1 Introduction
There are two binary operations on the class of ultrafilters commonly referred to as the product, each of which fails to live up to the name. The first, the Cartesian product, truly is a product in the category of filters. The subcategory of ultrafilters, however, is not closed under Cartesian products. The other, the tensor product, is at least an operation on ultrafilters. But it is not a product. It is not even commutative (up to isomorphism). This raises the following questions.
Question 1.1.
-
(1) When is the Cartesian product of two ultrafilters an ultrafilter?
-
(2) When is the tensor product of two ultrafilters commutative?
We note that the first question is equivalent to the question of when the Cartesian and tensor products coincide, and that if the Cartesian product of two ultrafilters is an ultrafilter, then their tensor product commutes.
For nonprincipal ultrafilters on
$\omega $
, the answer to the two questions above is never, as a consequence of Theorem 2.14.
The question for ultrafilters on arbitrary sets involves large cardinals. Namely, there is a pair of nonprincipal ultrafilters whose Cartesian product is an ultrafilter if and only if there is a pair of ultrafilters whose tensor product commutes if and only if there is a measurable cardinal.
We sketch the proof of one direction of this equivalence here since it is a key motivating example. Suppose
$\kappa $
is measurable, and let us exhibit a pair of nonprincipal ultrafilters whose Cartesian product is an ultrafilter. This is straightforward: fix any nonprincipal ultrafilter U on
$\omega $
and any nonprincipal
$\kappa $
-complete ultrafilter W on
$\kappa $
. Then
$U\times W$
is an ultrafilter; see Theorem 2.4.
The intuition behind this paper is that this basic example is not too far from the general situation. One aspect of this intuition is the conjecture that Question 1.1 (1) and (2) have the same answer.
Conjecture 1.2. The Cartesian product of two ultrafilters is an ultrafilter if and only if their tensor product is commutative.
We now define the two products in order to state this conjecture more precisely.
Definition 1.3. If F and G are filters on sets X and Y, the Cartesian product of F and G, denoted
$F\times G$
, is the filter on
$X\times Y$
generated by sets of the form
$A\times B$
where
$A\in F$
and
$B\in G$
.
The usual notation for the tensor product of ultrafilters is
$\otimes $
, but the following notation is more convenient here:
Definition 1.4. Suppose F and G are filters on sets X and Y. Their left and right tensor products are defined as follows:
$$ \begin{align*} F\ltimes G &= \{R\subseteq X\times Y : \{x\in X : (R)_x\in G\}\in F\}\\ F\rtimes G &= \{R\subseteq X\times Y : \{y\in Y : (R)^y\in F\}\in G\}. \end{align*} $$
Here
$(R)_x = \{y\in Y : (x,y)\in R\}$
and
$(R)^y = \{x\in X : (x,y)\in R\}$
.
Our conjecture now reads the following.
Conjecture 1.5. For any ultrafilters U and W, the following are equivalent:
-
(1)
$U\times W$
is an ultrafilter. -
(2)
$U\ltimes W = U\rtimes W$
.
The implication from (1) to (2) is easy enough to see:
$U\times W$
is certainly contained in both
$U\ltimes W$
and
$U\rtimes W$
, so if
$U\times W$
is an ultrafilter, its maximality implies that
$U\ltimes W = U\times W = U\rtimes W$
. So we really have just one open question.
Question 1.6. Suppose U and W are ultrafilters such that
$U\ltimes W = U\rtimes W$
. Must
$U\times W$
be an ultrafilter?
We will answer this question positively assuming the Generalized Continuum hypothesis.
Theorem 2.15 (GCH).
Suppose U and W are ultrafilters. Then
$U\ltimes W = U\rtimes W$
if and only if
$U \times W$
is an ultrafilter.
It is not true that if
$U\ltimes W$
and
$W\ltimes U$
are Rudin–Keisler equivalent, then
$U\times W$
is an ultrafilter. For example, take the case that
$U = W$
:
$U\times U$
is never an ultrafilter, but obviously
$U\ltimes U = U\ltimes U$
. So it is not enough that
$U\ltimes W\equiv _{\text {RK}} W\ltimes U$
; it is also necessary that this equivalence be witnessed by the coordinate-swapping bijection from
$X\times Y$
to
$Y\times X$
where X and Y are the underlying sets of U and W.
The theorem is proved by isolating a combinatorial criterion that arguably answers Question 1.1 under GCH.
Definition 1.7. A filter F is
$(\lambda ,\eta )$
-indecomposable if for any family of sets
$\mathcal P$
such that
$|\mathcal P| \leq \eta $
and
$\bigcup \mathcal P\in F$
, there is some
$\mathcal Q\subseteq \mathcal P$
such that
$|\mathcal Q| < \lambda $
and
$\bigcup Q\in F^+$
.
Theorem 2.13. Suppose U and W are ultrafilters such that for all cardinals
$\lambda $
, either U or W is
$(\lambda ,2^\lambda )$
-indecomposable. Then
$U\times W$
is an ultrafilter.
Theorem 2.14. Suppose U and W are ultrafilters such that
$U\ltimes W = U\rtimes W$
. Then for all cardinals
$\lambda $
, either U or W is
$(\lambda ,\lambda ^+)$
-indecomposable.
In addition, we will provide a third, somewhat more surprising condition equivalent to the statement that
$U \times W$
is an ultrafilter. This involves a variant of the Mitchell order called the internal relation. Recall that the Mitchell order is defined on countably complete ultrafilters U and W by setting
$U\lhd W$
if U belongs to the ultrapower of the universe by W, which will be denoted by
$V_W$
. The question of whether
$U\in V_W$
only makes sense when W is countably complete, so that
$V_W$
can be identified with a transitive class. The internal relation is a combinatorial proxy for the Mitchell order that in particular makes sense regardless of the completeness of the ultrafilters in question.
If M is a model of set theory, let us say that two elementary embeddings
${k_0 : M\to N_0}$
and
$k_1 : M\to N_1$
are isomorphic if there is an isomorphism
$i : N_0\to N_1$
such that
$i \circ k_0 = k_1$
.
Definition 1.8. The internal relation is defined on ultrafilters U and W by setting
$U\sqsubset W$
if there is a
$V_W$
-ultrafilter
$U_{*}\in V_W$
such that
$j_U\restriction V_W$
is isomorphic to
$(j_{U_{*}})^{V_W}$
.
Here
$j_U : V\to V_U$
denotes the ultrapower embedding, as it will throughout the paper.
One advantage of the internal relation over the Mitchell order is that the former relation is more amenable to algebraic or diagrammatic techniques. This is exploited in the author’s thesis to give a complete analysis of the internal relation on countably complete ultrafilters assuming the Ultrapower Axiom (UA). UA is a combinatorial principle motivated by inner model theory that enables one to develop a theory of countably complete ultrafilters that is far more detailed than the theory available assuming ZFC alone. (The precise statement of UA appears at the end of Section 2.1.) The theory of countably complete ultrafilters encompasses a major part of the theory of large cardinals, and so UA is a natural setting for large cardinal theory.Footnote 1
Conjecture 1.5 was originally motivated by the following consequence of UA.
Theorem (UA).
For any countably complete ultrafilters U and W,
$U\times W$
is an ultrafilter if and only if
$U\ltimes W = U\rtimes W$
if and only if
$U\sqsubset W$
and
$W\sqsubset U$
.
We show here that this theorem can be proved for arbitrary ultrafilters using GCH instead of UA.
Theorem 3.20 (GCH).
For any ultrafilters U and W,
$U\times W$
is an ultrafilter if and only if
$U\sqsubset W$
and
$W\sqsubset U$
.
The proof of Theorem 3.20 is motivated by ideas developed in the context of UA, and more specifically it makes use of the two main tools from [Reference Goldberg3], the Ketonen order and the factorization of ultrafilters. Arguably, the theorem could not have proved without the help of UA, even though UA makes no appearance in the proof.
2 Preliminaries
2.1 Ultrafilters and the Rudin–Keisler order
In this section, we briefly recall some facts about derived ultrafilters and the Rudin–Keisler order. A somewhat more leisurely account of similar material appears in [Reference Goldberg3, Sections 2.2 and 3.4.2].
