Let $\kappa$ be an infinite cardinal. We call a cardinal $\lambda \leq 2^\kappa$ intersecting if there is ${\cal C}\subseteq {\cal P}(\kappa)$ such that

  1. for every $A\in {\cal C}$ we have $|A|=\kappa$,
  2. $|A_0\cap A_1|<\lambda$ whenever $A_0\neq A_1\in {\cal C}$, and
  3. $|{\cal C}| > \kappa$.

We denote the smallest intersecting cardinal of $\kappa$ by $i(\kappa)$. For instance we have $i(\aleph_0) = \aleph_0$ (also see the concept of an almost disjoint family). By the comments of users bof and Alessandro Codenotti, we always have $i(\kappa) \leq \kappa$ for any infinite cardinal $\kappa$.

Question. If $\kappa$ is an infinite cardinal, is there a cardinal $\alpha\geq\kappa$ with $i(\alpha) < \alpha$?

  • 5
    $\begingroup$ For $\lambda>\kappa$, $[\kappa]^\kappa$ shows that $\lambda$ is intersecting. For $\lambda=\kappa$ we always have a mad family (and if $\kappa$ is regular 3 is satisfied), so the interesting case is $\lambda<\kappa$ unless I'm missing something $\endgroup$ – Alessandro Codenotti Jul 16 '20 at 20:57
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    $\begingroup$ You will probably be interested in Baumgartner's paper "Almost disjoint sets, the dense set problem and the partition calculus", Annals of Mathematical Logic 10 (1976) 401 - 439. The short answer is that this question is often independent of ZFC. You might find more recent information searching for "strongly almost disjoint families" or similar topics; see e.g. Koszmider's paper "On the existence of strong chains in $P(\omega_1)/fin$" from JSL Vol 63, No 3, Sept 1998. $\endgroup$ – Paul McKenney Jul 17 '20 at 3:40
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    $\begingroup$ It should be noted that $I(\alpha)\le\alpha$ for every infinite cardinal $\alpha$, so the answer to Qustion 3 is always "no". Moreover, if $\alpha$ is regular, and if $2^\lambda\le\alpha$ for every cardinal $\lambda\lt\alpha$, then $I(\alpha)=\alpha$, so the answer to Question 2 is "yes" if there is a cardinal $\beta$ such that $2^\beta=\beta^+\ge\kappa$, or if there is a strongly inaccessible cardinal $\alpha\ge\kappa$. In short, Question 1 is the interesting one. $\endgroup$ – bof Jul 17 '20 at 5:06
  • $\begingroup$ Thanks @bof and Alessandro for your comments. I will include them in the question $\endgroup$ – Dominic van der Zypen Jul 17 '20 at 5:23
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    $\begingroup$ In my previous comment the assumption "$\alpha$ is regular" was superfluous; $I(\alpha)=\alpha$ holds if $2^\lambda\le\alpha$ for all $\lambda\lt\alpha$, whether $\alpha$ is regular or singular. So the answer to your original question 2 is always "yes", since there is always a singular strong limit carsinal greater than $\kappa$. $\endgroup$ – bof Jul 17 '20 at 10:43

Since this question is still unanswered I thought I might write down some of what you can get out of Baumgartner's paper.

In Baumgartner's notation (see the beginning of section 2), $A(\kappa,\lambda,\mu,\nu)$ means that there exists a family of sets $F$ such that

  1. $F\subseteq P(\kappa)$,
  2. $|F| = \lambda$,
  3. $|X| = \mu$ for all $X\in F$, and
  4. $|X\cap Y| < \nu$ for all $X,Y\in F$ with $X\neq Y$.

Hence the connection is that $\lambda$ is intersecting (in your notation) if and only if $A(\kappa,\kappa^+,\kappa,\lambda)$ holds.

In Theorem 3.4(a) Baumgartner proves that, assuming GCH, for any cardinals $\nu \le \mu \le \kappa$, $A(\kappa,\kappa^+,\mu,\nu)$ holds if and only if $\mu = \nu$ and $cf(\mu) = cf(\kappa)$. Since we're only interested in the case where $\mu = \kappa$, this implies that, under GCH, $i(\kappa) = \kappa$ for all $\kappa$. Note that this conclusion already follows from bof's comments.

The other side is partly covered by Theorem 6.1, which says: assuming GCH holds in $V$, for any cardinals $\nu \le \kappa \le \lambda$ such that $\nu$ is regular, there is a forcing extension $V[G]$ which preserves the cofinalities (hence cardinals) of $V$, in which $A(\kappa,\lambda,\kappa,\nu)$ is true. Hence you can make $i(\kappa) = \omega$ true for any particular $\kappa$, starting from a model of GCH.

It remains to show the consistency of the statement in your question, i.e. for all $\kappa$ there is some $\alpha \ge \kappa$ such that $i(\alpha) < \alpha$. Maybe someone who knows about class forcing can step in.

  • $\begingroup$ Thank you Paul for your effort - beautiful answer! Consistency would be nice to know, but we can leave this for another question $\endgroup$ – Dominic van der Zypen Jul 21 '20 at 5:53

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