F_{1}\in\mathcal{F}_{1},\dots,F_{s+1}\in\mathcal{F}_{s+1}$. Using $|F_{1}\sqcup\ldots\sqcup F_{s+1}|\leq n$ and $|F_{s+1}|\geq n-d$ we infer $|F_{1}\sqcup\ldots\sqcup F_{s}|\leq d$, contradicting the $d$-cross-dependence property of $\mathcal{F}_{1},\dots,\mathcal{F}_{s}$. ∎

Now apply (1.12) to the $s+1$ families $\mathcal{F}_{1},\dots,\mathcal{F}_{s+1}$ to obtain $\|\mathcal{F}_{1}\|+\dots+\|\mathcal{F}_{s+1}\|\leq(n+1)s$. Subtracting $\|\mathcal{F}_{s+1}\|=d+1$ yields Theorem 2.2. ∎

Let us mention that analogous results were obtained in [FK1] for $|\mathcal{F}_{1}|+\dots+|\mathcal{F}_{s}|$ under the assumption of $d$-cross-dependence.

Let us conclude this section by another application of Theorem 1.6.

For families $\mathcal{F}_{1},\dots,\mathcal{F}_{s}\subset 2^{[n]}$ define the family $\mathcal{F}_{1}\square\ldots\square\mathcal{F}_{s}$ by\mathcal{F}_{1}\square\ldots\square\mathcal{F}_{s}=\bigl{\{}F\subset[n]: there exist pairwise disjoint F_{1}\in\mathcal{F}_{1},\dots,F_{s}\in\mathcal{F}_{s},\ F=F_{1}\sqcup\ldots\sqcup F_{s}\bigr{\}}.

In human language $\mathcal{F}_{1}\square\ldots\square\mathcal{F}_{s}$ consists of all the subsets that can be obtained as the disjoint union of $s$ members of the $\mathcal{F}_{i}$, one from each.

Theorem 2.4**.**

If $\mathcal{F}_{1},\dots,\mathcal{F}_{s}\subset 2^{[n]}$ and $\mathcal{F}_{1}$ is an up-set, then

[TABLE]

Proof.

Set \widetilde{\mathcal{F}}=\bigl{\{}[n]\setminus F:F\in\mathcal{F}_{1}\square\ldots\square\mathcal{F}_{s}\bigr{\}} and \mathcal{F}_{s+1}=2^{[n]}\bigm{/}\widetilde{\mathcal{F}}. Note that

[TABLE]

We claim that $\mathcal{F}_{1},\dots,\mathcal{F}_{s},\mathcal{F}_{s+1}$ are cross-dependent. Suppose the contrary and choose $F_{1}\in\mathcal{F}_{1},\dots,F_{s+1}\in\mathcal{F}_{s+1}$ that are pairwise disjoint. If $F_{1}\sqcup\ldots\sqcup F_{s+1}=[n]$, then $[n]\setminus F_{s+1}=F_{1}\square\ldots\square F_{s}$ which contradicts the definition of $\mathcal{F}_{s+1}$.

If the union is smaller, then we can use that $\mathcal{F}_{1}$ is an up-set and replace $F_{1}$ by its superset $[n]\setminus(F_{2}\sqcup\ldots\sqcup F_{s+1}]$ to obtain the same contradiction.

Now we may apply (1.12) to $\mathcal{F}_{1},\dots,\mathcal{F}_{s+1}$ and infer $\|\mathcal{F}_{1}\|+\dots+\|\mathcal{F}_{s+1}\|\leq s(n+1)$. Or equivalently,

[TABLE]

In view of (2.3) this is equivalent to (2.2). ∎

Let us show that there are many cases when equality holds in (2.2).

Example 2.5**.**

Let $t\geq 0$ and $T\in{[n]\choose t}$. Suppose that $t_{1},\dots,t_{s}$ are non-negative integers satisfying $t_{1}+\dots+t_{s}\leq t$. Define

[TABLE]

As we have shown before

[TABLE]

As to $\mathcal{F}_{1}\square\ldots\square\mathcal{F}_{s}$, it is equal to

[TABLE]

Thus $\varrho(\mathcal{F}_{1}\square\ldots\square\mathcal{F}_{s})=\varrho(\mathcal{F}_{1})+\dots+\varrho(\mathcal{F}_{s})-(s-1)$ showing that equality holds in (2.2).

