# PCF Theory and The Tukey Spectrum

Thomas Gilton  University of Pittsburgh Department of Mathematics. The
Dietrich School of Arts and Sciences, 301 Thackeray Hall, Pittsburgh, PA
15260, United States<EMAIL_ADDRESS>
###### Abstract.

In this paper, we investigate the relationship between the Tukey order and PCF
theory, as applied to sets of regular cardinals. We show that it is consistent
that for all sets $A$ of regular cardinals that the Tukey spectrum of $A$,
denoted $\operatorname{spec}(A)$, is equal to the set of possible cofinalities
of $A$, denoted $\operatorname{pcf}(A)$; this is to be read in light of the
$\mathsf{ZFC}$ fact that
$\operatorname{pcf}(A)\subseteq\operatorname{spec}(A)$ holds for all $A$. We
also prove results about when regular limit cardinals must be in the Tukey
spectrum or must be out of the Tukey spectrum of some $A$, and we show the
relevance of these for forcings which might separate $\operatorname{spec}(A)$
from $\operatorname{pcf}(A)$. Finally, we show that the _strong part_ of the
Tukey spectrum can be used in place of PCF-theoretic scales to lift the
existence of Jónsson algebras from below a singular to hold at its successor.
We close with a list of questions.

###### Key words and phrases:

PCF Theory, Tukey order, cofinality, weakly compact cardinals, Mahlo cardinals

###### 2020 Mathematics Subject Classification:

Primary 03E04, 03E05

##  1\. Introduction

The Tukey order has become a very useful tool for comparing directed,
partially ordered sets. The order is sufficiently coarse that is allows us to
compare many different partial orders, yet it is fine enough to preserve a
number of order-theoretic properties of interest (such as calibre properties;
see Proposition 2.4 below). The Tukey order works by comparing partial orders
in terms of what happens “eventually”, or more precisely, in terms of what
happens _cofinally_.

The Tukey order arose in the study of Moore-Smith convergence in topology
([21] and [35]), with [2] and [5] following shortly after. Schmidt ([25]) and
Isbell ([15] and [16]) later studied cofinal types among the class of directed
posets. Later, Todorčević ([32]) showed that it is consistent that there are
only five cofinal types of directed sets of size $\leq\aleph_{1}$. Todorčević
also showed that under the $\mathsf{CH}$, there are $2^{\mathfrak{c}}$-many
such cofinal types, and in [34] he extended this result to all transitive
relations on $\omega_{1}$.

Since then, there has been a tremendous amount of research on the Tukey order
in a variety of circumstances. Some results concern definable directed sets
([31], [30]). Others concern cofinal types of ultrafilters ([20], [6], [7],
[8], [24], [23], [18]). Additional research concerns the Tukey order on
various sets in topological spaces ([12], [13], [11]). Yet another batch of
results studies the number of cofinal types of partial orders of various sizes
below $\aleph_{\omega}$ ([19], [26]). See also [10] and [22].

Cofinal structure has been studied by set theorists coming from a different
angle. Especially important for us is Shelah’s theory of possible
cofinalities, or PCF theory for short. The main objects of study in PCF theory
are reduced products of sets of regular cardinals, modulo an ideal. Shelah has
developed the theory in a series of papers which culminated in the book [29].
PCF theory has had dramatic implications for our understanding of cardinal
arithmetic (see [1]), as well as plenty of applications both inside and
outside of set theory, such as [17]. See [28] for a discussion of further
applications.

These two ways of studying cofinal structure are related (and we will discuss
this more later): given a set $A$ of regular cardinals, we consider the Tukey
spectrum of $A$, which consists of all regular cardinals which are Tukey below
$(\prod A,<)$ (i.e., $\prod A$ with the pointwise domination ordering). We
denote this by $\operatorname{spec}(A)$. It follows quickly from the
definitions (which we give later) that for any set $A$ of regular cardinals,
$\operatorname{pcf}(A)\subseteq\operatorname{spec}(A)$. In this paper, we are
concerned with the following general question:

######  Question 1.1.

Does $\mathsf{ZFC}$ prove that for any set $A$ of regular cardinals,
$\operatorname{pcf}(A)=\operatorname{spec}(A)$?

