41

数理解析研究所講究録
第2081巻 2018年 41-56

On regular ultrafilters, Boolean ultrapowers, and Keisler’s order
Francesco Parente

School of Mathematics

University of East Anglia
Norwich NR4 7TJ, United Kingdom
f.parente@uea.ac.uk
Abstract

In this paper we analyse and compare two different notions of regularity for filters on complete
Boolean algebras. We also announce two results from a forthcoming paper in preparation, which
provide a characterization of Keisler’s order in terms of Boolean ultrapowers.

1

Introduction

Over the last decade, Malliaris and Shelah proved a striking sequence of results in the intersection

between model theory and set theory, settled affirmatively the question of whether \mathfrak{p}=\mathrm{t} , and developed
surprising connections between classification theory and cardinal characteristics of the continuum. The
starting point for their work is the study of Keisler’s order, introduced originally in 1967 as a device tocompare the complexity of complete theories by looking at regular ultrapowers of their models.
The intuitive idea behind Keisler’s order is simple: a theory T_{1} is“less complicated” than a theory T_{2}
if the ultrapowers of models of T_{1} are“more likely” to be saturated than the ultrapowers of models of T_{2}.

As Malliaris and Shelah [18] put it, Keisler’s order classifies “theories through the lens of ultrafilters”

It turns out that there is a specific class of ultrafilters which is particularly suitable for this classification

work, namely the regular ultrafilters.

Definition 1.1 (Keisler [10]). Let rc be an infinite cardinal. A filter F over a set I is $\kappa$ ‐regular iff there

exists a family \{X_{ $\alpha$} | $\alpha$< $\kappa$\}\subseteq F such that for every infinite I\subseteq \mathrm{K} we have \displaystyle \bigcap_{ $\alpha$\in I}X_{ $\alpha$}=\emptyset.

The importance of regular ultrafilters lies in the following theorem, which states that whether or not
the regular ultrapower of a model of a complete theory is saturated does not depend on the choice of theparticular model, but only on the theory itself.

Theorem 1.2 (Keisler [11, Corollary 2. 1\mathrm{a}Let $\kappa$ be an infinite cardinal; suppose U \dot{u} a $\kappa$ ‐regularultrafilter over a set I. If two L ‐structures \mathfrak{M} and \mathfrak{R} are elementarily equivalent, and |L| \leq $\kappa$ , then\mathfrak{M}^{I}/U is $\kappa$^{+} ‐saturated \Leftrightarrow ;\mathfrak{n}^{I}/U is $\kappa$^{+} ‐saturated.
This theorem then suggests a way of comparing theories according to the saturation of ultrapowers.

Definition 1.3 (Keisler [11]). Let T_{1} and T_{2} be complete countable theories and $\kappa$ an infinite cardinal.
We define T_{1} \underline{\triangleleft}_{ $\kap a$}T_{2} iff for every $\kappa$‐regular ultrafilter U over $\kappa$ and models \mathfrak{M}_{1} \models T_{1}, \mathfrak{M}_{2} \models T_{2} , if \mathfrak{M}_{2^{ $\kappa$}}/U
is $\kappa$^{+} ‐saturated then \mathfrak{M}_{1}^{ $\kappa$}/U is $\kappa$^{+} ‐saturated.
\triangleleft defined as follows: T_{1}\underline{\triangleleft}T_{2} iff for every infinite $\kappa$, T_{1} \underline{\triangleleft}_{ $\kap a$}T_{2}.
Keisler’s order is the preorder relation −As already mentioned, recent groundbreaking research has shed new light on the structure of Keisler’sorder. For example, Malliaris and Shelah [19] showed that there is an infinite strictly descending chain oftheories, and Ulrich [26] proved the existence of incomparable theories assuming a supercompact cardinal.
Although the definition of Keisler’s order makes use of regular ultrafilters over sets, Malliaris andShelah [17] developed the method of separation of variables, which involves a “paradigm shift” towards
building ultrafilters on complete Boolean algebras to classify theories. More specifically, suppose we42want to construct a regular ultrafilter over $\lambda$ with some specific saturation properties. We build instead
an ultrafilter U on a complete Boolean algebra \mathrm{B} , together with a suitable surjective homomorphismj:\mathcal{P}( $\lambda$)\rightar ow \mathrm{B} , in such a way that j^{-1}[U] will be a regular ultrafilter over $\lambda$ with the desired properties.
On the other hand, in a forthcoming paper by Raghavan and Shelah [23], Boolean ultrapowers offorcing iterations are used to force inequalities between cardinal invariants at and above $\omega$ . A common
thread is the construction of ultrafilters on complete Boolean algebras which will realize or omit some
types in the corresponding Boolean ultrapowers.
Motivated by the above results, in this paper we shall focus on the interaction between the combinat‐
orial properties of ultrafilters and the model‐theoretic properties of Boolean ultrapowers. More precisely,

we ask the following question: what kind of classification can arise when we compare theories according
to the saturation of Boolean ultrapowers of their models?
Since we have already seen the crucial role of regular ultrafilters in this context, the first step towards
an answer consists in finding the right definition of regularity for ultrafilters on complete Boolean algebras.
In fact, two such definitions have already appeared in the literature, both under the name “regular”. This
confusion motivates the results of Section 2, where those two notions are compared and shown not to beequivalent.
In Section 3, we analyse the model‐theoretic properties of regular ultrafilters in terms of Boolean ul‐
trapowers. In particular, we shall focus on cardinality, cofinality, and universality of Boolean ultrapowers.

In each case, one notion of regularity behaves as expected, while the other notion is not well behaved.
In the final section we shall announce two forthcoming results, which provide an answer to ourquestion: Keisler’s order can be equivalently characterized in terms of saturation of Boolean ultrapowers.
Hence, the model‐theoretic properties captured by ultrafilters on complete Boolean algebras are exactly
the same as for power‐set algebras, thus explaining the success of the“paradigm shift” mentioned above.

2

Two notions of regularity

In this section, we shall present and compare two different definitions of regularity for filters on com‐plete Boolean algebras. As we remarked earlier, both notions have appeared in the literature under thename “regular”. To avoid creating further confusion, we have decided to use the names “regular)’ and“quasiregular’) to distinguish them.
Before we present the first definition, we need to introduce some standard terminology.

Definition 2.1. Let $\kappa$ be a cardinal; a Boolean algebra \mathrm{B} is $\kappa$-c.c . iff every antichain in \mathrm{B} has cardinality
less than K. The saturation of \mathrm{B} , denoted by sat(B), is the least cardinal $\kappa$ such that \mathrm{B} is t￡‐c.c.

Theorem 2.2 (Erdõs and Tarski [3]). If \mathrm{B} is an infinite Boolean algebra, then sat(B) is an uncountable
regular cardinal.

The next remark is straightforward, but will be useful in the proof of Proposition 2.6.

Remark 2.3. Suppose \mathrm{B} is a complete Boolean algebra. Then, for every cardinal $\kappa$< sat(B) there exists
a maximal antichain A\subset \mathrm{B} with |A| = $\kappa$ . To prove this, we note that if $\kappa$< sat(B) then, by definition,\mathrm{B} has an antichain of cardinality \geq $\kappa$ . Using Zorn’s lemma, we may extend this antichain to a maximal
antichain W . Since $\kappa$ \leq |W| , it is possible to partition W into $\kappa$ many non‐empty disjoint pieces:
W=\displaystyle \bigcup_{i< $\kappa$}W_{i} . Then clearly A=\{\vee W_{i}|i< $\kappa$\} is a maximal antichain in \mathrm{B} such that |A| = $\kappa$.

We are now ready to state the first main definition, which is due to Shelah [24].
Deflnition 2.4. Let \mathrm{B} be a complete Boolean algebra and \mathrm{K} an infinite cardinal. We say that an filter
F on \mathrm{B} is $\kappa$ ‐regular iff there exist a family \{x_{ $\alpha$} | $\alpha$< $\kappa$\}\subseteq F and a maximal antichain A\subset \mathrm{B} such that:\bulletfor every $\alpha$< $\kappa$ and every a\in A , either a\leq x_{ $\alpha$} or a\wedge x_{ $\alpha$}=0 ;\bulletfor every a\in A , the set \{ $\alpha$< $\kappa$|a\leq x_{\mathfrak{a}}\} is finite.

It follows immediately from Definition 2.4 that if F is a $\kappa$‐regular filter and $\lambda$ \leq $\kappa$ , thenF is also$\lambda$‐regular.

Remark 2.5. Let F be a filter on B. If a family \{x_{ $\alpha$} | $\alpha$< $\kappa$\} \subseteq F and a maximal antichain A \subset \mathrm{B}
witness the $\kappa$‐regularity of F , then necessarily $\kappa$ \leq |A| . Indeed, for every $\alpha$ < $\kappa$ we can choose somea_{ $\alpha$} \in A such that a_{ $\alpha$} \leq x_{ $\alpha$} ; hence, by $\kappa$-‐regularity, the map $\alpha$\mapsto a_{ $\alpha$} is finite‐to‐one from $\kappa$ to A.

