Recall a stationary subset $S$ of a regular cardinal $\kappa$ is fat when for every $\alpha < \kappa$, and every club $C$, there is a closed set of order type $\alpha$ contained in $S \cap C$. It is a result of Stavi, proved [here](http://www.ams.org/mathscinet-getitem?mr=716625), that:


(1) For every regular cardinal $\kappa$ and every stationary $S \subseteq \kappa^+ \cap \mathrm{cf}(\kappa)$, $S \cup \mathrm{cf}(<\kappa)$ is fat.


(2) If $S \subseteq \kappa$ is fat, and $2^{<\alpha} < \kappa$ for all $\alpha< \kappa$, then there is a $<\kappa$-distributive forcing of size $2^{<\kappa}$ which forces a club $C \subseteq S$. Furthermore, the forcing preserves every stationary subset of $S$.


**Questions:** Suppose $\kappa$ is either (a) inaccessible or (b) the successor of singular cardinal. Is it true that there is a sequence of disjoint stationary sets $\langle S\_\alpha : \alpha < \kappa \rangle$ and some set $T$ disjoint from all $S\_\alpha$ with the following property? For all clubs $C$, and all $\alpha,\beta < \kappa$, there is a closed subset $p$ of $C \cap (T \cup S\_\alpha)$ with order type $\geq \beta$, and $\max p \in S\_\alpha$. An answer under additional combinatorial assumptions (known to be consistent) would be welcome.



Fat stationary sets
Suppose that $\lambda$ is a singular cardinal, $\square\_\lambda$ holds, and $2^\lambda=\lambda^+$. Then:


1. There exists a partition of $\lambda^+$ into $\lambda^+$ many pairwise disjoint fat stationary sets.
2. There exists a family of $2^{(\lambda^+)}$ pairwise almost-disjoint fat stationary sets.


Both clauses follow from Theorem D of <http://www.assafrinot.com/paper/11> . 


An application of Lemma 2.3 of the same paper shows that Clause (1) follows already from $\square\_\lambda$ (without the arithmetic hypothesis). 
For $\lambda$ regular, one obtains Clause (1) from $\square(\lambda^+)$ by feeding $\Gamma:=E^{\lambda^+}\_\lambda$ to Lemma 3.2 of <http://www.assafrinot.com/paper/18> . 


Update Jan/2017: <http://settheory.mathtalks.org/?p=7240>


