Nonregular ideals

TL;DR

This paper investigates degrees of regularity for ideals on successor cardinals, demonstrating that Taylor's theorem for nonregular ideals on  does not extend to higher cardinals like , and explores related properties of normal ideals.

Contribution

It establishes a dichotomy for degrees of regularity of - and -complete ideals and shows the limitations of Taylor's theorem at higher cardinals.

Findings

A nonregular ideal on  does not necessarily imply an -dense ideal.

Taylor's theorem for regularity does not generalize to higher cardinals.

Similar results hold for normal ideals on ppa(mbda).

Abstract

Generalizing Keisler's notion of regularity for ultrafilters, Taylor introduced degrees of regularity for ideals and showed that a countably complete nonregular ideal on $\omega_1$ must be somewhere $\omega_1$-dense. We prove a dichotomy about degrees of regularity for $\kappa$-complete ideals on successor cardinals $\kappa$ and apply this to show that Taylor's Theorem does not generalize to higher cardinals. In particular, the existence of a nonregular ideal on $\omega_2$ does not imply the existence of an $\omega_2$-dense ideal on $\omega_2$. We obtain similar results for normal ideals on $\mathcal P_\kappa(\lambda)$.

Equations

$$C_{\alpha }=B_{\alpha }\setminus \bigcup _{x\in [\alpha {]}^{\xi }}\bigcap _{\beta \in x}{B}_{\beta }.$$

$$\iota :\mathcal{B}(\mathbb{P}\ast \dot{\mathcal{P}(Z)}/\bar{I})\cong \mathcal{B}(\mathcal{P}(Z)/I\ast \dot{j(\mathbb{P})})$$

$$\left(\begin{array}[]{c}\lambda^{+}\\ \lambda^{<\kappa}\end{array}\right)\rightarrow\left(\begin{array}[]{c}\mu\\ \eta\end{array}\right)_{\nu}.$$

$$\left(\begin{array}[]{c}\theta^{+}\\ \theta\end{array}\right)\nrightarrow\left(\begin{array}[]{c}\mu\\ \theta\end{array}\right)_{2}.$$

Full text

An ideal on a set $X$ is a collection of subsets of $X$ closed under taking subsets and pairwise unions. If $\kappa$ is a cardinal, an ideal $I$ is called $\kappa$-complete if it is also closed under unions of size less than $\kappa$. An ideal $I$ on $X$ is called nonprincipal when for all $x\in X$, $\{x\}\in I$, and it is called proper when $X\notin I$. In this paper, we assume all our ideals are nonprincipal and proper. An ideal $I$ on $X$ gives a notion of a “negligible” subset of $X$, and members of $I$ are called $I$-measure-zero. Subsets of $X$ which are not in $I$ are called $I$-positive, and the collection of these is typically denoted by $I^{+}$. The dual filter to $I$, the collection of all complements of members of $I$, constitutes the collection $I$-measure-one sets and will be denoted by $I^{*}$. If an ideal $I$ renders every subset of $X$ either measure zero or measure one, then its dual filter is called an ultrafilter.

The notion of regularity of ultrafilters was introduced by Keisler [10] and has had many applications in set theory and model theory [3]. An ultrafilter $\mathcal{U}$ is called $(\alpha,\beta)$-regular when there is a sequence of sets $\langle A_{i}:i<\beta\rangle\subseteq\mathcal{U}$ such that for all $z\subseteq\beta$ of ordertype $\alpha$, $\bigcap_{i\in z}A_{i}=\emptyset$. Taylor [12] generalized this notion to arbitrary filters (or equivalently, ideals), defining an ideal $I$ to be $(\alpha,\beta)$-regular when for every sequence $\langle A_{i}:i<\beta\rangle\subseteq I^{+}$, there is a refinement $\langle B_{i}:i<\beta\rangle\subseteq I^{+}$, which means $B_{i}\subseteq A_{i}$ for each $i$, such that for all $z\subseteq\beta$ of ordertype $\alpha$, $\bigcap_{i\in z}B_{i}=\emptyset$. An ideal on a cardinal $\kappa$ is called simply regular when it is $(\omega,\kappa)$-regular. Taylor showed some connections between regularity properties of ideals and the structure of their associated quotient Boolean algebras, most notably the following:

Taylor also discussed degrees of regularity indexed by three ordinals. An ideal $I$ is said to be $(\alpha,\beta,\gamma)$-regular when for every sequence $\langle A_{i}:i<\gamma\rangle\subseteq I^{+}$, there is a refinement $\langle B_{i}:i<\gamma\rangle\subseteq I^{+}$ such that for every $x\subseteq\gamma$ of ordertype $\beta$, $|\bigcap_{i\in x}B_{i}|\leq\alpha$. We note the following easy relations between the regularity properties:

1. (1)

If $\alpha_{0}<\alpha_{1}$, then $(\alpha_{0},\beta,\gamma)$-regularity implies $(\alpha_{1},\beta,\gamma)$-regularity. 2. (2)

If $\beta_{0}<\beta_{1}$, then $(\alpha,\beta_{0},\gamma)$-regularity implies $(\alpha,\beta_{1},\gamma)$-regularity. 3. (3)

If $\gamma_{0}<\gamma_{1}$, then $(\alpha,\beta,\gamma_{1})$-regularity implies $(\alpha,\beta,\gamma_{0})$-regularity.

Taylor [12] showed that if $I$ is a $\kappa$-complete ideal on a regular uncountable cardinal $\kappa$, then $I$ is $(\omega,\kappa)$-regular if and only if it is $(2,\kappa)$-regular. The latter is known as the disjoint refinement property or Fodor’s property [1]. In [5], the author showed that under GCH, many more degrees of regularity are equivalent for $\kappa$-complete ideals on $\kappa$, where $\kappa$ is the successor of a regular cardinal, and this was used to examine the relationship between regularity and density of ideals on cardinals above $\omega_{1}$. In this paper as elsewhere, we only consider degrees of regularity of ideals on $\kappa$ for which the last index in the degree is at most $\kappa$. Under this restriction, we show that without any assumptions, there are only two possible flavors of two-variable regularity at successor cardinals, and with GCH, only two possible flavors of three-variable regularity:

1. (1)

$I$* is $(\operatorname{cf}(\mu)+1,\kappa)$-regular, $(1,\operatorname{cf}(\mu),\kappa)$-regular, and $(2,\delta)$-regular for $\delta<\kappa$.* 2. (2)

If $I$ is $(\operatorname{cf}(\mu),\kappa)$-regular, then $I$ is $(2,\kappa)$-regular. 3. (3)

If $I$ is $(\alpha,\beta,\kappa)$-regular for some $\alpha,\beta<\kappa$ such that $\mu^{\beta}=\mu$, then $I$ is $(2,\kappa)$-regular.

