Towers and clubs

TL;DR

This paper investigates club principles and the nonsaturation of the nonstationary ideal on regular uncountable cardinals, aiming to improve existing results by constructing towers of various lengths to demonstrate nonsaturation.

Contribution

It extends previous work by considering non-normal ideals and constructing towers of length possibly greater than ^+ to witness nonsaturation.

Findings

Established new conditions for nonsaturation of ideals.

Constructed towers of length exceeding ^+ in certain models.

Improved bounds related to club principles and nonstationary ideals.

Abstract

We revisit several results concerning club principles and nonsaturation of the nonstationary ideal, attempting to improve them in various ways. So we typically deal with a (non necessarily normal) ideal $J$ extending the nonstationary ideal on a regular uncountable (non necessarily successor) cardinal $\kappa$, our goal being to witness the nonsaturation of $J$ by the existence of towers (of length possibly greater than $\kappa^+$).

Full text

††footnotetext: MSC : 03E05††footnotetext: Keywords : club principle, saturated ideal, tower

1 INTRODUCTION

We will show that by modifying the proofs of some well-known results concerning non-saturation of the nonstationary ideal $NS_{\kappa}$, one may obtain towers in $(P(\kappa)/NS_{\kappa},\subseteq)$ or/and $(P(\kappa)/NS_{\kappa},\supseteq)$. Since these proofs usually involve one form or another of Club, we are led to revisit a number of results concerning this principle and its (many) variants.

2 DIAMOND LITE

2.1 Ideals and density

Let us first recall some definitions and facts that will be needed later. We start with ideals.

DEFINITION 2.1**.**

For a set $A$ and a cardinal $\rho$, we set $P_{\rho}(A)=\{a\subseteq A:|a|<\rho\}$ and $[A]^{\rho}=\{a\subseteq A:|a|=\rho\}$.

Throughout the paper $\kappa$ will denote a regular uncountable cardinal.

DEFINITION 2.2**.**

By an ideal on $\kappa$ we mean a nonempty collection $J$ of subsets of $\kappa$ such that

• •

$\kappa\notin J$.

• •

$\kappa\subseteq J$ ;

• •

$P(A)\subseteq J$ for all $A\in J$.

• •

$A\cup B\in J$ whenever $A,B\in J$.

Given an ideal $J$ on $\kappa$, we let $J^{+}=P(\kappa)\setminus J$, $J^{\ast}=\{A\subseteq\kappa:\kappa\setminus A\in J\}$, and $J|A=\{B\subseteq\kappa:B\cap A\in J\}$ for each $A\in J^{+}$. $J$ is prime if $J^{+}=J^{\ast}$, and nowhere prime if $J|A$ is prime for no $A\in J^{+}$. $J$ is $\kappa$-complete if $\bigcup Q\in J$ for every $Q\in P_{\kappa}(J)$. For a cardinal $\rho$ and $Y\subseteq P(\kappa)$, $J$ is $Y$-$\rho$-saturated if there is no $Q\subseteq J^{+}$ with $|Q|=\rho$ such that $A\cap B\in Y$ for any two distinct members $A,B$ of $Q$. $J$ is $\rho$-saturated if it is $J$-$\rho$-saturated.

$J$ is subnormal if $J\subseteq K$ for some normal ideal $K$ on $\kappa$.

DEFINITION 2.3**.**

We let $I_{\kappa}$ and $NS_{\kappa}$ denote, respectively, the noncofinal ideal on $\kappa$ and the nonstationary ideal on $\kappa$.

We let ${\cal C}_{\kappa}$ denote the collection of all closed unbounded subsets of $\kappa$.

For $A\subseteq\kappa$, we let $acc(A)=\{\alpha\in\kappa\setminus\{0\}:\sup(A\cap\alpha)=\alpha\}$.

DEFINITION 2.4**.**

Given a cardinal $\theta$, $E_{\theta}^{\kappa}$ (respectively, $E_{<\theta}^{\kappa}$, $E_{\geq\theta}^{\kappa}$) denotes the set of all $\alpha\in acc(\kappa)$ with $\mathrm{cf}(\alpha)=\theta$ (respectively, $\mathrm{cf}(\alpha)<\theta$, $\mathrm{cf}(\alpha)\geq\theta$).

DEFINITION 2.5**.**

Let $S$ be a stationary subset of $\kappa$. For $\gamma\in E^{\kappa}_{\geq\omega_{1}}$, $S$ reflects at $\gamma$ if $S\cap\gamma$ is stationary in $\gamma$.

REMARK 2.6*.*

If $S\subseteq E^{\kappa}_{\theta}$ reflects at $\gamma$, then $\mathrm{cf}(\gamma)>\theta$.

We next turn to density numbers and meeting numbers.

DEFINITION 2.7**.**

Given two cardinals $\tau\leq\sigma$ with $1\leq\tau$ and $\omega\leq\sigma$, $d(\tau,\sigma)$ (respectively, $m(\tau,\sigma)$) denotes the least cardinality of any $X\subseteq[\sigma]^{\tau}$ with the property that for any $e\in[\sigma]^{\tau}$, there is $x\in X$ with $x\subseteq e$ (respectively, $|x\cap e|=\tau$).

REMARK 2.8*.*

Thus $d(\tau,\sigma)=$ the cofinality of the poset $([\sigma]^{\tau},\supseteq)$.

DEFINITION 2.9**.**

Given two infinite cardinals $\tau\leq\sigma$, $\mathrm{MAD}_{\tau,\sigma}$ denotes the collection of all $Q\subseteq[\sigma]^{\tau}$ such that

• •

$|a\cap b|<\tau$ for any two distinct members $a,b$ of $Q$ ;

• •

for any $c\in[\sigma]^{\tau}$, there is $a\in Q$ with $|a\cap c|=\tau$.

FACT 2.10**.**

([21], [27], [29])* Let $\tau\leq\sigma$ be two infinite cardinals. Then the following hold :*

1. (i)

$\sigma\leq d(\tau,\sigma)\leq\sigma^{\tau}$. 2. (ii)

$d(\tau,\sigma)=\sigma$* if and only if $\mathrm{cf}(\sigma)\not=\mathrm{cf}(\tau)$ and $d(\tau,\chi)\leq\sigma$ for any cardinal $\chi$ with $\tau\leq\chi<\sigma$.* 3. (iii)

Suppose that $\chi^{<\tau}\leq\sigma<\chi^{\tau}$ for some cardinal $\chi$. Then $d(\tau,\sigma)=\sigma^{\tau}$. 4. (iv)

$d(\tau,\chi)\leq d(\tau,\sigma)$* for any cardinal $\chi$ with $\tau\leq\chi\leq\sigma$.* 5. (v)

$d(\tau,\sigma^{+})=\max\{d(\tau,\sigma),\sigma^{+}\}$. 6. (vi)

Suppose that $\sigma$ is a limit cardinal with $\mathrm{cf}(\sigma)\not=\mathrm{cf}(\tau)$. Then $d(\tau,\sigma)=\sup\{d(\tau,\chi):\tau\leq\chi<\sigma\}$. 7. (vii)

If $\sigma<\tau^{+\mathrm{cf}(\tau)}$, then $d(\tau,\sigma)=\max\{d(\tau,\tau),\sigma\}$. 8. (viii)

$d(\tau,\sigma)\geq|Q|$* for all $Q\in\mathrm{MAD}_{\tau,\sigma}$.* 9. (ix)

If $d(\tau,\tau)<d(\tau,\sigma)$, then $|Q|=d(\tau,\sigma)$ for all $Q\in\mathrm{MAD}_{\tau,\sigma}$. 10. (x)

$d(\tau,\sigma)=\max\{d(\tau,\tau),m(\tau,\sigma)\}$*. *

DEFINITION 2.11**.**

Shelah’s Strong Hypothesis (SSH) asserts that $\mathrm{pp}(\chi)=\chi^{+}$ for every singular cardinal $\chi$.

FACT 2.12**.**

([31])* The following are equivalent :*

1. (i)

Shelah’s Strong Hypothesis. 2. (ii)

Given two infinite cardinals $\tau<\sigma$, $m(\tau,\sigma)$ equals $\sigma$ if $\mathrm{cf}(\sigma)\not=\mathrm{cf}(\tau)$, and $\sigma^{+}$ otherwise.

FACT 2.13**.**

([45], [27])**

1. (i)

Let $\rho$ be an uncountable strong limit cardinal, and $\sigma\geq\rho$ be a cardinal. Then there is $\alpha<\rho$ such that for any infinite cardinal $\tau$ with $\alpha\leq\tau\leq\rho$, $d(\tau,\sigma)$ equals $\sigma$ if $\alpha\leq\mathrm{cf}(\tau)$, and $d(\mathrm{cf}(\tau),\sigma)$ otherwise. 2. (ii)

Suppose that $\rho<\kappa$ is an uncountable strong limit cardinal, and $\kappa$ is a limit cardinal. Then we may find $\chi<\rho$ with the following property : If $\tau$ is a regular cardinal with $\chi\leq\tau<\rho$, then $d(\tau,\sigma)<\kappa$ for every cardinal $\sigma$ with $\tau\leq\sigma<\kappa$.

2.2 J’enlève le haut

Our starting point is a result of Gregory on diamond star. The following guessing principles were introduced by Jensen [20].

DEFINITION 2.14**.**

Given a $\kappa$-complete ideal $J$ on $\kappa$, $\diamondsuit_{\kappa}^{\ast}[J]$ (respectively, $\diamondsuit_{\kappa}^{-}[J]$) asserts the existence of $t^{i}_{\alpha}\subseteq\alpha$ for $i<\alpha<\kappa$ such that $\{\alpha<\kappa:\exists i<\alpha(t^{i}_{\alpha}=A\cap\alpha)\}$ lies in $J^{\ast}$ (respectively, $J^{+}$) for every $A\subseteq\kappa$.

$\diamondsuit_{\kappa}[J]$ asserts the existence of $s_{\alpha}\subseteq\alpha$ for $\alpha<\kappa$ such that $\{\alpha<\kappa:s_{\alpha}=A\cap\alpha\}$ lies in $J^{+}$ for every $A\subseteq\kappa$.

REMARK 2.15*.*

1. (i)

If $\diamondsuit_{\kappa}^{\ast}[J]$ holds, then so does $\diamondsuit_{\kappa}^{\ast}[K]$ for any $\kappa$-complete ideal $K$ on $\kappa$ extending $J$. 2. (ii)

$\diamondsuit_{\kappa}^{\ast}[J]\Rightarrow\diamondsuit_{\kappa}^{-}[J]$. 3. (iii)

$\diamondsuit_{\kappa}[J]\Rightarrow\diamondsuit_{\kappa}^{-}[J]$.

FACT 2.16**.**

1. (i)

([25], [30])* Suppose that either $\kappa$ is a successor cardinal, or $J$ is normal. Then $\diamondsuit_{\kappa}^{-}[J]\Rightarrow\diamondsuit_{\kappa}[J]$.* 2. (ii)

([20], [30])* $\diamondsuit_{\kappa}^{-}[J]\Rightarrow 2^{<\kappa}=\kappa$.* 3. (iii)

(Folklore)* If $\diamondsuit_{\kappa}[J]$ holds, then $J$ is not $I_{\kappa}$-$2^{\kappa}$-saturated.*

Gregory’s result [18] asserted that if $\kappa=\nu^{+}=2^{\nu}$, then $\diamondsuit^{\ast}_{\kappa}[NS_{\kappa}|E^{\kappa}_{\theta}]$ holds for any regular infinite cardinal $\theta$ with $\nu^{\theta}=\nu$. It was later strengthened by Shelah ([40], [42]) and others ([38], [27]). Its present form (not necessarily the final one) reads as follows.

FACT 2.17**.**

Suppose that $\kappa=\nu^{+}=2^{\nu}$, and let $\theta$ be a regular infinite cardinal less than $\nu$ such that $d(\theta,\nu)=\nu$. Then $\diamondsuit^{\ast}_{\kappa}[NS_{\kappa}|E^{\kappa}_{\theta}]$ holds.

Let us discuss the requirement that $d(\theta,\nu)=\nu$. By Facts 2.10 (x) and 2.12, under SSH, it reduces to the condition that $d(\theta,\theta)\leq\nu$ (which will be satisfied if $\kappa$ is large enough) and $\theta\not=\mathrm{cf}(\nu)$. On the other hand, if there is a strong limit cardinal $\tau$ with $\theta<\tau<\kappa$, and $\theta$ is large enough, then by Fact 2.13, $d(\theta,\nu)=\nu$ will hold. So there are many cases when the condition $d(\theta,\nu)=\nu$ is satisfied. But what can be said when it is not ? Shelah has the following answer.

FACT 2.18**.**

([47])* Suppose that $\kappa=\nu^{+}=2^{\nu}$, and let $\theta$ be a regular infinite cardinal less than $\kappa$ with $\theta\not=\mathrm{cf}(\nu)$. Then $\diamondsuit_{\kappa}[J]$ holds for any $\kappa$-complete ideal on $\kappa$ extending $NS_{\kappa}|E^{\kappa}_{\theta}$.*

Notice that the result also applies to ideals that are not normal. Of course if $\diamondsuit_{\kappa}[K]$ holds for some normal ideal on $\kappa$, then so does $\diamondsuit_{\kappa}[J]$ for any $\kappa$-complete ideal $J$ on $\kappa$ included in $K$. So the ideals that would not be covered if the result were only stated for normal ideals are those that are not subnormal. The following result provides a description of these ideals.

FACT 2.19**.**

([3])* Given a $\kappa$-complete ideal $J$ on $\kappa$, the following are equivalent :*

1. (i)

$J$* is not subnormal.* 2. (ii)

There is a partition $\langle S_{\alpha}:\alpha<\kappa\rangle$ of $\kappa\setminus\{0\}$ into stationary sets $S_{\alpha}$ with $S_{\alpha}\cap(\alpha+1)=\emptyset$ such that $J$ extends the $\kappa$-complete ideal generated by $NS_{\kappa}\cup\{S_{\alpha}:\alpha<\kappa\}$.

Let us return to Shelah’s result. How does it look like if we go further and remove the remaining cardinal arithmetic hypothesis ? This paper originated in our desire to prove the following.

CONJECTURE 2.20**.**

Suppose that $\kappa=\nu^{+}$, and let $\theta$ be a regular infinite cardinal less than $\kappa$ with $\theta\not=\mathrm{cf}(\nu)$. Then no $\kappa$-complete ideal on $\kappa$ extending $NS_{\kappa}|E^{\kappa}_{\theta}$ is $I_{\kappa}$-$\kappa^{+}$-saturated.

Why only $\kappa^{+}$ ? Just to play it safe, since $2^{\kappa}$ would not be suitable (Foreman and Magidor [11] showed that if $V=L$ and $\sigma$ Cohen subsets of $\omega_{1}$ are added, where $\sigma\geq\kappa^{++}$, then in the extension, $NS_{\kappa}$ is $\kappa^{++}$-saturated).

2.3 J’enlève le bas

DEFINITION 2.21**.**

Given a $\kappa$-complete ideal $J$ on $\kappa$, $\clubsuit_{\kappa}[J]$) asserts the existence of $s_{\alpha}\subseteq\alpha$ with $\sup s_{\alpha}=\alpha$ for $\alpha\in acc(\kappa)$ such that $\{\alpha\in acc(\kappa):s_{\alpha}\subseteq A\}\in J^{+}$ for all $A\in[\kappa]^{\kappa}$.

The principle $\clubsuit_{\kappa}^{\ast}[J]$ asserts the existence of $s^{i}_{\delta}\subseteq\alpha$ with $\sup B^{i}_{\alpha}=\alpha$ for $i\in\alpha\in acc(\kappa)$ such that $\{\alpha<\kappa:\exists i<\alpha(s^{i}_{\alpha}\subseteq A)\}\in J^{\ast}$ for all $A\in[\kappa]^{\kappa}$.

$\clubsuit_{\omega_{1}}[NS_{\omega_{1}}]$ is usually denoted by $\clubsuit$ and known as Ostaszewski’s guessing principle.

It is easy to see that if $J$ extends $NS_{\kappa}$, then $\diamondsuit_{\kappa}[J]$ (respectively, $\diamondsuit^{\ast}_{\kappa}[J]$) implies $\clubsuit_{\kappa}[J]$ (respectively, $\clubsuit_{\kappa}^{\ast}[J])$. By a result of Devlin (see [35]), $\diamondsuit_{\omega_{1}}[NS_{\omega_{1}}]$ follows from CH + $\clubsuit$. This easily generalizes.

OBSERVATION 2.22**.**

Given a $\kappa$-complete ideal $J$ on $\kappa$ extending $NS_{\kappa}$, the following are equivalent :

1. (i)

$\diamondsuit_{\kappa}[J]$* holds.* 2. (ii)

$\clubsuit_{\kappa}[J]$* holds and $2^{<\kappa}=\kappa$.*

Proof. By the proof of Observation 3.4 below. $\square$

The starred version is established by a similar argument.

OBSERVATION 2.23**.**

Given a $\kappa$-complete ideal $J$ on $\kappa$ extending $NS_{\kappa}$, the following are equivalent :

1. (i)

$\diamondsuit^{\ast}_{\kappa}[J]$* holds.* 2. (ii)

$\clubsuit^{\ast}_{\kappa}[J]$* holds and $2^{<\kappa}=\kappa$.*

Observation 2.23 improves a result of [13] that asserts that if $\kappa=\nu^{+}$ and $\tau$ is a regular infinite cardinal less than $\nu$ such that $d(\tau,\sigma)\leq\nu$ for every cardinal $\sigma$ with $\tau\leq\sigma<\nu$, then for any $S\in NS^{+}_{\kappa}\cap P(E^{\kappa}_{\tau})$, $\diamondsuit^{\ast}_{\kappa}[NS_{\kappa}|S]$ holds just in case $\clubsuit^{\ast}_{\kappa}[NS_{\kappa}|S]$ holds and $2^{<\kappa}=\kappa$.

The consistency of $\clubsuit$ with the negation of the Continuum Hypothesis (and therefore with the negation of $\diamondsuit_{\omega_{1}}[NS_{\omega_{1}}]$) has been established by Shelah [41].

DEFINITION 2.24**.**

Given a $\kappa$-complete ideal $J$ on $\kappa$, $\clubsuit_{\kappa}^{\rm ev}[J]$ asserts the existence of $s_{\alpha}\subseteq\alpha$ with $\sup s_{\alpha}=\alpha$ for $\alpha\in acc(\kappa)$ such that $\{\alpha\in acc(\kappa):\exists\beta<\alpha(s_{\alpha}\setminus\beta\subseteq A)\}\in J^{+}$ for all $A\in[\kappa]^{\kappa}$.

Obviously, $\clubsuit_{\kappa}[J]\Rightarrow\clubsuit_{\kappa}^{\rm ev}[J]$. The principle $\clubsuit_{\omega_{1}}^{\rm ev}[NS_{\omega_{1}}]$ is denoted by $\clubsuit_{w}$ in [12], and by $\clubsuit^{1}$ in [8] where its consistency with the negation of $\clubsuit$ is established.

It is known (see [7], [12]) that for any $S\in NS^{+}_{\kappa}$, $\clubsuit_{\kappa}[NS^{+}_{\kappa}|S]$ holds if and only if there is $s_{\alpha}\subseteq\alpha$ with $\sup s_{\alpha}=\alpha$ for $\alpha\in acc(\kappa)$ such that $\{\alpha\in S\cap acc(\kappa):s_{\alpha}\subseteq A)\}\not=\emptyset$ for all $A\in[\kappa]^{\kappa}$. This works also for the eventual-guessing variant.

OBSERVATION 2.25**.**

Given $S\in NS^{+}_{\kappa}$, the following are equivalent :

1. (i)

$\clubsuit_{\kappa}^{\rm ev}[NS^{+}_{\kappa}|S]$. 2. (ii)

There is $s_{\alpha}\subseteq\alpha$ with $\sup s_{\alpha}=\alpha$ for $\alpha\in acc(\kappa)$ such that for any $A\in[\kappa]^{\kappa}$, $\{\alpha\in S\cap acc(\kappa):\exists\beta<\alpha(s_{\alpha}\setminus\beta\subseteq A)\}\not=\emptyset$.

Proof. (i) $\rightarrow$ (ii) : Trivial.

(ii)$\rightarrow$ (i) : Let $\langle s_{\alpha}:\alpha\in acc(\kappa)\rangle$ be as in (ii), and suppose toward a contradiction that (i) fails. Then there must be $A\in[\kappa]^{\kappa}$ and $C\in{\cal C}_{\kappa}\cap P(acc(\kappa))$ such that $(s_{\alpha}\setminus\beta)\setminus A\not=\emptyset$ whenever $\alpha\in C\cap S$ and $\beta<\alpha$. Set $B=\{\min(A\setminus\alpha):\alpha\in C\}$. We may find $\alpha\in S\cap acc(\kappa)$ and $\beta<\alpha$ such that $s_{\alpha}\setminus\beta\subseteq B\subseteq A$. But then $C$ is cofinal in $\alpha$, and consequently $\alpha\in C$. Contradiction. $\square$

REMARK 2.26*.*

Assuming GCH in $V$, Baumgartner [2] has constructed a cofinality-preserving generic extension in which there is $S_{i}\in[\kappa]^{\kappa}$ for $i<\kappa^{+}$ such that $|S_{i}\cap S_{j}|<\aleph_{0}$ whenever $i<j<\kappa^{+}$. In this extension, $\clubsuit_{\kappa}^{\rm ev}[NS_{\kappa}|E^{\kappa}_{\theta}]$ must fail for every regular infinite cardinal $\theta<\kappa$, since it is well-known (see e.g.[12]) that $\clubsuit_{\kappa}^{\rm ev}[NS_{\kappa}|E^{\kappa}_{\theta}]$ implies the existence of $X\subseteq[\kappa]^{\theta}$ with $|X|=\kappa$ such that for any $A\in[\kappa]^{\kappa}$, there is $x\in X$ with $x\subseteq A$.

Garti [14] observed that it follows from $\clubsuit$ that $NS_{\omega_{1}}$ is not $I_{\omega_{1}}$-$\omega_{2}$-saturated. An easy modification of his proof yields the following.