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We now present an existence result for regular ultrafilters; the argument is a simple modificationof the construction of Frayne, Morel and Scott [5, Theorem 1.17]. Also, we remark that more generalexistence results for regular ultrafilters will appear in Raghavan and Shelah [23].
Proposition 2.6. Let $\kappa$ be an infinite cardinal. For a complete Boolean algebra \mathrm{B} , the following condi‐
tions are equivalent:
1. \mathrm{B} is not $\kappa$-c.c.

2. there exists a $\kappa$ ‐regular ultrafilter on B.
r}
Proof. (2\Rightarrow 1) We already know from Remark 2.5 that if there exists a $\kappa$_{‐regularfilter on \mathrm{B} , then \mathrm{B}
has necessarily an antichain of cardinality \geq $\kappa$.
(1\Rightarrow 2) Suppose \mathrm{B} is not K‐c.c. By Remark 2.3, we can find a maximal antichain A=\{a_{i}|i< $\kappa$\}
in \mathrm{B} such that |A|= $\kappa$ . Let us fix an enumeration [ $\kap a$]^{<\mathrm{N}_{0} =\{S_{i} |i< $\kappa$\} and define for every $\alpha$< $\kappa$

x_{ $\alpha$}=\vee\{a_{i}| $\alpha$\in S_{i}\}.
Observe that for every $\alpha$_{1} , . . . , $\alpha$_{n}< $\kappa$ we havex_{$\alpha$_{1} \wedge\cdots\wedge x_{$\alpha$_{f1} =\vee\{a_{i} |\{$\alpha$_{1}, . . . , $\alpha$_{n}\}\subseteq S_{i}\}>0,
hence the family \{x_{ $\alpha$} | $\alpha$< $\kappa$\} has the finite intersection property, and so it generates an ultrafilter U
on B.

To show that U is $\kappa$‐regular, we just observe that for each $\alpha$ < $\kappa$ and every i < $\kappa$ we have the two
implicationsa_{i}\wedge x_{ $\alpha$} >0 \Rightarrow $\alpha$\in S_{i} \Rightarrow a_{i}\leq x_{ $\alpha$}.

From this, it follows immediately that the family \{x_{\mathfrak{a} | $\alpha$< $\kappa$\} \subseteq U and the maximal antichain A\subset \mathrm{B}\squaresatisfy the two conditions of Definition 2.4.

We shall now present a second definition of regularity, which can be found in Koppelberg and \mathrm{K}\mathrm{o}\trianglerightpelberg [13] and Huberich [7]. This is arguably the most natural generalization of Definition 1.1 to thelanguage of Boolean algebras; however, our choice of terminology “quasiregular” is motivated by the res‐ults in Section 3, which demonstrate that this natural generalization is in fact not well behaved model
theoretically.
Definition 2.7. Let \mathrm{B} be a complete Boolean algebra and $\kappa$ an infinite cardinal. We say that a filter F
on \mathrm{B} is $\kappa$ ‐quasiregular iff there exists a family \{x_{ $\alpha$} | $\alpha$< $\kappa$\} \subseteq F such that for every infinite I \subseteq $\kappa$ wehave \displaystyle \bigwedge_{ $\alpha$\in I}x_{ $\alpha$}=0.

Again, it follows from Definition 2.7 that if F is a Kr‐quasiregular filter and $\lambda$ \leq $\kappa$ , thenF is also$\lambda$‐quasiregular.

The next proposition is straightforward, and justifies our choice of terminology.
Proposition 2.8. Let rc be an infinite cardinal. For any complete Boolean algebra \mathrm{B} , every $\kappa$ ‐regularfilter on \mathrm{B} is also $\kappa$ ‐quasiregular.

Proof. Suppose F is a $\kappa$‐regular filter on \mathrm{B} ; this is witnessed by a family \{x_{ $\alpha$} | $\alpha$< $\kappa$\} \subseteq F and amaximal antichain A\subset \mathrm{B} . We shall prove that for every infinite I\subseteq $\kappa$ we have \displaystyle \bigwedge_{ $\alpha$\in I}x_{ $\alpha$}=0 . To obtaina contradiction, suppose this is not the case. Then \displaystyle \bigwedge_{ $\alpha$\in I}x_{ $\alpha$} > 0 , which implies the existence of somea\in A with a\displaystyle \wedge\bigwedge_{ $\alpha$\in I}x_{ $\alpha$} >0 , since A is maximal. Therefore, for every $\alpha$\in I we have a\wedge x_{ $\alpha$} >0 , which
implies a\leq x_{ $\alpha$} by the definition of $\kappa$‐regularity. Thus, we have shown that there exists a\in A such that\squarea\leq x_{ $\alpha$} for infinitely many $\alpha$' \mathrm{s} , a contradiction.

Lemma 2.9. Let $\kappa$ be an infinite cardinal and \mathrm{B} a complete Boolean algebra. If F is a $\kappa$ ‐regular filter
on \mathrm{B} , then the maximal antichain witnessing its regulamty can be chosen to have cardinality $\kappa$.

44

Proof. Suppose F is a $\kappa$‐regular filter on \mathrm{B} ; this is witnessed by a family \{x_{ $\alpha$}| $\alpha$< $\kappa$\} \subseteq

F

and a

maximal antichain A\subset \mathrm{B} . Consider the following antichain:

W=\displayst le\{ bigwedge_{$\alpha$\inI}x_{$\alpha$}\wedge\bigwedge_{$\alpha$\not\inI}\negx_{$\alpha$}|I\subset q$\kap a$\} backsla h\{0\}.
From the definition of W ,it follows that for every $\alpha$ <$\kappa$and every \mathrm{w} \inW,either w \leq

x_{ $\alpha$}orw\wedge x_{ $\alpha$}=0 . Furthermore, for every w\in W the set \{ $\alpha$< $\kappa$|w\leq x_{ $\alpha$}\} must be finite, otherwise it would

contradict the $\kappa$‐quasiregularity of F (Proposition 2.8).

To see that W is maximal, it suffices to observe for every a\in A there exists a set I\subseteq $\kappa$ such thata\displayst le\leq\bigwedge_{$\alpha$\inI}x_{$\alpha$}\wedge\bigwedge_{$\alpha$\not\inI}\negx_{$\alpha$}.
Hence, \mathrm{I}=A\leq\vee W and so W is maximal.

From Remark 2.5 we already know that $\kappa$\leq |W| . To see that |W|
infinite we must have \displaystyle \bigwedge_{ $\alpha$\in I}x_{ $\alpha$}=0 ; therefore we have the equality=$\kappa$, observe that whenever I is

W=\displaystyle\{ bigwedge_{$\alpha$\inI}x_{$\alpha$}\wedge\bigwedge_{$\alpha$\not\inI}\negx_{$\alpha$}|I\in[$\kap a$]^{<\aleph_{0} \} backsla h\{0\},which gives us |W| \leq$\kappa$^{<\aleph_{0} = $\kappa$.

\squareBefore the next result, let us recall first that a filter F on a complete Boolean algebra \mathrm{B} is \aleph_{1} ‐incomplete
iff there exists a countable subset X\subseteq F such that \wedge X\not\in F.

Proposition 2.10. Let \mathrm{B} be a complete Boolean algebra.

For an ultrafilter U on \mathrm{B} , the following

conditions are equivalent:

1. U is \aleph_{0} ‐regular;
2. U is \aleph_{0} ‐quasiregular;
3. U is \aleph_{1} ‐incomplete.

Proof. (1\Rightarrow 2) Follows immediately from Proposition 2.8.

(2\Rightarrow 3) Directly from the definition of \aleph_{0} ‐quasiregularity, we obtain the existence of some infinite

X\subseteq U with \wedge X=0\not\in U , as desired.

(3\Rightarrow 1) Suppose U is \aleph_{1} ‐incomplete; since U is an ultrafilter, this entails the existence of a countable
subset \{x_{n} |n< $\omega$\} \subseteq U such that \displaystyle \bigwedge_{n< $\omega$}x_{n}O. Without loss of generality, we may assume thatx_{n+1} <x_{n} for all n< $\omega$ , and x_{0}=\mathrm{I} . Let us define for every i< $\omega$=a_{i}=x_{i}\wedge\neg x_{i+1} ;it is clear that A=\{a_{i}|i< $\omega$\} is an antichain. Furthermore, A is maximal, because for all i< $\omega$a_{0}\vee\cdot \cdot \cdot\vee a_{i}=x_{0}\wedge\neg x_{i+1}=\mathrm{f}\mathrm{L}\wedge\neg x_{i+1}=\neg x_{i+1},
and therefore\displaystyle \ve A=\bigve _{i< $\omega$}(a_{0}\ve \cdots\ve a_{i})=\bigve _{i< $\omega$}\neg x_{i+1}=\neg\bigwedge_{i< $\omega$}x_{i+1}=1.
To show that U is \aleph_{0} ‐regular, it is sufficient to observe that for all i, n< $\omega$ we have the two implicationsa_{i}\wedge x_{n}>0 \Rightarrow n\leq i \Rightarrow a_{i}\leq x_{n}.

From this, we deduce that the family \{x_{n}|n< $\omega$\} and the maximal antichain A satisfy the two conditionsof Definition 2.4.

