Let $\kappa$ be an infinite ordinal, and suppose $E\subseteq {\cal P}(\kappa)$ with $|E| > \kappa$. Let us call $x\in \kappa$ popular if $$|\{e\in E: x\in e\}| > \kappa.$$
If $\text{Pop}(\kappa)$ is the collection of popular elements of $\kappa$, is it possible that $|\text{Pop}(\kappa)|<\kappa$?
Yes if there exists a cardinal $\gamma$ with $|\gamma|<|\kappa|<|2^\gamma|$. Just take $E=2^\gamma$.
No, otherwise. Throw out all popular elements from the sets in $E$; some of these sets become identical, but each equivalence class has at most $|\kappa|$ elements, since $|2^{\mathop{\rm Pop}(\kappa)}|\leq|\kappa|$. So wou still get a family of subsets of $\kappa$ of cardinality $>|\kappa|$ without popular elements, which is impossible.