OBSERVATION 2.27**.**

Let $J$ be a $\kappa$-complete ideal on $\kappa$ such that $\clubsuit_{\kappa}^{\rm ev}[J]$ holds, and $\rho$ be an infinite cardinal such that $I_{\kappa}$ is not $\rho$-saturated. Then $J$ is not $I_{\kappa}$-$\rho$-saturated.

Proof. Let $s_{\alpha}\subseteq\alpha$ with $\sup s_{\alpha}=\alpha$ for $\alpha\in acc(\kappa)$ be such that for any $A\in[\kappa]^{\kappa}$, $S_{A}\in J^{+}$, where

$S_{A}=\{\alpha\in acc(\kappa):\exists\beta<\alpha(s_{\alpha}\setminus\beta\subseteq A)\}$.

Pick $A_{i}\in[\kappa]^{\kappa}$ for $i<\rho$ so that $|A_{i}\cap A_{j}|<\kappa$ whenever $i<j<\rho$. We claim that $|S_{A_{i}}\cap S_{A_{j}}|<\kappa$ whenever $i<j<\rho$. Suppose otherwise, and fix $i<j<\kappa$ such that $|S_{A_{i}}\cap S_{A_{j}}|=\kappa$. Inductively define $\alpha_{\xi}\in S_{A_{i}}\cap S_{A_{j}}$ and $\gamma_{\xi}\in s_{\alpha_{\xi}}\cap A_{i}\cap A_{j}$ for $\xi<\kappa$ so that $\gamma_{\xi}>\sup\{\alpha_{\eta}:\eta<\xi\}$. Then $\{\gamma_{\xi}:\xi<\kappa\}$ is a size $\kappa$ subset of $A_{i}\cap A_{j}$. Contradiction. $\square$

Let us now introduce the kind of towers we will be working with.

DEFINITION 2.28**.**

Given an ideal $J$ on $\kappa$, $Y\subseteq P(\kappa)$ and an ordinal $\delta$, a descending (respectively, ascending) $(J,Y)$-tower of length $\delta$ is a sequence $\langle A_{\alpha}:\alpha<\delta\rangle$ such that

• •

$A_{\alpha}\in J^{+}$ for all $\alpha<\delta$ ;

• •

$A_{\beta}\setminus A_{\alpha}\in Y$ (respectively, $A_{\alpha}\setminus A_{\beta}\in Y$) and $A_{\alpha}\setminus A_{\beta}\in J^{+}$ (respectively, $A_{\beta}\setminus A_{\alpha}\in J^{+}$) whenever $\alpha<\beta<\delta$.

A descending (respectively, ascending) $(J,Y)$-tower is maximal if there is no descending (respectively, ascending) $(J,Y)$-tower properly extending it.

OBSERVATION 2.29**.**

Let $\tau\geq 2$ be a cardinal such that there exists a descending (respectively ascending) $(J,Y)$-tower of length $\tau$. Then $J$ is not $Y$-$\tau$-saturated.

Proof. Let $\langle A_{\alpha}:\alpha<\tau\rangle$ be an ascending $(J,Y)$-tower. For $\alpha<\tau$, set $S_{\alpha}=A_{\alpha+1}\setminus A_{\alpha}$. Then clearly $\{S_{\alpha}:\alpha<\tau\}\subseteq J^{+}$. Furthermore $S_{\gamma}\cap S_{\alpha}\subseteq A_{\gamma+1}\setminus A_{\alpha}$ whenever $\gamma<\alpha<\tau$. Descending towers are handled in a similar way. $\square$

DEFINITION 2.30**.**

We let $Depth([\kappa]^{\kappa},\nearrow)$ (respectively $Depth([\kappa]^{\kappa},\searrow)$) denote the least ordinal $\eta$ such that there is no ascending (respectively, descending) $(I_{\kappa},I_{\kappa})$-tower of length $\eta$.

THEOREM 2.31**.**

Let $J$ be a $\kappa$-complete ideal on $\kappa$ extending $NS_{\kappa}$, and $\eta$ be an infinite ordinal less than $Depth([\kappa]^{\kappa},\searrow)$ (respectively $Depth([\kappa]^{\kappa},\nearrow)$). Suppose that $\clubsuit^{\rm ev}_{\kappa}[J]$ holds. Then there exists a descending (respectively, ascending) $(J,I_{\kappa})$-tower of length $\eta$.

Proof. Select $s_{\alpha}\subseteq\alpha$ with $\sup s_{\alpha}=\alpha$ for $\alpha\in acc(\kappa)$ such that

$\{\alpha:\exists\beta<\alpha(s_{\alpha}\setminus\beta\subseteq A)\}\in J^{+}$

for all $A\in[\kappa]^{\kappa}$. For $A\in[\kappa]^{\kappa}$, let $S_{A}$ denote the set of all $\alpha\in acc(\kappa)$ such that $\sup(A\cap\alpha)=\alpha>\sup(s_{\alpha}\setminus A)$. Notice that $S_{A}\in J^{+}$. Now let $\langle A_{i}:i<\eta\rangle$ be an ascending $(I_{\kappa},I_{\kappa})$-tower. Fix $i<j<\eta$, and pick $\delta<\kappa$ so that $A_{i}\setminus\delta\subseteq A_{j}$. Then $S_{A_{i}}\setminus(\delta+1)\subseteq S_{A_{j}}$, and consequently $|S_{A_{i}}\setminus S_{A_{j}}|<\kappa$. Moreover, $S_{A_{j}\setminus A_{i}}\subseteq S_{A_{j}}\setminus S_{A_{i}}$. Thus $\langle S_{A_{i}}:i<\delta\rangle$ is an ascending $(J,I_{\kappa})$-tower. The descending case is left to the reader. $\square$

Thus $\clubsuit^{\rm ev}_{\kappa}[J]$ transmutes almost disjoint families of subsets of $\kappa$ into almost disjoint families of sets in $J^{+}$ of the same power, and descending (respectively, ascending) $(I_{\kappa},I_{\kappa})$-towers into descending (respectively, ascending) $(J,I_{\kappa})$-towers of the same length.

2.4 Depths

To give some background to Theorem 2.31, in this subsection we discuss the existence of ascending (respectively, descending) towers.

DEFINITION 2.32**.**

Given $f,g\in{}^{\kappa}\kappa$ , $f<^{\ast}g$ means that

$|\{\alpha<\kappa:f(\alpha)\geq g(\alpha\}|<\kappa$.

DEFINITION 2.33**.**

We let $\mathfrak{b}_{\kappa}$ denote the least cardinality of any $F\subseteq{}^{\kappa}\kappa$ with the property that there is no $g\in{}^{\kappa}\kappa$ such that $f<^{\ast}g$ for all $f\in F$.

REMARK 2.34*.*

By an argument that goes back to Rothberger, there exist $f_{\alpha}\in{}^{\kappa}\kappa$ for $\alpha<\mathfrak{b}_{\kappa}$ such that

• •

$f_{\alpha}<^{\ast}f_{\beta}$ for $\alpha<\beta<\mathfrak{b}_{\kappa}$ ;

• •

there is no $g\in{}^{\kappa}\kappa$ such that $f_{\alpha}<^{\ast}g$ for all $\alpha\in\mathfrak{b}_{\kappa}$.

Notice that it follows that $\mathfrak{b}_{\kappa}$ is regular.

FACT 2.35**.**

([1])* $\mathfrak{b}_{\kappa}$ is the least cardinality of any $F\subseteq{\cal C}_{\kappa}$ such that for any $A\in[\kappa]^{\kappa}$, there is $C\in F$ with $|A\setminus C|=\kappa$.*

OBSERVATION 2.36**.**

1. (i)

$\mathfrak{b}_{\kappa}$* is the least cardinality of any $F\subseteq{\cal C}_{\kappa}$ such that for any $D\in{\cal C}_{\kappa}$, there is $C\in F$ with $|D\setminus C|=\kappa$.* 2. (ii)

There is a maximal descending $(I_{\kappa},I_{\kappa})$-tower of length ${\mathfrak{b}}_{\kappa}$ consisting of closed unbounded subsets of $\kappa$. 3. (iii)

Let $S\in NS^{+}_{\kappa}$ be such that $NS_{\kappa}|S$ is not $\sigma$-saturated, where $\sigma$ is an infinite cardinal less than or equal to $\mathfrak{b}_{\kappa}$. Then $NS_{\kappa}|S$ is not $I_{\kappa}$-$\sigma$-saturated.

Proof. (i) : Let $F\subseteq{\cal C}_{\kappa}$ with $0<|F|\ <\mathfrak{b}_{\kappa}$. By Fact 2.35, there must be $A\in[\kappa]^{\kappa}$ such that $|A\setminus C|<\kappa$ for all $C\in F$. Set $D=acc(A)$. Then clearly, $D\in{\cal C}_{\kappa}$. Moreover, $|D\setminus C|<\kappa$ for all $C\in F$.

(ii) : By Fact 2.35, we may find $D_{i}\in{\cal C}_{\kappa}$ for $i<{\mathfrak{b}}_{\kappa}$ such that for any $A\in[\kappa]^{\kappa}$, there is $i<{\mathfrak{b}}_{\kappa}$ with $|A\setminus D_{i}|=\kappa$. We inductively define $C_{i}\in{\cal C}_{\kappa}$ as follows. Set $C_{0}=D_{0}$. Now suppose that $i>0$ and $C_{j}$ has been constructed for each $j<i$. By (i), there is $H\in{\cal C}_{\kappa}$ such that $|H\setminus C_{j}|<\kappa$ for all $j<i$. We let $C_{i}=H$ if $i$ is a limit ordinal, and $C_{i}=H\cap acc(C_{i-1})$ otherwise. It is easy to see that $\langle C_{i}:i<{\mathfrak{b}}_{\kappa}\rangle$ is a descending $(I_{\kappa},I_{\kappa})$-tower.

(iii) : Select $A_{\alpha}\in(NS_{\kappa}|S)^{+}$ for $\alpha<\sigma$ such that $A_{\beta}\cap A_{\alpha}\in NS_{\kappa}|S$ whenever $\beta<\alpha<\sigma$. For $\beta<\alpha<\sigma$, pick $C_{\beta\alpha}\in{\cal C}_{\kappa}$ such that $(A_{\beta}\cap S)\cap(A_{\alpha}\cap S)\cap C_{\beta\alpha}=\emptyset$. For each $\alpha<\sigma$, there is by (i) $D_{\alpha}\in{\cal C}_{\kappa}$ such that $|D_{\alpha}\setminus C_{\beta\alpha}|<\kappa$ for all $\beta<\alpha$. Then clearly, $|(A_{\beta}\cap S\cap D_{\beta})\cap(A_{\alpha}\cap S\cap D_{\alpha})|<\kappa$ whenever $\beta<\alpha<\sigma$. $\square$

OBSERVATION 2.37**.**

1. (i)

Let $J$ be a $\kappa$-complete ideal on $\kappa$. If there is a descending $(J,J)$-tower of length $\delta$, where $\delta\leq\kappa$, then there is a descending $(J,\{\emptyset\})$-tower of length $\delta$. 2. (ii)

Let $J$ be a normal ideal on $\kappa$. If there is a descending $(J,J)$-tower of length $\delta$, where $\delta\leq\kappa^{+}$, then there is a descending $(J,I_{\kappa})$-tower of length $\delta$. 3. (iii)

Let $J=NS_{\kappa}|S$ for some $S\in NS^{+}_{\kappa}$. If there is a descending $(J,J)$-tower of length $\delta$, where $\delta\leq\mathfrak{b}_{\kappa}$, then there is a descending $(J,I_{\kappa})$-tower of length $\delta$.

Proof. We prove (iii) and leave the similar proofs of (i) and (ii) to the reader. Thus suppose that $J=NS_{\kappa}|S$, where $S\in NS^{+}_{\kappa}$, and $\langle A_{i}:i<\delta\rangle$ is a descending $(J,J)$-tower of length $\delta$, where $0<\delta\leq\kappa$. We recursively define $B_{i}\in J^{+}\cap P(A_{i}\cap S)$ with $A_{i}\setminus B_{i}\in J$ for $i<\delta$ as follows. Put $B_{0}=A_{0}$. Now suppose that $i>0$, and $B_{j}$ has been constructed for every $j<i$. For each $j<i$, pick $C_{ji}\in{\cal C}_{\kappa}$ so that $(A_{i}\setminus B_{j})\cap S\cap C_{ji}=\emptyset$. By Fact 2.35, there must be $D_{i}\in{\cal C}_{\kappa}$ such that $|D_{i}\setminus C_{ji}|<\kappa$ for all $j<i$. We set $B_{i}=A_{i}\cap S\cap D_{i}$. Notice that given $j<i$, $B_{i}\setminus B_{j}\subseteq D_{i}\setminus C_{ji}$, and therefore $|B_{i}\setminus B_{j}|<\kappa$. Furthermore, $B_{j}\setminus B_{i}\subseteq(A_{j}\setminus A_{i})\setminus(A_{j}\setminus B_{j})$, and consequently $B_{j}\setminus B_{i}\in J^{+}$. $\square$

OBSERVATION 2.38**.**

Let $J$ be a $\kappa$-complete ideal on $\kappa$, and $\sigma$ be a cardinal with $2\leq\sigma\leq\kappa$. Then the following are equivalent :

1. (i)

$J$* is not $\sigma$-saturated.* 2. (ii)

There is a descending $(J,\{\emptyset\})$-tower of length $\sigma$. 3. (iii)

There is an ascending $(J,\{\emptyset\})$-tower of length $\sigma$.

Proof. (ii) $\rightarrow$ (i) and (iii) $\rightarrow$ (i) : By Observation 2.29.

(i) $\rightarrow$ (ii) and (iii) : Let $S_{i}\in J^{+}$ for $i<\sigma$ be such that $S_{i}\cap S_{j}\in J$ whenever $j<i<\sigma$. For $i<\sigma$, set $T_{i}=S_{i}\setminus(\bigcup_{j<i}S_{j})$. Now we can define an ascending $(J,\{\emptyset\})$-tower $\langle A_{i}:i<\sigma\rangle$ by $A_{i}=\bigcup_{j\leq i}T_{j}$, and a descending $(J,\{\emptyset\})$-tower $\langle B_{i}:i<\sigma\rangle$ by $B_{i}=\bigcup_{i\leq k<\sigma}T_{k}$. $\square$

OBSERVATION 2.39**.**

1. (i)

Let $J$ be a $\kappa$-complete ideal on $\kappa$, $\delta$ be a nonzero ordinal, and $\langle A_{i}:i<\delta\rangle$ be a maximal ascending $(J,Y)$-tower, where $Y\subseteq P(\kappa)$. Then $\delta$ is a successor ordinal. 2. (ii)

Let $J$ be a $\kappa$-complete, nowhere prime ideal on $\kappa$, $\delta$ be a nonzero ordinal, and $\langle A_{i}:i<\delta\rangle$ be a maximal descending $(J,Y)$-tower, where $Y$ is a subset of $P(\kappa)$ closed under subsets. Then $\delta$ is not a successor ordinal.

Proof. (i) : Suppose otherwise. Put $A_{\delta}=\bigcup_{i<\delta}A_{i}$. Then $\langle A_{j}:j\leq\delta\rangle$ is an ascending $(J,Y)$-tower. Contradiction.

(ii) : Suppose otherwise, and let $\delta=\xi+1$. Then $A_{\xi}$ can be written as the disjoint union of two members of $J^{+}$, say $B_{0}$ and $B_{1}$. Put $A_{\delta}=B_{0}$. Then $\langle A_{j}:j\leq\delta\rangle$ is a descending $(J,Y)$-tower. Contradiction. $\square$

OBSERVATION 2.40**.**

Let $J$ be a $\kappa$-complete ideal on $\kappa$, and $\sigma$ be a cardinal with $2\leq\sigma\leq\kappa$ such that $J$ is not $\sigma$-saturated. Then the following hold :

1. (i)

Suppose $\sigma\geq\omega$. Then there is a maximal descending $(J,\{\emptyset\})$-tower of length $\sigma$. 2. (ii)

There is a maximal ascending $(J,\{\emptyset\})$-tower of length $\delta$, where $\delta$ equals $\sigma$ if $\sigma<\omega$, and $\sigma+1$ otherwise.

Proof. Use (the proof of) Observations 2.38 and 2.39. $\square$

OBSERVATION 2.41**.**

Let $J$ be a normal ideal on $\kappa$ that is not $\kappa$-saturated. Then there is a maximal descending $(J,J)$-tower of length $\kappa$.

Proof. We use an argument that can be found in [49]. Pick a partition $\langle A_{\alpha}:\alpha<\kappa\rangle$ of $\kappa$ into members of $J^{+}$. Set $B_{0}=\bigcup_{0>\alpha<\kappa}(A_{\alpha}\cap(\alpha+1))$, and $B_{\alpha}=A_{\alpha}\setminus(\alpha+1)$ for $0<\alpha<\kappa$. Put $S_{i}=\bigcup_{\alpha\geq i}B_{\alpha}$ for each $i<\kappa$. Now let $S\subseteq\kappa$ be such that $S\setminus S_{i}\in J$. for all $i<\kappa$. For each $\alpha<\kappa$, we may find $C_{\alpha}\in J^{\ast}$ such that $S\cap B_{\alpha}\cap C_{\alpha}=\emptyset$. Then

$S\cap\bigtriangleup_{\alpha<\kappa}C_{\alpha}=\bigcup_{\alpha<\kappa}(S\cap(B_{\alpha}\cap\bigtriangleup_{\alpha<\kappa}C_{\alpha}))\subseteq 1$,

and therefore $S\in J$. Thus $\langle S_{i}:i<\kappa\rangle$ is a maximal descending $(J,J)$-tower. $\square$

The following is due to Moti Gitik [15].

THEOREM 2.42**.**

Let $J$ be a normal ideal on $\kappa$ that is not $\kappa^{+}$-saturated. Then there is a descending $(J,J)$-tower of length $\kappa^{+}$.

Proof. Pick $A_{\alpha}\in J^{+}$ for $\alpha<\kappa^{+}$ such that $A_{\beta}\cap A_{\alpha}\in J$ whenever $\beta<\alpha<\kappa^{+}$. For $\kappa\leq\alpha<\kappa^{+}$, select a bijection $j_{\alpha}:\kappa\rightarrow\alpha$ and put $B_{\alpha}=\bigtriangleup_{i<\kappa}(\kappa\setminus A_{j_{\alpha}(i)})$. Note that $|A_{\beta}\cap B_{\alpha}|<\kappa$ for all $\beta<\alpha$.

Claim 1. Let $\kappa\leq\beta<\kappa^{+}$. Then $A_{\beta}\setminus B_{\beta}\in J$.

Proof of Claim 1. Suppose otherwise. Define $f:A_{\beta}\setminus B_{\beta}\rightarrow\kappa$ by $f(\xi)=$ the least $i<\xi$ such that $\xi\in A_{j_{\beta}(i)}$. There must be $H\in J^{+}\cap P(A_{\beta}\setminus B_{\beta})$ such that $f$ is constant on $H$. This contradiction completes the proof of the claim.

Claim 2. Let $\kappa\leq\beta<\alpha<\kappa^{+}$. Then $B_{\beta}\setminus B_{\alpha}\in J^{+}$.

Proof of Claim 2. Clearly, $A_{\beta}\setminus(B_{\beta}\setminus B_{\alpha})$ is a subset of $(A_{\beta}\setminus B_{\beta})\cup(A\beta\cap B_{\alpha})$ which by Claim 1 lies in $J$. Hence $B_{\beta}\setminus B_{\alpha}\in J^{+}$, which completes the proof of the claim.

Claim 3. Let $\kappa\leq\beta<\alpha<\kappa^{+}$. Then $B_{\alpha}\setminus B_{\beta}\in J$.

Proof of Claim 3. Suppose otherwise. Define $g:B_{\alpha}\setminus B_{\beta}\rightarrow\kappa$ by $g(\xi)=$ the least $i<\xi$ such that $\xi\in A_{j_{\beta}(i)}$. We may find $G\in J^{+}\cap P(B_{\alpha}\setminus B_{\beta})$ and $i<\kappa$ such that $g$ takes the constant value $i$ on $G$. Let $k<\kappa$ be such that $j_{\beta}(i)=j_{\alpha}(k)$. Then $\xi\leq k$ for all $\xi\in G$. This contradiction completes the proof of the claim and that of the proposition. $\square$

REMARK 2.43*.*

Suppose that in the proof above, the family $\{A_{\alpha}:\alpha<\kappa^{+}\}$ has the additional property that for any $K\in J^{+}$, there is $\alpha$ with $K\cap A_{\alpha}\in J^{+}$. Then our $(J,J)$-tower $\langle B_{\beta}:\kappa\leq\beta<\kappa^{+}\rangle$ is maximal. To see this, recall that $|B_{\beta}\cap A_{\alpha}|<\kappa$ whenever $\alpha<\beta$ and $\kappa\leq\beta<\kappa^{+}$. It follows that if $W\subseteq\kappa$ is such that $W\setminus B_{\beta}\in J$ whenever $\kappa\leq\beta<\kappa^{+}$, then $W\in J$.

DEFINITION 2.44**.**

$Depth({}^{\kappa}\kappa)$ denotes the least ordinal $\eta$ such that there is no increasing sequence $\langle f_{i}:i<\eta\rangle$ in $({}^{\kappa}\kappa,<^{\ast})$.

DEFINITION 2.45**.**

$Depth({\cal C}_{\kappa})$ denotes the least ordinal $\delta$ such that there is no sequence $\langle C_{i}:i<\delta\rangle$ such that

• •

$C_{i}\in{\cal C}_{\kappa}$ ;

• •

$C_{i+1}\subseteq acc(C_{i})$ ;

• •

$|C_{i}\setminus C_{j}|<\kappa$ for all $j<i$.

OBSERVATION 2.46**.**

$Depth({}^{\kappa}\kappa)$* (respectively, $Depth({\cal C}_{\kappa})$) is not the successor of a successor ordinal.*

FACT 2.47**.**

([46])* $\mathfrak{b}_{\kappa}<Depth({\cal C}_{\kappa})\leq Depth({}^{\kappa}\kappa)\leq Depth({\cal C}_{\kappa})+1$.*

Proof. For the first inequality see the proof of Observation 2.36 (ii). To establish the second one, let $\langle C_{i}:i<\delta\rangle$ be such that

• •

$C_{i}\in{\cal C}_{\kappa}$ ;

• •

$C_{i+1}\subseteq acc(C_{i})$ ;

• •

$|C_{i}\setminus C_{j}|<\kappa$ for all $j<i$.