Part (1) already appeared in [12] for the case of $\mu$ regular, and though the extension to the general case does not require essentially new ideas, we include a proof here for completeness. We show similar results for $\kappa$-complete normal ideals on $\mathcal{P}_{\kappa}(\lambda)$.

We will say a normal ideal on $Z\subseteq\mathcal{P}(\lambda)$ is simply regular when it is $(2,\lambda)$-regular. Following Taylor [12], we show in Proposition 4 that $(2,\lambda)$-regularity is equivalent to $(\omega,\lambda)$-regularity for such ideals, so this definition accords with the original terminology of Keisler.

It is easy to see that a $\lambda$-dense normal ideal on $\mathcal{P}(\lambda)$ is nonregular. Taylor’s Theorem uses a result of Baumgarter-Hajnal-Máté [1], who showed that if a countably complete ideal on $\omega_{1}$ is nowhere $\omega_{1}$-dense, then it has the disjoint refinement property. This generalizes to normal ideals $I$ on $Z\subseteq\mathcal{P}_{\kappa}(\lambda)$ for $\kappa$ a successor cardinal, with an additional assumption about the quotient Boolean algebra $\mathcal{P}(Z)/I$ that is trivially satisfied for $\kappa=\omega_{1}$ (see [5]). However, it is possible to separate density and nonregularity above $\omega_{1}$:

By results in [5], the existence of a $\lambda$-dense, $\kappa$-complete, normal ideal on $\mathcal{P}_{\kappa}(\lambda)$, where $\kappa=\mu^{+}$, is consistent relative to an almost-huge cardinal, for any choice of regular cardinals $\mu<\kappa\leq\lambda$.

1. The regularity dichotomy

This section is devoted to a proof of Theorem 1. We will prove some more general facts about the regularity of normal ideals on $\mathcal{P}(\lambda)$ and show how they imply the desired results about $\kappa$-complete ideals on successor cardinals $\kappa$.

Our notations are mostly standard. By $\mathcal{P}_{\kappa}(\lambda)$ we mean $\{z\subseteq\lambda:|z|<\kappa\}$. If $x$ is a set of ordinals, then $\operatorname{ot}(x)$ denotes its ordertype. The notations $[\lambda]^{\kappa}$ and $[\lambda]^{<\kappa}$ stand for $\{z\subseteq\lambda:\operatorname{ot}(z)=\kappa\}$ and $\{z\subseteq\lambda:\operatorname{ot}(z)<\kappa\}$ respectively.

The following facts can be found in [6]. Recall that an ideal $I$ on $Z\subseteq\mathcal{P}(X)$ is normal when for all $x\in X$, $\hat{x}:=\{z\in Z:x\in z\}\in I^{*}$, and for all sequences $\langle A_{x}:x\in X\rangle\subseteq I$, the diagonal union $\nabla_{x\in X}A_{x}:=\bigcup_{x\in X}(A_{x}\cap\hat{x})\in I$. This is equivalent to the statement that for every $A\in I^{+}$ and every $f:A\to X$ such that $f(z)\in z$ for all $z\in A$, there is $B\in I^{+}$ such that $f$ is constant on $B$.

The smallest normal ideal on a set $Z$ is the nonstationary ideal on $Z$, which is the dual ideal to the club filter (closed-unbounded filter) generated by sets of the form $\{z\in Z:f[z^{<\omega}]\subseteq z\}$, where $f$ is a function from $X^{<\omega}$ to $X$. As the name suggests, positive sets for the nonstationary ideal are called stationary. Consequently, if there is a (proper) normal ideal on $Z\subseteq\mathcal{P}(X)$, then $Z$ is stationary.

A normal ideal $I$ on $Z\subseteq\mathcal{P}(X)$ is $\delta$-saturated for a cardinal $\delta$ if there is no sequence $\langle A_{\alpha}:\alpha<\delta\rangle$ such that $A_{\alpha}\cap A_{\beta}\in I$ for $\alpha<\beta$, and simply saturated if it is $|X|^{+}$-saturated. If $I$ is saturated, then $\mathcal{P}(Z)/I$ is a complete Boolean algebra, with suprema given by diagonal unions. If $\langle A_{x}:x\in X\rangle$ is an antichain, then we can use normality to refine it to a pairwise disjoint sequence of $I$-positive sets by replacing $A_{x}$ with $A_{x}\cap\hat{x}\setminus\bigcup_{y\not=x}(A_{y}\cap\hat{y})$.

The idea behind the following lemma is taken from [1].

Lemma 3**.**

Suppose $I$ is an ideal on $Z\subseteq\mathcal{P}(\lambda)$, $\delta\leq\lambda$, and $I$ is either normal or $\delta$-complete. If there is no $A\in I^{+}$ such that $I\restriction A$ is $\delta^{+}$-saturated, then $I$ is $(2,\delta)$-regular.

Proof.

Let $\langle A_{\alpha}:\alpha<\delta\rangle\subseteq I^{+}$, and for each $A_{\alpha}$, choose a sequence of $I$-positive sets $\langle B_{\alpha}^{\beta}:\beta<\delta^{+}\rangle$ such that each $B_{\alpha}^{\beta}\subseteq A_{\alpha}$ and $B_{\alpha}^{\beta}\cap B_{\alpha}^{\beta^{\prime}}\in I$ when $\beta<\beta^{\prime}<\delta^{+}$. For each $\alpha<\delta$, let $f(\alpha)\leq\alpha$ be the minimal ordinal such that $|\{\beta:A_{\alpha}\cap B_{f(\alpha)}^{\beta}\in I^{+}\}|=\delta^{+}$. We can find $\xi<\delta^{+}$ such that for all $\alpha<\delta$, all $\alpha^{\prime}<f(\alpha)$, and all $\beta\geq\xi$, $A_{\alpha}\cap B^{\beta}_{\alpha^{\prime}}\in I$.

