# Fat stationary sets

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.

• It seems that you want to use this sequence and $T$ to do some forcing. What's the endgame here? Apr 7, 2016 at 16:38
• Code some information into the stationarity/ nonstationarity of some sets. Apr 7, 2016 at 16:39
• Ah. So you essentially want a sequence of fat stationary sets, so you can shoot clubs into them without causing anything to collapse, and thus code information into which sets were shot in the club? Apr 7, 2016 at 16:41
• Yes. I think this kind of idea is due to Magidor... Apr 7, 2016 at 16:43
• In the work (with Sy Friedman, if my memory serves me right) on getting the number of normal measures to be anything, right? Apr 7, 2016 at 16:46

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

• How necessary is the square? Apr 8, 2016 at 4:31
• @asaf For Clause (1)? I just need a $\square(\lambda^+)$ sequence for which $\{ \alpha<\lambda^+\mid \text{otp}(C_\alpha)\ge\lambda\}$ is stationary.
– saf
Apr 8, 2016 at 5:14
• Correction: I just realized that MM implies the existence of $\aleph_2$ many pairwise disjoint fat subsets of $\aleph_2$.
– saf
Apr 8, 2016 at 5:50
• Okay, so we're back to square one. Pun semi intended. Apr 8, 2016 at 6:08
• @AsafKaragila I think I have a way of forcing this with a $\lambda^+$-closed forcing. This cannot introduce $\square_lambda$. Apr 8, 2016 at 14:26