For $S\subseteq\kappa$, let $e_{S}:o.t.(S)\rightarrow S$ be the increasing enumeration of $S$. For $i<\delta$, define $f_{i}\in{}^{\kappa}\kappa$ by $f_{i}(\alpha)=e_{C_{i}}(\alpha+1)$. Now fix $i<\delta$. Then for any $\beta<\kappa$, $e_{C_{i}}(\beta)\leq e_{acc(C_{i})}(\beta)$. Hence for each $\alpha<\kappa$,

$f_{i}(\alpha)<e_{C_{i}}(\alpha+\omega)\leq e_{acc(C_{i})}(\alpha+1)\leq f_{i+1}(\alpha)$.

Given $i+1<j<\delta$, we may find $\xi<\eta<\kappa$ such that $C_{j}\setminus C_{i+1}\subseteq\xi$ and $o.t.(C_{i+1}\cap\eta)=\eta=o.t.(C_{j}\cap\eta)$. Then clearly,

$f_{i}(\alpha)<f_{i+1}(\alpha)\leq f_{j}(\alpha)$

whenever $\eta\leq\alpha<\kappa$.

For the last inequality, given an increasing sequence $\langle f_{i}:i\leq\delta\rangle$ in $({}^{\kappa}\kappa,<^{\ast})$, let

$D$ denote the set of all $\alpha\in acc(\kappa)$ such that

• •

$f_{\delta}(\beta)<\alpha$ for all $\beta<\alpha$ ;

• •

$\omega^{\alpha}=\alpha$.

Now for each $i<\delta$, set

$C_{i}=\{\alpha+\omega^{f_{i}(\alpha)}\cdot\gamma:\alpha\in D$ and $\gamma<f_{i}(\alpha)<f_{\delta}(\alpha)\}$.

It is not difficult to see that

• •

$C_{i}\in{\cal C}_{\kappa}$ ;

• •

$C_{i+1}\subseteq acc(C_{i})$ ;

• •

$|C_{i}\setminus C_{j}|<\kappa$ for all $j<i$.

$\square$

DEFINITION 2.48**.**

For $f\in\kappa^{\kappa}$, let $M_{f}=\{(\alpha,\beta)\in\kappa\times\kappa:\beta\geq f(\alpha)\}$ and $m_{f}=\{(\alpha,\beta)\in\kappa\times\kappa:\beta\leq f(\alpha)\}$.

REMARK 2.49*.*

Since $f\subseteq m_{f}\cap M_{f}$, we have $\{m_{f},M_{f}\}\subset[\kappa\times\kappa]^{\kappa}$.

PROPOSITION 2.50**.**

Let $f,g\in\kappa^{\kappa}$ be such that $f<^{\ast}g$. Then the following hold :

1. (i)

$|M_{g}\setminus M_{f}|<\kappa$, and moreover $|M_{f}\setminus M_{g}|=\kappa$. 2. (ii)

$|m_{f}\setminus m_{g}|<\kappa$, and moreover $|m_{g}\setminus m_{f}|=\kappa$.

Proof. Let $\gamma<\kappa$ be such that $f(\alpha)<g(\alpha)$ for all $\alpha\geq\gamma$. Then the following is readily checked :

• •

$M_{g}\setminus M_{f}\subseteq\bigcup_{\alpha<\gamma}\{(\alpha,\beta):\beta<f(\alpha)\}$.

• •

$\{(\alpha,f(\alpha)):\alpha\geq\gamma\}\subseteq M_{f}\setminus M_{g}$.

• •

$m_{f}\setminus m_{g}\subseteq\bigcup_{\alpha<\gamma}\{(\alpha,\beta):g(\alpha)<\beta\leq f(\alpha)\}$.

• •

$\{(\alpha,g(\alpha)):\alpha\geq\gamma\}\subseteq m_{g}\setminus m_{f}$.

$\square$

COROLLARY 2.51**.**

1. (i)

([46])* $Depth({}^{\kappa}\kappa)\leq Depth([\kappa]^{\kappa},\nearrow)$.* 2. (ii)

$Depth({}^{\kappa}\kappa)\leq Depth([\kappa]^{\kappa},\searrow)$.

REMARK 2.52*.*

By Theorem 2.31, Fact 2.47 and Corollary 2.51, $\clubsuit^{\rm ev}_{\omega_{1}}[NS_{\omega_{1}}]$ implies the existence of a descending (respectively, ascending) $(NS_{\omega_{1}},I_{\omega_{1}})$-tower of length ${\mathfrak{b}}_{\omega_{1}}$. Let us mention in this connection that by a result of Baumgartner and Tall [49], $\diamondsuit_{\omega_{1}}[NS_{\omega_{1}}]$ (and hence $\clubsuit^{\rm ev}_{\omega_{1}}[NS_{\omega_{1}}]$) + $2^{\aleph_{1}}>\aleph_{2}$ + $P_{S}$ is consistent, where $P_{S}$ asserts the following :

For any family $F$ of less than $2^{\aleph_{1}}$ many stationary subsets of $\omega_{1}$ with the property that $\bigtriangleup_{\alpha<\omega_{1}}f(\alpha)\in NS^{+}_{\omega_{1}}$ for all $f:\omega_{1}\rightarrow F$, there is a stationary subset $T$ of $\omega_{1}$ such that $|T\setminus S|<\kappa$ for every $S\in F$.

Notice that it is immediate from Fact 2.35 that $P_{S}$ implies ${\mathfrak{b}}_{\omega_{1}}=2^{\aleph_{1}}$.

REMARK 2.53*.*

Suppose that the GCH holds in $V$, and let $\tau$ and $\sigma$ be two regular cardinals with $\kappa^{+}\leq\tau<\sigma$. Put ${\mathbb{Q}}=(\{0\}\times\tau)\cup(\{1\}\times\sigma)$. For $q,r\in{\mathbb{Q}}$, let $q<r$ just in case either $q=(0,\alpha)$ and $r=(0,\beta)$, where $\alpha<\beta<\tau$, or $q=(1,\gamma)$ and $r=(1,\delta)$, where $\gamma<\delta<\sigma$. Notice that $\tau=$ the least size of any unbounded subset of ${\mathbb{Q}}$. Furthermore ${\mathbb{Q}}$ is well-founded. Hence by a result of Cummings and Shelah [5], there is a $\kappa$-closed, $\kappa^{+}$-cc notion of forcing ${\mathbb{P}}$ such that in $V^{\mathbb{P}}$,

• •

${\mathfrak{b}}_{\kappa}=\tau$.

• •

There are $f_{q}\in{}^{\kappa}\kappa$ for $q\in{\mathbb{Q}}$ such that

1. (a)

for any $g\in{}^{\kappa}\kappa$, there is $q\in{\mathbb{Q}}$ with $g<^{\ast}f_{q}$ ; 2. (b)

for $q,r\in{\mathbb{Q}}$, $q<r$ if and only if $f_{q}<^{\ast}f_{r}$.

Thus in $V^{\mathbb{P}}$, $Depth({}^{\kappa}\kappa)>\sigma$.

REMARK 2.54*.*

Gitik observes that there is a natural forcing (let $P$ be the set of all $(c,F)$ such that $c$ is a closed subset of $\kappa$ of size less than $\kappa$, and $F\in P_{\kappa}({\cal C}_{\kappa})$, with the obvious ordering) that adds $C\in{\cal C}_{\kappa}$ such that $|C\setminus D|<\kappa$ for every $D$ in $({\cal C}_{\kappa})^{V}$. It can be iterated to any length, which tends to indicate that there is no nontrivial upper bound for $Depth({\cal C}_{\kappa})$.

2.5 Interdependent depths

Let us next discuss the following result of Shelah.

FACT 2.55**.**

([46])* Suppose that*

• •

$\rho$* is a singular cardinal of uncountable cofinality such that $\sigma^{\mathrm{cf}(\rho)}<\rho$ for every cardinal $\sigma<\rho$ ;*

• •

$\langle\rho_{i}:i<\mathrm{cf}(\rho)\rangle$* is an increasing, continuous sequence of infinite cardinals with supremum $\rho$ ;*

• •

$f$* is a function from $\mathrm{cf}(\rho)$ to the set of all regular infinite cardinals below $\rho$ such that $f(i)<Depth({\cal C}_{\rho_{i}^{+}})$ for all $i<\mathrm{cf}(\rho)$ ;*

• •

$I$* is a normal ideal on $\mathrm{cf}(\rho)$ ;*

• •

$\pi=tcf(\prod f/I)$.

Then $\pi<Depth({\cal C}_{\rho^{+}})$.

REMARK 2.56*.*

[46] contains more results of the same type.

Let us consider a concrete situation where Fact 2.55 can be applied. In [33] Merimovich constructs from large large cardinals a number of models where GCH massively fails. To be specific let us choose a model $V^{\mathbb{P}}$ in which there are an inaccessible cardinal $\theta$ and $C$ in ${\cal C}_{\theta}$ consisting of infinite cardinals such that for any infinite cardinal $\tau<\theta$, $2^{\tau}$ equals $\sigma^{+3}$ if there is $\sigma\in acc(C)$ such that $\sigma\leq\tau<\sigma^{+3}$, and $\tau^{+}$ otherwise. As pointed out by Gitik to the author, it can be arranged that in $V^{\mathbb{P}}$ , $\mathfrak{b}_{\sigma^{+}}=2^{\sigma}$ for every $\sigma\in acc(C)$. Now working in $V^{\mathbb{P}}$, let $\rho\in acc(C)$ be a singular cardinal of uncountable cofinality, and let $\langle\rho_{i}:i<\mathrm{cf}(\rho)\rangle$ be an increasing, continuous sequence of singular cardinals in $acc(C)$ with supremum $\rho$. Then $pp(\rho)=2^{\rho}=\rho^{+3}$. Hence by Lemma 9.2.9 in [19], there must be $D\in{\cal C}_{\mathrm{cf}(\rho)}$ and a function $f$ from $D$ to the set of all regular infinite cardinals below $\rho$ such that

• •

$tcf(\prod f/J)=\rho^{+3}$, where $J$ denotes the noncofinal ideal on $C$ ;

• •

for any $i\in D$,

$\rho_{i}<f(i)\leq pp_{\mathrm{cf}(\rho)}(\rho_{i})\leq 2^{\rho_{i}}=\rho_{i}^{+3}.$

Let $I$ be the nonstationary ideal on $C$. Then by Lemma 3.17 of [19], $tcf(\prod f/I)=tcf(\prod f/J)=\rho^{+3}$. Hence by Fact 2.55, $Depth({\cal C}_{\rho^{+}})>2^{(\rho^{+})}$.

The interpretation is that $Depth({\cal C}_{\rho^{+}})$ depends on the $Depth({\cal C}_{\rho_{i}^{+}})$’s . But what about $\mathfrak{b}_{\rho^{+}}$ in such a situation ? Does it also depend on the $\mathfrak{b}_{\rho_{i}^{+}}$’s ?

3 J’ENLEVE TOUT

3.1 Fromage ou dessert

Let us return to our starting point, when $\kappa=\nu^{+}$, $\theta$ is a regular cardinal less than $\nu$, and $J$ is a $\kappa$-complete ideal on $\kappa$ extending $NS_{\kappa}|E^{\kappa}_{\theta}$. We just saw that if $\clubsuit^{\rm ev}_{\kappa}[J]$ holds, then there is a descending (respectively, ascending) $(J,I_{\kappa})$-tower of length ${\mathfrak{b}}_{\kappa}$. But what if $\clubsuit^{\rm ev}_{\kappa}[J]$ fails ? To address this problem, we could weaken our club principle in a number of ways. We could for instance allow several guesses instead of just one.

DEFINITION 3.1**.**

Given a $\kappa$-complete ideal $J$ on $\kappa$ and a cardinal $\tau$ with $1\leq\tau<\kappa$, $\clubsuit_{\kappa}^{-/\tau}[J]$ asserts the existence of $B^{i}_{\delta}\subseteq\delta$ with $\sup B^{i}_{\delta}=\delta$ for $\delta\in acc(\kappa)$ and $i<\tau$ such that

$\{\delta\in acc(\kappa):\exists i<\tau(B^{i}_{\delta}\subseteq W)\}\in J^{+}$

for any $W\in[\kappa]^{\kappa}$.

$\clubsuit_{\kappa}^{-}[J]$ asserts the existence of $B^{i}_{\delta}\subseteq\delta$ with $\sup B^{i}_{\delta}=\delta$ for $\delta\in acc(\kappa)$ and $i<\delta$ such that

$\{\delta\in acc(\kappa):\exists i<\delta(B^{i}_{\delta}\subseteq W)\}\in J^{+}$

for any $W\in[\kappa]^{\kappa}$.

Note that $\clubsuit^{\rm ev}_{\kappa}[J]\Rightarrow\clubsuit_{\kappa}^{-}[J]$. Furthermore, if $E^{\kappa}_{\tau}\in J^{\ast}$, then $\clubsuit^{\rm ev}_{\kappa}[J]\Rightarrow\clubsuit_{\kappa}^{-/\tau}[J]$.

OBSERVATION 3.2**.**

1. (i)

Let $\langle B^{i}_{\delta}:\delta\in acc(\kappa)\rangle$ witness that $\clubsuit_{\kappa}^{-/\tau}[J]$ holds, where $J$ is a $\kappa$-complete ideal on $\kappa$, and $\tau$ a cardinal with $1\leq\tau<\kappa$. Then for any $W\in[\kappa]^{\kappa}$, there is $i<\tau$ such that $\{\delta\in acc(\kappa):B^{i}_{\delta}\subseteq W\}\in J^{+}$. 2. (ii)

Let $\langle B^{i}_{\delta}:\delta\in acc(\kappa)\rangle$ witness that $\clubsuit_{\kappa}^{-}[J]$ holds, where $J$ is a normal ideal on $\kappa$. Then for any $W\in[\kappa]^{\kappa}$, there is $i<\kappa$ such that $\{\delta\in acc(\kappa)\setminus(i+1):B^{i}_{\delta}\subseteq W\}\in J^{+}$.

OBSERVATION 3.3**.**

1. (i)

Suppose that $\clubsuit_{\kappa}^{-/\tau}[J]$ holds, where $J$ is a $\kappa$-complete ideal on $\kappa$, and $\tau$ a cardinal with $1\leq\tau<\kappa$, and let $\rho$ be a cardinal such that $\mathrm{cf}(\rho)>\tau$ and $I_{\kappa}$ is not $\rho$-saturated. Then $J$ is not $I_{\kappa}$-$\rho$-saturated. 2. (ii)

Suppose that $\clubsuit_{\kappa}^{-}[J]$ holds, where $J$ is a normal ideal on $\kappa$, and let $\rho$ be a cardinal such that $\mathrm{cf}(\rho)>\kappa$ and $I_{\kappa}$ is not $\rho$-saturated. Then $J$ is not $I_{\kappa}$-$\rho$-saturated.

Proof. By Observation 3.2 and the proof of Observation 2.27. $\square$

OBSERVATION 3.4**.**

Given a $\kappa$-complete ideal $J$ on $\kappa$ extending $NS_{\kappa}$, the following are equivalent :

1. (i)

$\diamondsuit_{\kappa}^{-}[J]$* holds.* 2. (ii)

$\clubsuit_{\kappa}^{-}[J]$* holds and $2^{<\kappa}=\kappa$.*

Proof. (i) $\rightarrow$ (ii) : Use Fact 2.16 (ii).

(ii)$\rightarrow$ (i) : Let $B^{i}_{\delta}\subseteq\delta$ with $\sup B^{i}_{\delta}=\delta$ for $\delta\in acc(\kappa)$ and $i<\delta$ be such that $\{\delta\in acc(\kappa):\exists i<\delta(B^{i}_{\delta}\subseteq W)\}\in J^{+}$ for any $W\in[\kappa]^{\kappa}$. For $\eta<\kappa$, define $\chi_{\eta}:P(\eta)\rightarrow{}^{\eta}2$ by : $\chi(\eta)(a)(\xi)=1$ if and only if $\xi\in a$. Select a bijection $F:\bigcup_{\eta<\kappa}{}^{\eta}2\rightarrow\kappa$. For $\delta\in acc(\kappa)$ and $i<\delta$, put

$s^{i}_{\delta}=\bigcup_{\eta<\kappa}(\bigcup\{a\subseteq\eta:F(\chi_{\eta}(a))\in B^{i}_{\delta}\})$.

Given $A\subseteq\kappa$, let $D$ be the set of all $\delta\in acc(\kappa)$ such that $\sup\{F(\chi_{\zeta}(A\cap\zeta)):\zeta\leq\alpha\}<\delta$ for all $\alpha<\delta$. Note that $D$ belongs to $NS_{\kappa}^{\ast}$ and hence to $J^{\ast}$. Now suppose that $\delta\in D$ and $i<\delta$ are such that $B^{i}_{\delta}\subseteq\{F(\chi_{\eta}(A\cap\eta)):\eta<\kappa\}$.

Claim. $s^{i}_{\delta}\cap\delta=A\cap\delta$.

Proof of the claim.

$\subseteq$ : Let $\alpha\in s^{i}_{\delta}\cap\delta$. We may find $\eta<\kappa$ and $a\subseteq\eta$ such that $\alpha\in a$ and $F(\chi_{\eta}(a))\in B^{i}_{\delta}$. Then clearly, $F(\chi_{\eta}(a))=F(\chi_{\eta}(A\cap\eta))$, and therefore $a=A\cap\eta$. Hence, $\alpha\in A\cap\delta$.

$\supseteq$ : Let $\alpha\in A\cap\delta$. There must be $\gamma\in B^{i}_{\delta}$ such that $\gamma>\sup\{F(\chi_{\zeta}(A\cap\zeta)):\zeta\leq\alpha\}$. Let $\eta<\kappa$ be such that $\gamma=F(\chi_{\eta}(A\cap\eta))$. Since $\eta>\alpha$, we have that $\alpha\in A\cap\eta$. It follows that $\alpha\in s^{i}_{\delta}$, which completes the proof of the claim and that of the observation. $\square$

There is another way to weaken $\clubsuit^{\rm ev}_{\kappa}[J]$. Instead of guessing eventually, we could content ourselves with guessing cofinally. But then we need an extra condition on our guess $B_{\delta}$, otherwise we would achieve success too easily with $B_{\delta}=\delta$.

DEFINITION 3.5**.**

Given a $\kappa$-complete ideal $J$ on $\kappa$ and a cardinal $\sigma<\kappa$, $\clubsuit^{\rm cof/\sigma}_{\kappa}[J]$ asserts the existence of $B_{\delta}\in P_{\sigma}(\delta)$ for $\delta<\kappa$ such that $\{\delta:\sup(W\cap B_{\delta})=\delta\}\in J^{+}$ for any $W\in[\kappa]^{\kappa}$.

$\clubsuit^{\rm cof}_{\kappa}[J]$ asserts the existence of $B_{\delta}\in P_{|\delta|}(\delta)$ for $0<\delta<\kappa$ such that $\{\delta:\sup(W\cap B_{\delta})=\delta\}\in J^{+}$ for any $W\in[\kappa]^{\kappa}$.

Note that if $E^{\kappa}_{\theta}\in J^{\ast}$, where $\theta^{+}<\kappa$, then $\clubsuit^{\rm ev}_{\kappa}[J]\Rightarrow\clubsuit^{\rm cof/\theta^{+}}_{\kappa}[J]$.

We finally settle for a doubly weaker principle.

DEFINITION 3.6**.**

Let $\sigma$ and $\tau$ be two cardinals with $\sigma<\kappa$ and $1\leq\tau<\kappa$, and $J$ be a $\kappa$-complete ideal on $\kappa$. The principle $\clubsuit_{\kappa}^{{\rm cof/\sigma},-/\tau}[J]$ asserts the existence of $B^{i}_{\delta}\in P_{\sigma}(\delta)$ for $\delta<\kappa$ and $i<\tau$ such that for any $W\in[\kappa]^{\kappa}$,

$\{\delta<\kappa:\exists i<\tau(\sup(W\cap B^{i}_{\delta})=\delta)\}\in J^{+}$.

$\clubsuit_{\kappa}^{{\rm cof/\sigma},-}[J]$ (respectively, $\clubsuit_{\kappa}^{{\rm cof},-}[J]$) asserts the existence of $B^{i}_{\delta}$ in $P_{\sigma}(\delta)$ (respectively, $P_{|\delta|}(\delta)$) for $i<\delta<\kappa$ such that for any $W\in[\kappa]^{\kappa}$,

$\{\delta<\kappa:\exists i<\delta(\sup(W\cap B^{i}_{\delta})=\delta)\}\in J^{+}$.

It is easy to see that if $\clubsuit_{\kappa}^{{\rm cof},-}[J]$ holds, then $NS^{\ast}_{\kappa}\subseteq J^{+}$. Notice that if $\kappa=\nu^{+}$, then $\clubsuit_{\kappa}^{{\rm cof/\sigma},-/\nu}[J]$ (respectively, $\clubsuit_{\kappa}^{{\rm cof/\nu},-}[J]$, $\clubsuit_{\kappa}^{-/\nu}[J]$) and $\clubsuit_{\kappa}^{{\rm cof/\sigma},-}[J]$ (respectively, $\clubsuit_{\kappa}^{{\rm cof},-}[J]$, $\clubsuit_{\kappa}^{-}[J]$) are equivalent.

OBSERVATION 3.7**.**

Suppose that $\kappa=\nu^{+}$ and $\clubsuit_{\kappa}^{{\rm cof},-}[J]$ holds. Then the following hold :

1. (i)

$\nu>\omega$. 2. (ii)

$E^{\kappa}_{<\nu}\in J^{+}$, and moreover $\clubsuit_{\kappa}^{{\rm cof},-}[J|E^{\kappa}_{<\nu}]$ holds. 3. (iii)

${\cal C}_{\kappa}\subseteq J^{+}$.

Assuming $\kappa$ is a successor cardinal, $\clubsuit_{\kappa}^{{\rm cof},-}[NS_{\kappa}|S]$ is denoted by $\clubsuit^{-}_{S}$ in [37] where, building on previous work by Džamonja and Shelah [7], Rinot proves that $\clubsuit^{-}_{S}$ implies that $NS_{\kappa}|S$ is not $\kappa^{+}$-saturated. Our presentation will closely follow his. We start with the following technical lemma.