Recursively choose a refinement $\langle C_{\alpha}:\alpha<\delta\rangle$ of $\langle A_{\alpha}:\alpha<\delta\rangle$ and an increasing sequence of ordinals $\langle\beta_{\alpha}:\alpha<\delta\rangle\subseteq\delta^{+}$ as follows. Let $C_{0}=B_{0}^{\xi}$ and $\beta_{0}=\xi$. Given $\langle C_{\alpha^{\prime}}:\alpha^{\prime}<\alpha\rangle$, let $C_{\alpha}$ be an $I$-positive set of the form $A_{\alpha}\cap B_{f(\alpha)}^{\beta_{\alpha}}$, where $\beta_{\alpha}\geq\sup_{\alpha^{\prime}<\alpha}(\beta_{\alpha^{\prime}}+1)$. Note that whenever $\alpha\not=\alpha^{\prime}$ are less than $\delta$, it is ensured that $C_{\alpha}\cap C_{\alpha^{\prime}}\in I$. This is because if $f(\alpha)=f(\alpha^{\prime})=\eta$, then $B^{\beta_{\alpha^{\prime}}}_{\eta}\cap B^{\beta_{\alpha}}_{\eta}\in I$ by construction, and if $f(\alpha)<f(\alpha^{\prime})$, then $B^{\beta_{\alpha}}_{f(\alpha)}\cap A_{\alpha^{\prime}}\in I$ since $\beta_{\alpha}\geq\xi$.

Finally, we refine $\langle C_{\alpha}:\alpha<\delta\rangle$, to a pairwise disjoint sequence $\langle D_{\alpha}:\alpha<\delta\rangle$. If $I$ is normal, we put $D_{\alpha}=C_{\alpha}\cap\hat{\alpha}\setminus\bigcup_{\alpha^{\prime}\not=\alpha}(C_{\alpha^{\prime}}\cap{\hat{\alpha}}^{\prime})$. If $I$ is $\delta$-complete, we put $D_{\alpha}=C_{\alpha}\setminus\bigcup_{\alpha^{\prime}<\alpha}C_{\alpha^{\prime}}$. ∎

Proposition 4**.**

A normal ideal on $\mathcal{P}(\lambda)$ is $(\omega,\lambda)$-regular if and only if it is $(2,\lambda)$-regular.

Proof.

Suppose $I$ is an $(\omega,\lambda)$-regular normal ideal on $\mathcal{P}(\lambda)$. Let $\langle A_{\alpha}:\alpha<\lambda\rangle\subseteq I^{+}$. We may assume that $A_{\alpha}\subseteq\hat{\alpha}$ for each $\alpha<\lambda$. Let $\langle B_{\alpha}:\alpha<\lambda\rangle\subseteq I^{+}$ be a refinement such that $\bigcap_{\alpha\in x}B_{\alpha}=\emptyset$ whenever $x\subseteq\lambda$ is infinite. For each $z\in\bigcup_{\alpha<\lambda}B_{\alpha}$, let $s(z)$ be the finite set $\{\alpha:z\in B_{\alpha}\}$. Note that $s(z)\subseteq z$.

Using the normality of $I$, for each $\alpha<\lambda$, there is an $I$-positive $C_{\alpha}\subseteq B_{\alpha}$ such that $s$ is constant on $C_{\alpha}$ with value $g(\alpha)\in[\lambda]^{<\omega}$. Note that if $g(\alpha)\not=g(\beta)$, then $C_{\alpha}\cap C_{\beta}=\emptyset$.

Now since $I$ is $(\omega,\lambda)$-regular and countably complete, there is no $A\in I^{+}$ such that the dual of $I\restriction A$ is a ultrafilter. Therefore every $I$-positive set can be partitioned into two disjoint $I$-positive sets, and thus infinitely many. Hence for every $x\in[\lambda]^{<\omega}$, the sequence $\langle C_{\alpha}:\alpha\in x\rangle$ has a disjoint refinement $\langle D^{x}_{\alpha}:\alpha\in x\rangle\subseteq I^{+}$ by Lemma 3. Finally, let $E_{\alpha}=D^{g(\alpha)}_{\alpha}$. Then $\langle E_{\alpha}:\alpha<\lambda\rangle$ is a pairwise disjoint refinement of $\langle A_{\alpha}:\alpha<\lambda\rangle$. ∎

The above argument can be generalized to show that for a $\kappa$-complete normal ideal on $\mathcal{P}(\lambda)$ and $\mu<\kappa$, $(\mu,\lambda)$-regularity implies $(2,\lambda)$-regularity if the ideal concentrates on $z\subseteq\lambda$ such that $|z|^{<\mu}=|z|$. However, we ultimately want to prove something more general about successor $\kappa$. The following lemma contains the key combinatorial idea.

Lemma 5**.**

Suppose $I$ is a normal ideal on $\mathcal{P}(\lambda)$, $\mu$ is a cardinal such that $\{z\subseteq\lambda:\operatorname{cf}(\sup z)\geq\mu\}\in I^{*}$, and for all $A\in I^{+}$ and $\delta<\lambda$, $I\restriction A$ is not $\delta^{+}$-saturated. If $I$ is $(\mu,\lambda)$-regular, then $I$ is regular.

Proof.

