# “Intersection number” of a cardinal

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 $$i(\alpha) < \alpha$$?

• 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 – Alessandro Codenotti Jul 16 '20 at 20:57
• 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. – Paul McKenney Jul 17 '20 at 3:40
• 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. – bof Jul 17 '20 at 5:06
• Thanks @bof and Alessandro for your comments. I will include them in the question – Dominic van der Zypen Jul 17 '20 at 5:23
• 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 $\kappa$. – 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.

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