LEMMA 3.8**.**

Suppose that $\kappa=\nu^{+}$, and $J$ is a $\kappa$-complete ideal on $\kappa$ such that $\clubsuit_{\kappa}^{{\rm cof},-}[J]$ holds. Then there exist $A^{i}_{\delta}\in P_{\nu}(\kappa\times\kappa)$ for $\delta<\kappa$ and $i<\nu$ such that for any $F:\kappa\rightarrow\kappa$,

$\{\delta<\kappa:\exists i<\nu\exists Z\subseteq\delta(\sup Z=\delta$ and $F|Z\subseteq A^{i}_{\delta})\}$

lies in $J^{+}$.

Proof. We follow the proof of Lemma 1.5 in [37]. Let ${\cal B}_{\delta}\in P_{\kappa}(P_{\nu}(\delta))$ for $\delta<\kappa$ witness that $\clubsuit_{\kappa}^{{\rm cof},-}[J]$ holds. Set $\chi=\mathrm{cf}(\nu)$. Let $\langle\xi_{j}:j<\chi\rangle$ be an increasing sequence of infinite ordinals with supremum $\nu$, and for each $\gamma\in E^{\kappa}_{\not=\chi}$, select $G^{j}_{\gamma}\subseteq\gamma\times\gamma$ for $j<\chi$ so that

• •

$|G^{j}_{\gamma}|\leq|\xi_{j}|$ ;

• •

$G^{k}_{\gamma}\subseteq G^{j}_{\gamma}$ for all $k<j$ ;

• •

$\bigcup_{j<\chi}G^{j}_{\gamma}=\gamma\times\gamma$.

For $\delta<\kappa$, let $\langle A^{i}_{\delta}:i<\nu\rangle$ be an enumeration of the set

$\{A:\exists B\in{\cal B}_{\delta}\exists j<\chi(A=\bigcup\{G^{j}_{\gamma}:\gamma\in B\cap E^{\kappa}_{\not=\chi}\})\}$.

Now fix $F:\kappa\rightarrow\kappa$. Put $X=\{\gamma\in E^{\kappa}_{\not=\chi}:\forall\alpha<\gamma(F(\alpha)<\gamma)\}$.

Claim 1. Let $\gamma\in X$. Then $F|Y\subseteq G^{j}_{\gamma}$ for some cofinal subset $Y$ of $\gamma$ and some $j<\chi$.

Proof of Claim 1. Pick a cofinal subset $e$ of $\gamma$ of order-type $\mathrm{cf}(\gamma)$. Define $h:e\rightarrow\chi$ by $h(\alpha)=\min\{j<\chi:(\alpha,F(\alpha))\in G^{j}_{\gamma}\}$. Then we may find $Y\subseteq e$ with $|Y|=\mathrm{cf}(\gamma)$, and $j<\chi$ such that $h(\alpha)\leq j$ for all $\alpha\in Y$. Clearly, $Y$ and $j$ are as desired, which completes the proof of the claim.

Using Claim 1, define $g:X\rightarrow\chi$ by $g(\gamma)=$ the least $j$ such that $F|Y\subseteq G^{j}_{\gamma}$ for some cofinal subset $Y$ of $\gamma$. Then we may find $j<\chi$ and a size $\kappa$ subset $W$ of $X$ such that $g$ takes the constant value $j$ on $W$. Then $S\in J^{+}$, where $S=\{\delta<\kappa:\exists B\in{\cal B}_{\delta}(\sup(W\cap B)=\delta)\}$.

Claim 2. Let $\delta\in S$, and let $B\in{\cal B}_{\delta}$ such that $\sup(W\cap B)=\delta$. Then there exists a cofinal subset $Z$ of $\delta$ such that $F|Z\subseteq\bigcup\{G^{j}_{\gamma}:\gamma\in B\cap E^{\kappa}_{\not=\chi}\}$.

Proof of Claim 2. For each $\gamma\in Z\cap B$, select a cofinal subset $Y_{\gamma}$ of $\gamma$ with $F|Y_{\gamma}\subseteq G^{j}_{\gamma}$. Then $W=\bigcup\{Y_{\gamma}:\gamma\in Z\cap B\}$ is as desired, which completes the proof of the claim and that of the lemma.

$\square$

THEOREM 3.9**.**

Suppose that

• •

$\kappa=\nu^{+}$* ;*

• •

$J$* is a $\kappa$-complete ideal on $\kappa$ such that $\clubsuit_{\kappa}^{{\rm cof},-}[J]$ holds ;*

• •

$\tau$* is a regular cardinal less than $Depth({}^{\kappa}\kappa)$.*

Then there exists either an ascending $(J,I_{\kappa})$-tower of length $\tau$, or a descending $(J,J)$-tower of length $\tau$.

Proof. The proof is a modification of that of Theorem 1.10 in [37]. Select $f_{\alpha}\in{}^{\kappa}\kappa$ for $\alpha<\tau$ such that $f_{\alpha}<^{\ast}f_{\beta}$ whenever $\alpha<\beta<\tau$. Let $A^{i}_{\delta}\in P_{\nu}(\kappa\times\kappa)$ for $\delta<\kappa$ and $i<\nu$ be as in the statement of Lemma 3.8. For each $\alpha<\tau$, pick $i_{\alpha}<\nu$ such that

$S_{\alpha}=\{\delta\in E^{\kappa}_{<\nu}:\exists W\subseteq\delta(\sup W=\delta$ and $f_{\alpha}|W\subseteq A^{i_{\alpha}}_{\delta})\}$

lies in $J^{+}$. By thinning out our sequence of functions, we may assume that there is $i<\nu$ such that $i_{\alpha}=i$ for all $\alpha<\tau$.

For $\delta\in E^{\kappa}_{<\nu}$, put

$D_{\delta}=\{j<\delta:\exists r((j,r)\in A^{i}_{\delta})\}$

and

$R_{\delta}=\{r<\delta:\exists j((j,r)\in A^{i}_{\delta})\}$.

For $\delta\in E^{\kappa}_{<\nu}$ and $\alpha<\tau$, define $f_{\alpha\delta}:D_{\delta}\rightarrow R_{\delta}$ by $f_{\alpha\delta}(j)=\min((R_{\delta}\cup\{\kappa\})\setminus f_{\alpha}(j))$. Finally, for $\alpha<\beta<\tau$, let

$S_{\alpha\beta}=\{\delta\in E^{\kappa}_{<\nu}:\sup\{j\in D_{\delta}:f_{\alpha\delta}(j)<f_{\beta\delta}(j)\}=\delta\}.$

Claim 1. Let $\alpha<\beta<\tau$. Then $|S_{\alpha}\setminus S_{\alpha\beta}|<\kappa$ (and hence $S_{\alpha\beta}\in J^{+}$).

Proof of Claim 1. Suppose not. Put $m=\sup\{j<\kappa:f_{\alpha}(j)\geq f_{\beta}(j)\}$, and pick $\delta\in S_{\alpha}\setminus S_{\alpha\beta}$ with $\delta>m$. Note that for any $j\in D_{\delta}$ with $j>m$, we have $f_{\alpha}(j)<f_{\beta}(j)$ and hence $f_{\alpha\delta}(j)\leq f_{\beta\delta}(j)$. Set $n=\sup\{j\in D_{\delta}:f_{\alpha\delta}(j)\not=f_{\beta\delta}(j)\}$. Since $\delta\notin S_{\alpha\beta}$, we have that $n<\delta$. On the other hand, $\delta\in S_{\alpha}$, so there is a cofinal subset $W$ of $\delta$ with $f_{\alpha}|W\subseteq A^{i}_{\delta}$. Now pick $j\in W$ with $j>\max(n,m)$. Then $(j,f_{\alpha}(j))\in A^{i}_{\delta}$, and consequently $j\in D_{\delta}$. Hence $f_{\alpha}(j)=f_{\alpha\delta}(j)=f_{\beta\delta}(j)\geq f_{\beta}(j)$. This contradiction completes the proof of the claim.

Claim 2. Let $\alpha<\beta<\gamma<\tau$. Then $|S_{\beta\gamma}\setminus S_{\alpha\gamma}|<\kappa$ and $|S_{\alpha\beta}\setminus S_{\alpha\gamma}|<\kappa$.

Proof of Claim 2. Let $k_{0}$ (respectively, $k_{1}$) in $\kappa$ be such that $f_{\alpha}(j)<f_{\beta}(j)$ (respectively, $f_{\beta}(j)<f_{\gamma}(j)$) for all $j$ greater than $k_{0}$ (respectively, $k_{1}$). Then for any $j$ greater than $k_{0}$ (respectively, $k_{1}$), $f_{\alpha\xi}(j)\leq f_{\beta\xi}(j)$ (respectively, $f_{\beta\xi}(j)\leq f_{\gamma\xi}(j)$) for all $\xi<\kappa$, and consequently $S_{\beta\gamma}\setminus S_{\alpha\gamma}\subseteq k_{0}$ (respectively, $S_{\alpha\beta}\setminus S_{\alpha\gamma}\subseteq k_{1}$), which completes the proof of the claim.

Since there is no ascending $(J,I_{\kappa})$-tower of length $\tau$, we may find, for each $\alpha<\tau$, $\alpha^{\ast}$ with $\alpha<\alpha^{\ast}<\tau$ such that $S_{\alpha\beta}\setminus S_{\alpha\alpha^{\ast}}\in J$ whenever $\alpha^{\ast}<\beta<\tau$.

Claim 3. Let $\alpha<\beta<\tau$. Then $S_{\beta\beta^{\ast}}\setminus S_{\alpha\alpha^{\ast}}\in J$.

Proof of Claim 3. Pick $\gamma$ with $\max(\alpha^{\ast},\beta^{\ast})<\gamma<\tau$. Then

$S_{\beta\beta^{\ast}}\setminus S_{\alpha\alpha^{\ast}}\subseteq(S_{\beta\beta^{\ast}}\setminus S_{\beta\gamma})\cup(S_{\beta\gamma}\setminus S_{\alpha\gamma})\cup(S_{\alpha\gamma}\setminus S_{\alpha\alpha^{\ast}})$,

which completes the proof of the claim.

Since there is no descending $(J,J)$-tower of length $\tau$, we may find $\gamma<\tau$ such that $S_{\gamma\gamma^{\ast}}\setminus S_{\beta\beta^{\ast}}\in J$ whenever $\gamma<\beta<\tau$. Select $T\in J^{+}\cap P(S_{\gamma\gamma^{\ast}})$ and $\theta<\nu$ such that $|A^{i}_{\delta}|=\theta$ for all $\delta\in T$. Inductively define $g:\theta^{+}\rightarrow\tau\setminus(\gamma+1)$ by : $g(\zeta)$ equals $\gamma+1$ if $\zeta=0$, and $(\sup\{g(\xi)^{\ast}:\xi<\zeta\})+1$ otherwise. Notice that if $\xi<\zeta<\theta^{+}$, then $\{S_{g(\xi)g(\xi)^{\ast}}\bigtriangleup S_{g(\xi)g(\zeta)},S_{g(\xi)g(\xi)^{\ast}}\bigtriangleup S_{\gamma\gamma^{\ast}}\}\subseteq J$, so we may find $C_{\xi\zeta}\in J^{\ast}$ such that $S_{g(\xi)g(\zeta)}\cap C_{\xi\zeta}=S_{\gamma\gamma^{\ast}}\cap C_{\xi\zeta}$. Set $C=\bigcap\{C_{\xi\zeta}:\xi<\zeta<\theta^{+}\}$. Then $T\cap C\subseteq S_{g(\xi)g(\zeta)}$ whenever $\xi<\zeta<\theta^{+}$. Put

$s=(\sup\bigcup_{\xi<\zeta<\theta^{+}}\{j<\kappa:f_{g(\xi)}(j)\geq f_{g(\zeta)}(j)\})+1$,

and pick $\delta\in T\cap C$ with $\delta>s$. Notice that since $\delta\in T$, we have $|R_{\delta}|\leq|A^{i}_{\delta}|<\theta^{+}$. For each $j\in\kappa\setminus s$, the sequence $\langle f_{g(\xi)}(j):\xi<\theta^{+}\rangle$ is strictly increasing. It follows that for any $j\in D_{\delta}\setminus s$, the sequence $\langle f_{g(\xi)\delta}(j):\xi<\theta^{+}\rangle$ is nondecreasing, and in fact eventually constant since $\{f_{g(\xi)\delta}(j):\xi<\theta^{+}\}\subseteq R_{\delta}\cup\{\kappa\}$. Thus we may find $\xi_{j}<\theta^{+}$ such that $f_{g(\xi)\delta}(j)=f_{g(\xi_{j})\delta}(j)$ whenever $\xi_{j}<\xi<\theta^{+}$. Put $\eta=\sup\{\xi_{j}:j\in D_{j}\setminus s\}$, and let $\eta<\xi<\zeta<\theta^{+}$. Then $f_{g(\xi)\delta}|(D_{j}\setminus s)=f_{g(\zeta)\delta}|(D_{j}\setminus s)$. However $\delta\in S_{g(\xi)g(\zeta)}$, and consequently $\ sup\{j\in D_{\delta}:f_{g(\xi)\delta}(j)<f_{g(\zeta)\delta}(j)\}=\delta$. Contradiction ! $\square$

3.2 Silly meeting

DEFINITION 3.10**.**

Given a regular infinite cardinal $\theta$, and a cardinal $\nu>\theta$, we let $M(\theta,\nu)=$ the least cardinality of any $X\subseteq P_{\nu}(\nu)$ with the property that for any $e\in[\nu]^{\theta}$, there is $x\in X$ with $|x\cap e|=\theta$.

OBSERVATION 3.11**.**

Let $\theta$ be a regular infinite cardinal, and $\nu>\theta$ be a cardinal with $\mathrm{cf}(\nu)\not=\theta$. Then $M(\theta,\nu)=\mathrm{cf}(\nu)$.

Proof. Select $x_{i}\in P_{\nu}(\nu)$ for $i<\mathrm{cf}(\nu)$ such that

• •

$x_{j}\subseteq x_{i}$ for all $j<i$.

• •

$\bigcup_{i<\mathrm{cf}(\nu)}x_{i}=\nu$.

Now given $e\in[\nu]^{\theta}$, define $f:e\rightarrow\mathrm{cf}(\nu)$ by $f(\alpha)=$ the least $j$ such that $\alpha\in x_{j}$. There must be $i<\mathrm{cf}(\nu)$ such that $|f^{-1}(i+1)|=\theta$. Then clearly $|x_{i}\cap e|=\theta$. $\square$

FACT 3.12**.**

([37])* Let $\theta$ be a regular infinite cardinal less than $\kappa$. Suppose that $\kappa=\nu^{+}$, where $\mathrm{cf}(\nu)\not=\theta$. Then there is ${\cal B}_{\delta}\subseteq P_{\nu}(\delta))$ with $|{\cal B}_{\delta}|\leq\mathrm{cf}(\nu)$ for $\delta<\kappa$ such that for any $A\in[\kappa]^{\kappa}$,*

$\{\delta<\kappa:\exists B\in{\cal B}_{\delta}(\sup(A\cap B)=\delta)\}\in(NS_{\kappa}|E^{\kappa}_{\theta})^{\ast}$.

Proof. Using Observation 3.11, for $\delta\in E^{\kappa}_{\theta}\setminus\nu$, pick ${\cal B}_{\delta}\subseteq P_{\nu}(\delta)$ with $|{\cal B}_{\delta}|\leq\mathrm{cf}(\nu)$ such that for any $e\in[\delta]^{\theta}$, there is $B\in{\cal B}_{\delta}$ with $|B\cap e|=\theta$. Given $A\in[\kappa]^{\kappa}$, set $C=\{\delta\in acc(\kappa)\setminus\nu:\sup(A\cap\delta)=\delta\}$. Now fix $\delta\in C\cap E^{\kappa}_{\theta}$. Select $e\subseteq A\cap\delta$ with $\sup e=\delta$ and o.t.$(e)=\theta$. We may find $B\in{\cal B}_{\delta}$ such that $|B\cap e|=\theta$. Then clearly, $\sup(B\cap A)=\delta$. $\square$

REMARK 3.13*.*

It obviously follows that if $\kappa=\nu^{+}$, where $\mathrm{cf}(\nu)\not=\theta$, then $\clubsuit_{\kappa}^{{\rm cof},-/cf(\nu)}[J]$ holds for any $\kappa$-complete ideal $J$ on $\kappa$ extending $NS_{\kappa}|E^{\kappa}_{\theta}$. For the case when $\theta=\mathrm{cf}(\nu)<\nu$ and $J=NS_{\kappa}|S$ for some $S\in NS^{+}_{\kappa}\cap P(E^{\kappa}_{\theta})$, see Theorem 2.6 in [37] which gives a condition in terms of approachability for $\clubsuit_{\kappa}^{{\rm cof},-/cf(\nu)}[J]$ to hold.

PROPOSITION 3.14**.**

Let $\theta$ be a regular infinite cardinal less than $\kappa$. Suppose that $\kappa=\nu^{+}$, where $\mathrm{cf}(\nu)\not=\theta$, and $J$ is a $\kappa$-complete ideal on $\kappa$ extending $NS_{\kappa}|E^{\kappa}_{\theta}$. Then for any regular cardinal $\tau<Depth({}^{\kappa}\kappa)$, there exists either an ascending $(J,I_{\kappa})$-tower of length $\tau$, or a descending $(J,J)$-tower of length $\tau$.

Proof. By Theorem 3.9 and Remark 3.13. $\square$

3.3 Slow train

We will now give a proof of Fact 2.18. Let us recall the setting : $\theta<\kappa$ is a regular uncountable cardinal, $\kappa=\nu^{+}=2^{\nu}$, where $\mathrm{cf}(\nu)\not=\theta$, and $J$ is a $\kappa$-complete ideal on $\kappa$ extending $NS_{\kappa}|E^{\kappa}_{\theta}$. By Remark 3.13, we already know that $\clubsuit_{\kappa}^{{\rm cof},-/cf(\nu)}[J]$ holds. We need to show that $\clubsuit_{\kappa}[J]$ holds. Just as Primavesi [36], we are not looking for a concise, beautiful proof. On the contrary, the more steps the better, as we would like to see in slow motion how $\clubsuit_{\kappa}^{{\rm cof},-/cf(\nu)}[J]$ gradually evolves into $\clubsuit_{\kappa}[J]$. The main component of the proof is assertion (i) in the following proposition.

PROPOSITION 3.15**.**

1. (i)

Suppose that $\clubsuit_{\kappa}^{{\rm cof}/\sigma,-/\tau}[J]$ holds, where $\sigma$ is an infinite cardinal with $\kappa^{\sigma}=\kappa$, $\tau$ is a cardinal with $1\leq\tau<\kappa$, and $J$ is a $\kappa$-complete ideal on $\kappa$. Then $\clubsuit_{\kappa}^{-/\tau}[J]$ holds. 2. (ii)

Suppose that $\clubsuit_{\kappa}^{{\rm cof}/\sigma,-}[J]$ holds, where $\sigma$ is an infinite cardinal with $\kappa^{\sigma}=\kappa$ and $J$ is a $\kappa$-complete ideal on $\kappa$. Then $\clubsuit_{\kappa}^{-}[J]$ holds.

Proof. (i) : We modify the proof (which we do not understand) of Theorem 3.5 in [36] that asserts that assertion (i) is valid for any ideal of the form $NS_{\kappa}|S$. Let $s^{i}_{\gamma}\in P_{\sigma}(\gamma)$ for $i<\tau$ and $\gamma\in acc(\kappa)$ witness that $\clubsuit_{\kappa}^{{\rm cof}/\sigma,-/\tau}[J]$ holds. Let $\langle v_{\delta}:\delta<\kappa\rangle$ be a one-to-one enumeration of ${}^{\sigma}\kappa$. Define $Bad:\sigma\times[\kappa]^{\kappa}\rightarrow P(\kappa)$ by $Bad(r,A)=\{\delta\in\kappa:v_{\delta}(r)\notin A\}$. Given $r<\sigma$ and $k:r\rightarrow[\kappa]^{\kappa}$, define $f_{k}:\tau\times acc(\kappa)\rightarrow P_{\sigma}(\kappa)$ by

$f_{k}(i,\gamma)=\{v_{\delta}(r):\delta\in s^{i}_{\gamma}\setminus(\bigcup_{q<r}Bad(q,k(q)))\}$.

Claim. There is $k\in\bigcup_{r<\sigma}{}^{r}(P(\kappa))$ such that

$\{\gamma\in acc(\kappa):\exists i<\tau(\sup f_{k}(i,\gamma)=\gamma\,{\rm and}\,f_{k}(i,\gamma)\subseteq T)\}\in J^{+}$

for all $T\in[\kappa]^{\kappa}$.

Proof of the claim. Suppose otherwise. For each $k\in\bigcup_{r<\sigma}{}^{r}([\kappa]^{\kappa})$, pick $T_{k}\in[\kappa]^{\kappa}$ and $C_{k}\in J^{\ast}\cap P(acc(\kappa))$ such that for any $\gamma\in C_{k}$ and any $i<\tau$, either $\sup f_{k}(i,\gamma)<\gamma$, or $f_{k}(i,\gamma)\setminus T_{k}\not=\emptyset$. Define $H:\sigma\rightarrow[\kappa]^{\kappa}$ so that $H(r)=T_{H|r}$ for all $r<\sigma$. For $r<\sigma$, let $\langle\xi^{r}_{\beta}:\beta<\kappa\rangle$ be the increasing enumeration of $H(r)$. Define $F:\kappa\rightarrow\kappa$ so that $v_{F(\beta)}(r)=\xi^{r}_{\beta}$ for every $\beta<\kappa$ and every $r<\sigma$. Notice that $F$ is one-to-one. Inductively define $\beta_{j}<\kappa$ for $j<\kappa$ so that $\sup\{\max(\beta_{l},F(\beta_{l})):l<j\}<\beta_{j}$. Put $\Delta=\{F(\beta_{j}):j<\kappa\}$. There must be $\gamma\in\bigcap_{r<\sigma}C_{H|r}$ and $i<\tau$ such that $\sup(s^{i}_{\gamma}\cap\Delta)=\gamma$.