Let $\langle A_{\alpha}:\alpha<\lambda\rangle\subseteq I^{+}$. We may assume that for all $\alpha<\lambda$ and all $z\in A_{\alpha}$, $\operatorname{cf}(\sup z)\geq\mu$ and $\alpha\in z$. Since $I$ is $(\mu,\lambda)$-regular, there is a sequence of $I$-positive sets $\langle B_{\alpha}:\alpha<\lambda\rangle$ such that $B_{\alpha}\subseteq A_{\alpha}$ for all $\alpha$, and for all $z$, $s(z):=\{\alpha:z\in B_{\alpha}\}$ has size $<\mu$. Note that $s(z)\subseteq z$. For all $z\in\bigcup_{\alpha<\lambda}B_{\alpha}$, since $\operatorname{cf}(\sup z)\geq\mu$, $s(z)$ is not cofinal in $z$. Thus let $f:\bigcup_{\alpha<\lambda}B_{\alpha}\to\lambda$ be such that $s(z)\subseteq f(z)\in z$. By normality, for all $\alpha<\lambda$, there is an $I$-positive $C_{\alpha}\subseteq B_{\alpha}$ on which $f$ is constant. Let $g(\alpha)$ be this constant value, and note that $g(\alpha)>\alpha$.

For each $\alpha<\lambda$, choose a pairwise disjoint refinement $\langle D^{\alpha}_{\beta}:\beta<\alpha\rangle\subseteq I^{+}$ of $\langle C_{\beta}:\beta<\alpha\rangle$, using Lemma 3. Then let $E_{\alpha}=D_{\alpha}^{g(\alpha)}$. If $g(\alpha_{0})=g(\alpha_{1})$, then $E_{\alpha_{0}}\cap E_{\alpha_{1}}=\emptyset$ by construction. If $g(\alpha_{0})\not=g(\alpha_{1})$, then $E_{\alpha_{0}}\cap E_{\alpha_{1}}=\emptyset$, since for $i<2$ and $z\in E_{\alpha_{i}}$, $f(z)=g(\alpha_{i})$. ∎

Lemma 6**.**

Suppose $\mu,\lambda$ are regular cardinals, and $I$ is a normal ideal on $\mathcal{P}(\lambda)$ such that $\{z\subseteq\lambda:\operatorname{cf}(\sup z)=\mu\}\in I^{*}$. Then $I$ is $(\mu+1,\lambda)$-regular. If the function $z\mapsto\sup z$ is $\leq\delta$ to one on a set in $I^{*}$, then $I$ is $(\delta,\mu,\lambda)$-regular.

Proof.

Let $\langle A_{i}:i<\lambda\rangle\subseteq I^{+}$. Let $Z=\{z\subseteq\lambda:\operatorname{cf}(\sup z)=\mu\}$. For each $z\in Z$, let $c_{z}\subseteq z$ be a cofinal subset of ordertype $\mu$. By induction, we build an increasing sequence $\langle\alpha_{i}:i<\lambda\rangle\subseteq\lambda$ and a refinement $\langle B_{i}:i<\lambda\rangle\subseteq I^{+}$ of $\langle A_{i}:i<\lambda\rangle$ as follows. Given $\langle\alpha_{i}:i<j\rangle$, $\sup z>\sup_{i<j}\alpha_{i}$ for $I$-almost all $z\in A_{j}$. For such $z$, let $\sup_{i<j}\alpha_{i}<\alpha_{j}(z)\in c_{z}$. Let $B_{j}\subseteq A_{j}$ be an $I$-positive set on which the function $z\mapsto\alpha_{j}(z)$ is constant, and let $\alpha_{j}$ be this constant value. For each $z$, let $s(z)=\{i<\lambda:z\in B_{i}\}$. Note that $z\in B_{i}$ implies $\alpha_{i}\in c_{z}$, so $\operatorname{ot}(s(z))\leq\mu$. This establishes the claim that $I$ is $(\mu+1,\lambda)$-regular. For the second claim, note that if $\operatorname{ot}(s(z))=\mu$, then $s(z)$ is cofinal in $c_{z}$ and thus in $z$. Thus, if $z\mapsto\sup z$ is $\leq\delta$ to one on a set in $I^{*}$, then we may take the sequence $\langle B_{i}:i<\lambda\rangle$ such that $|\bigcap_{i\in x}B_{i}|\leq\delta$ whenever $\operatorname{ot}(x)=\mu$. ∎

The following result was independently observed by Burke-Matsubara [2] and Foreman-Magidor [8]. Its proof uses deep results of Shelah [11] and Cummings [4].

Lemma 7**.**

Suppose $I$ is a normal saturated ideal on $\mathcal{P}(\lambda)$. Then $\{z:\operatorname{cf}(\sup z)=\operatorname{cf}(|z|)\}\in I^{*}$.

The following basic fact can be proved in multiple ways, for example via Ulam matrices or via generic ultrapowers (see [6]).

Lemma 8**.**

If $\kappa$ is a successor cardinal, then no $\kappa$-complete ideal on $\kappa$ is $\kappa$-saturated, and no $\kappa$-complete normal ideal on $\mathcal{P}_{\kappa}(\lambda)$ is $\lambda$-saturated.

Theorem 9**.**

Suppose $\kappa=\mu^{+}$ and $I$ is a $\kappa$-complete normal ideal on $\mathcal{P}_{\kappa}(\lambda)$. If $I$ is $(\operatorname{cf}(\mu),\lambda)$-regular, then $I$ is regular. If $\lambda$ is a regular cardinal, then $I$ is $(\operatorname{cf}(\mu)+1,\lambda)$-regular.

Proof.

Let $\langle A_{\alpha}:\alpha<\lambda\rangle\subseteq I^{+}$. We first separate the saturated and non-saturated parts. We choose an initial refinement by putting $B_{\alpha}=A_{\alpha}$ if there is no $B\subseteq A_{\alpha}$ such that $I\restriction B$ is saturated, and otherwise choose $B_{\alpha}\subseteq A_{\alpha}$ such that $I\restriction B_{\alpha}$ is saturated. Let $Y_{0}$ be the ordinals below $\lambda$ falling into the first case, and $Y_{1}$ those falling into the second. Note that whenever $\alpha\in Y_{0}$ and $\beta\in Y_{1}$, we have $B_{\alpha}\cap B_{\beta}\in I$. As in the proof of Lemma 3, we may refine to a sequence $\langle C_{\alpha}:\alpha<\lambda\rangle$ such that $C_{\alpha}\cap C_{\beta}=\emptyset$ whenever at least one of $\alpha,\beta$ is in $Y_{0}$. If we put $C=\nabla_{\alpha\in Y_{1}}C_{\alpha}$, then $I\restriction C$ is saturated, since if $\langle D_{\alpha}:\alpha<\lambda^{+}\rangle$ were an antichain in $\mathcal{P}(C)/I$, then for some $\beta<\lambda$, there would be $\lambda^{+}$-many $\alpha$ such that $C_{\beta}\cap D_{\alpha}\in I^{+}$, in contradiction to the fact that $I\restriction C_{\beta}$ is saturated.

