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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.

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  • $\begingroup$ It seems that you want to use this sequence and $T$ to do some forcing. What's the endgame here? $\endgroup$
    – Asaf Karagila ♦
    Commented Apr 7, 2016 at 16:38
  • $\begingroup$ Code some information into the stationarity/ nonstationarity of some sets. $\endgroup$
    – Monroe Eskew
    Commented Apr 7, 2016 at 16:39
  • $\begingroup$ 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? $\endgroup$
    – Asaf Karagila ♦
    Commented Apr 7, 2016 at 16:41
  • $\begingroup$ Yes. I think this kind of idea is due to Magidor... $\endgroup$
    – Monroe Eskew
    Commented Apr 7, 2016 at 16:43
  • $\begingroup$ In the work (with Sy Friedman, if my memory serves me right) on getting the number of normal measures to be anything, right? $\endgroup$
    – Asaf Karagila ♦
    Commented Apr 7, 2016 at 16:46

1 Answer 1

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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

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

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