# Lower bound for sets couples in an union-closed family such that $A \subset B$ or $B \subset A$

Given an union-closed family of sets $$\mathcal{F}$$, with $$n = \vert\mathcal{F}\vert$$ and thus $$n \choose 2$$ unordered couples of distinct sets $$\{A, B\}$$, $$A,B \in \mathcal{F}$$, I would like to compute a good lower bound for the number of couples such that $$A \subset B$$ or $$B \subset A$$ as a function of $$n$$, i.e.:

$$\text{Number of couples such that } A \subset B \text{ or } B \subset A \ge f(n)$$

Intuitively for a couple $$\{A, B\}$$ with $$A \not\subset B$$ and $$B \not\subset A$$, there are two couples $$\{A, C\}$$ and $$\{B, C\}$$, $$C = A \cup B$$, with $$A \subset C$$ and $$B \subset C$$, however two couples $$\{A, B_1\}$$ and $$\{A, B_2\}$$ may share the same $$\{A, C = A \cup B_1 = A \cup B_2\}$$, so this does not seem to help.

Any hint?

In terms of $$n$$ alone, and lacking any extra constraints, I think $$n-1$$ is the best lower bound you can get.
It is a lower bound, because if you take $$A = \bigcup {\cal F}$$, then for all $$B \in {\cal F} \setminus \{A\}$$ you have $$B \subset A$$, and this gives you $$n-1$$ pairs.
The bound is reached with the union-closed family $${\cal F} = A \;\cup\; \{B_i \;:\; i=1,\ldots,n-1\},$$ where $$A = \{1,2,\ldots,n-1\}$$, and $$B_i = A\setminus\{i\}$$.
• In what sense is $A$ a set of sets? Commented May 15 at 17:20