To our knowledge, the only result, so far, which addresses this question is
due to Gartside and Mamatelashvili ([11]) who have a proof showing that if $A$
is any _progressive_ set of regular cardinals, then
$\operatorname{pcf}(A)=\operatorname{spec}(A)$ (“progressive” is a common
assumption when doing PCF theory). However, there is a gap in their proof. We
address this gap later, observing that their argument rather shows that if $A$
is progressive, then
$\operatorname{spec}(A)\subseteq\operatorname{pcf}(A)\cup\lim(\operatorname{pcf}(A))$.111Since
$\operatorname{spec}(A)$ consists, by definition, of regular cardinals, if $A$
is progressive and $\operatorname{spec}(A)\neq\operatorname{pcf}(A)$, then
$\operatorname{pcf}(A)$ has a regular limit point. Additional assumptions on
$\operatorname{pcf}(A)$ then guarantee equality. However, the status of
Question 1.1 when $A$ is not progressive is far from clear.

In this work, we prove various results related to Question 1.1. After a review
of the basics of the Tukey order and PCF theory in Section 2, we turn in
Section 3 to the question of how much bigger $\operatorname{spec}(A)$ can be
than $\operatorname{pcf}(A)$. We review the theorem from [11] and address the
gap in their proof. Then we turn to showing that Question 1.1 has a consistent
positive answer. We also discuss circumstances under which, for all $A$,
$\operatorname{spec}(A)$ is no worse than $\operatorname{pcf}(A)\cup\lim(A)$.
In Section 4 we address the role that small large cardinals (Mahlo and weakly
compact) have in excluding a regular limit $\kappa$ from
$\operatorname{spec}(A)$ (where $A\subseteq\kappa$) or for ensuring that
$\kappa\in\operatorname{spec}(A)$. The upshot of these results is that they
may reduce the options for showing that Question 1.1 has a consistent negative
answer which is witnessed by a forcing separating $\operatorname{spec}(A)$ and
$\operatorname{pcf}(A)$ (if such exists). In the last main section, Section 5,
we show that a subset of the Tukey spectrum (what we call the “strong part” of
the Tukey spectrum) is sufficiently strong to be able to “lift” the existence
of Jónsson algebras; this generalizes Shelah’s celebrated result [27] that
scales in PCF theory can lift the existence of Jónsson algebras.

Acknowledgements We would like to thank Will Brian, James Cummings, Todd
Eisworth, and Paul Gartside for many helpful conversations about the Tukey
order and PCF theory and for suggesting ways of extending this line of
research.

##  2\. A quick overview of Tukey and PCF

In this section we review the basics of the Tukey order and PCF theory which
are relevant for this paper.

######  Remark 2.1.

Throughout the paper, all posets are assumed to be directed.

First we recall the definition of the Tukey order (see [12] for a detailed
development of these ideas).

######  Definition 2.2.

A poset $Q$ is said to be a _Tukey quotient_ of $P$ if there exists a function
$\varphi:P\longrightarrow Q$ which preserves cofinal sets. We denote this by
$P\geq_{T}Q$.

$P\geq_{T}Q$ is equivalent to the existence of a map $\psi:Q\longrightarrow P$
which preserves unbounded sets.

######  Definition 2.3.

Suppose that $\kappa\geq\lambda\geq\mu$ are cardinals. We say that a poset $P$
has _calibre_ $(\kappa,\lambda,\mu)$ if for all $\kappa$-sized
$P^{\prime}\subseteq P$ there is a $\lambda$-sized $R\subseteq P^{\prime}$ so
that every $\mu$-sized $B\subseteq R$ is bounded in $P$.

We say simply that $P$ has _calibre_ $\kappa$ if $P$ has calibre
$(\kappa,\kappa,\kappa)$.

Note that the definition of $P$ having calibre $\kappa$ simplifies to the
following: every $\kappa$-sized $P^{\prime}\subseteq P$ has a $\kappa$-sized
subset which is bounded in $P$.

The following item connects the ideas of $\geq_{T}$ and calibre.

######  Proposition 2.4.

Suppose that $\kappa$ is regular, that $P$ has calibre $(\kappa,\lambda,\mu)$,
and that $P\geq_{T}Q$. Then $Q$ has calibre $(\kappa,\lambda,\mu)$ too.

The next item is particularly relevant for us.

######  Proposition 2.5.

For a regular cardinal $\kappa$, a poset $P$ fails to have calibre $\kappa$
iff $P\geq_{T}\kappa$.

Now we define the Tukey spectrum of a poset.

