# 'stationary' almost disjoint families

Consider almost disjoint families on regular $$\kappa > \omega$$ consisting only of stationary sets.

My question: Is there consistently an upper bound $$<2^\kappa$$ on the size of such a 'stationary' almost disjoint family (under suitable large cardinal assumptions)?

E.g. a Woodin cardinal implies the consistency of '$$\text{NS}_{\aleph_1}$$ is $$\aleph_2$$-saturated', which implies that any s.a.d. family has size $$\leq \aleph_1$$. However, as $$X_i \cap X_j$$ is not only non-stationary but bounded in $$\kappa$$, maybe weaker assumptions also imply the consistency of 'Every s.a.d. family on $$\aleph_1$$ has size $$\leq \aleph_1$$' ? (solved)

EDIT: The following cases for $$\kappa$$ remain open:

Always require $$2^\kappa > \kappa^+$$:

• $$\kappa=\kappa^{<\kappa}$$ and $$\text{sad}< 2^\kappa$$ ?
• $$\text{sad} < \text{sat}(\text{NS}_\kappa)$$ ?
• and, of course, $$\text{sad} < \text{min} \{\text{sat}(\text{NS}_\kappa), \text{mad}\}$$ ?
• At least in the case for s.a.d. family of size $\aleph_2$, I think that you can argue that the non-saturation of $NS_{\aleph_1}$ implies that there is also a s.a.d. family of size $\aleph_2$ (for every $S_\alpha, S_\beta$ in the antichain take $D_{\alpha,\beta}$ club disjoint from $S_\alpha \cap S_\beta$. Now, take $E_\beta$ to be the diagonal intersection of $D_{\alpha,\beta}$ for $\alpha < \beta$, using some enumeration of $\beta$ of order type $\leq \omega_1$. The collection $S_\alpha \cap E_\alpha$ with be s.a.d.). This doesn't answer your question in the non-GCH case. – Yair Hayut Jan 23 '19 at 15:50
• Now I don’t understand the point of the question, after the edit. There is consistently (without large cardinals) an upper bound $<2^{\omega_1}$ on the size of an almost-dijsoint family of subsets of $\omega_1$. So in this model, such a cardinal also bounds the size of stationary almost-disjoint families. – Monroe Eskew Jan 25 '19 at 12:59
• I think that 'SAD family' would intuitively be a family of sets such that the symmetric difference of any two is non-stationary. An almost disjoint family of stationary sets would be just that. – Asaf Karagila Jan 25 '19 at 13:30
• @AsafKaragila symmetric difference being nonstationary means that they’re all in the same equivalence class mod NS, and in particular have stationary intersection. – Monroe Eskew Jan 25 '19 at 13:54
• And I would also be interested if the bound of the sad's can be strictly below the bound of the ad families, e.g if $\kappa=\kappa^{<\kappa}$ so there exists an ad familiy of size $2^\kappa$ – Johannes Schürz Jan 25 '19 at 14:56

First let us show that consistently there is no such bound, along with $$2^\kappa$$ larger than any prescribed cardinal. Assume $$\diamondsuit_\kappa$$. This is consistent with any large cardinal assumption and any value of $$2^\kappa$$, by forcing with $$Add(\kappa,\theta)$$. This principle states:
There is a sequence $$\langle a_\alpha : \alpha < \kappa \rangle$$ such that for every $$X \subseteq \kappa$$, $$\{ \alpha : X \cap \alpha = a_\alpha \}$$ is stationary.
It is easy to see that if $$X \not= Y$$, then the set points where the diamond sequence guesses $$X$$ is almost-disjoint from the set where it guesses $$Y$$. Thus under $$\diamondsuit_\kappa$$, there is an almost-disjoint family of stationary subsets of $$\kappa$$ of maximal size.
Second, let us show that consistently there is such a bound. This is inspired by exercise 23.11 in Jech. Suppose GCH holds in $$V$$. Force with $$Add(\omega,\omega_3)$$. In the extension, $$2^{\omega_1} = \omega_3$$. We will show that there is no almost-disjoint family of subsets of $$\omega_1$$ of size $$\omega_3$$. Suppose otherwise and let $$\langle \dot A_\alpha : \alpha < \omega_3 \rangle$$ be a name for a counterexample. For each pair $$\alpha<\beta$$, it is forced that $$A_\alpha \cap A_\beta$$ has size $$<\omega_1$$, and by the ccc, there is some ordinal $$\delta_{\alpha,\beta}$$ such that $$1 \Vdash \dot A_\alpha \cap \dot A_\beta \subseteq \check \delta_{\alpha,\beta}$$. By the Erdos-Rado theorem (and GCH), there is some $$X \subseteq \omega_3$$ of size $$\omega_2$$ and a $$\delta <\omega_1$$ such that $$\delta_{\alpha,\beta} = \delta$$ for all $$\alpha,\beta \in X$$. Thus it is forced that $$\{ A_\alpha \setminus \delta : \alpha \in X \}$$ is pairwise disjoint. This is impossible.
• I wonder if the existence of a bound is consistent with continuum $=\omega_2$. – Monroe Eskew Jan 25 '19 at 19:16