Recall our conventions on ultrapowers, briefly mentioned in the introduction: if U is an ultrafilter, then
$V_U$
denotes the ultrapower of the universe by U. We use this notation whether or not the ultrapower is well-founded, though we will usually identify the well-founded part with its transitive collapse.
If U is an ultrafilter on a set X and
$f : X\to Y$
is a function, the pushforward of U under f is the ultrafilter
$f_{*}(U) = \{A\subseteq Y : f^{-1}[A]\in U\}$
. If
$W = f_{*}(U)$
, then there is a unique elementary embedding
$k : V_W\to V_U$
such that
$k\circ j_W = j_U$
and
$k([\text {id}]_W) = [f]_U$
. Conversely, if there is an elementary embedding
$k : V_W\to V_U$
such that
$k\circ j_W = j_U$
, then
$W = f_{*}(U)$
for some f. (Note that these equivalences go through regardless of whether the ultrapowers in question are well-founded.)
Definition 2.1. We say that an ultrafilter W precedes U in the Rudin–Keisler order, and write
$W\leq _{\text {RK}} U$
, if W is the pushforward of U under some function.
Suppose M is a model of ZFC and
$X\in M$
. An M-ultrafilter on X is an ultrafilter over the Boolean algebra
$P^M(X)$
. Given an M-ultrafilter U on X, one can form the associated ultrapower, denoted
$M_U$
, by taking the quotient
$M^X\cap M$
under the equivalence relation of U-almost everywhere equality. The equivalence class of
$f\in M^X\cap M$
is denoted by
$[f]_U^M$
.
The associated ultrapower embedding
$$\begin{align*}j_U^M : M\to M_U\end{align*}$$
is given by
$j(s) = [c_s]_U^M$
where
$c_s : X\to \{s\}$
is the constant function.
Suppose
$j : M\to N$
is an elementary embedding. For any
$X\in M$
and
$a\in N$
such that
$N\vDash a\in j(X)$
, the M-ultrafilter on X derived from j using a is the ultrafilter
$$\begin{align*}D_X(j,a) = \{A\in P^M(X) : N\vDash a\in j(A)\}.\end{align*}$$
Letting
$D = D_X(j,a)$
, there is again a canonical factor embedding
$k : M_{D}\to N$
, denoted
$k_{a,j}$
, uniquely determined by requiring
$k\circ j_D = j$
and
$k([\text {id}]_D) = a$
. Note that if
$f_{*}(U) = W$
, then W is the ultrafilter derived from
$j_U$
using
$a= [f]_U$
, and
$k_{a,j_U}$
is the canonical factor embedding.
The UA states that for any countably complete ultrafilters U and W there exist
$W_{*}\in V_{U}$
and
$U_{*}\in V_{W}$
such that
$W_{*}$
is a countably complete
$V_{U}$
-ultrafilter,
$U_{*}$
is a countably complete
$V_{W_{*}}$
-ultrafilter,
$(V_{U})_{W_{*}} = (V_{W})_{U_{*}}$
, and
$j_{W_{*}}\circ j_{U} = j_{U_{*}}\circ j_{W}$
.
2.2 The symmetry of regularity
Part of the motivation for this work was to explain the connection between a theorem from Blass’s thesis [Reference Blass1, Theorem 3.6] and Kunen’s Commuting Ultrapowers Lemma [Reference Larson4, Lemma 1.1.25]. These lemmas concern two binary relations on ultrafilters that despite appearances turn out to be symmetric.
Definition 2.2. Suppose F and G are filters on sets X and Y.
-
• G is F-regular if there exist
$\langle B_x : x\in X\rangle \subseteq G$
such that for any
$A\in F^+$
,
$\bigcap _{x\in A}B_x = \emptyset $
. -
• G is F-closed if for any
$\langle B_x : x\in X\rangle \subseteq G$
, there is some
$A\in F$
such that
$\bigcap _{x\in A}B_x\in G$
.
Clearly a filter cannot be both F-regular and F-closed. In this section, we show that our main conjecture (Conjecture 1.5) reduces to the problem of whether, given an ultrafilter U, every ultrafilter is either U-regular or U-closed. (See the remark after Theorem 2.8.)
The concept of regularity is a natural generalization of the classical concept of a regular ultrafilter.
Definition 2.3. If
$\kappa \leq \lambda $
are cardinals, a filter G is
$(\kappa ,\lambda )$
-regular if there is a sequence
$\langle B_\alpha : \alpha < \lambda \rangle \subseteq G$
such that for any
$\sigma \subseteq \lambda $
with
$|\sigma |\geq \kappa $
,
$\bigcap _{\alpha \in \sigma }B_\alpha = \emptyset $
.
In other words, G is
$(\kappa ,\lambda )$
-regular if it is
$F_{\kappa }(\lambda )$
-regular where
$F_{\kappa }(\lambda )$
is the dual of the ideal
$P_\kappa (\lambda ) = \{\sigma \subseteq \lambda : |\sigma | < \kappa \}$
.
The concept of closure is extracted from the following theorem from Blass’s thesis.
Theorem 2.4 (Blass, [Reference Blass1, Theorem 3.6]).
Suppose U and W are ultrafilters on sets X and Y. Then the following are equivalent:
-
(1)
$U\times W$
is an ultrafilter. -
(2) W is U-closed.
-
(3) U is W-closed.
Although the symmetry expressed in Blass’s theorem does not extend to filters, one can deduce Blass’s Theorem from a more general filter-theoretic fact.
Lemma 2.5. Suppose F and G are filters. Then G is F-closed if and only if
$F\ltimes G = F\times G$
.
Proof. In general,
$F\times G\subseteq F\ltimes G$
, so it suffices to prove that the reverse inclusion is equivalent to the F-closure of G. This is a routine matter of applying the correspondence between binary relations
$R\subseteq X\times Y$
and X-indexed sequences
$\langle B_x : x\in X\rangle $
of subsets of Y.
Proof of Theorem 2.4.
It suffices to prove the equivalence of (1) and (2), since the obvious symmetry of (1) in U and W then immediately implies the equivalence of (2) and (3). If
$U\times W$
is an ultrafilter, then since
$U\times W$
is a maximal filter contained in
$U\ltimes W$
,
$U\times W = U\ltimes W$
, and hence W is U-closed by Lemma 2.5. Conversely if W is U-closed, then
$U\times W = U\ltimes W$
by Lemma 2.5, and so since
$U\ltimes W$
is an ultrafilter,
$U\times W$
is.
It turns out that regularity, like closure, is also a symmetric relation, but this time the symmetry does extend to all filters. Despite the extensive literature on regular ultrafilters, as far as the author knows, this symmetry has never been observed.
Lemma 2.6. Suppose F and G are filters. Then F is G-regular if and only if
$F\ltimes G$
is incompatible with
$F\rtimes G$
.
Proof. First assume F is G-nonregular. Fix
$R,S\subseteq X\times Y$
with
$R\in F\ltimes G$
and
$S\in F\rtimes G$
, and let us show that
$R\cap S$
is nonempty. Take
$A\in F$
such that for all
$x\in A$
,
$(R)_x\in G$
. Note that since
$S\in F\rtimes G$
, for G-almost all
$y\in Y$
,
$(S)^y\in F$
, and hence
$(S)^y\cap A\in F$
. Consider the sequence
$\langle (S)^y\cap A : y\in Y\rangle $
. Since F is G-nonregular, there is a G-positive set B such that
$\bigcap _{y\in B} (S)^y\cap A\neq \emptyset $
. Fix
$x\in \bigcap _{y\in B} (S)^y\cap A$
. Since
$x\in A$
,
$(R)_x\in G$
, and therefore
$(R)_x\cap B\neq \emptyset $
. Since
$x\in \bigcap _{y\in B} (S)^y$
, for any
$y\in (R)_x\cap B$
,
$x\in (S)^y$
; but then
$(x,y)\in R\cap S$
.