3 The binomial Harper Theorem and Katona Theorem

For two sets $F$ and $G$ let $F+G$ denote their symmetric difference (Boolean sum), i.e., $F+G=(F\setminus G)\sqcup(G\setminus F)$. For a family $\mathcal{F}\subset 2^{[n]}$ let $\partial\mathcal{F}$ denote its outer boundary:

[TABLE]

The usual Harper Theorem ([H], cf. [FF] or [F] for a simple proof) determines $\min|\partial\mathcal{F}|$ as a function of $|\mathcal{F}|$. For the binomial norm we have a much simpler result.

Theorem 3.1**.**

Suppose that $\emptyset\in\mathcal{F}$, $[n]\notin\mathcal{F}$. Then

[TABLE]

where the inequality is strict unless $\mathcal{F}=\{F\subset[n]:\,|F|\leq j\}$ for some $0\leq j<n$.

Proof.

Consider an arbitrary full chain $\emptyset=G_{0}\subset\dots\subset G_{n}=[n]$ where $|G_{i}|=i$. In view of $\emptyset\in\mathcal{F}$ and $[n]\notin\mathcal{F}$ one can find a $j$, $0\leq j<n$ such that $G_{j}\in\mathcal{F}$, $G_{j+1}\notin\mathcal{F}$. By definition, $G_{j+1}\in\partial\mathcal{F}$. Thus we proved that in all $n!$ full chains there is at least one member $G$ of $\partial\mathcal{F}$. Since the same $G$ occurs in $|G|!(n-|G|)!$ full chains

[TABLE]

Dividing both sides by $n!$ we infer

[TABLE]

Suppose that equality holds. By the above argument every full chain contains exactly one member of $\partial\mathcal{F}$. That is, $\partial\mathcal{F}$ is an antichain. Now equality in (3.1) implies that $\partial\mathcal{F}={[n]\choose j+1}$ for some $0\leq j<n$. This in turn easily entails $\mathcal{F}=\{F\subset[n]:\,|F|\leq j\}$. ∎

Remark 3.2**.**

One can conclude the proof without using the uniqueness for maximal antichains in the LYM inequality. Let us simply use that exactly one member of each full chain is in $\partial\mathcal{F}$. Let $G$ be such a set $|G|=j+1$. This implies that every $j$-subset $F$, $F\subset G$ must be in $\mathcal{F}$. For such an $F$ choose arbitrarily $H$, $F\subset H\subset[n]$, $|H|=j+1$. We claim that $H\notin\mathcal{F}$. Indeed, otherwise we could continue with the chain $F\subset H\subset G\cup H\subset\ldots$ and find a new member of $\partial\mathcal{F}$ containing $G$, a contradiction. Iterating we infer by connectivity of the containment graph between ${[n]\choose j+1}$ and ${[n]\choose j}$ that $\partial\mathcal{F}={[n]\choose j+1}$.

For a nonnegative integer $w<n$ we say that $\mathcal{F}\subset 2^{[n]}$ is $w$-union if $|F\cup F^{\prime}|\leq w$ for all $F,F^{\prime}\in\mathcal{F}$. Let us recall the following classical result.

Katona Theorem** ([Ka]).**

Let $0\leq w<n$ and suppose that $\mathcal{F}\subset 2^{[n]}$ is $w$-union. Then (i) or (ii) hold.

(i)* $w=2\ell$ and*

[TABLE]

(ii)* $w=2\ell-1$ and*

[TABLE]

The original proof of Katona gives the following binomial version.

Binomial Katona Theorem**.**

Let $0\leq w<n$ and suppose that $\mathcal{F}\subset 2^{[n]}$ is $w$-union. Then (i) or (ii) hold.

(i)* $w=2\ell$ and*

[TABLE]

(ii)* $w=2\ell-1$ and*

[TABLE]

Proof.