######  Definition 2.6.

The _Tukey spectrum_ of a poset $P$ is denoted by $\operatorname{spec}(P)$ and
defined to be
$\operatorname{spec}(P):=\left\\{\kappa:P\geq_{T}\kappa\wedge\kappa\text{ is
regular}\right\\}$.

When $A$ is a set of regular cardinals, we let $\operatorname{spec}(A)$
abbreviate $\operatorname{spec}\left(\prod A,<\right)$.

######  Remark 2.7.

$\operatorname{spec}(P)$ consists of all regular $\kappa$ so that $P$ does
_not_ have calibre $\kappa$.

It is helpful to get a better handle on what $\kappa\in\operatorname{spec}(A)$
means in the specific case that $A$ is a set of regular cardinals. Indeed,
$\kappa\in\operatorname{spec}(A)$ iff there exists a set $\cal{F}$ of
$\kappa$-many functions in $\prod A$ so that every
$\cal{F}_{0}\in[\cal{F}]^{\kappa}$ is unbounded in $(\prod A,<)$, i.e., $A$
with the pointwise domination ordering. That is to say, there is _at least
one_ coordinate $a\in A$ so that

$\left\\{f(a):f\in\cal{F}_{0}\right\\}$

is unbounded in the regular cardinal $a$. We give a name to these coordinates
in the next definition.

######  Definition 2.8.

Suppose that $A$ is a set of regular cardinals $\cal{F}\subseteq\prod A$. A
cardinal $a\in A$ is called an _unbounded coordinate_ of $\cal{F}$ if
$\left\\{f(a):f\in\cal{F}\right\\}$ is unbounded in $a$.

We let $\operatorname{ub}(\cal{F})$ denote the set of unbounded coordinates of
$\cal{F}$.

We address the question of how many coordinates are in
$\operatorname{ub}(\cal{F})$ in Section 5.

The following lemma is a standard part of Tukey-ology.

######  Lemma 2.9.

$\operatorname{spec}(P\times
Q)=\operatorname{spec}(P)\cup\operatorname{spec}(Q)$. Hence if $A$ and $B$ are
sets of regular cardinals, $\operatorname{spec}(A\cup
B)=\operatorname{spec}(A)\cup\operatorname{spec}(B)$.

This captures all of the basics of the Tukey order that we need. We now review
some of the central results in PCF theory, beginning with the central
definition (see [1] for a clear and detailed exposition of these and related
ideas).

######  Definition 2.10.

Let $A$ be a set of regular cardinals.

$\operatorname{pcf}(A):=\left\\{\operatorname{cf}\left(\prod
A/D\right):D\text{ is an ultrafilter on }A\right\\}.$

The following are routine facts about the pcf function.

######  Fact 2.11.

Suppose that $A$ and $B$ are sets of regular cardinals.

  1. (1)

$A\subseteq\operatorname{pcf}(A)$ (by using principal ultrafilters).

  2. (2)

If $A\subseteq B$, then $\operatorname{pcf}(A)\subseteq\operatorname{pcf}(B)$
(since any ultrafilter on $A$ can be extended to one on $B$).

  3. (3)

$\operatorname{pcf}(A\cup B)=\operatorname{pcf}(A)\cup\operatorname{pcf}(B)$
(since any ultrafilter on $A\cup B$ contains either $A$ or $B$).

The next lemma follows almost immediately from the definitions:

######  Lemma 2.12.

For any set $A$ of regular cardinals,
$\operatorname{pcf}(A)\subseteq\operatorname{spec}(A)$.

A very useful assumption when doing PCF theory is the following:

######  Definition 2.13.

A set $A$ of regular cardinals is _progressive_ if $|A|<\min(A)$.

We next define certain ideals which are naturally associated to the cardinals
in $\operatorname{pcf}(A)$. $A$ being progressive plays an important role in
the development of the ideas that follow.

######  Definition 2.14.

Let $A$ be a set of regular cardinals and $\lambda$ a cardinal (singular or
regular). Define the ideal $J_{<\lambda}[A]$ to consist of all $B\subseteq A$
so that for any ultrafilter $D$ on $A$ with $B\in D$, $\operatorname{cf}(\prod
A/D)<\lambda$.