Conversely, suppose F is G-regular, and let
$\langle A_y : y\in Y\rangle \subseteq F$
witnesses this. Let
$B_x = \{y\in Y : x\notin A_y\}$
. We claim that
$B_x\in G$
for all
$x\in X$
. This is because
$x\in \bigcap _{y\in Y\setminus B_x} A_y$
, so
$Y\setminus B_x\notin G^+$
, and hence
$B_x\in G$
. But let
$\mathcal A = \{(x,y) : x\in A_y\}$
and
$\mathcal B = \{(x,y) : y\in B_x\}$
. Then
$\mathcal A\in F\rtimes G$
,
$\mathcal B \in F\ltimes G$
, and
$\mathcal A = (X\times Y)\setminus \mathcal B$
. This shows
$F\rtimes G$
is incompatible with
$F\ltimes G$
.
Theorem 2.7. Suppose F and G are filters. Then F is G-regular if and only if G is F-regular.
For ultrafilters, we of course get more.
Theorem 2.8. If U and W are ultrafilters, then U is W-nonregular if and only if
$U\ltimes W = U\rtimes W$
.
As a consequence of Theorems 2.4 and 2.8, Conjecture 1.5 reduces to the question of whether, given an ultrafilter U, every ultrafilter is either U-regular or U-closed.
Let us just mention one more familiar characterization of regularity.
Definition 2.9. If F is a family of sets, the dual of F is the family
$F^*$
of sets of the form
$(\bigcup F)\setminus A$
for
$A\in F$
. The fine filter on the dual of F, denoted
$\mathbb F(F)$
, is the filter on
$F^*$
generated by sets of the form
$\{\sigma \in F^* : x\in \sigma \}$
where
$x\in \bigcup F$
.
The prototypical special case of this concept arises when
$F = F_{\kappa }(\lambda )$
is the filter of subsets of
$\lambda $
with complement of cardinality less than
$\kappa $
, so that
$\mathbb F(F)$
is the fine filter on
$P_\kappa (\lambda )$
. Note that if
$\bigcap F\neq \emptyset $
, then
$\mathbb F(F)$
is improper.
The Rudin–Keisler order can be generalized to filters in two different ways.
Definition 2.10. Suppose F is a filter on X and G is a filter on Y.
-
• If
$f: X\to Y$
is a function, the pushforward of F under f is the filter
$$\begin{align*}f_{*}(F) = \{B\subseteq Y : f^{-1}[B]\in F\}.\end{align*}$$
-
• The Rudin–Keisler order is defined by
$G\leq _{\text {RK}} F$
if
$G = f_{*}(F)$
for some
${f : X\to Y}$
. -
• The Katětov order is defined by
$G\leq _{\text {kat}} F$
if
$G\subseteq f_{*}(F)$
for some
$f : X\to Y$
.
Proposition 2.11. A filter F is G-regular if and only if
$\mathbb F(F)\leq _{\text {kat}} G$
.
Proof. Let X and Y be the underlying sets of F and G, respectively. Observe that a sequence of sets
$\langle A_y : y\in Y\rangle $
witnesses that F is G-regular if and only if the function
$f(y) = X\setminus A_y$
pushes G forward to a superset of
$\mathbb F(F)$
.
It follows immediately that if F is G-regular, then
$\mathbb F(F)\leq _{\text {kat}} G$
. Conversely, if
$\mathbb F(F)\leq _{\text {kat}} G$
via some function
$f : Y \to X$
, the observation of the previous paragraph implies that the sequence
$A_y = X\setminus f(y)$
witnesses F is G-regular.
An immediate consequence of Proposition 2.11 is that the F-regular filters are closed upwards in the Katětov order, or dually.
Lemma 2.12. Suppose
$F, G$
and H are filters such that F is G-regular and
$G\leq _{\text {kat}} H$
. Then F is H-regular.
Put another way, if F is H-nonregular, then the class of F-regular filters contains no Katětov predecessors of H.
2.3 Indecomposable ultrafilters
In this subsection, we prove the following theorems.
Theorem 2.13. Suppose U and W are ultrafilters such that for all cardinals
$\lambda $
, either U or W is
$(\lambda ,2^\lambda )$
-indecomposable. Then
$U\times W$
is an ultrafilter.
Theorem 2.14. Suppose U and W are ultrafilters such that
$U\ltimes W = U\rtimes W$
. Then for all cardinals
$\lambda $
, either U or W is
$(\lambda ,\lambda ^+)$
-indecomposable.
As an immediate corollary of Theorems 2.13 and 2.14, we obtain our main theorem.
Theorem 2.15 (GCH).
For any ultrafilters U and W,
$U\ltimes W = U\rtimes W$
if and only if
$U\times W$
is an ultrafilter.
Recall the notion of indecomposability introduced above (Definition 1.7).
Definition 2.16. If
$\lambda $
is a cardinal, F is
$\lambda $
-indecomposable if F is
$(\lambda ,\lambda )$
-indecomposable.
It is easy to see that U is
$(\lambda ,\eta )$
-indecomposable (see Definition 1.7) if and only if U is
$\gamma $
-indecomposable whenever
$\lambda \leq \gamma \leq \eta $
. We will use the term
$\lambda $
-decomposable in the obvious way. (We avoid the notion of
$(\lambda ,\eta )$
-decomposability in general, however, to avoid potential ambiguity.) We will use the following straightforward lemma.
Lemma 2.17. A filter F on X is
$\lambda $
-decomposable if and only if
$F_\lambda (\lambda )\leq _{\text {kat}} F$
where
$F_\lambda (\lambda )$
is the Fréchet filter on
$\lambda $
.
Proof. Suppose
$f : X\to \lambda $
witnesses that
$F_{\lambda }(\lambda )\leq _{\text {kat}} F$
. This means that for any
$\sigma \in P_\lambda (\lambda )$
,
$f^{-1}[\sigma ]$
is in the dual ideal of F. In other words, setting
$A_\alpha = f^{-1}[\{\alpha \}]$
and
$\mathcal P = \{A_\alpha : A_\alpha \neq \emptyset \}$
, we obtain a partition of X into
$\lambda $
pieces with the property that for all
$\mathcal Q\in P_\lambda (\mathcal P)$
,
$\bigcup Q$
is
$F\restriction S$
-null. This shows that F is
$\lambda $
-decomposable.
The converse is similar.
Corollary 2.18. If
$F\leq _{\text {kat}} G$
are filters and G is
$(\lambda ,\eta )$
-indecomposable, then F is
$(\lambda ,\eta )$
-indecomposable.
Proof. It suffices to show that F is
$\gamma $
-indecomposable for all
$\gamma $
such that
$\lambda \leq \gamma \leq \eta .$
But note that if F is
$\gamma $
-decomposable, then
$F_\gamma (\gamma )\leq _{\text {kat}} F\leq _{\text {kat}} G$
, and so G is
$\gamma $
-decomposable, contradicting that G is
$(\lambda ,\eta )$
-indecomposable.
We start with the proof of Theorem 2.13.
Proof of Theorem 2.13.
Suppose
$U\times W$
is not an ultrafilter. We will show that there is some cardinal
$\lambda $
such that neither U nor W is
$(\lambda ,2^\lambda )$
-decomposable. Let X be a set of minimal cardinality carrying an ultrafilter
$\tilde {U}\leq _{\text {RK}} U$
such that
$\tilde {U}\times W$
is not an ultrafilter. Let Y be a set of minimal cardinality carrying an ultrafilter
$\tilde {W}\leq _{\text {RK}} W$
such that
$\tilde {U}\times \tilde {W}$
is not an ultrafilter.
Let
$\langle A_y : y \in Y\rangle \subseteq \tilde {U}$
witness that
$\tilde {U}$
is not
$\tilde {W}$
-closed. Let
$f : Y \to P(X)$
be the function
$f(y) = A_y$
. Then
$\tilde {U}$
is not
$f_{*}(\tilde {W})$
-closed: indeed, setting
$A_{B} = B$
if
$B\in \text {ran}(f)$
and
$A_B = X$
otherwise, the sequence
$\langle A_B : B\in P(X)\rangle $
witnesses that
$\tilde {U}$
is not
$f_{*}(\tilde {W})$
-closed. The minimality of the cardinality of Y therefore implies
$|Y| \leq 2^{|X|}$
.