We may assume $\operatorname{cf}(\sup z)=\operatorname{cf}(\mu)$ for all $z\in C$. Since $I\restriction A$ is not $\lambda$-saturated for any $A\in I^{+}$, Lemma 5 implies that if $I$ is $(\operatorname{cf}(\mu),\lambda)$-regular, then there is a disjoint refinement of $\langle C_{\alpha}:\alpha\in Y_{1}\rangle$ into $I$-positive sets $\langle D_{\alpha}:\alpha\in Y_{1}\rangle$. Putting this together with $\langle C_{\alpha}:\alpha\in Y_{0}\rangle$, we have a disjoint refinement of the original sequence into $I$-positive sets.

If $\lambda$ is a regular cardinal, then by Lemma 6, there is a refinement $\langle E_{\alpha}:\alpha\in Y_{1}\rangle\subseteq I^{+}$ of $\langle C_{\alpha}:\alpha\in Y_{1}\rangle$ such that $\bigcap_{\alpha\in x}E_{\alpha}=\emptyset$ whenever $\operatorname{ot}(x)>\operatorname{cf}(\mu)$, showing that $I$ is $(\operatorname{cf}(\mu)+1,\lambda)$-regular. ∎

In order to prove Theorem 1, we use some results from [12] which allow a reduction to normal ideals:

Lemma 10** (Taylor).**

Let $I$ be a $\kappa$-complete ideal on $\kappa$.

1. (1)

Suppose every sequence $\langle A_{i}:i<\kappa\rangle\subseteq I^{+}$ has a refinement $\langle B_{i}:i<\kappa\rangle\subseteq I^{+}$ such that $I\restriction B_{i}$ is $(\alpha,\beta,\kappa)$-regular for each $i$. Then $I$ is $(\alpha,\beta,\kappa)$-regular. 2. (2)

If $\kappa=\mu^{+}$ and $I$ is $\kappa^{+}$-saturated, then there is $A\in I^{+}$ and a bijection $f:\kappa\to\kappa$ such that $\{f[X]:X\in I\restriction A\}$ is a normal ideal on $\kappa$.

Let $I$ be a $\kappa$-complete ideal on $\kappa=\mu^{+}$. First let us show part (1) of Theorem 1. By Lemmas 3 and 8, $I$ is $(2,\delta)$ regular for $\delta<\kappa$. For the other regularity properties, let $\langle A_{\alpha}:\alpha<\kappa\rangle\subseteq I^{+}$. Let $B_{\alpha}\subseteq A_{\alpha}$ be an $I$-positive set such that $I\restriction B_{\alpha}$ is $\kappa^{+}$-saturated if there is such a $B_{\alpha}$. In such a case, part (2) of Lemma 10 implies that we can find an $I$-positive $C_{\alpha}\subseteq B_{\alpha}$ such that $I\restriction C_{\alpha}$ is isomorphic to a normal ideal. By Lemmas 6 and 7, $I\restriction C_{\alpha}$ is $(\operatorname{cf}(\mu)+1,\kappa)$-regular and $(1,\operatorname{cf}(\mu),\kappa)$-regular whenever $C_{\alpha}$ is defined. If $C_{\alpha}$ is undefined, then $I\restriction A_{\alpha}$ is regular by Lemma 3. Part (1) of Lemma 10 then gives that $I$ is $(\operatorname{cf}(\mu)+1,\kappa)$-regular and $(1,\operatorname{cf}(\mu),\kappa)$-regular.

Now let us show part (2). Let $\langle A_{\alpha}:\alpha<\kappa\rangle\subseteq I^{+}$, and choose sets $C_{\alpha}$ exactly as above. If $I$ is $(\operatorname{cf}(\mu),\kappa)$-regular, then so is each $I\restriction C_{\alpha}$ when $C_{\alpha}$ is defined, and thus $I\restriction C_{\alpha}$ is regular by Theorem 9. Again by Lemma 3 and part (1) of Lemma 10, $I$ is regular in this case.

To show part (3) of Theorem 1, we introduce an extension of Taylor’s three-variable notion of regularity. Let us say an ideal $I$ is $(I,\alpha,\beta)$-regular if every sequence $\langle A_{i}:i<\beta\rangle\subseteq I^{+}$ has a refinement $\langle B_{i}:i<\beta\rangle\subseteq I^{+}$ such that $\bigcap_{i\in x}B_{i}\in I$ whenever $\operatorname{ot}(x)=\alpha$. If $I$ is a $\kappa$-complete ideal on $\kappa$, then $(I,\beta,\kappa)$-regularity is a weakening of $(\alpha,\beta,\kappa)$-regularity for every $\alpha<\kappa$.

Lemma 11**.**

Suppose $\kappa=\mu^{+}$ and $I$ is a $\kappa$-complete ideal on $\kappa$. If $I$ is $(I,\xi,\kappa)$-regular, where $\mu^{\xi}=\mu$, then $I$ is regular.

Proof.