Since $|s^{i}_{\gamma}|<\sigma$, we may find $r<\sigma$ such that

$(s^{i}_{\gamma}\setminus\bigcup_{q<r}Bad(q,H(q))\cap Bad(r,H(r))=\emptyset$.

Then $v_{\delta}(r)\in H(r)$ for any $\delta\in s^{i}_{\gamma}\setminus(\bigcup_{q<r}Bad(q,H(q))$, and therefore $f_{H|r}(i,\gamma)\subseteq H(r)$. Given $\alpha<\gamma$, pick $l<j<\kappa$ so that $\{F(\beta_{l}),F(\beta_{j})\}\subseteq s^{i}_{\gamma}$ and $F(\beta_{l})\geq\alpha$. Then $v_{F(\beta_{j})}(r)=\xi^{r}_{\beta_{j}}\geq\beta_{j}>F(\beta_{l})\geq\alpha$. Now clearly, $s^{i}_{\gamma}\cap\Delta\subseteq s^{i}_{\gamma}\setminus Bad(u,H(u))$ for all $u<\kappa$, so $s^{i}_{\gamma}\cap\Delta\subseteq s^{i}_{\gamma}\setminus(\bigcup_{q<r}Bad(q,H(q)))$. Hence, $v_{F(\beta_{j})}(r)\in f_{H|r}(i,\gamma)\setminus\alpha$. Thus $\sup f_{H|r}(i,\gamma)=\gamma$. This contradiction completes the proof of the claim and that of (i).

(ii) : The proof is a straightforward modification of that of (i). $\square$

Primavesi [36] established that if $\kappa^{\tau}=\kappa$ and $\clubsuit_{\kappa}^{-/\tau}[NS_{\kappa}|S]$ holds, where $S\in NS_{\kappa}^{+}$, then so does $\clubsuit_{\kappa}[NS_{\kappa}|S]$. This can be generalized as follows.

PROPOSITION 3.16**.**

Let $\tau$ be a cardinal with $1<\tau<\kappa$ and $\kappa^{\tau}=\kappa$, and $J$ be a $\kappa$-complete ideal on $\kappa$ extending $NS_{\kappa}$. Suppose that $\clubsuit_{\kappa}^{-/\tau}[J]$ holds. Then so does $\clubsuit_{\kappa}[J]$.

Proof. The proof is a modification of that of Theorem 6.2.3 in [36]. Thus let $s^{i}_{\alpha}\subseteq\alpha$ with $\sup s^{i}_{\alpha}=\alpha$ for $\alpha\in acc(\kappa)$ and $i<\tau$ be such that $\{\alpha\in acc(\kappa):\exists i<\tau(s^{i}_{\alpha}\subseteq A)\}\in J^{+}$ for every $A\in[\kappa]^{\kappa}$. Let $\langle e_{\gamma}:\gamma<\kappa\rangle$ be a $\kappa$-to-one enumeration of $[\tau\times\kappa]^{\tau}$. For $\alpha\in acc(\kappa)$ and $i<\tau$, put $t^{i}_{\alpha}=\bigcup_{\gamma\in s^{i}_{\alpha}}\{\delta<\kappa:(i,\delta)\in e_{\gamma}\}$.

We claim that there is $i<\tau$ such that for any $A\in[\kappa]^{\kappa}$, the set of all $\alpha\in acc(\kappa)$ such that $\sup t^{i}_{\alpha}=\alpha$ and $t^{i}_{\alpha}\subseteq A$ lies in $J^{+}$. Suppose otherwise. Then we may find $C_{i}\in J^{\ast}\cap P(acc(\kappa))$ and $A_{i}\in[\kappa]^{\kappa}$ for $i<\tau$ such that for any $i<\tau$ and any $\alpha\in C_{i}$, either $\sup t^{i}_{\alpha}\not=\alpha$, or $t^{i}_{\alpha}\setminus A_{i}\not=\emptyset$. For $i<\tau$, let $\langle a^{i}_{\beta}:\beta<\kappa\rangle$ be the increasing enumeration of $A_{i}$. We inductively define $\beta_{\xi},\gamma_{\xi}<\kappa$ for $\xi<\kappa$ as follows. We let $\beta_{0}=\gamma_{0}=0$. Assuming that $\xi>0$ and $\beta_{\zeta}$ and $\gamma_{\zeta}$ have been constructed for all $\zeta<\xi$, we let $\beta_{\xi}=$ the least $\beta>\sup\{\beta_{\zeta}:\zeta<\xi\}$ such that $\min\{a^{i}_{\beta}:i<\tau\}\geq\xi$, and $\gamma_{\xi}=$ the least $\gamma>\sup\{\gamma_{\zeta}:\zeta<\xi\}$ such that $e_{\gamma}=\{(i,a^{i}_{\beta_{\xi}}):i<\tau\}$. Let $D$ be the set of all $\alpha\in acc(\kappa)$ such that

• •

$\gamma_{\alpha}=\alpha$ ;

• •

$\alpha\in\bigcap_{i<\tau}C_{i}$ ;

• •

$\{\delta<\kappa:(i,\delta)\in e_{\gamma}\}\subseteq\alpha$ for any $\gamma<\alpha$ and and any $i<\tau$.

There must be $\alpha\in D$ and $i<\tau$ such that $s^{i}_{\alpha}\subseteq\{\gamma_{\xi}:\xi<\kappa\}$. It is readily checked that $t^{i}_{\alpha}\subseteq A_{i}\cap\alpha$. Given $i<\zeta<\alpha$, we may find $\xi<\kappa$ such that $\gamma_{\xi}\in s^{i}_{\alpha}\setminus\gamma_{\zeta}$. Since $i<\zeta\leq\xi\leq\beta_{\xi}$, we have $(i,a^{i}_{\beta_{\xi}})\in e_{\beta_{\xi}}$, and therefore $a^{i}_{\beta_{\xi}}\in t^{i}_{\alpha}$. Furthermore, $\zeta\leq\xi\leq a^{i}_{\beta_{\xi}}$. Thus $\sup t^{i}_{\alpha}=\alpha$. Contradiction. $\square$

Assertion (i) in the following proposition is due to Rinot [37] in the case when $J$ is a restriction of $NS_{\kappa}$.

PROPOSITION 3.17**.**

1. (i)

Suppose that $\kappa$ is a successor cardinal, and $J$ is a $\kappa$-complete ideal on $\kappa$ extending $NS_{\kappa}$. Then the following are equivalent :

1. (a)

$\diamondsuit_{\kappa}[J]$* holds.* 2. (b)

$\clubsuit_{\kappa}^{{\rm cof},-}[J]$* holds and $2^{<\kappa}=\kappa$.* 2. (ii)

Suppose that $\kappa$ is weakly inaccessible, and $J$ is a normal ideal on $\kappa$. Then for any infinite cardinal $\sigma<\kappa$, the following are equivalent :

1. (i)

$\diamondsuit_{\kappa}[J]$* holds.* 2. (ii)

$\clubsuit_{\kappa}^{{\rm cof}/\sigma,-}[J]$* holds and $2^{<\kappa}=\kappa$.*

Proof. (i) : By Observation 2.22 and Propositions 3.15 and 3.16.

(ii) : By Fact 2.16, Observation 3.4 and Propositions 3.15 and 3.16. $\square$

Proof of Fact 2.18. By Remark 3.13 and Proposition 3.17. $\square$

Mildenberger [34] showed that for any $S\in NS^{+}_{\omega_{1}}$, if CH and $\clubsuit_{\kappa}^{\rm ev}[NS_{\omega_{1}}|S]$ both hold, then $\diamondsuit_{\omega_{1}}[NS_{\omega_{1}}|S]$ holds (see [24] for more results of this type). This generalizes.

OBSERVATION 3.18**.**

Let $\theta$ be a regular infinite cardinal less than $\kappa$, and $J$ be a $\kappa$-complete ideal on $\kappa$ extending $NS_{\kappa}|E^{\kappa}_{\theta}$ such that $\clubsuit_{\kappa}^{\rm ev}[J]$ holds. Then the following hold :

1. (i)

Suppose that $\kappa^{\theta}=\kappa$. Then $\clubsuit_{\kappa}[J]$ holds. 2. (ii)

Suppose that $2^{<\kappa}=\kappa$. Then $\diamondsuit_{\kappa}[J]$ holds.

Proof. Observe that $\clubsuit_{\kappa}^{-/\theta}[J]$ holds, and appeal to Proposition 3.16 and Observation 2.22. $\square$

By considering other versions of the club principle, one can obtain variants of Proposition 3.16.

DEFINITION 3.19**.**

Given a cardinal $\tau$ with $1\leq\tau<\kappa$ and a $\kappa$-complete ideal $J$ on $\kappa$, $\clubsuit_{\kappa}^{\rm ev/-\tau}[J]$ asserts the existence of $s^{i}_{\alpha}\subseteq\alpha$ with $\sup s^{i}_{\alpha}=\alpha$ for $\alpha\in acc(\kappa)$ and $i<\tau$ such that $\bigcup_{i<\tau}\{\alpha\in acc(\kappa):\exists\beta<\alpha(s^{i}_{\alpha}\setminus\beta\subseteq A\}$ lies in $J^{+}$ for all $A\in[\kappa]^{\kappa}$.

OBSERVATION 3.20**.**

Let $\tau$ be a cardinal with $1<\tau<\kappa$ and $\kappa^{\tau}=\kappa$, and $J$ be a $\kappa$-complete ideal on $\kappa$ extending $NS_{\kappa}$. Then the following hold :

1. (i)

Suppose that $\clubsuit_{\kappa}^{\rm ev/-\tau}[J]$ holds. Then so does $\clubsuit_{\kappa}^{\rm ev}[J]$. 2. (ii)

Suppose that there are $s^{i}_{\alpha}\subseteq\alpha$ for $\alpha\in acc(\kappa)$ and $i<\tau$ such that $\bigcup_{i<\tau}\{\alpha\in acc(\kappa):\sup(s^{i}_{\alpha}\cap A)=\alpha\}\in J^{+}$ for all $A\in[\kappa]^{\kappa}$. Then there are $t_{\alpha}\subseteq\alpha$ with $|t_{\alpha}|\leq\max(\tau,|s^{i}_{\alpha}|)$ for $\alpha\in acc(\kappa)$ such that $\{\alpha\in acc(\kappa):\sup(t_{\alpha}\cap A)=\alpha\}\in J^{+}$ for all $A\in[\kappa]^{\kappa}$.

3.4 The case $\kappa=\nu^{+}$ with $\nu$ singular

We start by recalling the definition of covering numbers.

DEFINITION 3.21**.**

Given four cardinals $\rho_{1},\rho_{2},\rho_{3},\rho_{4}$ with $\rho_{1}\geq\rho_{2}\geq\rho_{3}\geq\omega$ and $\rho_{3}\geq\rho_{4}\geq 2$, $\mathrm{cov}(\rho_{1},\rho_{2},\rho_{3},\rho_{4})$ denotes the least cardinality of any $Z\subseteq P_{\rho_{2}}(\rho_{1})$ such that for any $a\in P_{\rho_{3}}(\rho_{1})$, there is $Q\in P_{\rho_{4}}(Z)$ with $a\subseteq\bigcup Q$.

OBSERVATION 3.22**.**

Let $\tau$, $\chi$ and $\sigma$ be three cardinals such that $1\leq\tau\leq\chi<\sigma<\kappa$ and $\mathrm{cov}(\kappa,\chi^{+},\tau^{+},2)=\kappa$. Suppose that $\clubsuit_{\kappa}^{{\rm cof}/\sigma,-/\tau}[J]$ holds, where $J$ is a $\kappa$-complete ideal on $\kappa$. Then $\clubsuit_{\kappa}^{{\rm cof}/\sigma}[J]$ holds.

Proof. The proof is similar to that of Proposition 3.16. Let $s^{i}_{\alpha}\in P_{\sigma}(\alpha)$ for $i<\tau$ and $\alpha\in acc(\kappa)$ witness that $\clubsuit_{\kappa}^{{\rm cof}/\sigma,-/\tau}[J]$ holds. Pick $Z\subseteq P_{\chi^{+}}(\kappa)$ such that $|Z|=\kappa$ and for any $a\in P_{\tau^{+}}(\kappa)$, there is $z\in Z$ with $a\subseteq z$. Let $\langle z_{\gamma}:\gamma<\kappa\rangle$ be a $\kappa$-to-one enumeration of $Z$. For $\alpha\in acc(\kappa)$ and $i<\tau$, put $t^{i}_{\alpha}=\alpha\cap(\bigcup_{\gamma\in s^{i}_{\alpha}}z_{\gamma})$. Note that $|t^{i}_{\alpha}|\leq\max\{|s^{i}_{\alpha}|,\chi\}<\sigma$.

We claim that there is $i<\tau$ such that for any $A\in[\kappa]^{\kappa}$, the set of all $\alpha\in acc(\kappa)$ such that $\sup(t^{i}_{\alpha}\cap A)=\alpha$ lies in $J^{+}$. Suppose otherwise. Then we may find $C_{i}\in J^{\ast}\cap P(acc(\kappa))$ and $A_{i}\in[\kappa]^{\kappa}$ for $i<\tau$ such that for any $i<\tau$ and any $\alpha\in C_{i}$, $\sup(t^{i}_{\alpha}\cap A_{i})<\alpha$. For $i<\tau$, let $\langle a^{i}_{\beta}:\beta<\kappa\rangle$ be the increasing enumeration of $A_{i}$. We inductively define $\beta_{\xi},\gamma_{\xi}<\kappa$ for $\xi<\kappa$ as follows. We let $\beta_{0}=\gamma_{0}=0$. Assuming that $\xi>0$ and $\beta_{\zeta}$ and $\gamma_{\zeta}$ have been constructed for all $\zeta<\xi$, we let $\beta_{\xi}=$ the least $\beta>\sup\{\beta_{\zeta}:\zeta<\xi\}$ such that $\min\{a^{i}_{\beta}:i<\tau\}\geq\xi$, and $\gamma_{\xi}=$ the least $\gamma>\sup\{\gamma_{\zeta}:\zeta<\xi\}$ such that $\{a^{i}_{\beta_{\xi}}:i<\tau\}\subseteq z_{\gamma}$. Let $D$ be the set of all $\alpha\in acc(\kappa)\cap\bigcap_{i<\tau}C_{i}$ such that $\bigcup_{\gamma<\alpha}z_{\gamma}\subseteq\alpha=\gamma_{\alpha}$. There must be $\alpha\in D$ and $i<\tau$ such that $\sup(s^{i}_{\alpha}\cap\{\gamma_{\xi}:\xi<\kappa\})=\alpha$. Given $i<\zeta<\alpha$, we may find $\xi<\kappa$ such that $\gamma_{\xi}\in s^{i}_{\alpha}\setminus\gamma_{\zeta}$. Then clearly $\xi<\alpha$ and $z_{\gamma_{\xi}}\subseteq\alpha$. Since $a^{i}_{\beta_{\xi}}\in z_{\gamma_{\xi}}$, it follows that $a^{i}_{\beta_{\xi}}\in t^{i}_{\alpha}$. Furthermore, $\zeta\leq\xi\leq a^{i}_{\beta_{\xi}}$. Thus $\sup(t^{i}_{\alpha}\cap A_{i})=\alpha$. Contradiction. $\square$

FACT 3.23**.**

Suppose that $\kappa=\nu^{+}$, where $\nu$ is singular. Then

$\mathrm{cov}(\nu,\nu,(\mathrm{cf}(\nu))^{+},2)=\mathrm{cov}(\kappa,\chi^{+},(\mathrm{cf}(\nu))^{+},2)$

for some cardinal $\chi<\nu$.

Proof. Set $\theta=\mathrm{cov}(\nu,\nu,(\mathrm{cf}(\nu))^{+},2)$. By [44, Observation 5.3 (10) p. 86], there must be a cardinal $\chi<\nu$ such that $\theta=\mathrm{cov}(\nu,\chi,(\mathrm{cf}(\nu))^{+},2)$. Then clearly, $\theta=\mathrm{cov}(\nu,\chi^{+},(\mathrm{cf}(\nu))^{+},2)$. On the other hand, by [44, Observation 5.3 (2) p. 86], $\mathrm{cov}(\kappa,\chi^{+},(\mathrm{cf}(\nu))^{+},2)=\max\{\theta,\kappa\}$. It remains to observe that by [44, Theorem 5.4 p. 88], $\theta\geq\mathrm{pp}(\nu)$. $\square$

PROPOSITION 3.24**.**

Let $\theta$ be a regular infinite cardinal less than $\kappa$. Suppose that $\kappa=\nu^{+}$, where $\mathrm{cf}(\nu)\in\nu\setminus\{\theta\}$ and $\mathrm{cov}(\nu,\nu,(\mathrm{cf}(\nu))^{+},2)=\kappa$, and $J$ is a $\kappa$-complete ideal on $\kappa$ extending $NS_{\kappa}|E^{\kappa}_{\theta}$. Then $\clubsuit_{\kappa}^{{\rm cof}}[J]$ holds.

Proof. $\clubsuit_{\kappa}^{{\rm cof},-/\tau}[J]$ holds by Remark 2.10, and hence so does $\clubsuit_{\kappa}^{{\rm cof}}[J]$ by Observation 3.22 and Fact 3.23. $\square$

Note that by results of Shelah ([44, Theorem 5.4 p. 87], [43]), SSH is equivalent to the statement that $\mathrm{cov}(\nu,\nu,(\mathrm{cf}(\nu))^{+},2)=\nu^{+}$ for any singular cardinal $\nu$.

3.5 Slow train II

This time we would like to retrace the path leading from Fact 3.12 to Fact 2.17.

DEFINITION 3.25**.**

Given a $\kappa$-complete ideal $J$ on $\kappa$ and a cardinal $\sigma<\kappa$, $\clubsuit^{\rm cof/\sigma,\ast}_{\kappa}[J]$ asserts the existence of $B^{i}_{\delta}\in P_{\sigma}(\delta))$ for $i<\delta<\kappa$ such that for any $A\in[\kappa]^{\kappa}$,

$\{\delta<\kappa:\exists i<\delta(\sup(A\cap B^{i}_{\delta})=\delta)\}\in J^{\ast}$.

We will follow Fuchs and Rinot who established [13] that if $\kappa=\nu^{+}=2^{\nu}$ and $\clubsuit^{\rm cof/\nu,\ast}_{\kappa}[NS_{\kappa}|S]$ holds for $S\in NS^{+}_{\kappa}|E^{\kappa}_{\theta}$, where $\theta$ is a regular infinite cardinal less than $\nu$ such that $d(\theta,\mu)\leq\nu$ for any cardinal $\mu$ with $\theta\leq\mu<\nu$, then $\diamondsuit^{\ast}_{\kappa}[NS_{\kappa}|S]$ holds. However we will make an extra stop at $\clubsuit_{\kappa}^{\ast}$.

PROPOSITION 3.26**.**

Let $\theta<\sigma$ be two infinite cardinals such that $\mathrm{cf}(\theta)=\theta$, $\sigma<\kappa$, and $d(\theta,\mu)<\kappa$ for any cardinal $\mu$ with $\theta\leq\mu<\sigma$. Further let $J$ be a $\kappa$-complete ideal on $\kappa$ extending $NS_{\kappa}|E^{\kappa}_{\theta}$ such that $\clubsuit_{\kappa}^{\rm cof/\sigma,\ast}[J]$ holds. Then $\clubsuit_{\kappa}^{\ast}[J]$ holds.

Proof. Let $B^{i}_{\delta}\in P_{\sigma}(\delta))$ for $i<\delta<\kappa$ witness that $\clubsuit_{\kappa}^{\rm cof/\sigma,\ast}[J]$ holds. For $i<\delta<\kappa$, we define $Z^{i}_{\delta}$ as follows. If $|B^{i}_{\delta}|<\theta$, put $Z^{i}_{\delta}=\emptyset$. Otherwise let $Z^{i}_{\delta}$ be a cofinal subset of $([B^{i}_{\delta}]^{\theta},\supseteq)$ of size $d(\theta,|B^{i}_{\delta}|)$. For $\delta<\kappa$, set ${\cal A}_{\delta}=\bigcup\{Z^{i}_{\delta}:i<\delta$   and  $|B^{i}_{\delta}|\geq\theta\}$. Note that if $\delta\geq\sup\{d(\theta,\mu):\theta\leq\mu<\sigma\}$, then $|{\cal A}_{\delta}|\leq|\delta|$. Given $A\in[\kappa]^{\kappa}$, we may find $C\in J^{\ast}$ such that for any $\delta\in C$, there is $i<\delta$ with $\sup(A\cap B^{i}_{\delta})=\delta$. Now fix $\delta\in C\cap E^{\kappa}_{\theta}$. Pick $i<\delta$ and $e\subseteq A\cap B^{i}_{\delta}$ so that $\sup e=\delta$ and o.t.$(e)=\theta$. There must be $z\in Z^{i}_{\delta}$ such that $z\subseteq e$. Then $z\in{\cal A}_{\delta}$. Moreover, $z\subseteq A$ and $\sup z=\delta$. $\square$

COROLLARY 3.27**.**

Suppose that $\kappa=\nu^{+}$, and $\theta$ is a regular infinite cardinal less than $\nu$ such that $d(\theta,\nu)=\nu$. Then $\clubsuit^{\ast}_{\kappa}[NS_{\kappa}|E^{\kappa}_{\theta}]$ holds.

Proof. Use Fact 3.12. $\square$

Proof of Fact 2.17. By Observation 2.23 and Corollary 3.27. $\square$

3.6 More clubbing

Corollary 3.27 can be easily generalized to weakly inaccessible cardinals, but it is not yet clear what these weak generalizations (already considered in [6] and [47]) are good for. The following strengthens a result of Džamonja [6].

OBSERVATION 3.28**.**

There is ${\cal B}_{\delta}\subseteq\{B\subseteq\delta:\sup B=\delta\}$ with $|{\cal B}_{\delta}|\leq d(\mathrm{cf}(\delta),|\delta|)$ for $\delta\in acc(\kappa)$ such that for any $A\in[\kappa]^{\kappa}$, there exists $C\in{\cal C}_{\kappa}$ with the property that $C\subseteq\{\delta\in acc(\kappa):\exists B\in{\cal B}_{\delta}(B\subseteq A)\}$.