Similarly,
$|X|\leq 2^{|Y|}$
. Let
$\lambda = \min \{|X|,|Y|\}$
. Then
$\tilde U$
and
$\tilde W$
are both uniform ultrafilters on sets whose cardinalities lie in the interval
$[\lambda ,2^\lambda ]$
. Since
$\tilde U\leq _{\text {RK}} U$
and
$\tilde W\leq _{\text {RK}} W$
, this implies (by Corollary 2.18) that neither U nor W is
$(\lambda ,2^\lambda )$
-indecomposable.
We now turn to Theorem 2.14, which follows from a simple combinatorial fact about the relationship between regularity and decomposability.
Corollary 2.19. Suppose F and G are filters such that F is G-nonregular. Then for any cardinal
$\lambda $
such that G is
$\lambda $
-decomposable, F is
$(\lambda ,\lambda )$
-nonregular.
Proof. Assume F is
$(\lambda ,\lambda )$
-regular, toward a contradiction. In other words, F is
$F_\lambda (\lambda )$
-regular where
$F_{\lambda }(\lambda )$
denotes the Fréchet filter on
$\lambda $
. By Lemma 2.17, the fact that G is
$\lambda $
-decomposable means that
$F_\lambda (\lambda )\leq _{\text {kat}} G$
. So by Lemma 2.12, F is G-regular, contrary to hypothesis.
Theorem 2.14 will be a consequence of Corollary 2.19 using the following theorem, considerably generalized by Lipparini [Reference Lipparini5, Theorem 2.2].
Lemma 2.20. If U is
$\lambda ^+$
-decomposable, either U is
$\text {cf}(\lambda )$
-decomposable or U is
$(\kappa ,\lambda ^+)$
-regular for some
$\kappa < \lambda $
. In particular, if
$\lambda $
is regular, then U is
$\lambda $
-decomposable.
Lemma 2.21. Suppose U is
$(\lambda ,\lambda )$
-nonregular. Then U is
$(\lambda ,\lambda ^+)$
-indecomposable.
Proof. Clearly U is
$\text {cf}(\lambda )$
-indecomposable (and in particular, U is
$\lambda $
-indecomposable). Therefore by Lemma 2.20, if U were
$\lambda ^+$
-decomposable, then U would
$(\gamma ,\lambda ^+)$
-regular for some
$\gamma < \lambda $
, but this of course would imply that U is
$(\lambda ,\lambda )$
-regular, contrary to assumption. Therefore U is
$\lambda ^+$
-indecomposable. It follows that U is
$\lambda $
-indecomposable and
$\lambda ^+$
-indecomposable, so U is
$(\lambda ,\lambda ^+)$
-indecomposable.
Note that a theorem of Lipparini [Reference Lipparini6] shows that U is
$(\lambda ,\lambda )$
-nonregular if and only if U is either
$\text {cf}(\lambda )$
-decomposable or
$\lambda ^+$
-decomposable.
Theorem 2.14 is now immediate.
Proof of Theorem 2.14.
We first note that either W or U is
$\lambda ^+$
-indecomposable. This is immediate from Corollary 2.19 since
$(\lambda ^+,\lambda ^+)$
-nonregularity is equivalent to
$\lambda ^+$
-indecomposability.
Therefore let us assume that W is
$\lambda ^+$
-indecomposable. If W is also
$\lambda $
-indecomposable, we are done, so assume instead that W is
$\lambda $
-decomposable.
Applying Corollary 2.19 again, U is
$(\lambda ,\lambda )$
-nonregular, and it follows from Lemma 2.21 that U is
$(\lambda ,\lambda ^+)$
-indecomposable.
3 The internal relation
3.1 Amalgamations
Our main questions concern the relationship between
$U\times W$
and its extensions
$U\ltimes W$
and
$U\rtimes W$
. More generally, one can consider arbitrary ultrafilters extending of
$U\times W$
, which we will call amalgamations of
$(U,W)$
. An amalgamation D of
$(U,W)$
is an upper bound of U and W in the Rudin–Keisler order, since
$U = (\pi _0)_{*}(D)$
and
$W= (\pi _1)_{*}(D)$
where
$\pi _n$
denotes projection to the nth coordinate.
There is a second way to view amalgamations: as iterated ultrapowers. Suppose U is an ultrafilter on X and W is an ultrafilter on Y. If D is an amalgamation of
$(U,W)$
, then there is a unique elementary embedding
$k : V_U\to V_D$
such that
$k\circ j_U = j_D$
and
$k([\text {id}]_U) = [\pi _0]_D$
. This embedding is isomorphic to the ultrapower of
$V_U$
by an
$V_U$
-ultrafilter, defined in this context as follows.
Definition 3.1. Suppose U and W are ultrafilters on sets X and Y and D is an amalgamation of
$(U,W)$
. Then
$(D)_0$
denotes the
$V_U$
-ultrafilter on
$j_U(Y)$
derived from k using
$[\pi _1]_D$
where
$k : V_U\to V_D$
is the canonical factor embedding.
The ultrafilter
$(D)_1$
is defined similarly, so
$(D)_1 = (\breve {D})_0$
where
$\breve {D}$
denotes the amalgamation of
$(W,U)$
given by D.
One can show that in this context, the canonical factor embedding
$k: V_U\to V_D$
is isomorphic to the ultrapower embedding of
$V_U$
associated with
$(D)_0$
. For this, one must show that the canonical factor embedding
$h : (V_U)_{{(D)_0}}\to V_D$
is an isomorphism. It is enough to show h is surjective, and this holds because every element of
$V_D$
is definable from parameters among
$j_D[V]$
,
$[\pi _0]_D$
, and
$[\pi _1]_D$
, and all these parameters lie in the range of h:
$$ \begin{align*} j_D[V] &= h\circ j_{(D)_0}^{V_U}\circ j_U[V]\\ [\pi_0]_D &= h(j_{(D)_0}^{V_U}([\text{id}]_U))\\ [\pi_1]_D &= h([\text{id}]_{(D)_0}^{V_U}). \end{align*} $$
One can also compute the following lemma.
Lemma 3.2. If D is an amalgamation of
$(U,W)$
, then
$$\begin{align*}(D)_0 = \{[f_R]_U : R\in D\}, \end{align*}$$
where
$f_R :X\to P(Y)$
is given by
$f_R(x) = (R)_x = \{y\in Y : (x,y)\in R\}$
.
3.2 Amalgamations and the internal relation
The internal relation is closely related to the ultrafilters
$U\ltimes W$
and
$U\rtimes W$
, viewed as amalgamations of
$(U,W)$
. This subsection is devoted to this relationship.
Lemma 3.3. If U and W are ultrafilters, then
$(U\ltimes W)_0 = j_U(W)$
.
The ultrafilter
$(U\ltimes W)_1$
is more interesting.
Definition 3.4. If U and W are ultrafilters, then
$s_W(U) = (U\ltimes W)_1$
.
The following lemma shows that
$s_W(U)$
is in a sense the canonical
$V_W$
-ultrafilter giving rise to the ultrapower embedding
$j_U\restriction V_W$
.
Lemma 3.5. Suppose U and W are ultrafilters on sets X and Y. Then
$s_W(U)$
is the
$V_W$
-ultrafilter on
$j_W(X)$
derived from
$j_U\restriction V_W : V_W\to j_U(V_W)$
using
$[j_W]_U$
, and the canonical factor embedding from
$(V_W)_{s_W(U)}$
to
$j_U(V_W)$
is an isomorphism. Finally,
$s_W(U)$
is given by the formula
$$\begin{align*}s_W(U) = \{A\in j_W(P(X)) : j_W^{-1}[A]\in U\}.\end{align*}$$
Proof. By definition,
$s_W(U)$
is the
$V_W$
-ultrafilter on
$j_W(X)$
derived from the canonical factor embedding
$i : V_W\to V_{U\ltimes W}$
using
$[\pi _0]_{U\ltimes W}$
. In general, the factor embedding
$p : (V_W)_{s_W(U)}\to V_{U\ltimes W}$
is an isomorphism; see the remarks following Lemma 3.2.