If the set $A$ is clear from context, we write $J_{<\lambda}$ instead of
$J_{<\lambda}[A]$. A crucial fact about the ideal $J_{<\lambda}$ is the
following:

######  Proposition 2.15.

Suppose that $A$ is progressive. Then for any cardinal $\lambda$, $\prod
A/J_{<\lambda}$ is $<\lambda$-directed (that is, any set of fewer than
$\lambda$-many functions in $\prod A$ has an upper bound mod $J_{<\lambda}$).

A major theorem in PCF theory is the existence of generators, which we define
now.

######  Definition 2.16.

Let $A$ be a set of regular cardinals and $\lambda\in\operatorname{pcf}(A)$. A
_generator_ for $\lambda$ is a set $B_{\lambda}\subseteq A$ in
$J_{<\lambda^{+}}$ so that $J_{<\lambda}\cup\left\\{B_{\lambda}\right\\}$
generates $J_{<\lambda^{+}}$. In other words, given $X\subseteq A$, $X\in
J_{<\lambda^{+}}$ iff $X\setminus B_{\lambda}\in J_{<\lambda}$.

Thus a generating set $B_{\lambda}$ is a maximal set in $J_{<\lambda^{+}}$,
modulo $J_{<\lambda}$; they are unique modulo $J_{<\lambda}$. The following is
often proven using universal cofinal sequences:

######  Proposition 2.17.

Suppose that $A$ is progressive. Then for any
$\lambda\in\operatorname{pcf}(A)$, there is a generator for $\lambda$.

Generators give an instance of compactness:

######  Proposition 2.18.

Suppose that $A$ is progressive and $B\subseteq A$. Let
$\left<B_{\lambda}:\lambda\in\operatorname{pcf}(A)\right>$ be a sequence of
generators. Then there exists a finite decreasing sequence
$\lambda_{0}>\dots>\lambda_{n}$ of elements of $\operatorname{pcf}(A)$ with
$\lambda_{0}=\max\operatorname{pcf}(B)$ so that

$B\subseteq\bigcup_{i\leq n}B_{\lambda_{i}}.$

Another application of generators and related ideas is the following fact:

######  Proposition 2.19.

Suppose that $A$ is progressive. Then $\operatorname{pcf}(A)$ has a maximum
element, and moreover,

$\max\operatorname{pcf}(A)=\operatorname{cf}\left(\prod A,<\right).$

The last collection of background facts to review concerns weak compactness.

######  Definition 2.20.

Let $\alpha$ be an inaccessible cardinal. We say that a transitive set $M$ is
an _$\alpha$ -model_ if $M\models\mathsf{ZFC}^{-}$, $M$ has size $\alpha$, $M$
is closed under $<\alpha$-sequences, and $\alpha\in M$.

Now we recall the following characterization of weak compactness (see [14]):

######  Fact 2.21.

An inaccessible cardinal $\kappa$ is weakly compact if and only if for any
$\kappa$-model $M$, there exist a $\kappa$-model $N$ and an elementary
embedding $j:M\to N$ with $\operatorname{crit}(j)=\kappa$.

In the context of the previous fact, note that $j:M\to N$ gives rise to an
$M$-normal ultrafilter $\cal{U}_{j}:=\left\\{X\in
M:X\subseteq\kappa\wedge\kappa\in j(X)\right\\}$ on $\cal{P}(\kappa)\cap M$
(recall that an $M$-ultrafilter $\cal{U}$ is $M$-normal if for any
$A\in\cal{U}$ and any regressive $f:A\to\kappa$ with $f\in M$, $f$ is constant
on a set in $\cal{U}$). Moreover, since $M$ is closed under
$<\kappa$-sequences (by definition of a $\kappa$-model), this ultrafilter is
$<\kappa$-closed in $V$ (but of course it only measures subsets of $\kappa$ in
$M$).

##  3\. How bad can Spec be?

In this section, we study various conditions which guarantee that either (a)
$\operatorname{spec}(A)$ is no worse than $\operatorname{pcf}(A)$ together
with (regular) limit points of $\operatorname{pcf}(A)$ or (b)
$\operatorname{spec}(A)$ is no worse than $\operatorname{pcf}(A)$ together
with regular limit points of $A$. These results, when coupled with anti large
cardinal hypotheses, give a consistent positive answer to Question 1.1.

We begin by looking at the theorem from [11] which we mentioned in the
introduction. Addressing a small gap in their argument, what their proof shows
is that for any progressive $A$, $\operatorname{spec}(A)$ can at worst add
regular limits of $\operatorname{pcf}(A)$. Additional assumptions on
$\operatorname{pcf}(A)$ then give the equality of $\operatorname{pcf}(A)$ and
$\operatorname{spec}(A)$. We first have some notation.

