TITLE: Fat stationary sets
QUESTION [6 upvotes]: 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, 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.

REPLY [8 votes]: Suppose that $\lambda$ is a singular cardinal, $\square_\lambda$ holds, and $2^\lambda=\lambda^+$. Then:

There exists a partition of $\lambda^+$ into $\lambda^+$ many pairwise disjoint fat stationary sets.
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