Let $f_{i}$ denote the number of $i$-sets in $\mathcal{F}$. The $w$-union condition implies $f_{i}=0$ for $i>w$. Katona [Ka] proves the inequalities

[TABLE]

For the case $w=2\ell-1$, $i=\ell$, the Erdős–Ko–Rado Theorem [EKR] yields

[TABLE]

Noting that ${n\choose i}<{n\choose w+1-i}$ for $1\leq i<\frac{w+1}{2}$, $\varphi(i)+\varphi(w+1-\ell)\leq 2$ follows. Summing these along with $\varphi(0)\leq 1$ gives (3.4) and (3.5). ∎

As one can guess from the easy proof, the inequalities (3.4) and (3.5) are considerably weaker than the bounds on $|\mathcal{F}|$.

4 Size versus binomial norm

Throughout this section let $\mathcal{F}\subset 2^{[n]}$ be a complex (down-set), that is, $E\subset F\in\mathcal{F}$ implies $E\in\mathcal{F}$.

The relation between $|\mathcal{F}|$ and $\|\mathcal{F}\|$ is very intricate even for complexes. However in certain cases one can deduce best possible lower bounds on $|\mathcal{F}|$ in terms of $\|\mathcal{F}\|$.

Let $\ell$ be an integer. Define ${[n]\choose<\ell}=\{E\subset[n],|E|<\ell\}$. This is a complex and it will remain a complex even if we add some $\ell$-element sets. Therefore the best we can hope for in terms of a relation between $\|\mathcal{F}\|$ and $|\mathcal{F}|$ is the following.

[TABLE]

Definition 4.1**.**

Fix a pair $(n,\ell)$, $n>\ell\geq 0$. If (4.1) holds for all $\mathcal{F}$ with $\|\mathcal{F}\|<n+1$, then we say that $(n,\ell)$ is perfect. If (4.1) holds for all $\mathcal{F}$ with $\|\mathcal{F}\|<\frac{n+1}{2}$, then we say that $(n,\ell)$ is quasi-perfect.

The main result of the present section is the following.

Proposition 4.2**.**

If $n\geq 3\ell$, then $(n,\ell)$ is quasi-perfect. Moreover, unless $n=3\ell$ and $2\leq\ell\leq 5$ it is perfect.

Recalling the fact $\|2^{[n-1]}\|_{n}=\frac{n+1}{2}$ one observes that in proving quasi-perfectness we need to consider only complexes without $(n-1)$-element sets $(\varphi(n-1)=0)$. As $\|\mathcal{F}\|<n+1$ in Definition 4.1, we are going to assume always that $\varphi(n)=0$.

The proof relies on the following inequality

Lemma 4.3**.**

Suppose that $n,k,\ell$ are positive integers, $n-1>k>\ell$, $n\geq 3\ell$. Then

[TABLE]

Moreover, except for $n=3\ell$, $2\leq\ell\leq 5$ (4.2) holds for $k=n-1$ as well.

Let us postpone the proof of (4.2) and prove Proposition 4.2 first.

Recall the definition of \varphi(i)=\left|\mathcal{F}\cap{[n]\choose i}\right|\Bigm{/}{n\choose i} and the monotonicity $\varphi(n)\leq\varphi(n-1)\leq\dots\leq\varphi(0)$ proved by Sperner [S]. This permits to define $\alpha(j)=\varphi(j)-\varphi(j+1)$ for $0\leq j\leq n-1$, where $\alpha(j)$ are nonnegative and by definition

[TABLE]

In the exceptional cases $n=3\ell$, $2\leq\ell\leq 5$, $\alpha(n-1)=0$ follows from $\varphi(n-1)=0$, as well. To prove all cases simultaneously, set $q=n-2$ in the exceptional cases (i) and $q=n-1$ for the rest. In view of (4.3) we have

[TABLE]