######  Notation 3.1.

Given functions $g_{0},\dots,g_{n}$ in some product $\prod A$ of regular
cardinals, we let $\max(g_{0},\dots,g_{n})$ be the function $h\in\prod A$
defined by

$h(a):=\max\left\\{g_{0}(a),\dots,g_{n}(a)\right\\}.$

######  Theorem 3.2.

(Almost entirely [11]) Suppose that $A$ is a progressive set of regular
cardinals. Then
$\operatorname{spec}(A)\subseteq\operatorname{pcf}(A)\cup\lim(\operatorname{pcf}(A))$.222The
original statement of their theorem applies to $A$ which are finite unions of
progressive sets and uses what we are calling Lemma 2.9 to reduce to the case
of a single progressive set.

In particular, if $\operatorname{pcf}(A)$ is closed under regular limits, or
if $\operatorname{pcf}(A)$ has no regular limits (for example, if
$\operatorname{pcf}(A)$ is itself progressive), then
$\operatorname{pcf}(A)=\operatorname{spec}(A)$.

###### Proof.

We begin with some set-up. Since $A$ is progressive, we may apply Proposition
2.17 to fix a sequence
$\left<B_{\lambda}:\lambda\in\operatorname{pcf}(A)\right>$ of generators for
$\operatorname{pcf}(A)$. For each $\lambda\in\operatorname{pcf}(A)$,
$B_{\lambda}$ is progressive, being a subset of the progressive set $A$. Thus
$\lambda=\operatorname{maxpcf}(B_{\lambda})=\operatorname{cf}\left(\prod
B_{\lambda},<\right)$, using $B_{\lambda}\in J_{<\lambda^{+}}\setminus
J_{<\lambda}$ for the first equality and applying Proposition 2.19 for the
second equality. Therefore, for each $\lambda\in\operatorname{pcf}(A)$, we may
choose a sequence
$\vec{f}^{\lambda}=\left<f^{\lambda}_{\alpha}:\alpha<\lambda\right>$ of
functions in $\prod A$ so that $\left<f^{\lambda}_{\alpha}\upharpoonright
B_{\lambda}:\alpha<\lambda\right>$ is cofinal in $\left(\prod
B_{\lambda},<\right)$.

Fix a cardinal $\kappa\in\operatorname{spec}(A)$, suppose that $\kappa$ is not
a limit point of $\operatorname{pcf}(A)$, and we will show that
$\kappa\in\operatorname{pcf}(A)$. Because $\kappa$ is not a limit point of
$\operatorname{pcf}(A)$, we know that
$|\operatorname{pcf}(A)\cap\kappa|<\kappa$; this will permit us to apply a
pigeonhole argument at a crucial step later in the proof.

Since $\kappa\in\operatorname{spec}(A)$, let
$\left<f_{\gamma}:\gamma<\kappa\right>$ enumerate a set $\cal{F}$ of
$\kappa$-many functions in $\prod A$ so that every $\kappa$-sized subset of
$\cal{F}$ is unbounded in $\left(\prod A,<\right)$. To show that
$\kappa\in\operatorname{pcf}(A)$, it suffices to show that $J_{<\kappa}\neq
J_{<\kappa^{+}}$. Since $\cal{F}$ is bounded in $\prod A/J_{<\kappa^{+}}$
(because this reduced product is $<\kappa^{+}$-directed, by Proposition 2.15),
it in turn suffices to show that $\cal{F}$ is unbounded in $\prod
A/J_{<\kappa}$.

We work by contradiction and suppose that $\cal{F}$ is bounded in $\prod
A/J_{<\kappa}$. Our goal is to define a set $\cal{X}$ of fewer than
$\kappa$-many functions in $\prod A$ so that every $f\in\cal{F}$ is pointwise
below some $g\in\cal{X}$. Supposing we can define such an $\cal{X}$, we obtain
a contradiction as follows: since $\kappa$ is regular, there is a single
$g\in\cal{X}$ which is pointwise above $\kappa$-many elements of $\cal{F}$,
and this contradicts the fact that $\cal{F}$ witnesses that
$\kappa\in\operatorname{spec}(A)$.

