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?


1 Answer 1


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 $B_i = A\setminus\{i\}$.


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