This question is motivated by [Frankl's union-closet sets conjecture][1]. 

Let $n\in\mathbb{N}$ and set $[n] = \{0,1,\ldots,n\}$. We say that a family ${\cal A} \subseteq {\cal P}([n])$ is *union-closed* if $\emptyset\notin{\cal A}$ and  $A,B\in {\cal A}$ implies $A\cup B\in{\cal A}$.

For $k\in[n]$ we define the *weight* of $k$ to be $$w(k) = |\{A\in {\cal A}: k\in A\}|.$$ For a set $A\in{\cal A}$ we set the *discrepancy* to be $$\text{disc}(A) = \max\{w(j):j\in A\} - \min\{w(j):j\in A\}.$$

Let $r\in ]0,1[$ be given. Is there $n\in\mathbb{N}$ and a union closed family ${\cal A}$ on $[n]$ and $M$ minimal amongst the members of ${\cal A}$ such that $$\frac{\text{disc}(M)}{|{\cal A}|} \geq r?$$

  [1]: https://en.wikipedia.org/wiki/Union-closed_sets_conjecture