Let $\kappa$ be an infinite cardinal. Suppose $E \subseteq {\cal P}(\kappa)$ has the following property: for $e\_1\neq e\_2\in E$ we have $|e\_1\cap e\_2|= 1$, and suppose $|E| = \kappa$.


Does this imply that at least one of the following statements is true?


1. there is $e\in E$ with $|e|=\kappa$;
2. there is $\alpha \in \kappa$ such that $|\{e\in E: \alpha \in e\}| = \kappa$.



On a set of sets intersecting in $1$ point
Consider two sets $e,f\in E$, assume that $\max(|f|,|e|)=:\mu<\kappa$, $\{x\}:=e\cap f$. Take arbitrary element $y\in f\setminus x$, it is contained in at most $\mu$ sets from $E$. Indeed, they all have a common element with $e$, and all those common elements are different. So, the elements of $f\setminus x$ are contained totally in at most $\mu\times \mu<\kappa$ sets, thus $x$ is contained in $\kappa$ sets.


