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This question is motivated by Frankl's union-closet 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 $w(x) = \kappa$?

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    $\begingroup$ Sorry for my ignorance, but why the down vote? $\endgroup$ – Dirk Feb 5 '16 at 19:27
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Counterexample. Identify the infinite cardinal $\kappa$ with its initial ordinal, and let $\mathcal A$ be the set of all final segments of $\kappa.$

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