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



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.

  • $\begingroup$ You probably meant $|g(x)| \leq \kappa$. $\endgroup$ – Ramiro de la Vega Nov 28 '16 at 14:10
  • 2
    $\begingroup$ Alternatively, note that each two element subset of $\frak S$ is covered at most once. $\endgroup$ – Fan Zheng Nov 28 '16 at 16:06

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.