# Counting sparse union-closed families

A family of sets $\mathcal{F} \subseteq 2^{[n]}$ is called union-closed if for any two sets $A,B \in \mathcal{F}$, $A \cup B \in \mathcal{F}$. We say that $\mathcal{F} \subseteq 2^{[n]}$ is sparse if the average size of a set in $\mathcal{F}$ is less than $n/2$. $\mathbb{E}_{A \in \mathcal{F}}|A| < n/2$.

Let's denote by $f(n)$ the number of sparse union-closed families of subsets of $[n]$. Any good bounds on $f(n)$?

Of course all subfamilies of $$2^{[n/2-1]}$$ are sparse.

For any $$k$$, the number of union-closed families $$\mathcal F\subseteq 2^{[k]}$$ is $$u(k)=2^{\binom{k}{\lfloor k/2\rfloor}(1+o(1))}$$ (link), where $$\binom{k}{[k/2]}\in\Theta(2^k/\sqrt{k})$$ (link).

Apply this with $$k=n/2$$ and we get some bounds:

$$\log_2 u(n) \in \Theta (2^n/\sqrt{n})$$,

$$\log_2 f(n) \in \Omega (2^{n/2}/\sqrt{n/2})$$, and $$%\log_2\log_2 U_n=n-\frac12\log_2 n+O(1)$$ $$%\log_2\log_2 S_n\ge \frac12 n -\frac12\log_2 n+O(1)$$ $$\frac12\le\lim_{n\to\infty}\frac{\log_2\log_2 f(n)}n\le \lim_{n\to\infty}\frac{\log_2\log_2 u(n)}n = 1.$$

• not all subfamilies of $2^{[n/2-1]}$ are union-closed Commented Oct 31, 2020 at 13:39
• @mathworker21 correct; so I'm substituting $[n/2-1]$ for $n$. Maybe I should write it more explicitly Commented Oct 31, 2020 at 15:17
• @mathworker21 updated now Commented Nov 1, 2020 at 0:44