A Note on the Frankl Conjecture

TL;DR

This paper explores variants of the Frankl conjecture, a well-known unsolved problem in combinatorics, by introducing and partially justifying new conjectural forms.

Contribution

It proposes and provides initial justification for several variants of the Frankl conjecture, expanding the scope of potential approaches.

Findings

Introduction of new variants of the Frankl conjecture

Partial theoretical justification for these variants

Potential implications for future research in combinatorics

Abstract

The Frankl conjecture (called also union-closed sets conjecture) is one of the famous unsolved conjectures in combinatorics of finite sets. In this short note, we introduce and to some extent justify some variants of the Frankl conjecture.

Equations

$$\mathcal{B}:=\{A\in \mathcal{A}:|A\cap S|\ge \ell \},$$

$${\mathcal{B}}^{\prime }:=\{A\in \mathcal{A}:\ell >|A\cap S|\ge 1\}.$$

$${\mathcal{A}}^{\prime }:=\{A\in \mathcal{A}:A\cap S=\varnothing \}.$$

$$|\mathcal{A}|=|\mathcal{B}|+|{\mathcal{B}}^{\prime }|+|{\mathcal{A}}^{\prime }|.$$

$$\begin{array}{rl}p& \ge |\mathcal{B}|+{\mathfrak{c}}_{k^{\prime }|\ell }|{\mathcal{A}}^{\prime }|(1+|{\mathcal{B}}^{\prime }|)\\ & =|\mathcal{B}|+{\mathfrak{c}}_{k^{\prime }|\ell }(|\mathcal{A}|-|\mathcal{B}|-|{\mathcal{B}}^{\prime }|)(1+|{\mathcal{B}}^{\prime }|)\\ & =|\mathcal{B}|+{\mathfrak{c}}_{k^{\prime }|\ell }|\mathcal{A}|-{\mathfrak{c}}_{k^{\prime }|\ell }|\mathcal{B}|+{\mathfrak{c}}_{k^{\prime }|\ell }(|\mathcal{A}|-|\mathcal{B}|-|{\mathcal{B}}^{\prime }|-1)\\ & \ge |\mathcal{B}|+{\mathfrak{c}}_{k^{\prime }|\ell }|\mathcal{A}|-{\mathfrak{c}}_{k^{\prime }|\ell }|\mathcal{B}|\\ & ={\mathfrak{c}}_{k|\ell }|\mathcal{A}|+r+{\mathfrak{c}}_{k^{\prime }|\ell }|\mathcal{A}|-{\mathfrak{c}}_{k^{\prime }|\ell }{\mathfrak{c}}_{k|\ell }|\mathcal{A}|-r{\mathfrak{c}}_{k^{\prime }|\ell }\\ & =({\mathfrak{c}}_{k|\ell }+{\mathfrak{c}}_{k^{\prime }|\ell }-{\mathfrak{c}}_{k|\ell }{\mathfrak{c}}_{k^{\prime }|\ell })|\mathcal{A}|+(1-{\mathfrak{c}}_{k^{\prime }|\ell })r\\ & \ge ({\mathfrak{c}}_{k|\ell }+{\mathfrak{c}}_{k^{\prime }|\ell }-{\mathfrak{c}}_{k|\ell }{\mathfrak{c}}_{k^{\prime }|\ell })|\mathcal{A}|\end{array}$$

$${\mathcal{A}}^{\prime \prime }:=\{A\setminus S:A\in \mathcal{A}\text{and}S\subseteq A\}\cup \{\varnothing \}.$$

$$\Bigg{[}2^{-(k+k^{\prime})}\sum_{i=\ell}^{k+k^{\prime}}\binom{k+k^{\prime}}{i}\Bigg{]}\geq\Bigg{[}2^{-k}\sum_{j=\ell}^{k}\binom{k}{j}\Bigg{]}+\Bigg{[}2^{-k^{\prime}}\sum_{j^{\prime}=\ell}^{k^{\prime}}\binom{k^{\prime}}{j^{\prime}}\Bigg{]}-\Bigg{[}2^{-k}\sum_{j=\ell}^{k}\binom{k}{j}\Bigg{]}\Bigg{[}2^{-k^{\prime}}\sum_{j^{\prime}=\ell}^{k^{\prime}}\binom{k^{\prime}}{j^{\prime}}\Bigg{]}$$

$$\Bigg{[}2^{-(k^{\prime\prime}+k)}\sum_{i=\ell}^{k^{\prime\prime}+k}\binom{k^{\prime\prime}+k}{i}\Bigg{]}\geq\Bigg{[}2^{-k^{\prime\prime}}\Bigg{]}\Bigg{[}2^{-k}\sum_{i=\ell}^{k}\binom{k}{i}\Bigg{]}$$

Full text

MSC 2010. Primary: 05D05. Secondary: 05A20.

Keywords. Union-closed sets conjecture, extremal set theory, combinatorics of finite sets, binomial expansion, combinatorics inequality.

