This question is motivated by [Frankl's union-closet sets conjecture](https://en.wikipedia.org/wiki/Union-closed_sets_conjecture).


Let $X$ be a non-empty set. We say that a family ${\cal A} \subseteq {\cal P}(X)$ is *union-closed* if $\emptyset\notin{\cal A}$ and $A,B\in {\cal A}$ implies $A\cup B\in{\cal A}$.


For $x\in X$ we define the *weight* of $x$ to be $$w(x) = |\{A\in {\cal A}: x\in A\}|.$$


Suppose that ${\cal A}$ is an infinite union-closed family and let $\kappa = |{\cal A}|$. Is there $x\in \bigcup{\cal A}$ such that $w(x) = \kappa$?



Frankl's union-closed sets conjecture for infinite families
**Counterexample.** Identify the infinite cardinal $\kappa$ with its initial ordinal, and let $\mathcal A$ be the set of all final segments of $\kappa.$


