TITLE: 'stationary' almost disjoint families
QUESTION [11 upvotes]: 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, $\text{sad} < \text{min} \{\text{sat}(\text{NS}_\kappa), \text{mad}\}$ ?

REPLY [3 votes]: 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.