Define $\mathcal{F}^{+}$ $(\mathcal{F}^{-})$ where a member $F\in\mathcal{F}$ is in $\mathcal{F}^{+}$ $(\mathcal{F}^{-})$ iff $|F|\geq\ell$ $(|F|<\ell)$, respectively. Obviously $\mathcal{F}=\mathcal{F}^{+}\sqcup\mathcal{F}^{-}$ is a partition. Multiplying both sides of (4.2) by $\alpha(k)$ we obtain:

[TABLE]

On the other hand (4.4) implies

[TABLE]

Comparing the RHS with (4.5) we infer

[TABLE]

For $\mathcal{F}^{-}$ define the nonnegative reals $\delta(i)=1-\varphi(i)$, $0\leq i<\ell$ and note that

[TABLE]

In view of ${n\choose\ell}>{n\choose i}$ for all $0\leq i<\ell$,

[TABLE]

Using $|\mathcal{F}|=|\mathcal{F}^{+}|+|\mathcal{F}^{-}|$ and $\|\mathcal{F}^{+}\|=\|\mathcal{F}\|-\|\mathcal{F}^{-}\|$, from (4.6) and (4.8) we derive

[TABLE]

Proof of (4.2).

Let us first note that (4.2) trivially holds for both $\ell=0$ and $\ell=1$.

For the case $\ell\geq 2$ we need the next formula:

[TABLE]

Let us prove

Claim 4.4**.**

For cases (i) $\sim$ (iii) one has

[TABLE]

(i)*  $n\geq 3\ell+2$,  $\ell\geq 2$,*

(ii)*  $n\geq 3\ell+1$,  $\ell\geq 4$,*

(iii)*  $n\geq 3\ell$,       $\ell\geq 6$.*

To prove (4.10) note that for fixed $\ell$ the LHS is an increasing function of $n$. Therefore it is sufficient to check the cases $n=3\ell+a$, $a=2,1,0$.

For $n=3\ell+2$, $n+1=3(\ell+1)$ and $\frac{n-\ell}{\ell+2}=2-\frac{2}{\ell+2}\geq\frac{3}{2}$ for $\ell\geq 2$. These prove (4.10).

For $n=3\ell+1$ after rearranging (4.10) is equivalent to

[TABLE]

which is true for $\ell\geq 4$.

For $n=3\ell$, (4.10) is equivalent to $2\ell^{2}-10\ell-8>0$ which is true for $\ell\geq 6$. ∎

The proof of Lemma 4.3.

First note that ${n\choose\ell}<{n\choose j}$ for $\ell<j<n-\ell$. Therefore (4.2) is obvious for $k\leq n-\ell$. This takes care of the cases $\ell=0$ and $\ell=1$.

Let us first consider the case $n\geq 3\ell+1$, $\ell\geq 4$. The terms ${n\choose j}$ that exceed ${n\choose\ell}$ are ${n\choose j}$ with $\ell+1\leq j\leq n-\ell-1$. This gives $n-2\ell-1\geq\ell$ choices for $j$. Noting that ${n\choose\ell+1}$ and ${n\choose\ell+2}$ are the smallest, (4.9) and (4.10) imply

[TABLE]

Adding ${n\choose\ell}$ and ${n\choose n-\ell}$ yields

[TABLE]

In the case $\ell=3$, (4.10) is valid for $n\geq 11$. Therefore the only remaining case is $n=10$, $\ell=3$. Then ${10\choose 4}+{10\choose 5}+{10\choose 6}=5.6{10\choose 3}$ which is less than $6{10\choose 3}$. However ${10\choose 8}+{10\choose 9}=55>0.4{10\choose 3}$, proving (4.2) for $k=8$ and $9$. Now let us consider the case $n=3\ell$. By Claim 4.4, (4.10) is true for $\ell\geq 6$. Consequently the above proof shows that $(3\ell,\ell)$ is perfect for $\ell\geq 6$.

Let $\ell=2$. For $n=7$ we have

[TABLE]

For $n\geq 8$ we can use Claim 4.4 (i) to conclude the proof.

The only remaining cases are $\ell=3,4,5$. The following equalities conclude the proof for $\ell=3,4$.