Let $X$ be a finite nonempty set and let $\mathcal{P}(X)$ denote the powerset of $X$. By a union-closed family on $X$ we mean a set $\mathcal{A}\subseteq\mathcal{P}(X)$ such that $\emptyset,X\in\mathcal{A}$, and such that if $A_{1},A_{2}\in\mathcal{A}$ then $A_{1}\cup A_{2}\in\mathcal{A}$. Frankl’s conjecture (FC for short), that is also called union-closed sets conjecture, is as follows.

For history of FC, some basic related results, and some equivalent formulations of the conjecture in terms of graphs and lattices, we refer the reader to [1, 2] and references therein. In this short note, we introduce some natural variants of FC.

Let us begin by some notations and definitions. For a set $S$, we denote by $|S|$ the cardinal of $S$. If $\mathcal{A}$ is a union-closed family on $X$, we denote $X$ by the symbol $\cup\mathcal{A}$. The class of all union-closed families on finite nonempty sets is denoted by $\mathfrak{A}$.

By definition, any union-closed family is $1|1$-separated; thus $\mathfrak{A}_{1|1}=\mathfrak{A}$. Also, $\mathfrak{A}_{k|k}$ is the class of all union-closed families $\mathcal{A}$ with $|\cup\mathcal{A}|\geq k$. As a trivial example, for every finite set $X$ with $|X|\geq k$, any union-closed family on $X$, containing the union-closed family $\{A\subseteq X:|A|\geq\ell\}\cup\{\emptyset\}$, is $k|\ell$-separated.

For every two natural numbers $k$ and $\ell$ as above, let $\mathfrak{c}_{k|\ell}$ be the supremum of all positive real numbers $\epsilon$ satisfying the following condition: For every $\mathcal{A}\in\mathfrak{A}_{k|\ell}$, there exists a set $S\subseteq\cup\mathcal{A}$ with $|S|=k$ such that there are at least $\epsilon|\mathcal{A}|$ members $A$ of $\mathcal{A}$ satisfying $|S\cap A|\geq\ell$. If there is no such an $\epsilon$, then $\mathfrak{c}_{k|\ell}$ is defined to be [math]. Note that the ‘supremum’ in the definition of $\mathfrak{c}_{k|\ell}$ is actually a ‘maximum’.

1. (1)

$0\leq\mathfrak{c}_{k|\ell}\leq 2^{-k}\sum_{i=\ell}^{k}\binom{k}{i}$** 2. (2)

$\mathfrak{c}_{k+k^{\prime}|\ell}\geq\mathfrak{c}_{k|\ell}+\mathfrak{c}_{k^{\prime}|\ell}-\mathfrak{c}_{k|\ell}\mathfrak{c}_{k^{\prime}|\ell}$** 3. (3)

$\mathfrak{c}_{k+k^{\prime\prime}|\ell+k^{\prime\prime}}\geq\mathfrak{c}_{k|\ell}\mathfrak{c}_{k^{\prime\prime}|k^{\prime\prime}}$** 4. (4)

$\mathfrak{c}_{k+k^{\prime}|k+k^{\prime}}\geq\mathfrak{c}_{k|k}\mathfrak{c}_{k^{\prime}|k^{\prime}}$**

(2) Let $\mathcal{A}\in\mathfrak{A}_{k+k^{\prime}|\ell}$. Then, it is easily checked that $\mathcal{A}$ also belongs to $\mathfrak{A}_{k|\ell}$. Thus, there exists $S\subset\cup\mathcal{A}$ with $|S|=k$, such that if the family $\mathcal{B}$ is defined by

[TABLE]

then $|\mathcal{B}|\geq\mathfrak{c}_{k|\ell}|\mathcal{A}|$. Let the real number $r\geq 0$ be such that $|\mathcal{B}|=\mathfrak{c}_{k|\ell}|\mathcal{A}|+r$. Suppose that the family $\mathcal{B}^{\prime}$ is defined to be the empty family if $\ell=1$, and otherwise,

[TABLE]

Also, let

[TABLE]

It is clear that $\mathcal{A}$ coincides with the disjoint union $\mathcal{B}\cup\mathcal{B}^{\prime}\cup\mathcal{A}^{\prime}$. Thus,

[TABLE]

On the other hand, it is easily checked that $\mathcal{A}^{\prime}$ is a union-closed family on the set $\cup\mathcal{A}\setminus S$ and $\mathcal{A}^{\prime}\in\mathfrak{A}_{k^{\prime}|\ell}$. Thus, there exists $S^{\prime}\subseteq\cup\mathcal{A}\setminus S$ with $|S^{\prime}|=k^{\prime}$ such that for at least $\mathfrak{c}_{k^{\prime}|\ell}|\mathcal{A}^{\prime}|$ members $A^{\prime}$ of $\mathcal{A}^{\prime}$, $|A^{\prime}\cap S^{\prime}|\geq\ell$. Let $p$ be the number of those members $A$ of $\mathcal{A}$ with the property $|A\cap(S\cup S^{\prime})|\geq\ell$. Then,

[TABLE]

The above inequality together with $|S\cup S^{\prime}|=k+k^{\prime}$ show that (2) is satisfied.

