Is there an infinite cardinal $\kappa$ and a set $\frak{E}$ of subsets of $\kappa$ with the following properties:


1. $|e\cap f| \leq 1$ for $e,f\in {\frak E}$ with $e\neq f$, and
2. $|\frak{E}| > \kappa$


?



"Strongly" almost disjoint subsets
No. For each element $x \in \kappa$, let $g(x)$ be the set of elements in $\frak{E}$ that contain $x$. By assumption, for all $x \in \kappa$, we have $|g(x)| \leq \kappa$. We may clearly assume $\emptyset \in \frak{E}$. But now, $\frak{E}$=$\{\emptyset\} \cup \bigcup\_{x \in \kappa} g(x)$, and so $|\frak{E}|$$\leq \kappa$, as required. 


The answer is somewhat surprisingly yes, if you weaken condition (1) so that $|e \cap f|$ is finite. See Andrés E. Caicedo's comment to my answer [here](https://mathoverflow.net/questions/59115/a-set-for-which-it-is-hard-to-determine-whether-or-not-it-is-countable/59181#59181). 