We first show
$s_W(U)$
is the
$V_W$
-ultrafilter on
$j_W(X)$
derived from
$j_U\restriction V_W$
using
$[j_W]_U$
. Let
$k : V_U\to V_{U\ltimes W}$
denote the canonical factor embedding. By Lemma 3.3, there is an isomorphism
$h : V_{U\ltimes W}\to (V_U)_{j_U(W)}$
such that
$h\circ k$
is equal to the ultrapower embedding of
$V_U$
associated with
$j_U(W)$
and
$h([\pi _1]_{U\ltimes W}) = [\text {id}]_{j_U(W)}$
.
We claim that
$$\begin{align*}h\circ i = j_U\restriction V_W.\end{align*}$$
Note that by elementarity
$j_U(V_W) = (V_U)_{j_U(W)}$
, so
$h\circ i$
and
$j_U\restriction V_W$
at least have the same codomain. It suffices to show that
$h\circ i$
and
$j_U$
agree on the parameters
$j_W[V]\cup \{[\text {id}]_W\}$
, from which all other elements of
$V_W$
are definable in
$V_W$
. First,
$$\begin{align*}h\circ i \circ j_W = h\circ j_{U\ltimes W} = j_{j_U(W)}\circ j_U = j_U\circ j_W\end{align*}$$
and this implies
$h\circ i\restriction j_W[V] = j_U\restriction j_W[V]$
. Second,
$$\begin{align*}h(i([\text{id}]_W)) = h([\pi_1]_{U\ltimes W}) = [\text{id}]_{j_U(W)} = j_U([\text{id}]_W).\end{align*}$$
This proves
$h\circ i = j_U\restriction V_W$
.
Now since
$s_W(U)$
is the
$V_W$
-ultrafilter on
$j_W(X)$
derived from the canonical factor embedding
$i : V_W\to V_{U\ltimes W}$
using
$[\pi _0]_{U\ltimes W}$
,
$s_W(U)$
is the
$V_W$
-ultrafilter on
$j_W(X)$
derived from
$h\circ i$
using
$h([\pi _0]_{U\ltimes W})$
. But
$h\circ i = j_U\restriction V_W$
and
$$\begin{align*}h([\pi_0]_{U\ltimes W}) = h(k([\text{id}]_U)) = j_U(j_W)([\text{id}]_U) = [j_W]_U.\end{align*}$$
This proves
$s_W(U)$
is the
$V_W$
-ultrafilter on
$j_W(X)$
derived from
$j_U\restriction V_W$
using
$[j_W]_U$
.
The canonical factor embedding
$p : (V_W)_{s_W(U)}\to V_{U\ltimes W}$
is an isomorphism, and the embedding
$h : V_{U\ltimes W}\to j_U(V_W)$
is an isomorphism, so
$h\circ p: (V_W)_{s_W(U)}\to j_U(V_W)$
is an isomorphism, and by uniqueness,
$h\circ p$
is the canonical factor embedding from
$(V_W)_{s_W(U)}$
to
$j_U(V_W)$
. This proves that the canonical factor embedding from
$(V_W)_{s_W(U)}$
to
$j_U(V_W)$
is an isomorphism.
The pushforward
$(j_W)_{*}(U) = \{A\subseteq j_W(X) : j_W^{-1}[A]\in U\}$
is as usual equal to the ultrafilter derived from
$j_U$
using
$[j_W]_U$
. Therefore
$s_W(U)$
, being the ultrafilter derived from
$j_U\restriction V_W$
using
$[j_W]_U$
, is equal to
$(j_W)_{*}(U)\cap V_W$
. In other words,
$$\begin{align*}s_W(U) = \{A\in j_W(P(X)) : j_W^{-1}[A]\in U\}.\\[-32pt] \end{align*}$$
From Lemma 3.5, we immediately obtain the following lemma.
Lemma 3.6. For any ultrafilters U and W,
$U\sqsubset W$
if and only if
$s_W(U)\in V_W$
.
Given this equivalence, the internal relation looks quite a bit like the Mitchell order, the difference being that instead of demanding that the V-ultrafilter U belong to
$V_W$
, one demands that the corresponding
$V_W$
-ultrafilter
$s_W(U)$
does.
Combining Lemma 3.5 and Theorem 2.8, we obtain a proof of Kunen’s Commuting Ultrapowers lemma.
Definition 3.7. If U is an ultrafilter, then
$\text {size}(U)$
denotes the minimum cardinality of a set in U.
Kunen’s Commuting Ultrapowers lemma is the following fact, which in a sense we have already considerably generalized.
Theorem 3.8 (Kunen, [Reference Larson4, Lemma 1.1.25]).
If U and W are countably complete ultrafilters such that W is
$\text {size}(U)^+$
-complete, then
$j_W(j_U) = j_U\restriction V_W$
and
$j_U(j_W) = j_W\restriction V_U$
.
Proof. By Lemma 3.5,
$j_U\restriction V_W$
is the ultrapower of
$V_W$
by
$s_W(U)$
, while
$j_W(j_U)$
is clearly the ultrapower of
$V_W$
by
$j_W(U)$
. So to show
$j_W(j_U) = j_U\restriction V_W$
, it suffices to show
$s_W(U) = j_W(U)$
, and similarly for
$j_U(j_W) = j_W\restriction V_U$
. But since W is
$\text {size}(U)^+$
-complete, W is U-closed. Hence
$U\times W$
is an ultrafilter, which implies
$U\ltimes W = U\rtimes W$
, and so by Theorem 2.8,
$s_W(U) = j_W(U)$
and
$s_U(W) = j_U(W)$
.
The basic relationship between the internal relation and products of ultrafilters is the following lemma.
Lemma 3.9. If U and W are ultrafilters, then the following are equivalent:
-
(1) U is W-nonregular.
-
(2)
$s_W(U) = j_W(U)$
.
Proof. Note that an amalgamation D of
$(U,W)$
is uniquely determined by the value of
$(D)_0$
: by Lemma 3.2,
$$\begin{align*}D = \{R\subseteq X\times Y : [f_R]_U\in (D)_0\}.\end{align*}$$
Similarly it is uniquely determined by
$(D)_1$
.
Now
$(U\rtimes W)_1 = (W\ltimes U)_0 = j_W(U)$
by Lemma 3.3 and
$(U\ltimes W)_1 = s_W(U)$
by definition, so
$s_W(U) = j_W(U)$
if and only if
$U\ltimes W = U\rtimes W$
.
3.3 The internal relation and the Ketonen order
The following theorem is almost proved in [Reference Goldberg2], but here we are interested in ultrafilters that may not be countably complete, so a new argument is needed.
Theorem 3.10. Suppose U and W are ultrafilters such that
$U\sqsubset W$
and
$W\sqsubset U$
. Then for any cardinal
$\delta $
, either U or W is
$\delta $
-indecomposable.
Theorem 3.19 below improves this theorem in the context of GCH, but we do not know how to prove Theorem 3.19 in ZFC.
To prove Theorem 3.10, we introduce a version of the Ketonen order on countably incomplete ultrafilters:
Definition 3.11. If X is a set, a normed ultrafilter on X is a pair
$(U,f)$
where U is an ultrafilter on a set X and
$f: X\to \text {Ord}$
is a function. If
$\delta $
is an ordinal,
$(U,f)$
is
$\delta $
-normed if
$f(x) < \delta $
for U-almost all
$x\in X$
.
Definition 3.12. The Ketonen order is defined on normed ultrafilters
$(U,f)$
and
$(W,g)$
by setting
$(U,f)<_{\Bbbk } (W,g)$
if there is some
$U_{*}\in V_W$
extending
$j_W[U]$
such that
$V_W$
satisfies that
$(U_{*},j_W(f))$
is a
$[g]_W$
-normed ultrafilter.
Equivalently,
$(U,f)<_{\Bbbk } (W,g)$
if there is a sequence
$\langle U_y : y\in Y\rangle $
of ultrafilters such that
$U = W\text {-}\lim _{y\in Y} U_y$
and for W-almost all y, for
$U_y$
-almost all x,
$f(x) < g(y)$
.