Proof. For $\delta\in acc(\kappa)$, pick ${\cal A}_{\delta}\subseteq[\delta]^{\mathrm{cf}(\delta)}$ with $|{\cal A}_{\delta}|=d(\mathrm{cf}(\delta),|\delta|)$ such that for any $e\in[\delta]^{\mathrm{cf}(\delta)}$, there is $B\in{\cal A}_{\delta}$ with $B\subseteq e$. Given $A\in[\kappa]^{\kappa}$, let $C=\{\delta\in acc(\kappa):\sup(A\cap\delta)=\delta\}$. Now fix $\delta\in C$. Select $e\subseteq A\cap\delta$ such that o.t.$(e)=\mathrm{cf}(\delta)$ and $\sup e=\delta$. There must be $B\in{\cal A}_{\delta}$ such that $B\subseteq e$. Then clearly $\sup B=\delta$, and moreover $B\subseteq A$.

$\square$

3.7 Order-type versus cardinality

To conclude this section let us mention the following result of Džamonja and Shelah that yields a variant of $\clubsuit_{\kappa}^{{\rm cof}}$ where the condition on $|s_{\alpha}|$ is replaced with one on o.t.$(s_{\alpha})$.

FACT 3.29**.**

([9])* Suppose that $\kappa=\nu^{+}$, where $\nu$ is regular, and $\theta$ and $\rho$ are two regular infinite cardinals with $\theta<\rho<\nu$. Suppose further that either $\theta>\omega$, or $\nu\geq 2^{\aleph_{0}}$, and let $S\in NS_{\kappa}^{+}\cap P(E^{\kappa}_{\theta})$ with the property that $S$ reflects at stationarily many $\delta\in E^{\kappa}_{\rho}$. Then there is $s_{\alpha}\subseteq\alpha$ for $\alpha\in S$ with o.t.$(s_{\alpha})<\nu^{+\omega}\cdot\rho$ such that $\{\alpha\in S:\sup(A\cap s_{\alpha})=\alpha\}\in NS^{+}_{\kappa}$ for all $A\in[\kappa]^{\kappa}$.*

4 GUESSING GENERALIZED CLUBS

In this section we revisit another result of Rinot where nonsaturation is derived from guessing generalized clubs. Let us start with some definitions.

DEFINITION 4.1**.**

Let $\sigma$ be an infinite cardinal, and $\delta$ be a limit ordinal greater than or equal to $\sigma$. A subset $C$ of $P_{\sigma}(\delta)$ is a generalized club if $\{x\in P_{\sigma}(\delta):F``P_{\omega}(x)\subseteq x\}\subseteq C$ for some $F:P_{\omega}(\delta)\rightarrow\delta$.

OBSERVATION 4.2**.**

1. (i)

If $\sigma=\omega$, then there is an empty generalized club subset of $P_{\sigma}(\delta)$. 2. (ii)

If $\sigma>\omega$, then any generalized club subset of $P_{\sigma}(\delta)$ is cofinal in $(P_{\sigma}(\delta),\subseteq)$.

DEFINITION 4.3**.**

Given an infinite cardinal $\sigma<\kappa$ and a $\kappa$-complete ideal $J$ on $\kappa$, $\curlywedge^{-}(\sigma,\kappa,J)$) asserts the existence of a cofinal subset $C^{i}_{\delta}$ of $(P_{\sigma}(\delta),\subseteq)$ for $i\in\delta\in acc(\kappa)\setminus\sigma$ such that $\{\delta:\exists i<\delta\,(C^{i}_{\delta}\subseteq D)\}\in J^{+}$ for every generalized club subset $D$ of $P_{\sigma}(\kappa)$.

$\curlywedge(\sigma,\kappa,J)$ asserts the existence of a generalized club subset $C_{\delta}$ of $P_{\sigma}(\delta)$ for $\delta\in acc(\kappa)\setminus\sigma$ such that $\{\delta:C_{\delta}\subseteq D\}\in J^{+}$ for every generalized club subset $D$ of $P_{\sigma}(\kappa)$.

Note that $\curlywedge(\sigma,\kappa,J)\Rightarrow\curlywedge^{-}(\sigma,\kappa,J)\Rightarrow\sigma>\omega$. $\curlywedge(\sigma,\kappa,NS_{\kappa}|S)$ (respectively, $\curlywedge^{-}(\sigma,\kappa,NS_{\kappa}|S)$) is identical with the principle $\curlywedge(\sigma,S)$ (respectively, $\curlywedge^{-}(\sigma,S)$) introduced in [39]. For the associated starred version see [22].

OBSERVATION 4.4**.**

The following are equivalent :

1. (i)

$\curlywedge^{-}(\sigma,\kappa,J)$* holds.* 2. (ii)

There is a subset $X^{i}_{\delta}$ of $P_{\sigma}(\delta)$ with the property that $\delta\subseteq\bigcup X^{i}_{\delta}$ for $i\in\delta\in acc(\kappa)\setminus\sigma$ such that $\{\delta:\exists i<\delta\,(X^{i}_{\delta}\subseteq D)\}\in J^{+}$ for every generalized club subset $D$ of $P_{\sigma}(\kappa)$.

Proof. It suffices to prove that (ii) implies (i) since the other direction is trivial. Thus let $X^{i}_{\delta}$ for $i\in\delta\in acc(\kappa)\setminus\sigma$ be as in (ii). For $\delta\in acc(\kappa)\setminus\sigma$, select a cofinal subset $Z_{\delta}$ of $P_{\sigma}(\delta)$, and $h:\delta\times\delta\rightarrow X^{i}_{\delta}$ such that $\alpha\in h(i,\alpha)$ for each $(i,\alpha)\in\delta\times\delta$. Set $C^{i}_{\delta}=\{\bigcup_{\alpha\in z}h(i,\alpha):z\in Z_{\delta}\}$. Now given a generalized club subset $D$ of $P_{\sigma}(\kappa)$, pick $F:P_{\omega}(\kappa)\rightarrow\kappa$ such that $D^{\prime}\subseteq D$, where $D^{\prime}=\{x\in P_{\rho}(\delta):F``P_{\omega}(x)\subseteq x\}$. Then clearly, $C^{i}_{\delta}\subseteq D^{\prime}\subseteq D$ for any $\delta\in acc(\kappa)\setminus\sigma$ with $X^{i}_{\delta}\subseteq D^{\prime}$. $\square$

COROLLARY 4.5**.**

If $\curlywedge^{-}(\sigma,\kappa,J)$ holds, then so does $\curlywedge^{-}(\sigma^{\prime},\kappa,J)$ for every cardinal $\sigma^{\prime}$ with $\sigma\leq\sigma^{\prime}<\kappa$.

The following is essentially due to Shelah.

OBSERVATION 4.6**.**

Suppose that $J$ extends $NS_{\kappa}$, and $\diamondsuit_{\kappa}[J]$ holds. Then for any ucountable cardinal $\sigma<\kappa$, $\curlywedge(\sigma,\kappa,J)$ holds.

Proof. Let $\langle s_{\delta}:\delta<\kappa\rangle$ witness that $\diamondsuit_{\kappa}[J]$ holds. Select a bijection $j:P_{\omega}(\kappa)\times\kappa\rightarrow\kappa$. For $\delta\in acc(\kappa)$, define $f_{\delta}:P_{\omega}(\delta)\rightarrow\delta$ by : $f_{\delta}(e)$ equals [math] if $\{\xi\in\delta:j(e,\xi)\in s_{\gamma}\}=\emptyset$, and $\min\{\xi\in\delta:j(e,\xi)\in s_{\gamma}\}$ otherwise. Now let $\sigma$ be a fixed uncountable cardinal less than $\kappa$. Set $C_{\delta}=\{x\in P_{\sigma}(\delta):f_{\delta}``P_{\omega}(x)\subseteq x\}$. Given a generalized club subset $D$ of $P_{\sigma}(\kappa)$ and $H\in J^{\ast}$, pick $F:P_{\omega}(\kappa)\rightarrow\kappa$ with $\{x\in P_{\omega}(\kappa):F``P_{\omega}(x)\subseteq x\}\subseteq D$. We may find $\delta\in H$ such that $F``P_{\omega}(\delta)\subseteq\delta=j``(P_{\omega}(\delta)\times\delta)$ and $s_{\delta}=\{j(e,F(e)):e\in P_{\omega}(\kappa)\}\cap\delta$. Then it is readily checked that $C_{\delta}\subseteq D$. $\square$

Rinot [39] established that if $\kappa=\nu^{+}$, where $\nu$ is regular, and $S\in NS^{+}_{\kappa}\cap P(E^{\kappa}_{\nu})$ is such that $\curlywedge^{-}(\sigma,\kappa,NS_{\kappa}|S)$ holds for some $\sigma<\kappa$, then $NS_{\kappa}|S$ is not $\kappa^{+}$-saturated. This can be extended as follows.

THEOREM 4.7**.**

1. (i)

Suppose that

• •

$\kappa=\nu^{+}$* ;*

• •

$\sigma$* is an uncountable cardinal less than $\nu$ ;*

• •

$J$* is a $\kappa$-complete ideal on $\kappa$ such that $E^{\kappa}_{>\sigma}\in J^{\ast}$ and $\curlywedge^{-}(\sigma,\kappa,J)$ holds ;*

• •

$\tau$* is a regular cardinal less than $Depth({}^{\kappa}\kappa)$.*

Then there exists either an ascending $(J,I_{\kappa})$-tower of length $\tau$, or a descending $(J,J)$-tower of length $\tau$. 2. (ii)

Suppose that

• •

$\kappa$* is weakly inaccessible ;*

• •

$\sigma$* is an uncountable cardinal less than $\kappa$ ;*

• •

$J$* is a normal ideal on $\kappa$ such that $E^{\kappa}_{>\sigma}\in J^{\ast}$ and $\curlywedge^{-}(\sigma,\kappa,J)$ holds ;*

• •

$\tau$* is a regular cardinal less than $Depth({}^{\kappa}\kappa)$.*

Then there exists either an ascending $(J,I_{\kappa})$-tower of length $\tau$, or a descending $(J,J)$-tower of length $\tau$.

Proof. We prove (ii) and leave the similar proof of (i) to the reader. The proof is a modification of that of Theorem 3.4 in [39]. Select $f_{\alpha}\in{}^{\kappa}\kappa$ for $\alpha<\tau$ such that $f_{\alpha}<^{\ast}f_{\beta}$ whenever $\alpha<\beta<\tau$. Let $C^{i}_{\delta}$ for $i\in\delta\in acc(\kappa)\setminus\sigma$ witness that $\curlywedge^{-}(\sigma,\kappa,J)$ holds. For $\delta\in E^{\kappa}_{>\sigma}$, select $h_{\delta}:\delta\times\delta\rightarrow P_{\sigma}(\delta)$ such that $j\in h_{\delta}(i,j)\in C^{i}_{\delta}$ for all $(i,j)\in\delta\times\delta$. For $\delta\in E^{\kappa}_{>\sigma}$ and $\alpha<\tau$, define $f^{\delta}_{\alpha}:\delta\times\delta\rightarrow\delta\cup\{\kappa\}$ by $f_{\alpha}^{\delta}(i,j)=\min((h_{\delta}(i,j)\cup\{\kappa\})\setminus f_{\alpha}(j))$. Notice that if $f_{\alpha}(j)\leq f_{\beta}(j)$, then $f_{\alpha}^{\delta}(i,j)\leq f_{\beta}^{\delta}(i,j)$. Finally, for $\alpha<\beta<\tau$ and $i<\kappa$, let

$S^{i}_{\alpha\beta}=\{\delta\in E^{\kappa}_{>\sigma}\setminus(i+1):\sup\{j\in\delta:f^{\delta}_{\alpha}(i,j)=f_{\beta}^{\delta}(i,j)\}<\delta\}.$

Claim 1. Let $\alpha<\beta<\gamma<\tau$ and $i<\kappa$. Then the following hold :

1. (i)

$|S^{i}_{\alpha\beta}\setminus S^{i}_{\alpha\gamma}|<\kappa$. 2. (ii)

$|S^{i}_{\beta\gamma}\setminus S^{i}_{\alpha\gamma}|<\kappa$.

Proof of Claim 1. Pick $\eta<\kappa$ so that $f_{\alpha}(j)<f_{\beta}(j)<f_{\gamma}(j)$ whenever $\eta\leq j<\kappa$. Let us first show that $S^{i}_{\alpha\beta}\setminus(\eta+1)\subseteq S^{i}_{\alpha\gamma}$. Given $\delta\in S^{i}_{\alpha\beta}\setminus(\eta+1)$, put $A=\{j<\delta:f^{\delta}_{\alpha}(i,j)=f^{\delta}_{\beta}(i,j)\}\cup\eta$. Then clearly, $f^{\delta}_{\alpha}(i,j)<f^{\delta}_{\beta}(i,j)\leq f^{\delta}_{\gamma}(i,j)$ for all $j\in\delta\setminus A$. Thus $\{j<\delta:f^{\delta}_{\alpha}(i,j)=f^{\delta}_{\gamma}(i,j)\}\subseteq A$, and consequently $\delta\in S^{i}_{\alpha\gamma}$. Next we show that $S^{i}_{\beta\gamma}\setminus(\eta+1)\subseteq S^{i}_{\alpha\gamma}$. Given $\delta\in S^{i}_{\beta\gamma}\setminus(\eta+1)$, put $B=\{j<\delta:f^{\delta}_{\beta}(i,j)=f^{\delta}_{\gamma}(i,j)\}\cup\eta$. Then clearly, $f^{\delta}_{\alpha}(i,j)\leq f^{\delta}_{\beta}(i,j)<f^{\delta}_{\gamma}(i,j)$ for all $j\in\delta\setminus B$. Hence $\{j<\delta:f^{\delta}_{\alpha}(i,j)=f^{\delta}_{\gamma}(i,j)\}\subseteq B$, and therefore $\delta\in S^{i}_{\alpha\gamma}$, which completes the proof of the claim.

Claim 2. There is $v:\tau\rightarrow\kappa$ such that $S^{v(\alpha)}_{\alpha\beta}\in J^{+}$ whenever $\alpha<\beta<\tau$.

Proof of Claim 2. Fix $\alpha<\kappa$. Set $k=(\sup\{j<\kappa:f_{\alpha}(j)\geq f_{\alpha+1}(j)\})+1$. Define $F:P_{\omega}(\kappa)\rightarrow\kappa$ by :

• •

$F(e)=0$ if $e=\emptyset$ ;

• •

$F(e)=\xi+1$ if $e=\{\xi\}$ :

• •

$F(e)=f_{\alpha}(\max e)$ if $|e|=2$ ;

• •

$F(e)=f_{\alpha+1}(\max e)$ if $|e|>2$.

Put $D=\{x\in P_{\sigma}(\kappa):F``P_{\omega}(x)\subseteq x\}$. Note that $\{f_{\alpha}(\zeta),f_{\alpha+1}(\zeta)\}\subseteq x$ whenever $x\in D$ and $\zeta\in x\setminus\omega$. By normality of $J$, there must be $i<\kappa$ such that

$T=\{\delta\in E^{\kappa}_{>\sigma}\setminus\max(i+1,k+1):C^{i}_{\delta}\subseteq D\}$

lies in $J^{+}$. If $\delta\in T$, then for any $j\in\delta$ with $j\geq\max(k,\omega)$, we have that $j\in h_{\delta}(i,j)\in D$, so $f^{\delta}_{\alpha}(i,j)=f_{\alpha}(j)<f_{\alpha+1}(j)=f^{\delta}_{\alpha+1}(i,j)$. Thus $T\subseteq S^{i}_{\alpha\alpha+1}$. Hence $S^{i}_{\alpha(\alpha+1)}$ lies in $J^{+}$, and by Claim 1 so does $S^{i}_{\alpha\beta}$ for every $\beta>\alpha$, which completes the proof of the claim.

There must be $i<\kappa$ such that $|v^{-1}(\{i\})|=\tau$. By thinning out our sequence of functions, we may assume that $v(\alpha)=i$ for all $\alpha<\tau$.

Since there is no ascending $(J,I_{\kappa})$-tower of length $\tau$, we may find, for each $\alpha<\tau$, $\alpha^{\ast}$ with $\alpha<\alpha^{\ast}<\tau$ such that $S^{i}_{\alpha\beta}\setminus S^{i}_{\alpha\alpha^{\ast}}\in J$ whenever $\alpha^{\ast}<\beta<\tau$.

Claim 3. Let $\alpha<\beta<\tau$. Then $S^{i}_{\beta\beta^{\ast}}\setminus S^{i}_{\alpha\alpha^{\ast}}\in J$.

Proof of Claim 3. Pick $\gamma$ with $\max(\alpha^{\ast},\beta^{\ast})<\gamma<\tau$. Then

$S^{i}_{\beta\beta^{\ast}}\setminus S{i}_{\alpha\alpha^{\ast}}\subseteq(S^{i}_{\beta\beta^{\ast}}\setminus S^{i}_{\beta\gamma})\cup(S^{i}_{\beta\gamma}\setminus S^{i}_{\alpha\gamma})\cup(S^{i}_{\alpha\gamma}\setminus S^{i}_{\alpha\alpha^{\ast}})$,

which completes the proof of the claim.

Since there is no descending $(J,J)$-tower of length $\tau$, we may find $\gamma<\tau$ such that $S_{\gamma\gamma^{\ast}}\setminus S_{\beta\beta^{\ast}}\in J$ whenever $\gamma<\beta<\tau$. Select $T\in J^{+}\cap P(S^{i}_{\gamma\gamma^{\ast}})$ and $\theta<\sigma$ such that $\sup\{j<\delta:|h_{\delta}(i,j)|=\theta\}=\delta$ for all $\delta\in T$. Inductively define $g:\theta^{+}\rightarrow\tau\setminus(\gamma+1)$ by : $g(\zeta)$ equals $\gamma+1$ if $\zeta=0$, and $(\sup\{g(\xi)^{\ast}:\xi<\zeta\})+1$ otherwise.

Notice that if $\xi<\zeta<\theta^{+}$, then $\{S^{i}_{g(\xi)g(\xi)^{\ast}}\bigtriangleup S^{i}_{g(\xi)g(\zeta)},S^{i}_{g(\xi)g(\xi)^{\ast}}\bigtriangleup S^{i}_{\gamma\gamma^{\ast}}\}\subseteq J$, so we may find $C_{\xi\zeta}\in J^{\ast}$ such that $S^{i}_{g(\xi)g(\zeta)}\cap C_{\xi\zeta}=S^{i}_{\gamma\gamma^{\ast}}\cap C_{\xi\zeta}$. Set $C=\bigcap\{C_{\xi\zeta}:\xi<\zeta<\theta^{+}\}$. Then $T\cap C\subseteq S^{i}_{g(\xi)g(\zeta)}$ whenever $\xi<\zeta<\theta^{+}$. Put

$s=(\sup\bigcup_{\xi<\zeta<\theta^{+}}\{j<\kappa:f_{g(\xi)}(j)\geq f_{g(\zeta)}(j)\})+1$,

and pick $\delta\in T\cap C$ with $\delta>s$. For $\xi<\theta^{+}$, set $W_{\xi}=\{j<\delta:f_{g(\xi}(j)>\sup h_{\delta}(i,j)\}$.

Claim 4. Let $\xi<\theta^{+}$. Then $\sup W_{\xi}<\delta$.

Proof of Claim 4. Suppose otherwise. Pick $\zeta$ with $\xi<\zeta<\theta^{+}$. Then $\kappa=f_{g(\xi}^{\delta}(i,j)\leq f_{g(\zeta}^{\delta}(i,j)\leq\kappa$, contradicting the fact that $\delta\in S^{i}_{g(\xi)g(\zeta)}$, which completes the proof of the claim.

Set $t=(\sup\bigcup_{\xi<\theta^{+}}W_{\xi})+1$, $u=\max(s,t)$ and $Q=\{j\in\delta\setminus u:|h_{\delta}(i,j)|=\theta\}$. For $j\in Q$, $\langle f^{\delta}_{g(\xi)}(i,j):\xi<\theta^{+}\rangle$ is a weakly increasing sequence of elements of $h_{\delta}(i,j)$, since the sequence $\langle f_{g(\xi)}(j):\xi<\theta^{+}\rangle$ is increasing, so there must be $\chi_{j}<\theta^{+}$ such that $f^{\delta}_{g(\xi)}(i,j)=f^{\delta}_{g(\chi_{j})}(i,j)$ whenever $\chi_{j}<\xi<\theta^{+}$. We may find $\chi<\theta^{+}$ and $M\subseteq Q$ with $\sup M=\delta$ such that $\chi_{j}=\chi$ for all $j\in M$. But now $f^{\delta}_{g(\xi)}|(\{i\}\times M)=f^{\delta}_{g(\xi)}|(\{i\}\times M)$ whenever $\chi<\xi<\zeta<\theta^{+}$. Contradiction ! $\square$

Let us now compare our principle for guessing generalized clubs with the principles considered in the previous section (a weak version of club) and the next section (club-guessing). The following extends Theorem 2.1 of [39].

OBSERVATION 4.8**.**

1. (i)

Suppose that $\kappa$ is weakly inaccessible and $\curlywedge^{-}(\sigma,\kappa,J)$ holds, where $J$ is a normal ideal on $\kappa$ such that $E^{\kappa}_{\geq\sigma}\in J^{\ast}$. Then there is a closed unbounded subset $X_{\delta}$ of $\delta$ for $\delta\in E^{\kappa}_{\geq\sigma}$ such that $\{\delta:\sup(X_{\delta}\setminus Y)<\delta\}\in J^{+}$ for every closed unbounded subset $Y$ of $\kappa$. 2. (ii)

Suppose that $\kappa$ is a successor cardinal and $\curlywedge^{-}(\sigma,\kappa,J)$ holds, where $J$ is a $\kappa$-complete ideal on $\kappa$ such that $E^{\kappa}_{\geq\sigma}\in J^{\ast}$. Then there is a closed unbounded subset $X_{\delta}$ of $\delta$ for $\delta\in E^{\kappa}_{\geq\sigma}$ such that $\{\delta:X_{\delta}\subseteq Y\}\in J^{+}$ for every closed unbounded subset $Y$ of $\kappa$.

