3
$\begingroup$

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

?

$\endgroup$

1 Answer 1

5
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ You probably meant $|g(x)| \leq \kappa$. $\endgroup$ Nov 28, 2016 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, 2016 at 16:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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