6
$\begingroup$

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 $i(\alpha) < \alpha$?

$\endgroup$
5
  • 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
  • 7
    $\begingroup$ 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
  • 5
    $\begingroup$ 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
  • 4
    $\begingroup$ 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
7
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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

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.