Proof. We prove (i) and leave the similar proof of (ii) to the reader. Let $C^{i}_{\delta}$ for $i\in\delta\in acc(\kappa)\setminus\sigma$ witness that $\curlywedge^{-}(\sigma,\kappa,J)$ holds. For $i\in\delta\in E^{\kappa}_{\geq\sigma}$, put $S^{i}_{\delta}=\{\sup x:x\in C^{i}_{\delta}\setminus\{\emptyset\}\}$. Note that $S^{i}_{\delta}\subseteq\delta$.

Claim 1. Let $Y$ be a closed unbounded subset of $\kappa$. Then $\{\delta\in E^{\kappa}_{\geq\sigma}:\exists i<\delta\,(S^{i}_{\delta}\subseteq Y)\}\in J^{+}$.

Proof of Claim 1. Define $F:P_{\omega}(\kappa)\rightarrow\kappa$ by $F(e)=\min(C\setminus\bigcup e)$, and let $D=\{x\in P_{\sigma}(\delta):F``P_{\omega}(x)\subseteq x\}$. Now suppose that $i$ and $\delta$ are such that $i\in\delta\in E^{\kappa}_{\geq\sigma}$ and $C^{i}_{\delta}\subseteq D$. Then clearly, $\sup x\in Y$ for every nonempty $x$ in $C^{i}_{\delta}$. Hence $S^{i}_{\delta}\subseteq Y$, which completes the proof of the claim.

For $i\in\delta\in E^{\kappa}_{\geq\sigma}$, put $T^{i}_{\delta}=\{\xi\in\delta\setminus\{0\}:\sup(\xi\cap T^{i}_{\delta})=\xi\}$. Then clearly, $T^{i}_{\delta}$ is a closed unbounded subset of $\delta$. Furthermore, $\{\delta\in E^{\kappa}_{\geq\sigma}:\exists i<\delta\,(T^{i}_{\delta}\subseteq Y)\}\in J^{+}$ for any closed unbounded subset $Y$ of $\kappa$.

Claim 2. There is $i<\kappa$ such that $\{\delta:T^{i}_{\delta}\setminus Y\subseteq i+1\}\in J^{+}$ for every closed unbounded subset $Y$ of $\kappa$.

Proof of Claim 2. Suppose otherwise. For $i<\kappa$, pick $Y_{i}\in NS_{\kappa}^{\ast}$ and $W_{i}\in J^{\ast}\cap P(E^{\kappa}_{\geq\sigma})$ such that $(T^{i}_{\delta}\setminus Y_{i})\setminus(i+1)\not=\emptyset$ whenever $i\in\delta\in W_{i}$. Set $Y=\bigtriangleup_{i<\kappa}Y_{i}$ and $W=\bigtriangleup_{i<\kappa}W_{i}$. We may find $\delta\in W$ and $i<\delta$ such that $T^{i}_{\delta}\subseteq Y$. Then $\delta\in W_{i}$, and moreover $Y^{i}_{\delta}\setminus(i+1)\subseteq Y_{i}\setminus(i+1)\subseteq Y$. This contradiction completes the proof of the claim and that of the observation.

$\square$

OBSERVATION 4.9**.**

Suppose that $\curlywedge^{-}(\sigma,\kappa,J)$ holds, where $J$ is a $\kappa$-complete ideal on $\kappa$ such that $E^{\kappa}_{\geq\sigma}\in J^{\ast}$. Then $NS^{\ast}_{\kappa}\cap J=\emptyset$.

Proof. By the proof of Observation 4.8. $\square$

OBSERVATION 4.10**.**

Suppose that $\curlywedge^{-}(\sigma,\kappa,J)$ holds and $E^{\kappa}_{\theta}\in J^{\ast}$, where $\theta$ is an infinite cardinal with $\theta^{+}<\kappa$. Then $\clubsuit_{\kappa}^{{\rm cof}/\rho,-}[J]$ holds, where $\rho=\max\{\sigma,\theta^{+}\}$.

Proof. Let $C^{i}_{\delta}$ for $i\in\delta\in acc(\kappa)\setminus\sigma$ witness that $\curlywedge^{-}(\sigma,\kappa,J)$ holds. For $\delta\in E^{\kappa}_{\theta}\setminus\sigma$, pick an increasing sequence $\langle\delta_{j}:j<\theta\rangle$ with supremum $\delta$, and $h:\delta\times\theta\rightarrow\delta$ such that $\delta_{j}\in h(i,j)\in C^{i}_{\delta}$ for every $(i,j)\in\delta\times\theta$. Set $s^{i}_{\delta}=\bigcup_{j<\theta}h(i,j)$ for every $i<\delta$. Now given $A\in[\kappa]^{\kappa}$, define $F:P_{\omega}(\kappa)\rightarrow\kappa$ by $F(e)=\min(A\setminus\cup e)$, and put $D=\{x\in P_{\sigma}(\delta):F``P_{\omega}(x)\subseteq x\}$. Then clearly, $\sup(A\cap s^{i}_{\delta})=\delta$ whenever $\delta\in E^{\kappa}_{\theta}\setminus\sigma$ and $i<\delta$ are such that $C^{i}_{\delta}\subseteq D$. $\square$

The following is yet another indication of the strength of $\curlywedge^{-}$.

OBSERVATION 4.11**.**

Given a regular uncountable cardinal $\sigma<\kappa$, the following hold :

1. (i)

Suppose that $\curlywedge(\sigma,\kappa,NS_{\kappa}|S)$ holds, where $S\in NS_{\kappa}^{+}\cap P(E^{\kappa}_{<\sigma})$. Then there is a stationary subset $X$ of $P_{\sigma}(\kappa)$ such that

• •

$\{\sup x:x\in X\}=S$* ;*

• •

the sup-function is one-to-one on $X$. 2. (ii)

Suppose that $\curlywedge^{-}(\sigma,\kappa,NS_{\kappa}|S)$ holds, where $S\in NS_{\kappa}^{+}\cap P(E^{\kappa}_{<\sigma})$. Then there is a stationary subset $X$ of $P_{\sigma}(\kappa)$ such that

• •

$\{\sup x:x\in X\}=S$* ;*

• •

$|\{x\in X:\sup x=\alpha\}|\leq|\alpha|$* for all $\alpha\in S$.*

Proof. We prove (i) and leave the similar proof of (ii) to the reader. Let $\langle C_{\delta}:\delta\in acc(\kappa)\setminus\sigma\rangle$ witness that $\curlywedge(\sigma,\kappa,NS_{\kappa}|S)$ holds. We define $x_{\delta}$ for $\delta\in S$ as follows. Given $\delta\in S$, select $e_{\delta}\subseteq\delta$ so that o.t.$(e_{\delta})=\mathrm{cf}(\delta)$ and $\sup e_{\delta}=\delta$. Inductively define $x^{n}_{\delta}\in C_{\delta}$ for $n<\omega$ so that

• •

$x^{0}_{\delta}=e_{\delta}$ ;

• •

$x^{n}_{\delta}\cup\sup(x^{n}_{\delta}\cap\kappa)\subseteq x^{n+1}_{\delta}$.

Finally, $X=\{\bigcup_{n<\omega}x^{n}_{\delta}:\delta\in S\}$ is as desired. $\square$

COROLLARY 4.12**.**

Suppose that $\curlywedge^{-}(\sigma,\kappa,NS_{\kappa}|S)$ holds, where $\sigma$ is a regular uncountable cardinal less than $\kappa$, and $S\in NS_{\kappa}^{+}\cap P(E^{\kappa}_{<\sigma})$. Then $\mathrm{cov}(\kappa,\sigma,\sigma,2)=\kappa$.

Notice that in the other direction, the following is known.

FACT 4.13**.**

([17], [32])* Let $S\in NS^{+}_{\kappa}\cap P(E^{\kappa}_{<\sigma})$. Suppose that there is a stationary subset $X$ of $P_{\sigma}(\kappa)$ such that*

• •

$\{\sup x:x\in X\}\subseteq S$* ;*

• •

the sup-function is one-to-one on $X$ (respectively, $|\{x\in X:\sup x=\alpha\}|\leq|\alpha|$ for all $\alpha\in S$).

Then $\clubsuit_{\kappa}^{{\rm cof}/\sigma}[NS_{\kappa}|S]$ (respectively, $\clubsuit_{\kappa}^{{\rm cof}/\sigma,-}[NS_{\kappa}|S]$) holds.

5 GITIK-SHELAH ON NONSATURATION

We are looking for another result on ideal nonsaturation where towers might be involved. A natural candidate is the following result of Gitik and Shelah.

FACT 5.1**.**

([17])* Suppose that $\max(\omega_{2},\theta^{+})<\kappa$, where $\theta$ is a regular infinite cardinal. Then $NS_{\kappa}|E^{\kappa}_{\theta}$ is not $\kappa^{+}$-saturated.*

REMARK 5.2*.*

Notice that in general the conclusion will not remain valid if $NS_{\kappa}|E^{\kappa}_{\theta}$ is replaced with $NS_{\kappa}|S$ for some stationary subset $S$ of $E^{\kappa}_{\theta}$ (see [17]).

The result was already revisited by Krueger [23], and we will follow his reading.

5.1 Club-guessing

We start with an easy generalization of Shelah’s club guessing principle (see [17]).

Throughout Subsections 5.1 - 5.4, $\theta$ and $\rho$ will denote two regular infinite cardinals with $\theta<\rho<\kappa$.

PROPOSITION 5.3**.**

Let $K$ be a $\kappa$-complete ideal on $\kappa$ extending $NS_{\kappa}|E^{\kappa}_{\theta}$. Then there is $c_{\alpha}\subseteq E^{\kappa}_{\geq\rho}\cap\alpha$ with $\sup c_{\alpha}=\alpha$ for $\alpha\in E^{\kappa}_{\theta}\cap acc(E^{\kappa}_{\geq\rho})$ such that $\{\alpha\in E^{\kappa}_{\theta}\cap acc(E^{\kappa}_{\geq\rho}):c_{\alpha}\subseteq C\}\in K^{+}$ for every $C\in{\cal C}_{\kappa}$.

Proof. The proof follows that of Hirata given in [48]. For each $\beta\in acc(\kappa)$, select a cofinal subset $d_{\beta}$ of $\beta$ of order-type $\mathrm{cf}(\beta)$. Given $\alpha\in E^{\kappa}_{\theta}\cap acc(E^{\kappa}_{\geq\rho})$ and $D\in{\cal C}_{\kappa}\cap P(acc(\kappa))$, we inductively define $x^{D}_{\alpha,n}\subseteq D$ with $|x^{D}_{\alpha,n}|<\rho$ for $n<\omega$ as follows :

• •

$x^{D}_{\alpha,0}=\{\sup(D\cap\gamma):\gamma\in d_{\alpha}\}$ ;

• •

$x^{D}_{\alpha,n+1}=\{\sup(D\cap\gamma):\exists\beta\in x^{D}_{\alpha,n}\cap E^{\kappa}_{<\rho}(\gamma\in d_{\beta})\}$.

Finally, we let $x^{D}_{\alpha}=(\bigcup_{n<\omega}x^{D}_{\alpha,n})\setminus\{0\}$.

Claim. There is $D\in{\cal C}_{\kappa}\cap P(acc(\kappa))$ such that $\{\alpha\in E^{\kappa}_{\theta}\cap acc(E^{\kappa}_{\geq\rho}):x^{D}_{\alpha}\subseteq C\}\in K^{+}$ for every $C\in{\cal C}_{\kappa}$.

Proof of the claim. Suppose otherwise. Inductively define $C_{\xi}\in{\cal C}_{\kappa}\cap P(acc(\kappa))$ and $H_{\xi}\in K^{\ast}$ for $\xi<\rho$ such that

• •

$C_{\xi+1}\subseteq C_{\xi}$ ;

• •

$H_{\xi+1}\cap\{\alpha\in E^{\kappa}_{\theta}\cap acc(E^{\kappa}_{\geq\rho}):x^{C_{\xi}}_{\alpha}\subseteq C_{\xi+1}\}=\emptyset$.

Select $\alpha$ in $E^{\kappa}_{\theta}\cap acc(E^{\kappa}_{\geq\rho})\cap\bigcap_{\xi<\rho}C_{\xi}\cap\bigcap_{\xi<\rho}H_{\xi}$. Since the map $\xi\rightarrow\sup(C_{\xi}\cap\gamma)$ is nonincreasing for every $\gamma<\kappa$, and $|x^{C_{\xi}}_{\alpha,n}|<\rho$ for all $\xi<\rho$ and all $n<\omega$, we may inductively define $\xi_{n}$ for $n<\omega$ so that

• •

$\xi_{n}\leq\xi_{n+1}$ ;

• •

$x^{C_{\xi}}_{\alpha,n}=x^{C_{\xi_{n}}}_{\alpha,n}$ for $\xi_{n}\leq\xi<\rho$.

Set $\xi=\sup\{\xi_{n}:n<\omega\}$. Then $x^{C_{\xi}}_{\alpha}=x^{C_{\xi+1}}_{\alpha}\subseteq C_{\xi+1}$. This contradiction completes the proof of the claim.

By the claim, for any $C\in{\cal C}_{\kappa}$,

$\{\alpha\in E^{\kappa}_{\theta}\cap acc(E^{\kappa}_{\geq\rho})\cap acc(D):x^{D}_{\alpha}\setminus acc(x^{D}_{\alpha})\subseteq C\cap acc(D)\}\in K^{+}$.

So it remains to observe the following. Let $\alpha\in E^{\kappa}_{\theta}\cap acc(E^{\kappa}_{\geq\rho})\cap acc(D)$ be such that $x^{D}_{\alpha}\subseteq acc(D)$. Then $x^{D}_{\alpha,0}$ is cofinal in $\alpha$, and hence so are $x^{D}_{\alpha}$ and $x^{D}_{\alpha}\setminus acc(x^{D}_{\alpha})$. Furthermore for any $n<\omega$ and any $\beta\in x^{D}_{\alpha,n}\cap E^{\kappa}_{<\rho}$, $\beta$ lies in $acc(D)$ and therefore in $acc(x^{D}_{\alpha,n+1})$. It follows that $x^{D}_{\alpha}\setminus acc(x^{D}_{\alpha})\subseteq E^{\kappa}_{\geq\rho}$. $\square$

5.2 Strong guessing

Let $J$ be a $\kappa$-complete ideal on $\kappa$ extending $NS_{\kappa}|E^{\kappa}_{\theta}$. By Proposition 5.3, for any $A\in J^{+}\cap P(E^{\kappa}_{\theta}\cap acc(E^{\kappa}_{\geq\rho}))$, we may find $c^{A,J}_{\alpha}\subseteq E^{\kappa}_{\geq\rho}\cap\alpha$ for $\alpha\in A$ such that

• •

for any $\alpha\in A$, $\sup c^{A,J}_{\alpha}=\alpha$ and o.t.$(c^{A,J}_{\alpha})=\theta$ ;

• •

$\{\alpha\in A:c^{A,J}_{\alpha}\subseteq C\}\in J^{+}$ for every $C\in{\cal C}_{\kappa}$.

PROPOSITION 5.4**.**

Let $J$ be a $\kappa$-complete ideal on $\kappa$ extending $NS_{\kappa}|E^{\kappa}_{\theta}$. Suppose that there is no descending $(J,I_{\kappa})$-tower of length ${\mathfrak{b}}_{\kappa}$. Then for any $A\in J^{+}\cap P(E^{\kappa}_{\theta}\cap acc(E^{\kappa}_{\geq\rho}))$, there is $B\in J^{+}\cap P(A)$ such that

$\{\alpha\in A:\exists\beta<\alpha(c^{A,J}_{\alpha}\setminus\beta\subseteq C)\}\in(J|B)^{\ast}$

for every $C\in{\cal C}_{\kappa}$.

Proof. We follow the proof of Claim 2.1 in [17]. Fix $A\in J^{+}\cap P(E^{\kappa}_{\theta}\cap acc(E^{\kappa}_{\geq\rho}))$. Let $\Phi:{\cal C}_{\kappa}\rightarrow J^{+}\cap P(A)$ be defined by $\Phi(C)=\{\alpha\in A:\exists\beta<\alpha(c^{A,J}_{\alpha}\setminus\beta\subseteq C)\}$. Notice that

• •

$\Phi(D)\subseteq\Phi(C)$ for all $D\in{\cal C}_{\kappa}\cap P(C)$ ;

• •

$|\Phi(D)\setminus\Phi(C)|<\kappa$ for all $D\in{\cal C}_{\kappa}$ such that $|D\setminus C|<\kappa$.

Now suppose toward a contradiction that for any $B\in J^{+}\cap P(A)$, there is $C\in{\cal C}_{\kappa}$ such that $B\setminus\Phi(C)\in J^{+}$. We inductively define $C_{i}\in{\cal C}_{\kappa}$ for $i<{\mathfrak{b}}_{\kappa}$ so that for $j<i<{\mathfrak{b}}_{\kappa}$, $\Phi(C_{j})\setminus\Phi(C_{i})\in J^{+}$ and $|\Phi(C_{i})\setminus\Phi(C_{j})|<\kappa$. We put $C_{0}=\kappa$. Now suppose that $i>0$, and $C_{j}$ has been constructed for every $j<i$. By Observation 2.36 (i), we may find $D\in{\cal C}_{\kappa}$ such that $|D\setminus C_{j}|<\kappa$ for all $j<i$. By assumption, there must be $H\in{\cal C}_{\kappa}$ such that $\Phi(D)\setminus\Phi(H)\in J^{+}$. We let $C_{i}=D\cap H$. Finally, $\langle\Phi(C_{i}):i<{\mathfrak{b}}_{\kappa}\rangle$ is a descending $(J,I_{\kappa})$-tower of length ${\mathfrak{b}}_{\kappa}$, which yields the desired contradiction. $\square$

5.3 Club-guessing ideals

DEFINITION 5.5**.**

Given a $\kappa$-complete ideal $J$ on $\kappa$ extending $NS_{\kappa}|E^{\kappa}_{\theta}$, we let ${\cal X}_{J}$ denote the collection of all $B\subseteq\kappa$ such that either $B\in J$, or $B\in J^{+}$ and there is $d_{\alpha}\subseteq E^{\kappa}_{\geq\rho}\cap\alpha$ for $\alpha\in B\cap E^{\kappa}_{\theta}\cap acc(E^{\kappa}_{\geq\rho})$ such that

• •

for any $\alpha\in B\cap E^{\kappa}_{\theta}\cap acc(E^{\kappa}_{\geq\rho})$, $\sup d_{\alpha}=\alpha$ and o.t.$(d_{\alpha})=\theta$ ;

• •

$\{\alpha\in B\cap E^{\kappa}_{\theta}\cap acc(E^{\kappa}_{\geq\rho}):\exists\beta<\alpha(d_{\alpha}\setminus\beta\subseteq C)\}\in(J|B)^{\ast}$ for every $C\in{\cal C}_{\kappa}$.

OBSERVATION 5.6**.**

1. (i)

Let $B\in{\cal X}_{J}$. Then $P(B)\subseteq{\cal X}_{J}$. 2. (ii)

Let $A,B\in{\cal X}_{J}$. Then $A\cup B\in{\cal X}_{J}$.

REMARK 5.7*.*

Suppose that there is no descending $(J,I_{\kappa})$-tower of length ${\mathfrak{b}}_{\kappa}$. Then by Proposition 5.4, ${\cal X}_{J}\cap P(H)\cap J^{+}\not=\emptyset$ for any $H\in J^{+}$.

PROPOSITION 5.8**.**

Let $J$ be a normal ideal on $\kappa$ with $E^{\kappa}_{\theta}\in J^{\ast}$. Suppose that there is no descending $(J,I_{\kappa})$-tower of length ${\mathfrak{b}}_{\kappa}$, and no ascending $(J,I_{\kappa})$-tower of length $\kappa^{+}$. Then $\kappa\in{\cal X}_{J}$.

Proof. The proof is a modification of that of Lemma 2 in [17]. Let ${\cal T}$ denote the collection of those ascending $(J,I_{\kappa})$-towers that have all their members in ${\cal X}_{J}$. Given $T,W\in{\cal T}$, put $T<W$ just in case $W$ is a proper extension of $T$. By Zorn’s Lemma, $({\cal T},<)$ has a maximal element, say $T=\langle B_{\eta}:\eta<\delta\rangle$.

Case 1 : $\delta$ is a successor ordinal, say $\delta=\xi+1$.

Claim 1. $\kappa\setminus B_{\xi}\in J$.

Proof of Claim 1. Suppose otherwise. Then by Remark 5.7, we may find $R$ in ${\cal X}_{J}$ with $R\in J^{+}\cap P(\kappa\setminus B_{\xi})$. Set $B_{\delta}=B_{\xi}\cup R$ and $W=\langle B_{\gamma}:\gamma\leq\delta\rangle$. Then by Observation 5.6, $W\in{\cal T}$, and moreover $T<W$. This contradiction completes the proof of the claim.

It follows from the claim and Observation 5.6 that $\kappa\in{\cal X}_{J}$.

Case 2 : $\delta$ is a limit ordinal.

Put $\sigma=\mathrm{cf}(\delta)$, and let $\langle\delta_{i}:i<\sigma\rangle$ be an increasing sequence of ordinals with supremum $\delta$. For $i<\sigma$, set $K_{i}=B_{\delta_{i+1}}$. Then it is simple to see that $\langle K_{i}:i<\sigma\rangle$ is also a maximal element of ${\cal T}$. Note that $\sigma\leq\kappa$. Set $H_{0}=K_{0}\setminus 1$, and for each $i$ with $0<i<\sigma$, $H_{i}=K_{i}\setminus((\bigcup_{j<i}K_{j})\cup(i+1))$. Notice that $H_{m}\cap H_{n}=\emptyset$ whenever $m<n<\sigma$.

Claim 2. Let $i<\sigma$. Then $H_{i}\in J^{+}$.

Proof of Claim 2. This is obvious for $i=0$. For $i>0$, it suffices to observe that

$(B_{\delta_{i+1}}\setminus B_{\delta_{i}})\setminus\bigcup_{j<i}(B_{\delta_{j+1}}\setminus B_{\delta_{i}})\subseteq H_{i}$,

which completes the proof of the claim.

