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Let $X\neq \emptyset$ be a set. We say $E\subseteq {\cal P}(X)$ is a linear set system if for all $a\neq b\in X$ there is exactly one $e\in E$ with $\{a,b\}\subseteq e$.

Is there an infinite cardinal $\kappa$ and a linear set system $E\subseteq {\cal P}(\kappa)$ with $\kappa \notin E$ such that for all $x\in \kappa$ we have $|\{e\in E: x\in e\}| < \kappa$?

EDIT. Had to exclude cases like $E = \{\kappa\}$ which Noah Schweber made me aware of (thanks!).

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    $\begingroup$ What prevents $E=\{X\}$? $\endgroup$
    – Noah Schweber
    Commented Aug 14, 2022 at 19:50
  • $\begingroup$ Right - will exclude that case. Are there other examples? $\endgroup$
    – Dominic van der Zypen
    Commented Aug 15, 2022 at 9:32

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Denote $\deg x =|\{e\in E\colon x\in e\}|$.

Take any $e\in E$; choose $x\notin e$. Then $e$ meets any $e'\ni x$ by at most one element, whence $|e|\leq \deg x<\kappa$.

Now, any $x$ is contained in $\deg x<\kappa$ sets of cardinality $<\kappa$ each; moreover (if this is needed), any of them except one has cardinality at most $\deg y<\kappa$ (for an arbitrary $y\neq x$). All this implies that their union is of cardinality $<\kappa$; but their union is $\kappa$.

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