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Is it possible to find a union closed family $\mathcal{F}$, $\emptyset \notin \mathcal{F}$, with $|\mathcal{F}| = n$ sets, such that there are at most:

$$\left(1-\frac{1}{\left\lfloor \frac{n-1}{2} \right\rfloor}\right)\frac{n^2}{2}$$

unordered couples of sets in it with at least one element in common?

If that is not possible, and if we build a graph such that every vertex represents a set and any edge exists if and only if the two corresponding sets have an element in common, by Turán's theorem there will be a $K_{\lceil \frac{n}{2} \rceil}$ complete subgraph in it, and therefore a subfamily of $\mathcal{F}$ of size $\lceil \frac{n}{2} \rceil$ where each unordered couple of sets has an element in common. Maybe from that subfamily it will be then possible to build another subfamily of size $\lceil \frac{n}{2} \rceil$ with the same element in common among all sets.

Note that every union closed family with $n$ sets and without the empty set has at least $\frac{2}{3}\binom{n}{2}$ unordered couples of sets with at least one element in common, because for every couple without an element in common $\{A,B\}$ we can build two unique other couples $\{A,A \cup B\}$ and $\{B,A \cup B\}$ with non-empty intersection. However $\frac{2}{3}\binom{n}{2}$ is bigger than the above value only for $n \lt 7$.

Since a family as required by the question would imply that there are $\frac{n^2}{2\lfloor \frac{n-1}{2} \rfloor} \gt n$ unordered couples of sets with empty intersection, maybe it is possible to show that these couples would generate more than $n$ sets, or that to build those couples we need more than the maximum number of elements ($\frac{n-1}{4}$) required for a possible counterexample of the union closed sets conjecture.

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Well, thinking more carefully I have found the simple counterexample given by the powerset on $n$ elements without the empty set, $\mathcal{P}([n]) \setminus \emptyset$, for $n \ge 4$.

I have decided not to delete the question just in case someone finds a useful restriction on the family properties.

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