For each $i<\sigma$, $H_{i}\in{\cal X}_{J}$ by Observation 5.6, so we may find $d^{i}_{\alpha}\subseteq E^{\kappa}_{\geq\rho}\cap\alpha$ for $\alpha\in H_{i}\cap E^{\kappa}_{\theta}\cap acc(E^{\kappa}_{\geq\rho})$ such that

• •

for any $\alpha\in H_{i}\cap E^{\kappa}_{\theta}\cap acc(E^{\kappa}_{\geq\rho})$, $\sup d^{i}_{\alpha}=\alpha$ and o.t.$(d^{i}_{\alpha})=\theta$ ;

• •

$\{\alpha\in H_{i}\cap E^{\kappa}_{\theta}\cap acc(E^{\kappa}_{\geq\rho}):\exists\beta<\alpha(d^{i}_{\alpha}\setminus\beta\subseteq C)\}\in(J|H_{i})^{\ast}$ for every $C\in{\cal C}_{\kappa}$.

Put $A=\bigcup_{i<\sigma}H_{i}$.

Claim 3. $A\in{\cal X}_{J}$.

Proof of Claim 3. For $\alpha\in A$, set $c_{\alpha}=d^{i}_{\alpha}$, where $\alpha\in H_{i}$. Now let $C\in{\cal C}_{\kappa}$. For each $i<\sigma$, we may find $D_{i}\in J^{\ast}$ such that for any $\alpha\in D_{i}\cap H_{i}\cap E^{\kappa}_{\theta}\cap acc(E^{\kappa}_{\geq\rho})$, there is $\beta<\alpha$ such that $c_{\alpha}\setminus\beta\subseteq C$. Put $D=\{\alpha<\kappa:\forall i\in\sigma\cap\alpha(\alpha\in D_{i})\}$. Notice that $D\in J^{\ast}$ by normality of $J$. Clearly, $H_{i}\cap D\subseteq D_{i}$ for all $i<\sigma$. It follows that for any $\alpha\in D\cap A\cap E^{\kappa}_{\theta}\cap acc(E^{\kappa}_{\geq\rho})$, there is $\beta<\alpha$ such that $c_{\alpha}\setminus\beta\subseteq C$, which completes the proof of the claim.

Claim 4. $\kappa\setminus A\in J$.

Proof of Claim 4. Suppose not. We proceed as in the proof of Claim 1. By Remark 5.7, there must be $R\in{\cal X}_{J}$ such that $R\in J^{+}\cap P(\kappa\setminus A)$. Set $K_{\delta}=A\cup R$ and $W=\langle K_{i}:i\leq\sigma\rangle$. Now for each $i\leq\sigma$, $K_{i}\setminus A\subseteq i+1$, and therefore $|K_{i}\setminus A|<\kappa$. By Observation 5.6, it follows that $W\in{\cal T}$. However, $T<W$. This contradiction completes the proof of the claim.

By Claim 4 and Observation 5.6, $\kappa\in{\cal X}_{J}$. $\square$

5.4 Full reflection

DEFINITION 5.9**.**

Given a stationary subset $S$ of $\kappa$, and a stationary subset $T$ of $E^{\kappa}_{\geq\omega_{1}}$, $S$ reflects fully in $T$ if there is $G\in{\cal C}_{\kappa}$ such that $S$ reflects at every $\gamma\in G\cap T$.

OBSERVATION 5.10**.**

Let $\sigma<\kappa$ be a regular uncountable cardinal, and $S$ be a stationary subset of $\kappa$ that reflects fully in $E^{\kappa}_{\sigma}$. Then for any regular cardinal $\tau$ with $\sigma\leq\tau<\kappa$, $S$ reflects fully in $E^{\kappa}_{\tau}$.

FACT 5.11**.**

([23])* Suppose that $\kappa$ is either weakly inaccessible, or the successor of a singular cardinal. Then for any regular uncountable cardinal $\chi<\kappa$, and any stationary subset $S$ of $\kappa$, $S$ reflects fully in $E^{\kappa}_{\geq\chi}$ if and only if there are $C\in{\cal C}_{\kappa}$ and $\eta<\kappa$ such that $C\setminus S$ has no closed subset of order-type $\eta$.*

THEOREM 5.12**.**

Let $S$ be a stationary subset of $E^{\kappa}_{\theta}$ that reflects fully in $E^{\kappa}_{\geq\rho}$. Suppose that either $\theta>\omega$, or $\rho\geq\omega_{2}$. Then, setting $J=NS_{\kappa}|S$, there is either a descending $(J,I_{\kappa})$-tower of length ${\mathfrak{b}}_{\kappa}$, or an ascending $(J,I_{\kappa})$-tower of length $\kappa^{+}$.

Proof. We closely follow the proof of Lemma 3 in [17]. Suppose that the conclusion fails. Then by Proposition 5.4, there is $c_{\alpha}\subseteq E^{\kappa}_{\geq\rho}\cap\alpha$ for $\alpha\in S\cap acc(E^{\kappa}_{\geq\rho})$ such that

• •

for any $\alpha\in S\cap acc(E^{\kappa}_{\geq\rho})$, $\sup c_{\alpha}=\alpha$ and o.t.$(c_{\alpha})=\theta$ ;

• •

$\{\alpha\in S\cap acc(E^{\kappa}_{\geq\rho}):\exists\beta<\alpha(c_{\alpha}\setminus\beta\subseteq C)\}\in J^{\ast}$ for any $C\in{\cal C}_{\kappa}$.

Pick $G\in{\cal C}_{\kappa}$ such that $S$ reflects at every $\gamma\in G\cap E^{\kappa}_{\geq\rho}$.

Case 1 : $\theta>\omega$.

Inductively define $C_{n}\in{\cal C}_{\kappa}$ for $n<\omega$ as follows. Set $C_{0}=G\cap acc(E^{\kappa}_{\geq\rho})$. Now suppose that $C_{n}$ has been constructed. There must be $H\in{\cal C}_{\kappa}$ with the property that for any $\alpha\in H\cap S\cap acc(E^{\kappa}_{\geq\rho})$, there is $\beta<\alpha$ such that $c_{\alpha}\setminus\beta\subseteq acc(C_{n})$. Put $C_{n+1}=acc(C_{n})\cap H$. Finally, set $C=\bigcap_{n<\omega}C_{n}$ and $\alpha=\min(C\cap S)$. Since $\alpha\in\bigcap_{n<\omega}C_{n+1}$, we may find $\eta<\alpha$ such that $c_{\alpha}\setminus\eta\subseteq\bigcap_{n<\omega}acc(C_{n})$. Pick $\gamma\in c_{\alpha}\cap\bigcap_{n<\omega}acc(C_{n})$. Then $\gamma\in E^{\kappa}_{\geq\rho}$. Moreover, $C_{n}\cap\gamma$ is a closed unbounded subset of $\gamma$ for every $n<\omega$. Hence $C\cap\gamma$ is a closed unbounded subset of $\gamma$, and therefore $(C\cap S)\cap\gamma\not=\emptyset$, which contradicts the minimality of $\alpha$.

Case 2 : $\theta=\omega$.

In the same spirit as in Case 1, we define $C_{\xi}\in{\cal C}_{\kappa}$ for $\xi<\omega_{1}$ so that

• •

$C_{0}=G\cap acc(E^{\kappa}_{\geq\rho})$ ;

• •

$C_{\zeta}\subseteq C_{\xi}$ for all $\zeta<\xi$ ;

• •

for any $\alpha\in C_{\xi+1}\cap S$, there is $\beta<\alpha$ such that $c_{\alpha}\setminus\beta\subseteq acc(C_{\xi})$.

Set $C=\bigcap_{\xi<\omega_{1}}C_{\xi}$ and $\alpha=\min(C\cap S)$. Since $\alpha\in\bigcap_{\xi<\omega_{1}}C_{\xi+1}$ and o.t.$(c_{\alpha})=\omega$, there must be $\eta<\alpha$ such that $c_{\alpha}\setminus\eta\subseteq\bigcap_{\xi<\omega_{1}}acc(C_{\xi})$. Pick $\gamma\in c_{\alpha}\cap\bigcap_{\xi<\omega_{1}}acc(C_{\xi})$. Then $\mathrm{cf}(\gamma)\geq\rho>\omega_{1}$, and moreover $C\cap\gamma$ is a closed unbounded subset of $\gamma$. Hence $(C\cap S)\cap\gamma\not=\emptyset$. Contradiction. $\square$

Full reflection is a sufficient condition, but in general not a necessary one, as the following shows.

FACT 5.13**.**

([23])* Suppose that one of the following conditions holds :*

• •

$\kappa$* is weakly inaccessible.*

• •

$\kappa=\nu^{+}$, where $\nu$ is singular.

• •

$\kappa=\nu^{+}$, where $\nu$ is regular and $\square_{\nu}$ holds.

Then any stationary subset $T$ of $\kappa$ has a stationary subset $S$ with the property that for every regular uncountable cardinal $\sigma<\kappa$, $S$ does not reflect fully in $E^{\kappa}_{\sigma}$.

5.5 Good points

Let $A$ be an infinite set of regular cardinals such that $|A|<\min A$ and $\sup A<\kappa$, and $I$ be an ideal on $A$ such that $\{A\cap a:a\in A\}\subseteq I$.

DEFINITION 5.14**.**

We let $\prod A=\prod_{a\in A}a$. For $f,g\in\prod A$, we let $f<_{I}g$ if $\{a\in A:f(a)\geq g(a)\}\in I$.

DEFINITION 5.15**.**

Let $\vec{f}=\langle f_{\alpha}:\alpha<\kappa\rangle$ be an increasing, cofinal sequence in $(\prod A,<_{I})$. An infinite limit ordinal $\delta<\kappa$ is a good point for $\vec{f}$ if there is a cofinal subset $X\subseteq\delta$, and $Z_{\xi}\in I$ for $\xi\in X$ such that $f_{\beta}(a)<f_{\xi}(a)$ whenever $\beta<\xi$ are in $X$ and $a\in A\setminus(Z_{\beta}\cup Z_{\xi})$.

We let $G(\vec{f})$ denote the set of good points for $\vec{f}$.

FACT 5.16**.**

1. (i)

(Folklore)* If $\delta\in G(\vec{f})$, then $\mathrm{cf}(\delta)<\sup A$.* 2. (ii)

([4], [28])* Let $\delta<\pi$ be an infinite limit ordinal such that $I$ is $\mathrm{cf}(\delta)$-complete. Then $\delta\in G(\vec{f})$.*

The following is due to Shelah.

FACT 5.17**.**

For $i=0,1$, let $\vec{f_{i}}=\langle f^{i}_{\alpha}:\alpha<\kappa\rangle$ be an increasing, cofinal sequence in $(\prod A,<_{I})$. Then $G(\vec{f_{0}})\bigtriangleup G(\vec{f_{1}})\in NS_{\kappa}$.

Proof. Let $D$ be the set of all $\delta\in acc(\kappa)$ with the property that for any $\xi<\delta$, there are $\beta,\gamma<\delta$ such that $f^{0}_{\xi}<_{I}f^{1}_{\beta}$ and $f^{1}_{\xi}<_{I}f^{0}_{\gamma}$. Let us show that $D\cap G(\vec{f_{0}})=D\cap G(\vec{f_{1}})$. Thus fix $i<1$ and $\delta\in D\cap G(\vec{f_{i}})$. Let $X\subseteq\delta$ and $Z_{\xi}\in I$ for $\xi\in X$ witness that $\delta$ is a good point for $\vec{f_{i}}$. Define two increasing sequences $\langle\beta_{j}:j<\mathrm{cf}(\delta)\rangle$ and $\langle\gamma_{j}:j<\mathrm{cf}(\delta)\rangle$ so that

• •

$\beta_{j}\in X$ and $\gamma_{j}<\delta$ ;

• •

$f^{i}_{\beta_{j}}<_{I}f^{1-i}_{\gamma_{j}}<_{I}f^{i}_{\beta_{j+1}}$.

For $j<\mathrm{cf}(\delta)$, set

$W_{j}=\{a:f^{i}_{\beta_{j}}(a)\geq f^{1-i}_{\gamma_{j}}(a)\}\cup Z_{\beta_{j}}\cup\{a:f^{1-i}_{\gamma_{j}}(a)\geq f^{i}_{\beta_{j+1}}(a)\}\cup Z_{\beta_{j+1}}$.

Then

$f^{1-i}_{\gamma_{j}}(a)<f^{i}_{\beta_{j+1}}(a)\leq f^{i}_{\beta_{k}}(a)<f^{1-i}_{\gamma_{k}}(a)$

whenever $j<k<\mathrm{cf}(\delta)$ and $a\in A\setminus(W_{j}\cup W_{k})$. $\square$

FACT 5.18**.**

([26])* Suppose that there exists an increasing, cofinal sequence $\vec{f}=\langle f_{\alpha}:\alpha<\kappa\rangle$ in $(\prod A,<_{I})$. Then there exists an increasing, cofinal sequence $\vec{g}=\langle g_{\alpha}:\alpha<\kappa\rangle$ in $(\prod A,<_{I})$ such that for any regular cardinal $\sigma$ with $|A|<\sigma<\sup A$, and any ordinal $\eta$ with $\sigma\leq\eta<\sigma^{+}$, the following holds. For any $\beta\in E^{\kappa}_{\sigma^{+3}}$, and any closed unbounded subset $C$ of $\beta$, there is a closed subset $H$ of $C$ of order-type $\eta$ with $H\subseteq G(\vec{g})$.*

PROPOSITION 5.19**.**

Let $\vec{f}=\langle f_{\alpha}:\alpha<\kappa\rangle$ be an increasing, cofinal sequence in $(\prod A,<_{I})$, and $\sigma$ be a regular cardinal with $|A|<\sigma<\sup A$. Then, setting $J=NS_{\kappa}|(G(\vec{f})\cap E^{\kappa}_{\sigma})$, there is either a descending $(J,I_{\kappa})$-tower of length ${\mathfrak{b}}_{\kappa}$, or an ascending $(J,I_{\kappa})$-tower of length $\kappa^{+}$.

Proof. Set $\rho=\sigma^{+3}$, and let $\vec{g}=\langle g_{\alpha}:\alpha<\kappa\rangle$ be as in the statement of Fact 5.18.

Claim. Let $\gamma\in E^{\kappa}_{\geq\rho}$. Then $G(\vec{f})\cap E^{\kappa}_{\sigma}\cap\gamma$ is stationary in $\gamma$.

Proof of the claim. Let $C$ be a closed unbounded subset of $\gamma$. Then we may find $\beta\in E^{\kappa}_{\rho}\cap(\gamma+1)$ such that $C\cap\beta$ is cofinal in $\beta$. There must be a closed subset $H$ of $C\cap\beta$ of order-type $\sigma+1$ with $H\subseteq G(\vec{g})$. Then $\max H\in C\cap G(\vec{f})\cap E^{\kappa}_{\sigma}$, which completes the proof of the claim.

Note that by the claim, $G(\vec{g})\cap E^{\kappa}_{\sigma}$ is stationary in $\kappa$. By Fact 5.17, so is $G(\vec{f})\cap E^{\kappa}_{\sigma}$, and moreover $NS_{\kappa}|(G(\vec{g})\cap E^{\kappa}_{\sigma})=J$. The desired conclusion is now immediate from Theorem 5.12. $\square$

5.6 Robustness of Diamond

Unlike Club (see e.g. [8]), Diamond is remarkably robust, in the sense that a small modification in its definition will often yield an equivalent principle. For a striking example of this, consider the following result which was first established by Primavesi [36] for any $J$ of the form $NS_{\omega_{1}}|S$.

OBSERVATION 5.20**.**

Given a $\kappa$-complete ideal $J$ on $\kappa$ extending $NS_{\kappa}$, the following are equivalent :

1. (i)

$\diamondsuit_{\kappa}[J]$* holds.* 2. (ii)

There is $s_{\alpha}\subseteq\alpha$ for $\alpha<\kappa$ such that $\{\alpha:s_{\alpha}=C\cap\alpha\}\in J^{+}$ for all $C\in{\cal C}_{\kappa}$.

Proof. (i) $\rightarrow$ (ii) : Trivial.

(ii) $\rightarrow$ (i) : The proof is a modification of that of Theorem 3.0.10 in [36]. Let $\langle s_{\alpha}:\alpha<\kappa\rangle$ be as in (ii).

Claim 1. $\clubsuit_{\kappa}[J]$ holds.

Proof of Claim 1. For $\alpha\in acc(\kappa)$, put $t_{\alpha}=s_{\alpha}\setminus acc(s_{\alpha})$. Now fix $A\in[\kappa]^{\kappa}$. Put $C=A\cup acc(A)$,

$D=\{\alpha\in acc(\kappa):\sup((C\setminus acc(C))\cap\alpha)=\alpha\}$

and $S=\{\alpha<\kappa:s_{\alpha}=C\cap\alpha\}$. Clearly for any $\alpha\in D\cap S$, $t_{\alpha}\subseteq C\setminus acc(C)\subseteq A$, and moreover $\sup t_{\alpha}\geq\sup((C\setminus acc(C))\cap\alpha)=\alpha$, which completes the proof of the claim.

Claim 2. $2^{<\kappa}=\kappa$.

Proof of Claim 2. Let $\tau$ be an infinite cardinal less than $\kappa$. Put $c=acc(\kappa)\cap\tau$, and for any $a\subseteq\tau\setminus acc(\kappa)$, $C_{a}=c\cup a\cup(\kappa\setminus\tau)$ and $T_{a}=\{\alpha<\kappa:s_{\alpha}=C_{a}\cap\alpha\}$. Then clearly, $s_{\alpha}\cap(\tau\setminus acc(\kappa))=a$ for all $\alpha\in T_{a}$. It follows that $2^{\tau}\leq\kappa$, which completes the proof of the claim.

Finally, by Claims 1 and 2 and Observation 2.22, $\diamondsuit_{\kappa}[J]$ holds. $\square$

The starred version is established in the same way :

OBSERVATION 5.21**.**

Given a $\kappa$-complete ideal $J$ on $\kappa$ extending $NS_{\kappa}$, the following are equivalent :

1. (i)

$\diamondsuit_{\kappa}^{\ast}[J]$* holds.* 2. (ii)

There is $s^{i}_{\alpha}\subseteq\alpha$ for $i<\alpha<\kappa$ such that $\{\alpha:\exists i<\alpha(s^{i}_{\alpha}=C\cap\alpha)\}\in J^{\ast}$ for all $C\in{\cal C}_{\kappa}$.

Not so surprisingly, the situation is different with Club. Suppose for instance that $\kappa=\nu^{+}$, where $\nu$ is singular, and $J$ is a normal $\kappa^{+}$-saturated ideal on $\kappa$ (by work of Foreman [10], this is consistent relative to a huge cardinal). Then by Observation 2.27, $\clubsuit_{\kappa}^{\rm ev}[J]$ fails. However by Proposition 5.3, we may find $s_{\alpha}\subseteq\alpha$ with $\sup s_{\alpha}=\alpha$ for $\alpha\in acc(\kappa)$ such that $\{\alpha:s_{\alpha}\subseteq C\}\in J^{+}$ for all $C\in{\cal C}_{\kappa}$.

6 Embarrassing questions

In this section we attempt to probe the depth of the author’s ignorance. Unfortunately, as will shortly be seen, no lower bounds were found.

QUESTION 6.1**.**

Let $J$ be a normal ideal on $\kappa$ that is not $\kappa^{+}$-saturated. Does there then exist an ascending $(J,J)$-tower of length $\kappa^{+}$ ?

QUESTION 6.2**.**

Does $\diamondsuit_{\kappa}[J]$ imply the existence of a descending (respectively, ascending) $(J,J)$-tower of length $2^{\kappa}$ ?

Is it always true that any nontrivial club-like principle for $J$ implies some degree of nonsaturation for $J$ ? Let us consider the following test case. For a $\kappa$-complete ideal $J$ on $\kappa$ extending $NS_{\kappa}$, and a cardinal $\sigma$ with $2\leq\sigma\leq\kappa$, let $\clubsuit_{\kappa}^{\rm ev,\sigma}[J]$ assert the existence of $s_{\alpha}\subseteq\alpha$ with $\sup s_{\alpha}=\alpha$ for $\alpha\in acc(\kappa)$ with the property that for any $f:\kappa\rightarrow\sigma$, there is $i<\sigma$ such that

$\{\alpha\in acc(\kappa):\exists\beta<\alpha((s_{\alpha}\setminus\beta)\cap f^{-1}(\{i\})=\emptyset)\}\in J^{+}$.

$\clubsuit_{\kappa}^{\rm ev,2}[NS_{\omega_{1}}]$ is the principle $\clubsuit_{w^{2}}$ studied in [12]. Notice that $\clubsuit_{\kappa}^{\rm ev,\sigma}[J]$ gets weaker as $\sigma$ increases.

OBSERVATION 6.3**.**

Suppose that $J$ is normal and $\clubsuit_{\kappa}^{\rm ev,\kappa}[J]$ holds. Then $J$ is not prime.

Proof. Suppose otherwise. Let $s_{\alpha}\subseteq\alpha$ for $\alpha\in acc(\kappa)$ witness that $\clubsuit_{\kappa}^{\rm ev,\kappa}[J]$ holds. By a standard argument we may find $S\subseteq\kappa$ such that $T=\{\alpha\in acc(\kappa):s_{\alpha}=S\cap\alpha\}$ lies in $J^{\ast}$. Notice that $|S|=\kappa$. Select $f:\kappa\rightarrow\kappa$ so that for any $i<\kappa$, $|f^{-1}(\{i\})|=\kappa$, and moreover $|f^{-1}(\{i\})\setminus S|\leq 1$. Now fix $i<\kappa$. Then clearly $\sup(f^{-1}(\{i\})\cap s_{\alpha})=\alpha$ for any $\alpha\in T$ such that $\sup(f^{-1}(\{i\})\cap\alpha)=\alpha$. Contradiction. $\square$

QUESTION 6.4**.**

Does $\clubsuit_{\kappa}^{\rm ev,2}[J]$ imply that $J$ is not $\kappa$-saturated ?

7 Turrology / Clubology

We think that behind each result on nonsaturation, there is a tower (respectively, a club). If we do not see it right away, it does not mean that it is not there, just that more research is needed to find it. It is our opinion that such research is socially useful, as there should be a tower (respectively, a club) for everyone, not just for higher-ups.

Acknowledgements. The author would like to thank Assaf Rinot for fruitful discussions. He is deeply indebted to Moti Gitik for his contribution to this paper.