A *Frankl family* is a nonempty finite family $\mathcal F$ of nonempty finite sets such that $A,B\in\mathcal F\implies A\cup B\in\mathcal F.$ Define $d\_\mathcal F(x)=|\{A\in\mathcal F:x\in A\}|$ and $\Delta(\mathcal F)=\max\_xd\_\mathcal F(x).$ Frankl's union-closed sets conjecture says that $\Delta(\mathcal F)\gt\frac12|\mathcal F|,$ or in other words $|\mathcal F|\le2\Delta(F)-1,$ for any Frankl family $\mathcal F.$ Has the following stronger conjecture been considered, and is a counterexample known?



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> For any Frankl family $\mathcal F$ there is an element $x$ such that $\Delta(\{A\in\mathcal F:x\notin A\})\le\frac12\Delta(\mathcal F),$
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or better yet,



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> for any Frankl family $\mathcal F$ and any element $x,$ if $d\_\mathcal F(x)=\Delta(\mathcal F)$ then $\Delta(\{A\in\mathcal F:x\notin A\})\le\frac12\Delta(\mathcal F)$?
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A strengthening of Frankl's union-closed sets conjecture?
It seems that here is a counterexample to the stronger statement. Perhaps, it can be modified to provide one for the weaker statement?


Our universe is $\{1,\dots,7\}$. Each set of the family contains either $6$, or $7$, or both. We asume that the elements $1,\dots,5$ are arranged into a *cycle* $1-2-3-4-5-1$.


The sets containing $6$ but not $7$ are obtained from $\{6\}$ by adding either (2-6) two consecutive elements from the cycle, or (3-6) three consecutive elements, or $\{1,2,4\}$, or $\{2,3,5\}$, or $\{1,3,4\}$, or (4-6) arbitrary four or five elements from the cycle. 19 sets in all.


The sets containing $7$ but not $6$ are obtained from $\{7\}$ by adding either (2-7) two non-consecutive elements from the cycle, or (3-6) three non-consecutive elements, or $\{3,4,5\}$, or $\{5,1,2\}$, or (4-6) arbitrary four or five elements from the cycle. $18$ sets in all.


The sets containing both $6$ and $7$ are obtained from $\{6,7\}$ by adding at least three elements from the cycle. $16$ sets in all.


These sets form a Frankl family. Now, $6$ is contained in $35$ sets, $7$ --- in $34$ sets, any other element --- in at most $34$ sets. Thus the unique element with the largest degree is $6$ (this was the aim for which we have added extra triples), and this degree is $35.$ But all the sets not containing $6$ do contain $7$, and there are $18>35/2$ of them.


Thus the conjectured property may fail even for *every* element of the largest degree.