\squareThus, when $\kappa$=\aleph_{0} , both regularity properties coincide with \aleph_{1} ‐incompleteness. When \mathrm{K} is arbitrary,an additional distributivity assumption on \mathrm{B} will also make the two properties coincide.

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Definition 2.11 (Smith and Tarski [25]). Let $\kappa$ and $\lambda$ be cardinals. A complete Boolean algebra \mathrm{B} is\{ $\kappa$, $\lambda$\rangle ‐distributive iff for every function b : $\kappa$\times $\lambda$\rightar ow \mathrm{B} we have\displayst le\bigwedge_{$\alpha$<$\kap a$}\bigve _{$\beta$<$\lambda$}b($\alpha$, \beta$)=\bigve _{f\in^{$\kap a$} \lambda$}\bigwedge_{$\alpha$<$\kap a$}b($\alpha$,f($\alpha$).

Proposition 2.12. Let $\kappa$ be an infinite cardinal. If \mathrm{B} is a \{ $\kappa$, 2\rangle ‐distributive complete Boolean algebra,
then every $\kappa$ ‐quasiregular filter on \mathrm{B} is $\kappa$ ‐regular.
Proof. Suppose F is a $\kappa$‐quasiregular filter on \mathrm{B} ; by definition, there exists a family \{x_{ $\alpha$} | $\alpha$< $\kappa$\} \subseteq
such that for every infinite I\subseteq $\kappa$ we have \displaystyle \bigwedge_{ $\alpha$\in I}x_{ $\alpha$} =0.

F

Let us define

A=\displayst le\{ bigwedge_{$\alpha$\inI}x_{$\alpha$}\wedge\bigwedge_{$\alpha$\not\inI}\negx_{$\alpha$}|I\subset q$\kap a$\} backsla h\{0\})first of all, it is clear that A is an antichain of B. Furthermore, since each subset I \subseteq $\kappa$ corresponds to
its characteristic function f : $\kappa$\rightarrow 2 , we can apply \{ $\kappa$, 2\rangle ‐distributivity to conclude that\displayst le\ve A=\ve \{ bigwedge_{$\alpha$\inI}x_{$\alpha$}\wedge\bigwedge_{$\alpha$\not\inI}\negx_{$\alpha$}|I\subset q$\kap a$\}=\bigwedge_{$\alpha$<$\kap a$}(x_{$\alpha$}\ve \negx_{$\alpha$})=\bigwedge_{$\alpha$<$\kap a$}\mathrm{I}=1.
This shows that A is a maximal antichain. By definition of A , it follows that for every $\alpha$< $\kappa$ and every
a \in A , either a \leq x_{ $\alpha$} or a\wedge x_{ $\alpha$}
O. Furthermore, for every a \in A the set \{ $\alpha$< $\kappa$|a\leq x_{ $\alpha$}\} is finite,\squareotherwise we would contradict the $\kappa$‐quasiregularity of F . This shows that F is $\kappa$ ‐regular.
=

The Cohen algebra
We now focus on a specific complete Boolean algebra which will provide many examples of quasiregularultrafilters which are not regular.
Definition 2.13. For an infinite cardinal $\kappa$ , let \mathb {P}_{$\kap a$} be the set of finite partial functions from $\kappa$ to 2.
Given p, q\in \mathbb{P}_{ $\kappa$} , we define q\leq p if and only if p\subseteq q . Thus, \langle \mathb {P}_{ $\kap a$}, \leq\rangle is the forcing notion that adjoins rc
Cohen reals.

As usual, we say that two conditions p, q\in \mathbb{P}_{ $\kappa$} are compatible iff there exists r\in \mathbb{P}_{ $\kappa$} such that r\leq p
and r\leq q (otherwise, p and q are incompatible).

By a standard result (see Jech [8, Corollary 14.12]), there exists a unique complete Boolean algebra\mathb {C}_{ $\kap a$} , usually referred to as the Cohen algebra, with a function e:\mathbb{P}_{ $\kappa$}\rightar ow \mathbb{C}_{ $\kappa$}\backslash \{0\} such that:\bulletif q\leq p then e(q)\leq e(p) ;\bulletp and q are compatible in \mathb {P}_{ $\kap a$} if and only if e(p)\wedge e(q) >0 ;\bullete[\mathb {P}_{ $\kap a$}] is dense in \mathbb{C}_{ $\kappa$}\backslash \{0\}.

The following fact is well known and we do not prove it here; a proof can be found, for example,

in Jech [8].
Fact 2.14. The Cohen algebra \mathb {C}_{$\kap a$} is an \aleph_{1} ‐c. c . complete Boolean algebra of cardinality \mathrm{K}^{\mathrm{N}_{0}In particular, Proposition 2.6 implies that no filter on \mathb {C}_{ $\kap a$} is \aleph_{1} ‐regular. On the other hand, thefollowin \mathrm{g} lemma will provide plenty of $\kappa$ ‐quasiregular ultrafilters on \mathb {C}_{ $\kap a$}.

Lemma 2.15 (Koppelberg and Koppelberg [13]). On the Cohen algebra \mathb {C}_{$\kap a$} , every ultrafilter is $\kappa$quasiregular.

Proof. For each $\alpha$< $\kappa$ , let us define in \mathb {P}_{$\kap a$}

p_{ $\alpha$,0}=\{\langle $\alpha$, 0\rangle\},p_{ $\alpha$,1}=\{\{ $\alpha$ , 1} \}.

46

Firstly, we prove that for every $\alpha$< $\kappa$\neg e(p_{ $\alpha$,0})=e(p_{ $\alpha$,1}).

Clearly, since p_{ $\alpha$,0} and p_{ $\alpha$,1} are incompatible, we have e(p_{ $\alpha$,0}) \wedge e(p_{ $\alpha$,1})
O. Furthermore, we have
1 , because otherwise, by density of e[\mathb {P}_{ $\kap a$}] , we could find some p \in \mathb {P}_{ $\kap a$} such thate(p_{ $\alpha$,0}) \vee e(p_{ $\alpha$,1})
e(p)\wedge(e(p_{ $\alpha$,0})\vee e(p_{ $\alpha$,1}))=0 . But then, as a consequence, p and p_{ $\alpha$,0} are incompatible, which means thatp( $\alpha$)=1 , but also p and p_{ $\alpha$,1} are incompatible, which means that p( $\alpha$)=0 , a contradiction.
==

Secondly, we prove that whenever I\subseteq $\kappa$ is infinite, we have\displaystyle\bigwedge_{$\alpha$\inI}e(p_{$\alpha$_{)}0 )=0 and \displaystyle\bigwedge_{$\alpha$\inI}e(p_{$\alpha$,1})=0 .

(1)

To obtain a contradiction, suppose that \displaystyle \bigwedge_{ $\alpha$\in I}e(p_{ $\alpha$,0}) > O. Then, by density of e[\mathb {P}_{ $\kap a$}] , there exists somep\in \mathbb{P}_{ $\kappa$} such thate(p)\displaystyle \leq\bigwedge_{ $\alpha$\in I}e(p_{ $\alpha$,0}) .

(2)

We now distinguish two cases, and derive a contradiction in each case. If p\leq p_{ $\alpha$},0 for every $\alpha$\in I , then\displaystyle \bigcup_{ $\alpha$\in I}p_{ $\alpha$,0} \subseteq p , which is impossible as p is a finite function. On the other hand, if p\not\leq p_{ $\alpha$,0} for some $\alpha$\in I,
then \langle $\alpha$, 0 ) \not\in p , therefore there is some q\leq p such that q( $\alpha$)=1 . Hence q\leq p_{ $\alpha$,1} , but then using (2) wederivee(q) \leq e(p_{ $\alpha$,1})\wedge e(p)=0,
which is another contradiction. Of course, the same argument also shows that, if I\subseteq $\kappa$ is infinite, then\displaystyle \bigwedge_{ $\alpha$\in I}e(p_{ $\alpha$,1})=0 . This completes the proof of (1).
Now let U be any ultrafilter on \mathb {C}_{$\kap a$} . If we define

G=\{p_{ $\alpha$,0} | $\alpha$< $\kappa$\},it follows from what we proved so far that the set

X=(G\cap U)\cup\{\neg g|g\in G\backslash U\}
is a subset of U with |X|$\kap a$_{ $\G‐quasiregular.
amma$}=$\kappa$, such that wheneverY \subseteq

X

is infinite we have \wedge Y=

O. Thus, U is\squareOK ultrafilters

While the main focus in this paper is on regular ultrafilters, we conclude this section with a digression
on OK ultrafilters. Our motivation here is to show that if an ultrafilter is \aleph_{1} ‐incomplete and tc‐OK, thenit is $\kappa$‐regular in the sense of Definition 2.4.
OK ultrafilters were originally defined by Kunen [14] in the context of the topology of $\beta \omega$ , the Stone‐Čech compactification of the set of natural numbers. Five years later, Dow [2] rephrased Kunen’s definitionin terms of existence of multiplicative functions: this is the definition we present below.
Definition 2.16. Let X be any set, \mathrm{B} a Boolean algebra and f : [X]^{<\mathrm{N}_{0} \rightarrow \mathrm{B}.
\bulletf is monotonically decreasing iff for all S, T\in [X]^{<\mathrm{N}_{0} , S\subseteq T implies f(T) \leq f(S) .