(3) Let $\mathcal{A}\in\mathfrak{A}_{k+k^{\prime\prime}|\ell+k^{\prime\prime}}$. Then, $\mathcal{A}\in\mathfrak{A}_{k^{\prime\prime}|k^{\prime\prime}}$, and there exists a set $S\subset\cup\mathcal{A}$ with $|S|=k^{\prime\prime}$ such that $S$ is contained in at least $\mathfrak{c}_{k^{\prime\prime}|k^{\prime\prime}}|\mathcal{A}|$ members of $\mathcal{A}$. Let

[TABLE]

Therefore, we have $|\mathcal{A}^{\prime\prime}|\geq\mathfrak{c}_{k^{\prime\prime}|k^{\prime\prime}}|\mathcal{A}|$. On the other hand, it is easily seen that $\mathcal{A}^{\prime\prime}\in\mathfrak{A}_{k|\ell}$. Thus, there exists a set $S^{\prime\prime}\subseteq\cup\mathcal{A}^{\prime\prime}$ with $|S^{\prime\prime}|=k$ such that there are at least $\mathfrak{c}_{k|\ell}$ members $A^{\prime\prime}$ of $\mathcal{A}^{\prime\prime}$ satisfying $|A^{\prime\prime}\cap S^{\prime\prime}|\geq\ell$. Therefore, $S\cup S^{\prime\prime}$ has $k+k^{\prime\prime}$ elements and there exists at least $\mathfrak{c}_{k|\ell}\mathfrak{c}_{k^{\prime\prime}|k^{\prime\prime}}|\mathcal{A}|$ members $A$ of $\mathcal{A}$ satisfying $|A\cap(S\cup S^{\prime\prime})|\geq\ell+k^{\prime\prime}$. This implies that (3) is satisfied.

(4) follows directly from (3). ∎

Now, we introduce our variant of the Frankl conjecture.

It is clear that FC of order $1|1$, is the original Frankl conjecture. We have suggested Conjecture 4, from inequality (1) in Theorem 3, and special cases of the conjecture of orders $k|1$ and $k|k$.

1. (1)

Validity of FC of orders $k|1$ and $k^{\prime}|1$ imply validity of FC of order $k+k^{\prime}|1$. In particular, validity of the Frankl conjecture implies validity of FC of order $k|1$. 2. (2)

Validity of FC of orders $k|k$ and $k^{\prime}|k^{\prime}$ imply validity of FC of order $k+k^{\prime}|k+k^{\prime}$. In particular, validity of the Frankl conjecture implies validity of FC of order $k|k$.

It is easily seen that the following two inequalities are satisfied:

[TABLE]

[TABLE]

Hence, Conjecture 4 is compatible with inequalities (2) and (3) of Theorem 3. This is another justification for Conjecture 4.

We suggest that FC of order $k|\ell$ is also satisfied in a bigger class of union-closed families than $\mathfrak{A}_{k|\ell}$: Let $\mathcal{A}$ be a union-closed family such that $|\cup\mathcal{A}|\geq k$. We say that $\mathcal{A}$ is weakly $k|\ell$-separated if the following condition is satisfied: For any $k$ distinct elements $x_{1},\ldots,x_{k}\in\cup\mathcal{A}$, there exists $A\in\mathcal{A}$ such that $\{x_{1},\ldots,x_{\ell}\}\cap A\neq\emptyset$ and $\{x_{\ell+1},\ldots,x_{k}\}\cap A=\emptyset$. we denote the class of all weakly $k|\ell$-separated union-closed families by $\widetilde{\mathfrak{A}}_{k|\ell}$. It is clear that $\mathfrak{A}_{k|\ell}\subset\widetilde{\mathfrak{A}}_{k|\ell}$. We let the real number $\widetilde{\mathfrak{c}}_{k|\ell}\geq 0$ be exactly defined as $\mathfrak{c}_{k|\ell}$ except that this time the supremum being taken over $\widetilde{\mathfrak{A}}_{k|\ell}$. Obviously, we have $\widetilde{\mathfrak{c}}_{k|\ell}\leq\mathfrak{c}_{k|\ell}$.

Note that the statements and proofs of Theorems 3 and 5 hold if the symbols $\mathfrak{c},\mathfrak{A}$ are replaced by $\widetilde{\mathfrak{c}},\widetilde{\mathfrak{A}}$, and the phrase ‘FC of order’ by ‘strong FC of order’. Since $\widetilde{\mathfrak{A}}_{k|1}=\mathfrak{A}_{k|1}$ (resp. $\widetilde{\mathfrak{A}}_{k|k}=\mathfrak{A}_{k|k}$), strong FC and FC of order $k|1$ (resp. $k|k$) coincide.