# A strengthening of Frankl's union-closed conjecture?

Frankl's union-closed conjecture states that if $F$ is a finite union-closed family of sets (i.e. a family that is closed under taking unions), then there must be an element that belongs to at least half the sets.

1) Does anyone know an example of a finite union-closed family $F$ such that the set $\mathcal{A}(F)$ of elements that belong to at least half the sets of $F$ is not a member of $F$? [edit: this is answered below by Thomas Bloom]

2) Does anyone know an example where no member of $F$ is a subset of $\mathcal{A}(F)$?

The simplest example I found is the family

$$\emptyset, \{1\}, \{1,2,3\}, \{1,2,4\}, \{1,2,3,4\}$$

This is a union-closed family with 5 elements. 1 appears in 4, 2 appears in 3, 3 and 4 both appear in 2, but $\{1,2\}$ is not in the family.

• Thanks a lot! I was too optimistic! Yet, I'm wondering if there is an example of a finite union-closed family $F$ such that the set of elements that belong to at least half the sets of $F$ does not contain any member of $F$? – MassiveJack Apr 7 '14 at 14:35

As for question 2), I just found the following example:

Let $\mathcal{P}$ be the family of all non-empty subsets of $\{1,2,3,4\}$, and $\mathcal{F}$ be the family whose elements are all the sets $P\cup\{5,6,7,8,9\}$ with $P\in\mathcal{P}$ and the 4 sets

$$\{1,5,6,7,8\},\{2,5,6,7,9\},\{3,5,6,8,9\},\{4,5,7,8,9\}.$$

This union-closed family has 19 members. 1,2,3 and 4 belong to 9 sets, 6,7,8,9 to 18 sets and 5 belongs to 19 sets. Thus $\mathcal{A}(F)=\{5,6,7,8,9\}$ and no member of the family is a subset of $\mathcal{A}(F)$.

What if you take $F$ to be all subsets of size $>k$ of $[n]$, add to it the sets $\{1...k-1\}$ and $\{2...k\}$ ? This family $F$ contains all sets of size $>k$ and two sets of size exactly $k-1$ (thus it is union-closed), and I would say that the most frequent elements are $\{2,...,k-1\}$, i.e. a set of size $k-2$ that does not belong to $F$.

Nathann

• I think you mean $\ge k$. If so, then for $k=3, n=4$ this is not an example. $4 \in \lbrace 1,2,3,4 \rbrace,\lbrace 1,2,4 \rbrace, \lbrace 1,3,4\rbrace, \lbrace 2,3,4\rbrace$ so $4$ is in $4$ out of $7$ sets. $1$ and $3$ are in $5$ out of $7$. $2$ is in $6$ out of $7$. – Douglas Zare Apr 7 '14 at 11:25
• I ran a couple of tests in Sage, which convinced me that the idea does not work :-) – Nathann Cohen Apr 7 '14 at 11:37
• (I mean that the most frequent elements are those I want and they do not form a set of $F$, but all other elements will all appear more than 50%) – Nathann Cohen Apr 7 '14 at 11:40
• For other parameters, every element is in the set, since there is an injection from sets not containing $x$ to sets containing $x$ by adding $x$. – Douglas Zare Apr 7 '14 at 11:57