Let $\{A_1,B_1\},\ldots,\{A_k,B_k\}$ be all the distinct unordered couples of subsets, with $A_i \cap B_i = \emptyset, 1 \le i \le k$, that can formed from a set $\{C_1,\ldots,C_q\}$, $q \le \binom{n}{2}$, of subsets of $[n]=\{1,2,\ldots,n\}$ of size $2$, $A_i = \{a_{i,1},a_{i,2}\}, B_i = \{b_{i,1},b_{i,2}\}, 1 \le i \le k$.
Let $\mathcal{E}_j, 1 \le j \le n$ be the families of sets such that any set $E_j \in \mathcal{E}_j$ has the following properties:
- $j \in E_j$
- $a_{i,1} \in E_j \lor a_{i,2} \in E_j \iff b_{i,1} \in E_j \lor b_{i,2} \in E_j$
Let $f_j$ be the minimum size of a set in $\mathcal{E}_j$.
Finally define:
$$f = \max_{1 \le j \le n} f_j$$
I would like to prove that, for $3 \le m \le n-1$:
$$k \ge \frac{\binom{m+1}{2}\binom{m-1}{2}}{2} \implies f \ge m \tag{1}$$
And I am particularly interested in the case $m = \lceil n/2 \rceil$.
I think the conjecture is true if we assign to $\{A_1,B_1\},\ldots,\{A_k,B_k\}$ all the possible couples of disjoint subsets of size $2$ of $[m+1]=\{1,2,\ldots,m+1\}$, which are exactly $k=\frac{\binom{m+1}{2}\binom{m-1}{2}}{2}$, because, due to symmetry, we have $f_1 = f_2 = \cdots = f_{m+1}$, and we can use induction on $m$.
But how to prove it for a general choice of $\{C_1,\ldots,C_q\}$?
Motivation: this conjecture together with this lower bound, if both proved correct, could be used to prove the union-closed sets conjecture.