\bulletf is multiplicative iff for all S, T\in[X]^{<\aleph_{0}}, f(S\cup T)=f(S)\wedge f(T) .

Definition 2.17. Let $\kappa$ be an infinite cardinal. A filter F on a complete Boolean algebra \mathrm{B} is rc‐OK iff for|T| implies f(S) =f(T) , there
every monotonically decreasing function f : [ $\kap a$]^{<\mathrm{N}_{0} \rightarrow F such that |S|exists a multiplicative function g : [ $\kap a$]^{<\aleph_{0} \rightarrow F with the property that g(S) \leq f(S) for all S\in[ $\kappa$]^{<\aleph_{\mathrm{O} }=Although it is not completely obvious from Definition 2.17, it is an easy exercise to verify that if F isa tc‐OK filter and $\lambda$\leq $\kappa$ , thenF is also $\lambda$-\mathrm{O}\mathrm{K}.

The model‐theoretic relevance of OK ultrafilters lies in a property called flexibility, first isolatedby Malliaris [16]. For more details about the connection between OK ultrafilters and Keisler’s order werefer the reader to the work of Malliaris and Shelah [18].

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Theorem 2.18 (Ulrich [27, Theorem 5.5]). Let \mathrm{K} be an infinite cardinal; suppose U is an \aleph_{1} ‐incomplete$\kappa$-OK ultrafilter on a complete Boolean algebra B.

Then \mathrm{B} is not $\kappa$-c.c.

The purpose of the next proposition is to show that Ulrich’s argument, which follows the proofof Mansfield [20, Theorem 4.1], can be slightly adapted to obtain a stronger result.
Proposition 2.19. Let rc be an infinite cardinal; suppose U is an \aleph_{1} ‐incomplete $\kappa$-OK ultrafilter on acomplete Boolean algebra B. Then U is $\kappa$ ‐regular.

Proof. Since U is an \aleph_{1} ‐incomplete ultrafilter, there exists a countable subset \{a_{n} |n< $\omega$\} \subseteq U suchthat \displaystyle \bigwedge_{n< $\omega$}a_{n}=0 . Without loss of generality, we may assume that a_{n+1} <a_{n} for all n< $\omega$ , and a_{0}=\mathrm{I}.
Using this sequence, we can define a monotonically decreasing function as follows:f : [ $\kap a$]^{<\mathrm{N}_{0} \rightarrow U

S\mapsto a_{|S|}Since U is K‐OK, we can find a multiplicative function g : [ $\kap a$]^{<\aleph_{0} \rightarrow U such that g(S) \leq f(S) for all
S\in [ $\kap a$]^{<\aleph_{\mathrm{O} Note that f(\emptyset)=a_{0}=1 , so we may assume without loss of generality that g(\emptyset)=1 as well.
Now, for every $\alpha$< $\kappa$ define

x_{ $\alpha$}=g(\{ $\alpha$\})For everyn<\mathrm{w}.

, if $\alpha$_{1} , . . . , $\alpha$_{n}< $\kappa$ are all distinct, then by the multiplicativity of gx_{ $\alpha$ 1}\wedge\cdots\wedge x_{$\alpha$_{n}} =g(\{$\alpha$_{1}\})\wedge\cdots\wedge g(\{$\alpha$_{n}\})=g(\{$\alpha$_{1}, \ldots, $\alpha$_{n}\}) \leq f(\{$\alpha$_{1) )}$\alpha$_{n}\})=a_{n}.
This shows that, whenever I\subseteq $\kappa$ is infinite, we have \displaystyle \bigwedge_{ $\alpha$\in I}x_{ $\alpha$} =0.

To conclude the proof, we need to find a maximal antichain A \subset \mathrm{B} such that for every $\alpha$ < rc andevery a\in A , either a\leq x_{ $\alpha$} or a\wedge x_{ $\alpha$}=0 . In order to do so, it is sufficient to prove that the set
D= { d\in \mathrm{B}\backslash \{0\} | for every $\alpha$< $\kappa$ , either

d\leq x_{ $\alpha$} or d\wedge x_{ $\alpha$}=0 }is dense in \mathrm{B}\backslash \{0\} . Then, every maximal antichain A \subseteq D will have the desired property. So let
b\in \mathrm{B}\backslash \{0\} ; we need to find some d\in D such that d\leq b . For every n< $\omega$ , definec_{n}=\vee\{g(S) |S\in[ $\kappa$]^{n} \}.
1 . It follows that there exists
It is easy to verify that c_{n+1} \leq c_{n} for all n < $\omega$ and that c_{0}g(\emptyset)
0, \mathrm{a}some i < $\omega$ such that b A c_{i}\wedge\neg c_{i+1} 〉 0 (otherwise, we would have b \leq \displaystyle \bigwedge_{n<$\omega$^{C_{n} \leq \displaystyle \bigwedge_{n< $\omega$}a_{n}contradiction). Therefore, by definition of \mathrm{c}_{i} , there exists S\in[ $\kappa$]^{i} such that===

d=b\wedge g(S)\wedge\neg c_{i+1} > O.
Clearly d\leq b , so we just need to show that d\in D . Let $\alpha$< $\kappa$ ; if $\alpha$\in S then

d=b\wedge g(S)\wedge\neg c_{i+1}\leq g(S) \leq g(\{ $\alpha$\})=x_{ $\alpha$}.
Otherwise, if $\alpha$\not\in S , then by the multiplicativity of gd\wedge x_{ $\alpha$}=b\wedge g(S)\wedge g(\{ $\alpha$\})\wedge\neg c_{i+1}=b\wedge g(S\cup\{ $\alpha$\})\wedge\neg c_{i+1}\leq b\wedge c_{i+1}\wedge\neg c_{i+1}=0.
Therefore d\in D , as desired.

3\squareModel‐theoretic properties

In this third section we shall analyse the model‐theoretic properties of regular and quasiregular ultrafilters.

The natural tool for this analysis is the Boolean ultrapower construction, which dates back to Foster [4].
The standard reference for Boolean ultrapowers is Mansfield [20]; however, since we shall use a slightly

different (but equivalent) formulation, the details will be spelled out in the first part of the section.
Even though we have been working until now with filters on complete Boolean algebras, from now on
only ultrafilters will be considered; this is due to the relevance of Theorem 3.9 which we shall be using
essentially.

48

The Boolean ultrapower construction
Before we present the details of the Boolean ultrapower construction, we need some terminology.
Definition 3.1. Let A and W be maximal antichains of a complete Boolean algebra B. We say that W
is a refinement of A iff for every w \in W there exists a \in A such that w \leq a . Note that this elementa\in A is unique.

Definition 3.2 (Hamkins and Seabold [6]). Let X be any set, A a maximal antichain, and f : A\rightarrow X.
If W is a refinement of A , the reduction of f to W is the function(f\downarrow W) : W\rightarrow X
w\mapsto f(a)’where a is the unique element of A such that w\leq a.

Remark 3.3. Finitely many maximal antichains A_{1} , . . . , A_{n} always admit a common refinement, which is
the maximal antichain\{a_{1}\wedge\cdots\wedge a_{n} |a_{i}\in A_{i}\}\backslash \{0\}.
After these preliminary definitions, we proceed to define a Boolean‐valued semantics. The first Defin‐
ition 3.4 deals with the interpretations of the symbols in the language.
Definition 3.4. Let \mathfrak{M} be an L‐structure and \mathrm{B} a complete Boolean algebra. We define first the set ofnamesM^{[\mathrm{B}]}\bullet={ $\tau$ : A\rightarrow M|A\subset \mathrm{B} is a maximal antichain}.

We now define the Boolean value of the equality symbol: if $\tau$, $\sigma$\in M^{[\mathrm{B}]} , choose a common refinement
W of \mathrm{d}\mathrm{o}\mathrm{m}( $\tau$) and \mathrm{d}\mathrm{o}\mathrm{m}( $\sigma$) , and define[ $\tau$= $\sigma$ \mathrm{J}^{\mathfrak{M}^{| \mathrm{B}|} =\vee\{w\in W| ( $\tau$\downarrow W)(w)=( $\sigma$\downarrow W)(w)\}.
\bulletThe Boolean values of the symbols in L are defined aô follows:‐ if R\in L is an $\eta$,‐ary function symbol and $\tau$_{1} , . . . , $\tau$_{n} \in M^{[\mathrm{B}]} , choose a common refinement W
of \mathrm{d}\mathrm{o}\mathrm{m}($\tau$_{1}) , . . . , dom ($\tau$_{n}) , and define[R($\tau$_{1}, \ldots, $\tau$_{n})\mathrm{I}^{\mathfrak{M}^{|\mathrm{K}|} =\vee\{w\in W|\mathfrak{M}\models R(($\tau$_{1}\downarrow W)(w), \ldots, ($\tau$_{n}\downarrow W)(w) \};‐ if f\in L is an n‐ary function symbol and $\tau$_{1} , . . . , $\tau$_{n}, $\sigma$\in M^{[\mathrm{B}]} , choose a common refinement W
of dom ($\tau$_{1}) , . . . , dom ($\tau$_{n}) , dom ( $\sigma$) , and define[f($\tau$_{1}, \ldots, $\tau$_{n})= $\sigma$ \mathrm{I}^{\mathfrak{M}^{\mathrm{l}\mathrm{R}|} =\vee\{\mathrm{w}\in W|\mathfrak{M}\models f( $\tau$_{1}\downarrow W)(w), \ldots, ($\tau$_{n}\downarrow W)(w) =( $\sigma$\downarrow W)(w)\};‐ if c\in L is a constant symbol, its interpretation is the namec^{9\mathrm{J}l^{[1\mathrm{R}] } : \{\mathrm{I}\}\rightarrow M\mathrm{I}\mapsto c^{9\mathrm{J}\mathfrak{l}Following Mansfield [20], the definition of the Boolean values is extended to all formulae in thelanguage, not necessarily atomic: if $\varphi$(x_{1}, \ldots, x_{n}) is an L-‐formula and $\tau$_{1} , . . . , $\tau$_{n} \in M^{|\mathrm{B}]} , the Boolean
value[ $\varphi$($\tau$_{1) )}$\tau$_{n})\mathrm{I}^{y}n^{[\mathrm{R}]}can be defined recursively. From now on, when there is no danger of confusion, the superscript \mathfrak{M}^{[\mathrm{B}] will
be omitted.

