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 $\mathcal{A}(F)$?
 A: 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.
A: 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 $\mathcal{A}(F)$. 
A: 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