Let $\langle A_{\alpha}:\alpha<\kappa\rangle\subseteq I^{+}$, and let $\langle B_{\alpha}:\alpha<\kappa\rangle\subseteq I^{+}$ be a refinement such that $B_{\alpha}\subseteq\hat{\alpha}$ for all $\alpha$, and $\bigcap_{\alpha\in x}B_{\alpha}\in I$ whenever $\operatorname{ot}(x)\geq\xi$. For every $\alpha<\kappa$ we can define an $I$-positive $C_{\alpha}\subseteq B_{\alpha}$ by

[TABLE]

If $x$ is a subset of $\kappa$ of ordertype $\xi+1$, then let $\alpha=\max(x)$. If $\beta\in C_{\alpha}$, then $\beta\notin\bigcap_{\gamma\in x\cap\alpha}C_{\gamma}$. This shows that $I$ is $(\xi+1,\kappa)$-regular. Since $\mu^{\xi}=\mu$ implies $\xi<\operatorname{cf}(\mu)$, $I$ is regular by part (2) of Theorem 1. ∎

2. Consistency results

This section is devoted to a proof of Theorem 2. If $V\subseteq W$ are models of set theory and $I\in V$ is an ideal, then in $W$ we can generate an ideal $\bar{I}$ from $I$ by taking all sets which are covered by a set from $I$. Let us first show the preservation of nonregular ideals by forcings with a strong enough chain condition, as a consequence of Theorem 9.

Lemma 12**.**

Suppose $\kappa=\mu^{+}$, $\lambda\geq\kappa$, and $I$ is a nonregular, $\kappa$-complete, normal ideal on $Z\subseteq\mathcal{P}_{\kappa}(\lambda)$. If $\mathbb{P}$ is $\operatorname{cf}(\mu)$-c.c., then in $V^{\mathbb{P}}$, the ideal $\bar{I}$ generated by $I$ is nonregular.

Proof.

Let us show the contrapositive, that if $\bar{I}$ is regular in a $\mathbb{P}$-generic extension, then $I$ is regular in $V$. Let $\langle A_{\alpha}:\alpha<\lambda\rangle\subseteq I^{+}$ be in $V$. If $p\Vdash\bar{I}$ is regular, then there is a $\mathbb{P}$-name for a refinement $\langle\dot{B}_{\alpha}:\alpha<\lambda\rangle$ such that each $z\in Z$ is forced by $p$ to be in at most one $B_{\alpha}$. In $V$, for each $\alpha$ let $C_{\alpha}=\{z\in A_{\alpha}:(\exists q\leq p)q\Vdash z\in\dot{B}_{\alpha}\}$. Since $p\Vdash\dot{B}_{\alpha}\subseteq\check{C}_{\alpha}$, each $C_{\alpha}$ is $I$-positive. By the chain condition, for each $z$, the set $s(z):=\{\alpha:(\exists q\leq p)q\Vdash z\in\dot{B}_{\alpha}\}=\{\alpha:z\in C_{\alpha}\}$ has size $<\operatorname{cf}(\mu)$. This shows that $I$ is $(\operatorname{cf}(\mu),\lambda)$-regular in $V$, and thus regular by Theorem 9. ∎

If $I$ is a $\kappa$-complete normal ideal and $\mathbb{P}$ is a $\kappa$-c.c. forcing, then it is easy to show that the ideal generated by $I$ is also $\kappa$-complete and normal in $V^{\mathbb{P}}$. If $I$ is saturated, then Foreman’s Duality Theorem [7] allows us to say much more. This is connected to the forcing properties of the quotient algebra and generic elementary embeddings.

The following facts can be found in[6]. If $I$ is an ideal on $Z$ and $G\subseteq\mathcal{P}(Z)/I$ is generic, then in $V[G]$, we can form the ultrapower embedding $j:V\to V^{Z}/G$. If $Z\subseteq\mathcal{P}(\lambda)$ and $I$ is normal, then the pointwise image of $\lambda$ under $j$ is represented in the ultrapower by the identity function on $Z$, i.e. $[\operatorname{id}]_{G}=j[\lambda]$. If $I$ is $\kappa$-complete, $\kappa=\mu^{+}$, and $Z\subseteq\mathcal{P}_{\kappa}(\lambda)$, then $\kappa$ is the critical point of $j$, and $V^{Z}/G\models|j[\lambda]|<j(\kappa)$. Consequently, $V[G]\models|\lambda|=|\mu|$. This implies that there is no condition $A\in I^{+}$ such that $I\restriction A$ is $\lambda$-saturated. Thus in this context, $I$ being saturated is the same as $\mathcal{P}(Z)/I$ having the best possible chain condition. If this occurs, then $I$ is precipitous, meaning that whenever $G\subseteq\mathcal{P}(Z)/I$ is generic, $V^{Z}/G$ is well-founded and thus isomorphic to a transitive class $M\subseteq V[G]$.

Theorem 13** (Foreman [7]).**

Suppose $I$ is a $\kappa$-complete precipitous ideal on $Z$, and $\mathbb{P}$ is a $\kappa$-c.c. forcing. In $V^{\mathbb{P}}$, let $\bar{I}$ denote the ideal generated by $I$, and let $j$ denote a generic ultrapower embedding obtained from forcing with $\mathcal{P}(Z)/I$. Then there is an isomorphism

[TABLE]

given by $\iota(p,\dot{A})=||[\operatorname{id}]\in j(\dot{A})||\wedge(1,\dot{j(p)})$.

The next proposition shows the relevance of the cardinal arithmetic assumption in Lemma 11. For example, we can produce a model in which CH fails and there is a nonregular ideal $I$ on $\omega_{2}$ which is $(I,\omega,\omega_{2})$-regular.

Proposition 14**.**

Suppose $\kappa=\mu^{+}$, $\nu\leq\mu$ is such that $\nu^{<\nu}=\nu$, and $I$ is a saturated, nonregular, $\kappa$-complete ideal on $\kappa$. If $G\subseteq\operatorname{Add}(\nu,\kappa)$ is generic, then in $V[G]$, $\bar{I}$ is $(\bar{I},\nu,\kappa)$-regular.

Proof.