This Boolean‐valued semantics is made explicit in the next proposition, which could be taken as adefinition:

49

Proposition 3.5 (Mansfield [20, Theorem 1.1]). Let \mathfrak{M} be an L ‐structure and \mathrm{B} a complete Boolean
algebra. Let $\varphi$(x_{1}, \ldots, x_{n}) be an L ‐formula and $\tau$_{1} , . . . , $\tau$_{n} \in M^{[\mathrm{B}]} . If W is any common refinement ofdom ($\tau$_{1})) . . . , dom ($\tau$_{n}) , then

[ $\varphi$($\tau$_{1)}\ldots, $\tau$_{n})\mathrm{J}=\vee\{w\in W|\mathfrak{M}\models $\varphi$( $\tau$_{1}\downarrow W)(w), )($\tau$_{n}\downarrow W)(w) \}.
The following result, sometimes called “mixing property” in the literature, will be useful in the proofof Proposition 3.16.

Proposition 3.6 (Mansfield [20, Theorem 1.3]). Let \mathfrak{M} be an L ‐structure and \mathrm{B} a complete Boolean
algebra. If A\subset \mathrm{B} is an antichain and \{$\tau$_{a} |a\in A\}\subseteq M^{[\mathrm{B}]} , then there is $\tau$\in M^{[1\mathrm{B}]} such that a\leq [ $\tau$=$\tau$_{a}\mathrm{I}for all a\in A.

We are now ready to present the main definition.

Definition 3.7. Let \mathfrak{M} be an L‐structure, \mathrm{B} a complete Boolean algebra, and U an ultrafilter on B. The
Boolean ultrapower of \mathfrak{M} by U , denoted by \mathfrak{M}^{[\mathrm{B}]}/U , is the L ‐structure defined as follows:\bulletIts domain, denoted by M^{[\mathrm{B}]}/U , is the quotient of M^{|\mathrm{B}]} by the equivalence relation \equiv U defined as$\tau$\equiv U $\sigma$ \Leftrightar ow^{\mathrm{d}\mathrm{e}\mathrm{f} [ $\tau$= $\sigma$ \mathrm{J} \in U.
The \equiv U ‐equivalence class of a name $\tau$\in M^{[\mathrm{B}]} is denoted by [ $\tau$]_{U}.
\bulletThe interpretations of the symbols in L are defined in the natural way; for example, if R\in L is ann ‐ary relation symbol, thenR^{\mathrm{g}\mathfrak{n}^{1\mathb {R}\mathrm{J} /U}=\{\langle[$\tau$_{1}]_{U}

,

.

.

.

,

[$\tau$_{n}]_{U}\rangle \in n(M^{[\mathrm{B}]}/U)| [R($\tau$_{1,}$\tau$_{n})\mathrm{I}\in U\})and similarly for function and constant symbols.
Remark 3.8. Suppose U is an ultrafilter over a set I . Then, for every structure \mathfrak{M}\mathfrak{M}^{[\mathcal{P}(I)]}/U\cong \mathfrak{M}^{I}/U ;hence, Boolean ultrapowers are indeed a generalization of ultrapowers.

The following is the analogue for Boolean ultrapowers of a well‐known theorem of Loś [15].
Theorem 3.9 (Mansfield [20, Theorem 1.5]). Let EM be an L ‐structure, \mathrm{B} a complete Boolean algebra,
and U an ultrafilter on B. For every L ‐formula $\varphi$ (x_{1}, \ldots , x_{n}) and $\tau$_{1} , . . . , $\tau$_{n}\in M^{[\mathrm{B}]} we have\mathfrak{M}^{[\mathrm{B}]}/U\models $\varphi$ ([$\tau$_{1}]_{U}, . . . [$\tau$_{n}]_{U}) \Leftrightar ow [ $\varphi$($\tau$_{1}, . . . $\tau$_{n})\mathrm{I}\in U.
In particular, if for every m\in M we define the name\check{m} :\{\mathrm{I}\}\rightarrow M
1\mapsto m )then we obtain the following corollary.

Corollary 3.10. Let \mathfrak{M} be an L ‐structure, \mathrm{B} a complete Boolean algebra, and U an ultrafilter on B.
Then the natural embedding, defined asd:M\rightarrow M^{\mathrm{B}\mathrm{J} /U

m\mapsto[\check{m}]_{U} )is an elementary embedding of \mathfrak{M} ínto \mathfrak{M}^{[\mathrm{B}]}/U.

50

Cardinality
The problem of determining the possible cardinalities of the ultrapowers of a given structure starts withan simple observation: if U is an ultrafilter over I , then for every structure \mathfrak{M}|M|\leq |M^{I}/U| \leq|M|^{|I|}(3)

Of course, if U is principal then |M|= |M^{I}/U| , hence the lower bound in (3) can be attained. Therefore,
it is natural to ask whether or not the upper bound in (3) can be attained for some ultrafilter U over I.
This question led Frayne, Morel and Scott to consider regular ultrafilters in [5].

Theorem 3.11 (FYayne, Morel and Scott [5]). Let rc be an infinite cardinal; suppose U is a $\kappa$ ‐regularultrafilter over a set I. For every infinite structure \mathfrak{M}_{f} we have|M|^{ $\kappa$}\leq |M^{I}/U|.
In particular, if|I|= $\kappa$ then the upper bound 1 is attained.
Motivated by this result, we can ask whether the same is true for Boolean ultrapowers. As we shallsee, the parallel of Theorem 3.11 is true for regular ultrafilters on complete Boolean algebras, but canfail for quasiregular ultrafilters. First, we need to establish a bound analogous to (3).
Lemma 3.12. Let U be an ultrafilter on a complete Boolean algebra B. For every structure \mathfrak{M}_{f} we have

|M|\leq |M^{[\mathrm{B}] /U| \leq|M|^{<\mathrm{s}\mathrm{a}\mathrm{t}(\mathrm{B}) +|\mathrm{B}|^{<\mathrm{s}\mathrm{a}\mathrm{t}(\mathrm{B})(4)

Proof. The inequality |M| \leq |M^{[\mathrm{B}]}/U| follows immediately from Corollary 3.10. On the other hand,|M^{[\mathrm{B}]}/U| \leq |M^{[\mathrm{B}] |=| { $\tau$ : A\rightarrow M| A\subset \mathrm{B} is a maximal antichain} |\leq |\cup\{x_{M} |X\in[\mathrm{B}]^{<\mathrm{s}\mathrm{a}\mathrm{t}(\mathrm{B}) \}| =|M|^{<\mathrm{s}\mathrm{a}\mathrm{t}(\mathrm{B}) +|\mathrm{B}|^{<\mathrm{s}\mathrm{a}\mathrm{t}(\mathrm{B}) ,\squareas desired.

We now show that regular ultrafilters produce Boolean ultrapowers of large cardinality; the proof ofthis result is just a minor modification of the proof of Theorem 3.11.

Proposition 3.13. Let rc be an infinite cardinal; suppose U is a $\kappa$ ‐regular ultrafilter on a complete
Boolean algebra B. For every infinite structure \mathfrak{M} , we have|M|^{ $\kappa$}\leq|M^{ $\beta$]}/U| .

(5)

In particular, if \mathrm{B} is a $\kappa$^{+} ‐c. c . Boolean algebra of size\leq 2^{ $\kappa$} , then the upper bound in (4) is attained.
Proof. Since |^{< $\iota$ v}M|
|M| , it is sufficient to find an injective function i : $\kappa$ M \rightarrow (^{< $\omega$}M)^{ $\beta$]}/U . Let theamma$}
family \{x_{ $\alpha$} | $\alpha$< $\kappa$\} \subseteq U and the maximal antichain A \subset \mathrm{B} witness the $\kap a$_{$\G‐regularityof U . Hence, for=every a\in A the set

S(a)=\{ $\alpha$< $\kappa$|a\leq x_{ $\alpha$}\}is finite.

