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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. $|\frak{E}| > \kappa$

?

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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.

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  • $\begingroup$ You probably meant $|g(x)| \leq \kappa$. $\endgroup$ – Ramiro de la Vega Nov 28 '16 at 14:10
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    $\begingroup$ Alternatively, note that each two element subset of $\frak S$ is covered at most once. $\endgroup$ – Fan Zheng Nov 28 '16 at 16:06

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