If M and N are models of set theory and
$j : M\to N$
is an elementary embedding, then j is close to M if for all
$A\in N$
, there is some
$\bar A\in M$
such that
$$\begin{align*}j^{-1}[A] = \{x\in M : M\vDash x\in \bar A\}.\end{align*}$$
Lemma 3.13. For normed ultrafilters
$(U,f)$
and
$(W,g)$
,
$(U,f)<_{\Bbbk } (W,g)$
if and only if there is a model of set theory N admitting an elementary embedding
$k : V_U\to N$
and a close embedding
$j : V_W\to N$
such that
$k\circ j_U = j\circ j_W$
and
$k([f]_U) < j([g]_W)$
.
Sketch.
Assuming
$(U,f)<_{\Bbbk } (W,g)$
, let
$j : V_W\to N$
be the ultrapower embedding of
$V_W$
associated with
$U_{*}$
, and let
$k : V_U\to N$
map
$[h]_U$
to
$[j_W(h)]_{U_{*}}$
.
Conversely, given such embeddings j and k, the ultrafilter
$U_{*}$
derived from j using
$k([f]_U)$
witnesses
$(U,f)<_{\Bbbk } (W,g)$
.
The Ketonen order is a strict partial order of the normed ultrafilters. We leave the proof of transitivity to the reader, but we do show that it is irreflexive.
Theorem 3.14.
$<_{\Bbbk }$
is irreflexive.
Proof. Suppose
$(U,\varphi )$
is a normed ultrafilter on X, and suppose
$\langle U_x : x\in X\rangle $
satisfies
$U = U\text {-}\lim _{x\in X} U_x$
. We must show that for U-almost all x, for
$U_x$
-almost all y,
$\varphi (x) \leq \varphi (y)$
.
Let B be the set of
$x\in X$
such that for
$U_x$
-almost all
$y\in Y$
,
$\{y\in X : \varphi (y) < \varphi (x)\}\in U_x$
. Let
$\prec $
be the wellfounded relation on X defined by setting
$y\prec x$
if
$\varphi (y) < \varphi (x)$
. We define a set
$A\subseteq B$
by recursion on
$\prec $
, putting
$x\in A$
if and only if
$\{y\in X:y\prec x\text { and }y\in A\}\notin U_x$
. By the definition of B, for all
$x\in B$
,
$\{y\in X : y\prec x\}\in U_x$
, and therefore by the definition of A,
$x\in A$
if and only if
$A\notin U_x$
. That is,
$A = \{x\in B : A\notin U_x\}$
. As a consequence,
$$ \begin{align} A\in U&\iff \{x\in X: A\in U_x\}\in U\nonumber\\ &\iff \{x\in X: A\notin U_x\}\notin U\nonumber\\ &\,\hspace{.02cm}\implies \{x\in B: A\notin U_x\}\notin U\\ &\iff A\notin U.\nonumber \end{align} $$
The implication (1) cannot be reversed on pain of contradiction, so
$B\notin U$
.
The following corollary seems to have been unknown.
Corollary 3.15. Suppose U is an ultrafilter,
$j,k:V_U\to N$
are elementary embeddings with
$j\circ j_U = k\circ j_U$
, and j is close to
$V_U$
. Then for every ordinal a of
$V_U$
,
$j(a) \leq k(a)$
.
The analog of Theorem 3.14 proved in [Reference Goldberg2] applies to arbitrary directed systems of ultrafilters whose limit ultrapower is wellfounded. Theorem 3.14, on the other hand, applies to a single ultrafilter but requires no assumption of wellfoundedness, or equivalently of countable completeness. The two theorems cannot be mutually generalized: it simply is not true that the Ketonen order on arbitrary directed systems of ultrafilters, defined following [Reference Goldberg2], is irreflexive. For example, there is a directed system of ultrafilters E whose limit ultrapower
$V_E$
admits a nontrivial elementary self embedding
$k : V_E\to V_E$
such that
$k\circ j_E = j_E$
, yet k is isomorphic to an internal ultrapower embedding of
$V_E$
. This means Corollary 3.15 does not generalize to E, and hence Theorem 3.14 cannot either.
We now explain the connection between the Ketonen order and the internal relation.
Definition 3.16. If
$(W,g)$
is a normed ultrafilter, then
$\delta (W,g)$
denotes the least ordinal
$\delta $
such that
$(W,g)$
is
$\delta $
-normed.
Theorem 3.17. If
$U\sqsubset W$
,
$\delta = \delta (W,g)$
is a limit ordinal, and
$(U,f)$
is
$\delta $
-normed, then
$(U,f)<_{\Bbbk } (W,g)$
.
Proof. For any
$\epsilon < \delta $
, for W-almost all y,
$g(y)> \epsilon $
. Therefore in
$V_U$
, for any
$\epsilon < j_U(\delta )$
, for
$j_U(W)$
-almost all y,
$j_U(g)(y)> \epsilon $
. It follows that for
$j_U(W)$
-almost all y,
$j_U(g)(y)> [f]_U$
. In other words, for U-almost all x, for W-almost all y,
$f(x) < g(y)$
. This means
$\{x : j_W(f(x)) < [g]_W\}\in U$
, or equivalently,
$\{x : j_W(f)(x) < [g]_W\}\in s_W(U)$
. This means that
$(s_W(U),j_W(f))$
is a
$[g]_W$
-normed ultrafilter in
$V_W$
, and hence
$s_W(U)$
witnesses that
$(U,f)<_{\Bbbk } (W,g)$
.
We now prove our first indecomposability theorem for the internal relation.
Proof of Theorem 3.10.
Assume toward a contradiction that the theorem fails, and neither U nor W is
$\delta $
-indecomposable. Let X and Y be the underlying sets of U and W. The
$\delta $
-decomposability of U yields a function
$f : X\to \delta $
such that f is not bounded below
$\delta $
on a set in U. In other words,
$\delta (U,f) = \delta $
. Similarly, for some
$g : X \to \delta $
,
$\delta (W,g) = \delta $
. By Theorem 3.17, since
$U\sqsubset W$
and
$(U,f)$
is
$\delta $
-normed,
$(U,f)<_{\Bbbk } (W,g)$
. The same logic shows
$(W,g)<_{\Bbbk } (U,f)$
. By the transitivity of the Ketonen order,
$(U,f)<_{\Bbbk } (U,f)$
, and this contradicts Theorem 3.14.
The following corollary is proved more directly in [Reference Goldberg3, Corollary 5.5.16].
Corollary 3.18. Any ultrafilter U such that
$U\sqsubset U$
is principal.
Proof. Suppose U is nonprincipal. Let
$\kappa $
be the completeness of U, the supremum of all cardinals
$\delta $
such that U is
$\delta $
-complete. Then
$\kappa $
is regular and U is
$\kappa $
-decomposable. But
$U\sqsubset U$
, so Theorem 3.10 implies U is not
$\kappa $
-decomposable, and this is a contradiction.
3.4
$(\lambda ,\lambda ^+)$
-indecomposability for mutually internal ultrafilters
This final subsection is devoted to the following theorem.
Theorem 3.19 (GCH).
Suppose U and W are ultrafilters such that
$U\sqsubset W$
and
$W\sqsubset U$
. Then for all cardinals
$\lambda $
, either U or W is
$(\lambda ,\lambda ^+)$
-indecomposable.
Combining this with Theorem 2.13 immediately yields the last of our main results.
Theorem 3.20 (GCH).
Suppose U and W are ultrafilters such that
$U\sqsubset W$
and
$W\sqsubset U$
. Then
$U\times W$
is an ultrafilter.
Under UA, one can easily prove that if
$U\sqsubset W$
and
$W\sqsubset U$
, then U is W-nonregular. This fact is unique among UA results in that it applies to arbitrary ultrafilters (including countably incomplete ones), so we sketch a proof for the reader familiar with [Reference Goldberg3].
Theorem 3.21 (UA).
Suppose U and W are ultrafilters such that
$U\sqsubset W$
and
$W\sqsubset U$
. Then U is W-nonregular.