Given a function f : $\kappa$\rightarrow

M , we define $\tau$_{f} : A \rightarrow< $\omega$ M as follows.

S(a) increasingly as $\alpha$_{1} <\cdots<$\alpha$_{n} and define$\tau$_{f}(a)=\langle f($\alpha$_{1})

,

.

.

.

,

f($\alpha$_{n})\rangle.

We now prove that the function
i : $\kappa$ M\rightarrow(^{< $\omega$}M)^{\mathrm{F}]}/U

f\mapsto [$\tau$_{f}]_{u}Fix a \in A ; list all the elements of

51

is injective. Let f, g : $\kappa$\rightarrow M ; if f \neq g then there exists some $\alpha$ < $\kappa$ such that f( $\alpha$) \neq g( $\alpha$) . For all
a\in A , if a\leq x_{ $\alpha$} then $\alpha$\in S(a) and therefore, by construction, $\tau$_{f}(a)\neq$\tau$_{g}(a) . It follows that[$\tau$_{f}\neq$\tau$_{g}\mathrm{J}=\vee\{a\in A|$\tau$_{f}(a)\neq$\tau$_{g}(a)\}\geq\vee\{a\in A|a\leq x_{ $\alpha$}\}=x_{ $\alpha$} \in U,
hence [$\tau$_{f}\neq$\tau$_{g}\mathrm{I}\in U , as required. This shows that i : $\kappa$ M\rightar ow(^{< $\omega$}M)^{[\mathrm{B}]}/U is injective, establishing (5).
Now, if we assume further that \mathrm{B} is a $\kappa$^{+}-\mathrm{c}.\mathrm{c} . Boolean algebra of size \leq 2^{ $\kappa$} , then for every infinite
structure \mathfrak{M}|M|^{ $\kap a$}\leq |M^{|\mathrm{B}] /U| \leq |M|^{<\mathrm{s}\mathrm{a}\mathrm{t}(\mathrm{B}) +|\mathrm{B}|^{<\mathrm{s}\mathrm{a}\mathrm{t}(\mathrm{B}) \leq|M|^{ $\kap a$}+(2^{ $\kap a$})^{ $\kap a$}=|M|^{ $\kap a$},\squarehence we have equality throughout.

Using the Cohen algebra, we can find a counterexample for quasiregular ultrafilters.

Proposition 3.14. Let $\kappa$ be an uncountable cardinal. Then there exists a complete Boolean algebra \mathrm{B}and a $\kappa$ ‐quasiregular ultrafilter U on \mathrm{B} such that, for some infinite structure SErt,|M^{[\mathrm{B}]}/U| <|M|^{ $\kappa$}
mma$}
Proof. Let \mathb {C}_{$\kap a$} be the Cohen algebra and let U be an ultrafilter on \mathb {C}_{$\kap a$} ; we know that U is $\kap a$_{$\Ga‐quasiregularby Lemma 2.15.

Let $\lambda$ be a cardinal such that$\kap a$\leq$\lambda$^{\mathrm{N}_{0} <$\lambda$^{ $\kap a$}(6)Note that it is always possible to find such a cardinal: for example, if $\lambda$ \geq $\kappa$ is a strong limit cardinalwith \mathrm{c}\mathrm{f}( $\lambda$)=\aleph_{1} , then $\lambda$ satisfies (6). Now, if \mathfrak{M} is a structure with |M|= $\lambda$ then Lemma 3.12 gives us|M^{[\mathb {C}_{ $\kap a$}] /U| \leq$\lambda$^{<\aleph_{1} +($\kap a$^{\aleph_{0} )^{<\aleph_{1} =$\lambda$^{\mathrm{N}_{0} +$\kap a$^{\mathrm{N}_{\mathrm{O} =$\lambda$^{\aleph_{0} <$\lambda$^{ $\kap a$},\squareas desired.

Cofinality
An important feature of regular ultrafilters is that they produce ultrapowers of large cofinality. We
shall now investigate whether the same is true in the context of complete Boolean algebras and Boolean
ultrapowers. Again, our results show that regular ultrafilters behave as expected, while quasiregularultrafilters are not well behaved.

Proposition 3.15. Let $\kappa$ be an infinite cardinal; suppose U is a $\kappa$ ‐regular ultrafilter over a set I. For
every infinite cardinal $\lambda$ , the ultrapower \langle $\lambda$, <\rangle^{I}/U has cofinality > $\kappa$.

The above result can be found in Benda and Ketonen [1, Theorem 1.3], where it is referred to as a(‘standard fact” It appears also in Koppelberg [12, Lemma 2].
By adapting the usual proof of Proposition 3.15, and using the mixing property of Proposition 3.6,
we can establish the corresponding result for Boolean ultrapowers.

Proposition 3.16. Let $\kappa$ be an infinite cardinal; suppose U is a $\kappa$ ‐regular ultrafilter on a complete
Boolean algebra B. For every infinite cardinal $\lambda$ , the Boolean ultrapower \langle $\lambda$, <\rangle^{[\mathrm{B}]}/U has cofinality > $\kappa$.

Proof. Let the family \{x_{ $\alpha$} | $\alpha$< $\kappa$\} \subseteq U and the maximal antichain A \subset \mathrm{B} witness the $\kappa$‐regularity ofU.

In particular, this means that for every a\in A the set

S(a)=\{ $\alpha$< $\kappa$|a\leq x_{ $\alpha$}\}is finite.

Given any \{$\tau$_{ $\alpha$} | $\alpha$< $\kap a$\}\subset$\lambda$^{[\mathrm{B}]} , we show that the sequence \{[$\tau$_{ $\alpha$}]_{U} | $\alpha$< $\kappa$\} is not cofinal in \langle $\lambda$, <\rangle^{\mathbb{R}]}/U

by finding some $\sigma$\in$\lambda$^{[\mathrm{B}]} such that [$\tau$_{ $\alpha$} \leq $\sigma$ \mathrm{I}\in U for all $\alpha$< $\kappa$.

52

For every a\in A we wish to define a name $\sigma$_{a} \in$\lambda$^{[\mathrm{m}]} such that\displaystyle\bigwedge_{$\alpha$\inS(a)}[$\tau$_{$\alpha$}\leq$\sigma$_{a}\mathrm{I}=1.

(7)

To do so, consider the finitely many names \{$\tau$_{ $\alpha$} | $\alpha$\in S(a)\} . Bring their domains to a common refinement
W_{a} and define $\sigma$_{a} : W_{a}\rightarrow $\lambda$ as follows: for all w\in W_{a}$\sigma$_{a}(w)=\displaystyle \max\{($\tau$_{ $\alpha$}\downarrow W_{a})(w) | $\alpha$\in S(a)\}.
Clearly $\sigma$_{a} will bound each $\tau$_{ $\alpha$)} for $\alpha$\in S(a) , with Boolean value fl, and so (7) is proved.
Now, use Proposition 3.6 to obtain a name $\sigma$ \in $\lambda$^{[\mathrm{B}] such that a \leq [ $\sigma$=$\sigma$_{a}\mathrm{I} for each a \in A . Sincex_{ $\alpha$}\in U for every $\alpha$< $\kappa$ , to complete the proof it is sufficient to show thatx_{ $\alpha$} \leq[$\tau$_{ $\alpha$}\leq $\sigma$ \mathrm{I} .

(8)

For all a\in A , if a\leq x_{ $\alpha$} then $\alpha$\in S(a) , hence [$\tau$_{ $\alpha$}\leq$\sigma$_{a}\mathrm{I}=1 anda\leq[ $\sigma$=$\sigma$_{a}\mathrm{I}=[ $\sigma$=$\sigma$_{a}\mathrm{J}\wedge 1=[ $\sigma$=$\sigma$_{a}\mathrm{J}\wedge[$\tau$_{ $\alpha$} \leq$\sigma$_{a}\mathrm{J}\leq [$\tau$_{ $\alpha$} \leq $\sigma$ \mathrm{J}.
Thus we have shown that, for all a\in A , if a\leq x_{ $\alpha$} then a\leq[$\tau$_{ $\alpha$} \leq $\sigma$ \mathrm{I} . Now (8) follows: for every $\alpha$< $\kappa$[$\tau$_{ $\alpha$} \leq $\sigma$ \mathrm{J}\geq\vee\{a\in A|a\leq[$\tau$_{ $\alpha$} \leq $\sigma$ \mathrm{J}\}\geq\vee\{a\in A|a\leq x_{ $\alpha$}\}=x_{ $\alpha$}\in U,\squarethus showing that [$\tau$_{ $\alpha$} \leq $\sigma$ \mathrm{I}\in U.

Since the cofinality of an ordered set is not greater than its cardinality, from the estimate of Lemma 3.12
we already obtain a counterexample for quasiregular ultrafilters. To see this, let $\kappa$ be a cardinal suchthat $\kap a$^{\mathrm{N}_{0} = $\kappa$ . If U is any ultrafilter on \mathb {C}_{$\kap a$} , then U is $\kappa$‐quasiregular, however by Lemma 3.12
\mathrm{c}\mathrm{f}(\langle$\kap a$_{)} <\rangle^{[\mathb {C}_{ $\kap a$}] /U) \leq |$\kap a$^{[\mathb {C}_{$\kap a$}] /U| \leq $\kappa$^{<\aleph_{1} ($\kap a$^{\aleph_{0} )^{<\aleph_{1}+=$\kap a$^{\mathrm{N}_{0}=$\kappa$.

