# Conjecture about separating union-closed families

Consider a separating union-closed family $$\mathcal{F}$$ of $$n$$ finite sets with $$\mathcal{F} \not = \{ \emptyset \}$$.

Let $$U(\mathcal{F})$$ be the union of all sets in $$\mathcal{F}$$ (the universe). Let $$q = |U(\mathcal{F})|$$.

Separating means that for any two elements $$x,y$$ in the universe there exist $$A \in \mathcal{F}$$ such that $$x \in A$$ and $$y \not\in A$$, or $$x \not\in A$$ and $$y \in A$$.

Let $$S = \{ |A \Delta B| : A,B \in \mathcal{F}, A \not= B \}$$ be the set of all sizes of symmetric differences between any two sets in $$\mathcal{F}$$ (Hamming distances).

I have run many tests on example families and always found $$S = [m] = \{1, 2, \ldots, m-1, m\}$$ with $$m \le q$$.

I have thought about $$|A \Delta B| = |A \Delta (A \cup B)|+|B \Delta (A \cup B)|$$, but that doesn't seem to help much.

Can we prove that $$S = [m]$$, or find a counterexample?

Checking more carefully, I have found that this answer can be used as a counterexample here as well, since $$S = [21] \setminus \{18\}$$.