Since $\nu^{<\nu}=\nu$, $\operatorname{Add}(\nu,\kappa)$ is $\nu^{+}$-c.c. By Theorem 13, in $V[G]$, there is an isomorphism $\sigma:\mathcal{P}(\kappa)/\bar{I}\cong\mathcal{B}(\mathcal{P}(\kappa)^{V}/I\times\operatorname{Add}(\nu,\kappa^{+}))$. If $\langle A_{\alpha}:\alpha<\kappa\rangle\subseteq\bar{I}^{+}$, choose for each $\alpha$ some $(B_{\alpha},p_{\alpha})\leq\sigma(A_{\alpha})$. Let $\beta<\kappa^{+}$ be such that $\operatorname{dom}p_{\alpha}\subseteq\beta\times\nu$ for all $\alpha$. Let $q_{\alpha}=\{((\beta+\alpha,0),0)\}$ for $\alpha<\kappa$, and choose $C_{\alpha}\leq\sigma^{-1}(B_{\alpha},p_{\alpha}\wedge q_{\alpha})$. The intersection of any $\nu$-many $C_{\alpha}$ is in $\bar{I}$, since there is no lower bound to $\nu$-many $q_{\alpha}$. ∎

Lemma 15**.**

Suppose $I$ is a normal ideal on $Z\subseteq\mathcal{P}(X)$. Then $I$ is $|X|^{+}$-saturated if and only if every normal $J\supseteq I$ is equal to $I\restriction A$ for some $A\subseteq Z$.

Proof.

Suppose $I$ is $|X|^{+}$-saturated. Let $Y\subseteq X$ and $\{A_{x}:x\in Y\}$ be such that $\{[A_{x}]_{I}:x\in Y\}$ is an antichain in $\mathcal{P}(Z)/I$ of $I$-positive sets that are $J$-measure-zero, and is maximal among all such collections. Then $[\nabla A_{x}]_{I}$ is the $\subseteq_{I}$-largest element of $J\cap I^{+}$, so $J=I\restriction(Z\setminus\nabla A_{x})$. Now suppose $I$ is not $|X|^{+}$-saturated, and let $\{A_{\alpha}:\alpha<\delta\}$ be a maximal antichain where $\delta\geq|X|^{+}$. Let $J$ be the ideal generated by $\bigcup\{\Sigma_{\alpha\in Y}[A_{\alpha}]:Y\in\mathcal{P}_{|X|^{+}}(\delta)\}$. Then $J$ is a proper normal ideal extending $I$. $J$ cannot be equal $I\restriction A$ for some $A\in I^{+}$ because if this were so, there would some $\alpha$ such that $A\cap A_{\alpha}\in I^{+}$. $A\cap A_{\alpha}\in J$ by construction, but every $I$-positive subset of $A$ is $(I\restriction A)$-positive. ∎

A partial order is said to be $\kappa$-dense if it has a dense subset of size $\leq\kappa$. It is said to be nowhere $\kappa$-dense if it is not $\kappa$-dense below any condition. An ideal is said to be $\lambda$-dense or nowhere $\lambda$-dense when its associated Boolean algebra has these properties.

Lemma 16**.**

Suppose $\mu^{+}=\kappa\leq\lambda$, $\nu\leq\mu$ is such that $\nu^{<\nu}=\nu$, and $Z\subseteq\mathcal{P}_{\kappa}(\lambda)$ is stationary. Let $\mathbb{P}=\operatorname{Add}(\nu,\theta)$ for some $\theta\geq\kappa$. Then in $V^{\mathbb{P}}$, there are no normal, $\kappa$-complete, $\lambda$-dense ideals on $Z$.

Proof.

Since a $\lambda$-dense ideal is $\lambda^{+}$-saturated, it suffices to consider $\mathbb{P}$-names for $\lambda^{+}$-saturated normal ideals on $Z$. Suppose $p\Vdash\dot{J}$ is a $\kappa$-complete, $\lambda^{+}$-saturated, normal ideal on $Z$. Let $I=\{X\subseteq Z:p\Vdash X\in\dot{J}\}$. It is easy to check that $I$ is normal and $\kappa$-complete. The map $\sigma:\mathcal{P}(Z)/I\to\mathcal{B}(\mathbb{P}\restriction p)*\mathcal{P}(Z)/\dot{J}$ that sends $X$ to $(||\check{X}\in\dot{J}^{+}||,\dot{[X]_{J}})$ is order-preserving and antichain-preserving. Since $\nu^{<\nu}=\nu$, $\mathbb{P}$ is $\kappa$-c.c., so the two-step iteration $\mathbb{P}\restriction p*\mathcal{P}(Z)/\dot{J}$ is $\lambda^{+}$-c.c. Thus $I$ is $\lambda^{+}$-saturated.

Let $H$ be $\mathbb{P}$-generic over $V$ with $p\in H$. Since $\mathbb{P}$ is $\kappa$-c.c., $\bar{I}$ remains normal. By Theorem 13, the map $e:q\mapsto(1,\dot{j(q)})$ is a regular embedding of $\mathbb{P}$ into $\mathcal{P}(Z)/I*\dot{j(\mathbb{P})}$. Thus in $V[H]$, $\mathcal{P}(Z)/\bar{I}\cong\mathcal{P}^{V}(Z)/I*\operatorname{Add}(\nu,\dot{\eta})$, where $\Vdash\dot{\eta}=\operatorname{ot}(j(\theta)\setminus j[\theta])$. By the saturation of $I$, $\Vdash j(\kappa)=\lambda^{+}$, so $\Vdash\dot{\eta}\geq\lambda^{+}$.

$\bar{I}$ is normal and $\lambda^{+}$-saturated, and $\bar{I}\subseteq J$. By Lemma 15, there is $A\in\bar{I}^{+}$ such that $J=\bar{I}\restriction A$. Since $\operatorname{Add}(\nu,\eta)$ is nowhere $\lambda$-dense, $\mathcal{P}(Z)/\bar{I}$ is nowhere $\lambda$-dense. Thus $J$ is not $\lambda$-dense. ∎

Thus we may rid the universe of dense ideals that concentrate on $\mathcal{P}_{\kappa}(\lambda)^{V}$. This finishes the job if $\kappa=\lambda$, but not necessarily in other cases. For example, Gitik [9] showed that if $V\subseteq W$ are models of set theory, $\kappa<\lambda$ are regular in $W$, and there is a real number in $W\setminus V$, then $\mathcal{P}_{\kappa}(\lambda)^{W}\setminus\mathcal{P}_{\kappa}(\lambda)^{V}$ is stationary. In order to take care of such problems, we use some arguments of Laver and Hajnal-Juhasz that are reproduced in [6].

