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, $\text{sad} < \text{min} \{\text{sat}(\text{NS}_\kappa), \text{mad}\}$ ?
  • 1
    $\begingroup$ 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. $\endgroup$
    – Yair Hayut
    Jan 23, 2019 at 15:50
  • $\begingroup$ 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. $\endgroup$ Jan 25, 2019 at 12:59
  • $\begingroup$ 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. $\endgroup$
    – Asaf Karagila
    Jan 25, 2019 at 13:30
  • $\begingroup$ @AsafKaragila symmetric difference being nonstationary means that they’re all in the same equivalence class mod NS, and in particular have stationary intersection. $\endgroup$ Jan 25, 2019 at 13:54
  • 1
    $\begingroup$ 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 $2^\kappa$ $\endgroup$ Jan 25, 2019 at 14:56

1 Answer 1


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.

  • $\begingroup$ I wonder if the existence of a bound is consistent with continuum $=\omega_2$. $\endgroup$ Jan 25, 2019 at 19:16

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