[TABLE]

For $\ell=5$ using (4.9) yields

[TABLE]

Let us mention that for every positive $\varepsilon$ and $\ell>\ell_{0}(\varepsilon)$ one can show that $(n,\ell)$ is perfect for $n>(2+\varepsilon)\ell$.

5 Applications and outlook

Let us first use (1.15) and Proposition 4.2 to deduce the following important result of Kleitman.

Kleitman Theorem** ([Kl]).**

Let $n\geq s\geq 3$. Suppose that $\mathcal{K}\subset 2^{[n]}$ contains no $s$ pairwise disjoint members. Then (i) and (ii) hold.

(i)* If $n=sk-1$, then*

[TABLE]

(ii)* If $n=sk$, then*

[TABLE]

Proof.

Note that if $\mathcal{K}$ contains no $s$ pairwise disjoint members, then the up-set $\mathcal{K}^{*}=\{H\subset[n]:\exists K\in\mathcal{K},K\subset H\}$ has the same property as well. Hence when proving (i) and (ii) we may assume that $\mathcal{K}$ itself is an up-set. Define $\mathcal{F}=2^{[n]}\setminus\mathcal{K}$. Since $\mathcal{K}$ is an up-set, $\mathcal{F}$ is a complex. In view of (1.15), $\|\mathcal{K}\|\leq\frac{(s-1)(n+1)}{s}$. Consequently, $\|\mathcal{F}\|\geq\frac{n+1}{s}$.

In the case $n=sk-1$ we have $\|\mathcal{F}\|\geq k$. Applying Proposition 4.2 with $\ell=k-1$ we infer

[TABLE]

In the case $n=sk$ we have $\|\mathcal{F}\|\geq k+\frac{1}{s}$.

Applying Proposition 4.2 with $\ell=k$ yields

[TABLE]

What happens in the case $n\not\equiv 0$ or $-1$ $(\text{\rm mod }s)$? As mentioned in the introduction to find the exact value of $\max|\mathcal{F}|$ in general appears to be very difficult. The case of $n\equiv-2$ $(\text{\rm mod }s)$ was solved by Quinn [Q] for $s=3$ and recently by Kupavskii and the author (cf. [FK2], [FK3]) for all $s\geq 4$.

However for the remaining congruence classes we are still in the search of the right conjectures.

Many of the constructions fail to be of the form $\{$all subsets of size less than $k\}$ plus some $k$-element sets. This suggests that (1.15) in itself is not sufficiently strong to establish their optimality.

In a sense this suggests that for cross-dependence and for families without $s$ pairwise disjoint members the binomial norm is the right setting.

As another application let us give a short proof for a recent result of Kupavskii and the author.

Theorem 5.1** ([FK1]).**

Suppose that $n=s\ell+r$ where $0\leq r<s$, $s\geq 3$. Let $\mathcal{F}_{1},\dots,\mathcal{F}_{s}\subset 2^{[n]}$ be cross-dependent. Then

[TABLE]

Considering the families $\mathcal{H}_{1}=\dots=\mathcal{H}_{r+1}=\{H\subset[n]:|H|>\ell\}$ and $\mathcal{H}_{r+2}=\dots=\mathcal{H}_{s}=\{H\subset[n]:|H|\geq\ell\}$ shows that (5.3) is best possible.

Proof of (5.3).

Setting $\mathcal{G}_{i}=2^{[n]}\setminus\mathcal{F}_{i}$, $1\leq i\leq s$, (5.3) is equivalent to

[TABLE]

Since $\|\mathcal{G}_{i}\|=(n+1)-\|\mathcal{F}_{i}\|$, (1.12) yields

[TABLE]

Supposing that $(n,\ell)$ is perfect, using (4.1) we infer

[TABLE]

proving (5.3).

For $s\geq 4$ or for $s=3$ and $r>0$, $n\geq 3\ell+1$ and the perfectness of the pair $(n,\ell)$ follows from Proposition 4.2. The only remaining cases are $s=3$, $r=0$, $n=3\ell$. By the same proposition $(3\ell,\ell)$ is perfect, unless $2\leq\ell\leq 5$. For these four cases one can use quasi-perfectness to conclude the proof. ∎