Proof. At least one of U and W is countably complete by Theorem 3.10. So without loss of generality, assume that U is.
Applying UA in
$V_W$
, the ultrapower embeddings associated with
$s_W(U)$
and
$j_W(U)$
admit an internal ultrapower comparison. In other words, there exist internal ultrapower embeddings
$k : j_U(V_W)\to N$
and
$\ell : j_U(V_W)\to N$
such that
$k\circ j_U\restriction V_W = \ell \circ j_W(j_U)$
. To conclude that
$s_W(U) = j_W(U)$
(and hence that U is W-nonregular), it suffices to show that
$k([\text {id}]_{s_W(U)}) = \ell ([\text {id}]_{j_W(U)})$
.
We have that
$k([\text {id}]_{s_W(U)}) = k(j_U(j_W)([\text {id}]_U))$
while
$\ell ([\text {id}]_{j_W(U)}) = \ell (j_W([\text {id}]_U))$
. Now
$\ell \circ j_W$
and
$k\circ j_U(j_W)$
are both internal ultrapower embeddings from
$V_U$
to N, and moreover
$\ell \circ j_W \circ j_U = k\circ j_U(j_W)\circ j_U$
. By the uniqueness of such embeddings (Corollary 3.15),
$\ell \circ j_W=k\circ j_U(j_W)$
and in particular
$\ell \circ j_W([\text {id}]_U) =k\circ j_U(j_W)([\text {id}]_U)$
. It follows that
$k([\text {id}]_{s_W(U)}) = \ell ([\text {id}]_{j_W(U)})$
, and this proves the theorem.
We now turn to the proof of Theorem 3.19 itself, where we seem to need a completely different argument. This will require some lemmas on the interaction between the Rudin–Keisler order and the internal relation:
Theorem 3.22. Suppose
$\bar U\leq _{\text {RK}} U$
and W are ultrafilters.
-
(1) If
$\bar U = f_{*}(U)$
, then
$s_W(\bar U) = j_W(f)_{*}(s_W(U))$
. In particular, if
$U\sqsubset W$
, then
$\bar U\sqsubset W$
. -
(2) If
$k : V_{\bar U}\to V_U$
is an elementary embedding such that
$k\circ j_{\bar U} = j_U$
, then
$s_{\bar U}(W) = k^{-1}[s_U(W)]$
. In particular, if
$W\sqsubset U$
and
$s_U(W)\in \text {ran}(k)$
, then
$W\sqsubset \bar U$
.
Proof. For (1), recall that for any Z,
$s_W(Z) = (j_W)_{*}(Z)\cap V_W$
and observe:
$$\begin{align*}&j_W(f)_{*}((j_W)_{*}(U))\\&\quad = (j_W(f)\circ f)_{*}(U) = (j_W\circ f)_{*}(U)= (j_W)_{*}(f_{*}(U)) = (j_W)_{*}(\bar U). \end{align*}$$
For (2), note that
$A\in k^{-1}[s_{U}(W)]$
if and only if
$j_{U}^{-1}[k(A)]\in W$
, but
$j_{U}^{-1}[k(A)] = j_{\bar U}^{-1}[A]$
, so
$j_{U}^{-1}[k(A)]\in W$
if and only if
$j_{\bar U}^{-1}[A]\in W$
or equivalently
$A\in s_{\bar U}(W)$
.
Corollary 3.23. If
$U\ltimes W = U\rtimes W$
, then for any
$\bar U\leq _{\text {RK}} U$
and
$\bar W \leq _{\text {RK}} W$
,
$\bar U\ltimes \bar W = \bar U\rtimes \bar W$
.
Proof. It suffices to show that
$\bar U\ltimes W = \bar U\rtimes W$
. By Lemma 3.9, it suffices to show that
$j_{\bar U}(W) = s_{\bar U}(W)$
. Since
$\bar U \leq _{\text {RK}} U$
, there is an elementary embedding
$k : V_{\bar U}\to V_U$
such that
$k\circ j_{\bar U} = j_U$
. Therefore
$s_U(W) = j_U(W) = k(j_{\bar U}(W))$
. Hence
$j_{\bar U}(W) = k^{-1}[s_U(W)] = s_{\bar U}(W)$
, applying Theorem 3.22. Now by Lemma 3.9,
$\bar U\ltimes W = \bar U\rtimes W$
.
A model M has the
$\lambda $
-cover property if for all
$\sigma \in P_\lambda (M)$
, there is some
$\tau \in M$
such that
$\sigma \subseteq \tau $
.
Lemma 3.24. Suppose
$\lambda $
is a strong limit cardinal, M is a model with the
$\lambda $
-cover property, and U is an M-
$\lambda $
-complete M-ultrafilter. Then U generates a
$\lambda $
-complete filter.
Proof. Suppose
$\sigma \subseteq U$
and
$|\sigma | < \lambda $
. We will show that there is some
$A\in U$
with
$A\subseteq \bigcap \sigma $
.
Using the
$\lambda $
-cover property, fix
$\tau \in M$
such that
$\sigma \subseteq \tau $
and
$|\tau | < \lambda $
. We claim that
$\tau \cap U\in M$
. First, the M-
$\lambda $
-completeness of U implies
$j_U[\tau \cap U] = \{A\in j_U(\tau ) : [\text {id}]_U\in A\} \in M_U$
. But since
$2^{|\tau |} < \lambda $
, the M-
$\lambda $
-completeness of U implies
$j_U : P^M(\tau )\to j_U(P^M(\tau ))$
is an isomorphism, and so
$\tau \cap U = j_U^{-1}[j_U[\tau \cap U]]\in M$
.
Since U is M-
$\lambda $
-complete,
$A = \bigcap (\tau \cap U)\in U$
. Since
$\sigma \subseteq (\tau \cap U)$
,
$A\subseteq \bigcap \sigma $
.
Lemma 3.25 (Kunen).
Suppose W is
$(\kappa ,\delta )$
-regular and F is a
$\kappa $
-complete filter generated by at most
$\delta $
-many sets. Then
$F\leq _{\text {kat}} W$
.
Proof. Let
$\mathcal B$
be a basis for F with
$|\mathcal B| \leq \delta $
. Since W is
$(\kappa ,\delta )$
-regular, the filter G on
$P_\kappa (\mathcal B)$
generated by sets of the form
$\{\sigma \in P_\kappa (\mathcal B) : A\in \sigma \}$
for
$A\in \mathcal B$
lies below W in the Katětov order. Therefore it suffices to show that
$F\leq _{\text {kat}} G$
. But let
$f : P_\kappa (\mathcal B) \to X$
be any function such that
$f(\sigma )\in \bigcap \sigma $
. Easily,
$F\subseteq f_{*}(G)$
, as desired.
Finally, we need a theorem due essentially to Silver.
Definition 3.26. Suppose U and W are ultrafilters on sets X and Y. Then
$U\leq _{\text {RK}}^\gamma W$
if there is a function
$p :Y\to X$
such that
$p_{*}(W) = U$
and for all
$\beta < \gamma $
, for all
$f : Y\to \beta $
, f factors through p modulo W: there is a
$g : X\to \beta $
such that
$[f]_W = [g\circ p]_W$
.
We have the following characterization of this ordering in terms of ultrapowers.
Proposition 3.27.
$U\leq _{\text {RK}}^\gamma W$
if and only if there is an elementary embedding
$k : V_U\to V_W$
such that
$k\circ j_U = j_W$
and
$j_W(\beta )\subseteq \text {ran}(k)$
for all
$\beta < \gamma $
.
Proof. Suppose
$U\leq _{\text {RK}}^\gamma W$
as witnessed by
$p : Y\to X$
. Let
$k : V_U \to V_W$
be defined by
$k([g]_U) = [g\circ p]_W$
. Note that for any
$\beta < \gamma $
and any
$\alpha < j_W(\beta )$
,
$\alpha \in \text {ran}(k)$
: there is some
$f : Y\to \text {Ord}$
such that
$[f]_W = \alpha $
, so by the definition of
$\leq _{\text {RK}}^\gamma $
, there is some g such that
$[f]_W = [g\circ p]_W = k([g]_U)$
.
