# How to find an example of a union-closed family with two given properties

Consider a union-closed family $$\mathcal{F}$$ of $$n=\vert \mathcal{F} \vert$$ sets, $$n$$ odd, and its family $$\mathit{J}(\mathcal{F})$$ of $$m = \vert\mathit{J}(\mathcal{F})\vert$$ basis sets (or $$\cup$$-irreducible sets, also called generators), and define $$\mathcal{F_a} = \{A \in \mathcal{F}: a \in A \}$$, $$\mathit{J_a}(\mathcal{F}) = \{A \in \mathit{J}(\mathcal{F}) : a \in A \}$$.

It is easy to prove by contradiction that if $$\exists a$$ such that $$\vert\mathit{J_a}(\mathcal{F})\vert \ge m-\Bigl\lfloor\log_2{\frac{n+1}{2}}\Bigr\rfloor$$, then $$\vert \mathcal{F_a} \vert \ge \frac{n}{2}$$.

I have tried the two examples in "The journey of the union-closed sets conjecture" by Bruhn and Schaudt on section 6.3 "Limits of averaging" and both seem to satisfy the above requirement.

I understand it might not be easy, but any idea on how to construct an example of a union-closed family with average set size less than half of the universe and such that $$\forall a$$ $$\vert\mathit{J_a}(\mathcal{F})\vert \lt m-\Bigl\lfloor\log_2{\frac{n+1}{2}}\Bigr\rfloor$$?

Not a good example, but it might hint for a better one.

Take two integers $$N=2k+1,M$$. Let the basis sets be $$\{x,x+1,...,x+k\},1\leq x\leq N$$, taken modulo $$N$$ (identify $$N\equiv 0$$) and $$\{1,2,...,N,a_1,a_2,...,a_M\}$$.

The union-closed family $$\mathcal{F}$$ contains all set of the form $$\{x,x+1,...,x+l\},1\leq x\leq N,k\leq l\leq N-1$$ and $$\{1,2,...,N,a_1,a_2,...,a_M\}$$. So it's easy to see that $$m=N+1,n=\frac{N(N-1)}{2}+2,J_i(\mathcal{F})=\frac{N+1}{2},J_{a_j}(\mathcal{F})=1$$. The average set size is $$f(N)+\frac{M}{\frac{N(N-1)}{2}+2}$$ ($$f(N)$$ is quite complicative to write, but it's independent of $$M$$).

Now we choosing $$N,M$$ such that $$\frac{N-1}{2}$$ is odd, $$log_2(\frac{N(N-1)}{2}+2)<\frac{N+1}{2}, f(N)+\frac{M}{\frac{N(N-1)}{2}+2}<\frac{N+M}{2}$$ then $$\mathcal{F}$$ is the union-close family that we want.

• Since you want $k = \frac{N-1}2$ odd, take $k = 2k_0 + 1$. Then the average set size is $$\frac{24k_0^3 + 46k_0^2 + 37k_0 + M + 12}{2(4k_0^2 + 7k_0 + 4)}$$ and the constraint that the average set size is less than half of the universe simplifies to $M > \frac{2k_0^2}{k_0+1}$ or, if you prefer, $M > \frac{(N-3)^2}{2(N+1)}$ Commented Nov 4, 2022 at 8:54
• Thank you very much. I accept it, although now another more restrictive question would be if there exists an example with a separating family and/or a family with less than $n/4$ elements... Commented Nov 4, 2022 at 10:02
• @BillyJoe, the additional requirement for a family with fewer than $n/4$ elements comes down to the constraint $8k_0^2 - 6k_0 - 7 > 4M$. Combine that with the previous constraint on $M$ to get $8k_0^2 - 6k_0 - 7 > 4M > \frac{8k_0^2}{k_0 + 1}$. Since the upper bound is $\Theta(k_0^2)$ and the lower bound is $\Theta(k_0)$, there's plenty of margin to find values of $k_0$ and $M$ which satisfy both inequalities simultaneously. Commented Nov 4, 2022 at 11:16
• @PeterTaylor yes, sorry, I didn't check carefully. Commented Nov 4, 2022 at 11:27