Actually, we can prove a more general result.

Proposition 3.17. Let $\kappa$ be a regular uncountable cardinal and \mathrm{B} a is‐c. c . complete Boolean algebra. For
every ultrafilter U on \mathrm{B} , the Boolean ultrapower \langle $\kap a$, <\}^{[\mathrm{B}]}/U has cofinality $\kappa$.
Proof. We observe first that for every $\tau$\in $\kap a$^{[\mathrm{B}] there exists some $\alpha$ < $\kappa$ such that [ $\tau$\leq\dot{ $\alpha$}\mathrm{J}I. Indeed,
given a name $\tau$ , the $\kappa$_{r}\mathrm{c}.\mathrm{c} . implies that |\mathrm{d}\mathrm{o}\mathrm{m}( $\tau$)| < $\kappa$ . Since $\kappa$ is a regular cardinal, there exists an $\alpha$< $\kappa$
such that $\tau$(a)\leq $\alpha$ for all a\in \mathrm{d}\mathrm{o}\mathrm{m}( $\tau$) , as required.
Consequently, the natural embedding=

d:$\kappa$\rightar ow$\kappa$^{[\mathrm{B}\} /U$\alpha$\mapsto[\check{ $\alpha$}]_{U}

is strictly increasing and cofinal in \langle $\kap a$, <\rangle^{[\mathrm{B}]}/U . Hence, the cofinality of \langle $\kap a$, <\rangle^{[\mathrm{B}]}/U is $\kappa$.

\squareWe conclude by mentioning a related result for Boolean ultrapowers of \langle $\omega$, <\rangle.

Proposition 3.18 (Koppelberg and Koppelberg [13, Lemma 3]). Let $\kappa$ be a regular cardinal with $\kap a$^{\mathrm{N}_{\mathrm{O} = $\kappa$.
Then there exists an ultrafilter U on \mathb {C}_{ $\kap a$} such thatcf(\{ $\omega$, <)^{[\mathb {C}_{ $\kap a$}] /U) =|$\omega$^{[\mathb {C}_{ $\kap a$}] /U|= $\kappa$.

Starting from Proposition 3.18, the topic of the possible cardinality and cofinality of a Boolean ultra‐

power of \langle $\omega$, <\rangle was further explored by Koppelberg [12] and Jin and Shelah [9].

53

Universality
The third model‐theoretic property we consider for Boolean ultrapowers is universality. Let us recall firstthe definition of universal structure.

Definition 3.19 (Morley and Vaught [21]). Let $\lambda$ be a cardinal. An L ‐structure \mathfrak{M} is $\lambda$ ‐universal iff forevery L ‐structure \mathfrak{R} , if |N| < $\lambda$ and 0\mathrm{t}\equiv \mathfrak{M} then there is an elementary embedding j : \mathfrak{N}\rightarrow \mathfrak{M}.

The following characterization of regularity is implicit in Frayne, Morel and Scott [5] and appears
explicitly in Keisler [11, Theorem 1. 5\mathrm{a} ].
Theorem 3.20. Let \mathrm{K} be an infinite cardinal; for an ultrafilter U over a set I , the following conditions
are equivalent:
1. U is $\kappa$ ‐regular;
2. for every L ‐structure \mathfrak{M} , with |L| \leq$\kappa$_{f} the ultrapower \mathfrak{M}^{I}/U is $\kappa$^{+} ‐universal.

Again, we can adapt the proof of Theorem 3.20 to establish a similar characterization of regularity
for ultrafilters on complete Boolean algebras.

Proposition 3.21. Let \mathrm{B} be a complete Boolean algebra and $\kappa$ an infinite cardinal. For an ultrafilter U
on \mathrm{B} , the following conditions are equivalent:
1. U is $\kappa$ ‐regular;
2. for every L ‐structure \mathfrak{M} , with |L| \leq $\kappa$ , the Boolean ultrapower \mathfrak{M}^{[\mathrm{B}]}/U is $\kappa$^{+} ‐universal.

Proof. (1\Rightarrow 2) Suppose U is $\kappa$ ‐regular; this is witnessed by a family \{x_{ $\alpha$} | $\alpha$< $\kappa$\}\subseteq U and a maximal
antichain A\subset \mathrm{B} . In particular, this means that for every a\in A the set

S(a)=\{ $\alpha$< $\kappa$|a\leq x_{ $\alpha$}\}is finite.

Let \mathfrak{N} be an L ‐structure such that |N| \leq $\kappa$ and \mathfrak{N}\equiv \mathfrak{M}^{[\mathrm{B}]}/U ; we need to find an elementary embedding
j:\mathfrak{R}\rightar ow \mathfrak{M}^{[\mathrm{B}]}/U . Let L(N)=L\cup\{c_{n}|n\in N\} be the language obtained from L by adding a new constantsymbol c_{n} for each n\in N . We may expand \mathfrak{R} to L(N) in a natural way: the interpretation of the symbolc_{n} is simply n ; this expansion is denoted by \mathfrak{R}_{N} . Let Th (\mathfrak{R}_{N}) be the set of all L(N) ‐sentences $\varphi$ suchthat \Re_{N} \models $\varphi$ . Since |\mathrm{T}\mathrm{h}(\mathfrak{R}_{N})| \leq |L|+|N|\leq $\kappa$ , we can enumerate this theory as
Th (\mathfrak{R}_{N})=\{$\varphi$_{ $\alpha$} | $\alpha$< $\kappa$\}.

For each a\in A , we proceed to define a sequence \{$\tau$_{n}(a)|n\in N\rangle of elements of M in the following way:
let $\varphi$ be the finite conjunction \displaystyle \bigwedge_{ $\alpha$\in S(a)}$\varphi$_{ $\alpha$} . Let n_{1} , . . . , n_{k} be the elements of N appearing as parameters
in $\varphi$ , so that we can write it as $\varphi$ (c_{n_{1}}, \ldots , c_{n_{k}}) .

Since $\varphi$(c_{n_{1}}, \ldots, c_{n_{k}}) \in \mathrm{T}\mathrm{h}(\mathfrak{R}_{N}) , clearly we have\mathfrak{N}\models\exists x_{1}\ldots\exists x_{k} $\varphi$(x_{1, )}x_{k}),where x_{1} , . . . , x_{k} are new variables. But \mathfrak{R}\equiv \mathfrak{M}^{[\mathrm{N}]}/U\equiv \mathfrak{M} , and therefore\mathfrak{M}\models\exists x_{1}\ldots\exists x_{k} $\varphi$(x_{1}, \ldots, x_{k}).

This allows us to choose $\tau$_{ $\tau \iota$_{1} (a) , . . . , $\tau$_{n_{k}}(a) in M such that\mathfrak{M}\models $\varphi$($\tau$_{n_{1}}(a), \ldots {}_{)}T_{n_{k}}(a)).

On the other hand, if n \in N does not appear as a parameter in $\varphi$ , then we are free to define $\tau$_{n}(a)arbitrarily. This completes the definition of the sequence \langle $\tau$_{n}(a)|n\in N\rangle.
Note that for every n\in N we have defined a name $\tau$_{n} : A\rightarrow M . We claim that the function
j : N\rightarrow M^{\mathrm{F}]}/U

n\mapsto[$\tau$_{n}]_{U}54

is an elementary embedding. Given any formula $\varphi$_{ $\alpha$}(c_{n_{1} , \ldots, c_{n_{k} )\in\mathrm{T}\mathrm{h}(\Re_{N}) , we need to show that\mathfrak{M}^{[\mathrm{E}]}/U\models$\varphi$_{ $\alpha$}([$\tau$_{n_{1} ]_{U}, \ldots, [$\tau$_{n_{k} ]_{U}) . For all a\in A , if a\leq x_{ $\alpha$} then $\alpha$\in S(a) and therefore, by construction,\mathfrak{M}\models$\varphi$_{ $\alpha$}($\tau$_{n_{1} (a), \ldots)$\tau$_{n_{k} (a) . It follows that[$\varphi$_{ $\alpha$}($\tau$_{n_{1} , \ldots, $\tau$_{n_{k} )\mathrm{J}=\vee\{a\in A|\mathfrak{M}\models$\varphi$_{ $\alpha$}($\tau$_{n_{1} (a) \ldots, $\tau$_{n}k(a) \}\geq\vee\{a\in A|a\leq x_{ $\alpha$}\}=x_{ $\alpha$}\in U
which implies by Theorem 3.9 that j is an elementary embedding.
\subseteq,
Let \mathfrak{M}be the structure in the language(2 \Rightarrow 1)=\langle[$\kap a$]^{<\aleph_{\mathrm{O} , \{\{ $\alpha$\}| $\alpha$< $\kap a$\rangle\rangleLwith a binary

relation symbol for the inclusion and $\kappa$ many constant symbols for the singletons \{ $\alpha$\}\in [ $\kap a$]^{<\aleph_{0} , for $\alpha$< $\kappa$.
We now define a set of L ‐formulae$\Sigma$(x)=\{\{ $\alpha$\}\subseteq x| $\alpha$< $\kappa$\},and we show that $\Sigma$(x) is realized in \mathfrak{M}^{[\mathrm{B}]}/U.

