TITLE: Difficult examples for Frankl's union-closed conjecture
QUESTION [46 upvotes]: Frankl's well-known union-closed conjecture states that if F is a finite family of sets that is closed under taking unions (that is, if A and B belong to the family then so does $A\cup B$), then there must be an element that belongs to at least half the sets.
I know that pretty well any naive approach one takes to this conjecture is known to fail. By "naive approach" I suppose I mean something like an observation that it would follow from such-and-such a stronger conjecture -- it seems that all sensible stronger conjectures one thinks of are false. A very simple example of a stronger conjecture would be that if you pick a random element then on average it will belong to at least half the sets. That is completely false: take the family that consists of the empty set, {1}, and {1,2,3,4,5,6,7,8,9,10}, for example. One can try to "correct" this strengthening by devices such as insisting that for any two elements there is a set that contains one and not the other (which WLOG is the case), but such corrections don't get one very far.
What I am asking for is examples, either small ones or ones that are constructed theoretically, of union-closed families that defeat more sophisticated strengthenings of the original conjecture. I'm fairly sure they are out there but I am not an expert on this problem so I don't know them myself.
Apologies in advance if this resembles an existing question (which it feels as thought it easily might). But I've looked and not found anything.

REPLY [16 votes]: Here is a nice example due to Bjorn Poonen, which I have taken from this survey paper of Bruhn and Schaudt.  It is motivated by the following observations.  Let $\mathcal{A}$ be a union-closed family. Call an element $x$ abundant if $x$ is contained in at least half of the members of $\mathcal{A}$.
Observation. If $\mathcal{A}$ contains a singleton $\{x\}$, then $x$ is abundant. 
Proof. Partition $\mathcal{A}$ as $\mathcal{A}_x$ and $\overline{A}_x$ where $\mathcal{A}_x$ consists of the members of $\mathcal{A}$ containing $x$ and $\overline{A}_x$ are the members of $\mathcal{A}$ not containing $x$.  Since $\{x\} \in \mathcal{A}$ and $\mathcal{A}$ is union-closed, the map $S \mapsto S \cup \{x\}$ is an injection from $\overline{A}_x$ to $\mathcal{A}_x$.  Thus, $|\overline{A}_x| \leq |\mathcal{A}_x|$, as required. $\square$
Similarly, we also have the following.
Observation. If $\mathcal{A}$ contains a $2$-set $\{x,y\}$, then $x$ or $y$ is abundant.
This might lead one to conjecture the following false strengthening of Frankl's Union-closed Conjecture.
False Strengthening. Let $S$ be a non-empty member of $\mathcal{A}$ with the smallest number of elements.  Then $S$ contains an abundant element. 
Disproof. For two families $\mathcal{A}$ and $\mathcal{B}$ we let 
$$
\mathcal{A} \uplus \mathcal{B}=\{S \cup T: S \in \mathcal{A}, T \in \mathcal{B}\}.
$$
Fix $k \geq 3$ and for each $i \in [k]$ define
$$
\mathcal{B}^i=\{[k+1,3k] \setminus \{2i+k-1\}, [k+1,3k] \setminus \{2i+k\}\}.
$$
Finally, define
$$
\mathcal{A}^k=\{[k]\} \cup \bigcup_{i=1}^k(\{\emptyset, \{i\}, [k]\} \uplus \mathcal{B}^i) \cup (2^{[k]} \uplus [k+1,3k]).
$$
It is easy to check that $\mathcal{A}^k$ is union-closed and that $[k]$ is the unique smallest set in $\mathcal{A}^k$.  However, $|\mathcal{A}^k|=1+6k+2^k$ and each $i \in [k]$ is contained in only $1+(2k+2)+2^{k-1}$ members of $\mathcal{A}^k$.  Thus, no element of $[k]$ is abundant. $\square$
This example highlights one of the obstacles in proving the Union-closed Conjecture; it is difficult to predict where abundant elements will be.