The notation \left(\begin{array}[]{c}\alpha\\ \beta\end{array}\right)\rightarrow\left(\begin{array}[]{c}\gamma\\ \delta\end{array}\right)_{\eta} stands for the assertion that for every $f:\alpha\times\beta\to\eta$, there is $A\in[\alpha]^{\gamma}$ and $B\in[\beta]^{\delta}$ such that $f$ is constant on $A\times B$. As usual with arrow notations, if ordinals on the left side are increased and ordinals on the right side are decreased, then we get a weaker statement.

Lemma 17**.**

Suppose there is a $\lambda$-dense, $\kappa$-complete, normal ideal $I$ on $\mathcal{P}_{\kappa}(\lambda)$ such that every $I$-positive set has cardinality $\geq\eta$. Then for $\mu,\nu<\kappa$,

[TABLE]

Proof.

Let $\theta=\lambda^{<\kappa}$, and enumerate $\mathcal{P}_{\kappa}(\lambda)$ as $\langle z_{\alpha}:\alpha<\theta\rangle$. Let $f:\lambda^{+}\times\theta\to\nu$. By $\kappa$-completeness, for each $\alpha<\lambda^{+}$, there is $\gamma<\nu$ such that $X_{\alpha}:=\{z_{\beta}:f(\alpha,\beta)=\gamma\}\in I^{+}$. By $\lambda$-density, there is a set $S\in[\lambda^{+}]^{\lambda^{+}}$, a set $D\in I^{+}$, and a $\gamma^{*}<\nu$ such that for all $\alpha\in S$, $D\subseteq_{I}X_{\alpha}$ and $f(\alpha,\beta)=\gamma^{*}$ for $z_{\beta}\in X_{\alpha}$. Let $A\subseteq S$ have size $\mu$. Since $\bigcap_{\alpha\in A}X_{\alpha}$ is $I$-positive, there is a set $B\subseteq\theta$ of size $\geq\eta$ such that for all $\alpha\in A$ and all $\beta\in B$, $f(\alpha,\beta)=\gamma^{*}$. ∎

Lemma 18**.**

Suppose $\theta$ is regular and $\mu<\theta$ is such that $\mu^{<\mu}=\mu$. If $G\subseteq\operatorname{Add}(\mu,\theta)$ is generic, then in $V[G]$,

[TABLE]

Proof.

In $V$, choose an almost-disjoint family $\{X_{\alpha}:\alpha<\theta^{+}\}\subseteq\mathcal{P}(\theta)$, and for each $\alpha$, let $\langle\gamma^{\alpha}_{\beta}:\beta<\theta\rangle$ enumerate $X_{\alpha}$ in increasing order. In $V[G]$, let $f:\theta^{+}\times\theta\to 2$ be defined by $f(\alpha,\beta)=G(\gamma^{\alpha}_{\beta},0)$. Let $A\subseteq\theta^{+}$ be a set of size $\mu$ in $V[G]$. By the chain condition, there is a $\zeta<\theta$ such that $G=G_{0}\times G_{1}$, where $G_{0}$ is $\operatorname{Add}(\mu,\zeta)$-generic, and $A\in V[G_{0}]$. In $V[G_{0}]$, $\delta$ be such that $\zeta<\delta<\theta$ and $\{X_{\alpha}\setminus\delta:\alpha\in A\}$ is pairwise disjoint. For any $p\in\operatorname{Add}(\mu,\theta\setminus\zeta)$ and any $\eta\geq\delta$, there are $q\leq p$ and $\alpha,\beta\in A$ such that $q(\gamma_{\eta}^{\alpha},0)\not=q(\gamma_{\eta}^{\beta},0)$. Since $G_{1}$ is generic, we have that for all $\eta\geq\delta$, there are $\alpha,\beta\in A$ such that $f(\alpha,\eta)\not=f(\beta,\eta)$. Thus there is no $B\subseteq\theta$ of size $\theta$ such that $f$ is constant on $A\times B$. ∎

We can now prove Theorem 2. Suppose that in $V$, $I$ is a nonregular, $\kappa$-complete, normal ideal on $\mathcal{P}_{\kappa}(\lambda)$, where $\kappa=\mu^{+}$ and $\operatorname{cf}(\mu)$ is uncountable. Let $\theta\geq\lambda^{\mu}$ be regular and such that $\theta^{\mu}=\theta$. Let $G\subseteq\operatorname{Add}(\omega,\theta)$ be generic. By Lemma 12, $\bar{I}$ is nonregular in $V[G]$. Suppose $Z\subseteq\mathcal{P}_{\kappa}(\lambda)$ has cardinality $<\theta$. Then there is $\zeta<\theta$ such that $G=G_{0}\times G_{1}$, where $G_{0}$ is $\operatorname{Add}(\omega,\zeta)$-generic, and $Z\in V[G_{0}]$. By Lemma 16, there is no $\lambda$-dense, $\kappa$-complete, normal ideal concentrating on $Z$ in $V[G]$. For $\kappa$-complete normal ideals on $\mathcal{P}_{\kappa}(\lambda)$ in $V[G]$ that do not have any positive set of size $<\theta$, Lemma 17 implies that if such an ideal were $\lambda$-dense, then we would have \left(\begin{array}[]{c}\lambda^{+}\\ \theta\end{array}\right)\rightarrow\left(\begin{array}[]{c}\mu\\ \theta\end{array}\right)_{\mu}, since $\lambda^{\mu}=\theta$ in $V[G]$. But Lemma 18 implies that the weaker relation \left(\begin{array}[]{c}\theta^{+}\\ \theta\end{array}\right)\rightarrow\left(\begin{array}[]{c}\omega\\ \theta\end{array}\right)_{2} fails in $V[G]$. Thus $V[G]$ has no $\lambda$-dense, $\kappa$-complete, normal ideal on $\mathcal{P}_{\kappa}(\lambda)$.