Conversely, suppose such an embedding k exists. Fix a function
$p : Y \to X$
such that
$[p]_W = k([\text {id}]_U)$
. Easily
$p_{*}(W) = U$
, and moreover
$k([g]_U) = [g\circ p]_W$
.
Fix
$\beta < \gamma $
and
$f : Y\to \beta $
. Since
$[f]_W < j_W(\beta )$
,
$[f]_W\in \text {ran}(k)$
, and so there is some
$g: X\to \beta $
be such that
$[f]_W = k([g]_U) = [g\circ p]_W$
.
A key ingredient in the proof of Theorem 3.19 is the following theorem, which is a minor generalization of a result of Silver’s on indecomposable ultrafilters. The generalization required here is proved as [Reference Goldberg3, Theorem 7.5.26]. There the theorem is stated only for countably complete ultrafilters, but as the author notes just below the theorem, this assumption is used only for notational convenience.
Theorem 3.28 (Silver, [Reference Silver7]).
Suppose
$\lambda $
is a regular cardinal and U is a
$(\lambda ,\eta )$
-indecomposable ultrafilter and
$2^\lambda \leq \eta $
. Then there is an ultrafilter D on a cardinal less than
$\lambda $
such that
$D\leq _{\text {RK}}^{\eta } U$
.
Proof of Theorem 3.19.
We assume by induction that the theorem holds for all
$\bar \lambda < \lambda $
, and toward a contradiction that there exist ultrafilters U and W such that
$U\sqsubset W$
,
$W\sqsubset U$
, but neither U nor W is
$(\lambda ,\lambda ^+)$
-indecomposable. By Lemma 2.20, we may assume without loss of generality that
$\lambda $
is singular, W is
$\lambda $
-decomposable, and U is
$(\kappa ,\lambda ^+)$
-regular for some
$\kappa < \lambda $
.
We claim one can reduce to the case that W is a uniform ultrafilter on
$\lambda $
. Note that W is
$(\lambda ^+,\lambda ^{+\omega })$
-indecomposable since W is
$\lambda ^+$
-indecomposable (by Lemma 2.20). Applying Theorem 3.28 and the GCH, let
$\tilde {W}$
be a uniform ultrafilter on
$\lambda $
such that
$\tilde {W}\leq _{\text {RK}}^{\lambda ^{+\omega }} W$
. Let p witness that
$\tilde {W}\leq _{\text {RK}}^{\lambda ^{+\omega }} W$
, so
$p_{*}(W) = \tilde {W}$
. Let
$\tilde {U}$
be the ultrafilter on
$\beta (\lambda )\times P_{\kappa }(\lambda )$
derived from
$j_U$
using
$(s_U(\tilde W),\sigma )$
where
$\sigma \in j_U(P_\kappa (\lambda ^+))$
covers
$j_U[\lambda ^+]$
. (Here
$\beta (\lambda )$
denotes the set of ultrafilters on
$\lambda $
.) Thus by Theorem 3.22,
$\tilde W\sqsubset \tilde U$
and by Lemma 2.12,
$\tilde U$
is
$(\kappa ,\lambda ^+)$
-regular. Let f be such that
$[f]_W = s_W(\tilde {U})$
, and assume without loss of generality that the range of f has cardinality strictly less than
$\lambda ^{+\omega }$
. Since
$\tilde {W} \leq _{\text {RK}}^{\lambda ^{+\omega }} W$
, there is a function
$\tilde {f}$
such that
$\tilde {f}\circ p = f$
. By Theorem 3.22,
$[\tilde {f}]_{\tilde {W}} = s_{\tilde {W}}(\tilde {U})$
. So
$\tilde U \sqsubset \tilde W$
. Replacing U and W with
$\tilde U$
and
$\tilde W$
, we reduce to the case that W is a uniform ultrafilter on
$\lambda $
.
Since U is
$(\kappa ,\lambda ^+)$
-regular, U is
$\delta $
-decomposable for every regular cardinal
$\delta $
such that
$\kappa \leq \delta < \lambda $
. (This is a well-known fact, but it is also an immediate consequence of Lemmas 2.17 and 3.25.) Therefore by our induction hypothesis, W is
$\eta $
-indecomposable for all cardinals
$\eta $
such that
$\kappa \leq \eta < \lambda $
: we know at least one of U and W is
$(\eta ,\eta ^+)$
-indecomposable, and U is not (since U is
$\eta ^+$
-decomposable since
$\eta ^+$
is regular). We are assuming GCH, so we can apply Silver’s lemma (Theorem 3.28) to obtain an ultrafilter D on a cardinal
$\gamma $
less than
$\kappa $
such that
$D\leq _{\lambda } W$
. Equivalently, there is an elementary embedding
$$\begin{align*}k : V_D\to V_W\end{align*}$$
such that
$k\circ j_D = j_W$
and
$j_W(\eta )\subseteq \text {ran}(k)$
for all
$\eta < \lambda $
.
By our induction hypothesis, for any cardinal
$\bar \lambda < \lambda $
, either U or W is
$(\bar \lambda ,(\bar \lambda )^+)$
-indecomposable. Since
$D\leq _{\text {RK}} W$
and
$\text {size}(D) < \lambda $
, for any cardinal
$\bar \lambda < \lambda $
, either U or D is
$(\bar \lambda ,(\bar \lambda )^+)$
-indecomposable. Applying Theorem 2.13,
$D\times U$
is an ultrafilter. In particular,
$j_D(U) = s_D(U)$
. Let
$W_{*}$
be the
$V_D$
-ultrafilter on
$j_D(\lambda )$
derived from k using
$[\text {id}]_{W}$
. Then
$V_{W}$
is isomorphic to the ultrapower of
$V_D$
by
$W_{*}$
and k to the associated ultrapower embedding. We will identify the two models via the unique isomorphism between k and
$(j_{W_{*}})^{V_D}$
. Therefore,
$V_W = (V_D)_{W_{*}}$
and
$k = (j_{W_{*}})^{V_D}$
.
Let F be the filter generated by
$W_{*}$
. By Lemma 3.24, F is
$\lambda $
-complete, and since
$|j_D(P(\lambda ))|\leq (\lambda ^+)^\gamma = \lambda ^+$
, F is generated by at most
$\lambda ^+$
-many sets. Therefore
$F\leq _{\text {kat}} U$
by Lemma 3.25, and in particular, F extends to an ultrafilter G preceding U in the Rudin–Keisler order. Since
$G\leq _{\text {RK}} U \sqsubset D$
, Theorem 3.22 yields that
$s_D(G)\leq _{\text {RK}} s_D(U)$
in
$V_D$
, and in particular
$G\sqsubset D$
. Note that
$W_{*}$
is the ultrafilter derived from
$j_G\restriction V_D$
using
$[\text {id}]_G$
. Therefore
$W_{*}\in V_D$
and
$W_{*}\leq _{\text {RK}} s_D(G)$
in
$V_D$
.
So far we have shown that in
$V_D$
,
$$\begin{align*}W_{*}\leq_{\text{RK}} s_D(G)\leq_{\text{RK}} s_D(U).\end{align*}$$
Since
$U\sqsubset W$
,
$s_W(U)\in V_W$
. But
$s_W(U) = s_{W_{*}}(s_D(U))$
, so since
$s_{W_{*}}(s_D(U))\in V_W = (V_D)_{W_{*}}$
,
$s_D(U)\sqsubset W_{*}$
in
$V_D$
. Now in
$V_D$
,
$W_{*}\leq _{\text {RK}} s_D(U)\sqsubset W_{*}$
. Applying Theorem 3.22 one last time,
$W_{*}\sqsubset W_{*}$
in
$V_D$
, and it follows (by Corollary 3.18) that
$W_{*}$
is a principal ultrafilter of
$V_D$
. Thus
$j_D = j_{W_{*}}^{V_D}\circ j_D = j_W$
, or in other words,
$D\equiv _{\text {RK}} W$
, contradicting that W is
$\lambda $
-decomposable, while
$\text {size}(D) < \lambda $
.