Since every finite subset of $\Sigma$(x) is realized in \mathfrak{M} , by compactness there exists a model \mathfrak{R} of the theoryof \mathfrak{M} in which $\Sigma$(x) is realized. Since |L| = $\kappa$ , by Löwenheim‐Skolem we may assume that |N| = $\kappa$ . We
have \mathfrak{R}\equiv \mathfrak{M}\equiv \mathfrak{M}^{[\mathrm{B}]}/U , and \mathfrak{M}^{[\mathrm{B}]}/U is $\kappa$^{+} ‐universal by hypothesis, therefore there exists an elementaryembedding j:\mathfrak{R}\rightar ow \mathfrak{M}^{[\mathrm{B}]}/U . So, if n\in N realizes $\Sigma$(x) in \mathfrak{N} , then by elementarity j(n) realizes $\Sigma$(x) in\mathfrak{M}^{[\mathrm{B}]}/U . This completes the proof that $\Sigma$(x) is realized in \mathfrak{M}^{[\mathrm{B}]}/U.
Now, let $\tau$:A\rightar ow[ $\kappa$]^{<\aleph_{0} be such that [ $\tau$]_{U} realizes $\Sigma$(x) in \mathfrak{M}^{[\mathrm{B}]}/U . For each $\alpha$< $\kappa$ define

x_{ $\alpha$} =\vee\{a\in A|\mathfrak{M}\models\{ $\alpha$\}\subseteq $\tau$(a)\})and note that x_{ $\alpha$}\in U . To show that U is $\kappa$‐regular, we just observe that for each $\alpha$< $\kappa$ and every a\in A
we have the two implications

a\wedge x_{ $\alpha$}>0 \Rightarrow $\alpha$\in $\tau$(a) \Rightarrow a\leq x_{ $\alpha$}.
From this, we conclude that the family \{x_{ $\alpha$} | $\alpha$< $\kappa$\} and the maximal antichain A satisfy the two
conditions of Definition 2.4.

\squareKoppelberg and Koppelberg [13] showed the existence of a $\kappa$‐quasiregular ultrafilter U on \mathb {C}_{$\kap a$} suchthat, for some L ‐structure \mathfrak{M} , with |L| =\aleph_{1} , the Boolean ultrapower \mathfrak{M}^{[\mathb {C}_{ $\kap a$}]}/U is not \aleph_{2} ‐universal.
With Proposition 3.21 available to us, we can give a very simple proof of this fact. Let $\kappa$ be any

uncountable cardinal and let U be an ultrafilter on \mathb {C}_{$\kap a$} . We already know (Lemma 2.15) that U is $\kappa$quasiregular, however U cannot be \aleph_{1} ‐regular, due to the \aleph_{1^{-} \mathrm{c}.\mathrm{c} . Therefore, by Proposition 3.21 there
exists some L ‐structure \mathfrak{M} , with |L| \leq\aleph_{1} , such that the Boolean ultrapower \mathfrak{M}^{[\mathb {C}_{ $\kap a$}]}/U is not \aleph_{2} ‐universal.

4

Keisler’s order via Boolean ultrapowers

In this final section we announce two results from a forthcoming paper in preparation [22]. As we discussed
in the introduction, regular ultrafilters play an important role in the classification of theories due to the
crucial Theorem 1.2. The first result we announce here is a generalization of Theorem 1.2 to regularultrafilters on complete Boolean algebras.

Theorem 4.1. Let $\kappa$ be an infinite cardinal. Suppose \mathrm{B} is a \{ $\kappa$, 2\rangle ‐distributive complete Boolean algebra
and U is a $\kappa$ ‐regular ultrafilter on B. If two L ‐structures SM and or are elementarily equivalent, and|L| \leq $\kappa$ , then\mathfrak{M}^{[\mathrm{B}]}/U is $\kappa$^{+} ‐saturated \Leftrightarrow \Re^{[\mathrm{B}]}/U is $\kappa$^{+} ‐saturated.

While all the model‐theoretic properties considered in Section 3 were generalized smoothly to thecontext of arbitrary complete Boolean algebras, the analogue of Theorem 1.2 was established under an
additional distributivity assumption on B. We conjecture that, without this assumption, Theorem 4.1
can be false in general.

Conjecture 4.2. There exists a cardinal $\kappa$ , a complete Boolean algebra \mathrm{B} , and a $\kappa$ ‐regular ultrafilter U on
with |L|\leq $\kappa$ , the Boolean ultrapower \mathfrak{M}^{\mathrm{F}]}/U is $\kappa$^{+} ‐saturated\mathrm{B} such that, for some L ‐structures \mathfrak{M}\equiv \mathfrak{R}_{f}but \mathfrak{R}^{\mathrm{F}\mathrm{J} /U is not $\kappa$^{+} ‐saturated.

55

The above Conjecture 4.2 will be addressed, among other things, in our future work in preparation [22].
For the moment, we just observe that this conjecture has an immediate positive answer if we replace $\iota$_{\mathrm{K}-}regulaf’ with “‘ $\kappa$\acute{} quasiregular”
Proposition 4.3. Let $\kappa$ be a cardinal with $\kap a$^{\aleph 0} = $\kappa$ . There are two elementarily equivalent \emptyset ‐structures\mathfrak{M} \equiv \mathfrak{R} such that, for every ultrafilter U on \mathb {C}_{ $\kap a$} , the Boolean ultrapower \mathfrak{M}^{[\mathb {C}_{ $\kap a$}]}/U is $\kappa$^{+} ‐saturated, but

9\mathrm{t}^{[\mathb {C}_{ $\kap a$}] /U is not $\kappa$^{+} ‐saturated.

Proof. First, observe that an infinite structure \mathfrak{M} in the empty language L\mathrm{o}\mathrm{n}]\mathrm{y} if $\kappa$< |M|.
Now, let \Re= $\kappa$ . By Lemma 3.12, for each ultrafilter U on \mathb {C}_{$\kap a$} we have=

\emptyset is $\kappa$^{+} ‐saturated if and|$\kap a$^{[\mathb {C}_{ $\kap a$}] /U| \leq$\kap a$^{<\mathrm{N}_{1} +($\kap a$^{\mathrm{N}_{0} )^{<\mathrm{N}_{1} =$\kap a$^{\aleph_{0} = $\kap a$,hence \mathfrak{R}^{[\mathb {C}_{ $\kap a$}] /U is not $\kappa$^{+} ‐saturated.

On the other hand, let \mathfrak{M} be a structure of cardinality at least $\kappa$^{+} such that \mathfrak{M}\equiv \mathfrak{R} . Then$\kappa$<|M| \leq
which means that \mathfrak{M}^{[\mathb {C}_{ $\kap a$}] /U is $\kappa$^{+} ‐saturated.

|M^{[\mathb {C}_{ $\kap a$}] /U|\squareThus, not only Theorem 4.1 fails for quasiregular ultrafilters, but also this failure is due trivially tothe cardinality of the Boolean ultrapowers and not to their saturation properties.
We now announce the second result, which answers the question we asked in the introduction, namely:
what kind of classification can arise when we compare theories according to the saturation of Boolean
ultrapowers of their models?
When trying to define a Boolean‐algebraic analogue of Keisler’s order and compare it with the usual
one, the first obstacle is that, as far as we know, regular ultrafilters on complete Boolean algebras may
not satisfy the generalization of Theorem 1.2. In other words, given a complete theory T , whether or not
the Boolean ultrapower of a model of T is saturated may depend on the choice of a particular model.
However, the next definition is designed to work also in this context.
Definition 4.4. Let $\lambda$ be a cardinal and \mathrm{B} a complete Boolean algebra. Suppose U is an ultrafilter on\mathrm{B} ; we say that U $\lambda$ ‐saturates a complete theory T iff for every $\lambda$ ‐saturated model \mathfrak{M}\models T , the Boolean

ultrapower \mathfrak{M}^{[\mathrm{B}]}/U is $\lambda$ ‐saturated.

Using the techniques developed by Malliaris and Shelah [17] and Shelah [24], we can establish thefollowing characterization, due to appear in [22].
Theorem 4.5. Let $\kappa$ be an infinite cardinal and T_{1}, T_{2} complete countable theories. Then the following
are equivalent:\bullet\bulletT_{1} \underline{\triangleleft}{ _{ $\kap a$}T_{2} ;for every $\kappa$^{+} ‐c. c . complete Boolean algebra \mathrm{B} of size\leq 2^{ $\kappa$} , and every $\kappa$ ‐regular ultrafilter U on \mathrm{B},if U $\kappa$^{+} ‐saturates T_{2} then U $\kappa$^{+} ‐saturates T_{1}.

In conclusion, this characterization explains the shift towards constructing regular ultrafilters oncomplete Boolean algebras: indeed, those ultrafilters are able to detect exactly the same properties oftheories as ultrafilters on power‐set algebras.

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