We begin the construction of $\cal{X}$ as follows. Since $\cal{F}$ is bounded
in $\prod A/J_{<\kappa}$, let $g$ be a bound. For each $n\in\omega$, each
sequence $\vec{\lambda}=\lambda_{0}>\dots>\lambda_{n}$ in
$\operatorname{pcf}(A)\cap\kappa$, and each sequence
$\vec{\alpha}=\left<\alpha_{0},\dots,\alpha_{n}\right>$ with
$\alpha_{i}\in\lambda_{i}$ for all $i\leq n$, we let
$h(\vec{\alpha},\vec{\lambda})$ be the function
$\max(f^{\lambda_{0}}_{\alpha_{0}},\dots,f^{\lambda_{n}}_{\alpha_{n}},g)$; see
Notation 3.1. We let $\cal{X}$ be the set of such $h$. Since by assumption
$\operatorname{pcf}(A)\cap\kappa$ is bounded below $\kappa$, we see that
$\cal{X}$ consists of fewer than $\kappa$-many functions.

Now we verify that $\cal{X}$ has the desired property. Recalling that
$\left<f_{\alpha}:\alpha<\kappa\right>$ enumerates $\cal{F}$, fix some
$\alpha<\kappa$. Since $f_{\alpha}<_{J_{<\kappa}}g$, we know that

$z_{\alpha}:=\left\\{a\in A:f_{\alpha}(a)\geq g(a)\right\\}$

is in $J_{<\kappa}$. Thus $\max\operatorname{pcf}(z_{\alpha})<\kappa$, and so
by Proposition 2.18, we may find a sequence $\lambda_{0}>\dots>\lambda_{n}$ in
$\operatorname{pcf}(A)\cap\kappa$ so that $z_{\alpha}\subseteq\bigcup_{i\leq
n}B_{\lambda_{i}}$. Then, for each $i\leq n$, pick $\alpha_{i}<\lambda_{i}$ so
that

$f_{\alpha}\upharpoonright
B_{\lambda_{i}}<f^{\lambda_{i}}_{\alpha_{i}}\upharpoonright B_{\lambda_{i}}$

(i.e., this holds pointwise on $B_{\lambda_{i}}$). Then we see that

$f_{\alpha}<\max(f^{\lambda_{0}}_{\alpha_{0}},\dots,f^{\lambda_{n}}_{\alpha_{n}},g).$

Indeed, if $a\notin z_{\alpha}$, then $f_{\alpha}(a)<g(a)$. On the other hand,
if $a\in z_{\alpha}$, there is some $i\leq n$ so that $a\in B_{\lambda_{i}}$,
and hence $f_{\alpha}(a)<f^{\lambda_{i}}_{\alpha_{i}}(a)$. Since the function
$\max(f^{\lambda_{0}}_{\alpha_{0}},\dots,f^{\lambda_{n}}_{\alpha_{n}},g)$ is
in $\cal{X}$, this completes the main part of the proof.

For the “in particular” part of the theorem, observe that if
$\operatorname{pcf}(A)$ is progressive, then there are no regular limit
$\kappa$ in $\lim(\operatorname{pcf}(A))$. Otherwise,
$|\operatorname{pcf}(A)|\geq\kappa>\min(\operatorname{pcf}(A))$, contradicting
that $\operatorname{pcf}(A)$ is progressive. ∎

In the rest of this section, we work to isolate a condition (Theorem 3.5)
which is consistent with $\mathsf{ZFC}$ and which implies that
$\operatorname{spec}(A)\subseteq\operatorname{pcf}(A)\cup\lim(\operatorname{pcf}(A))$
for all $A$. In fact, it implies the stronger result that
$\operatorname{spec}(A)\subseteq\operatorname{pcf}(A)\cup\lim(A)$ for all $A$.
We have some preliminary lemmas. The first of these illustrates a “dropping”
phenomenon in the Tukey spectrum; we will use this lemma as part of a later
inductive argument.

######  Lemma 3.3.

Suppose that $A$ is a set of regular cardinals. Let
$\kappa\in\operatorname{spec}(A)\cap\sup(A)$. Then
$\kappa\in\operatorname{spec}(A\cap(\kappa+1))$.

###### Proof.

Set $\mu:=\sup(A)$; note that we are making no assumption about whether or not
$\mu\in A$. Since $\kappa\in\operatorname{spec}(A)$, let $\cal{F}$ be a set of
$\kappa$-many functions in $\prod A$ so that for all
$\cal{F}_{0}\in[\cal{F}]^{\kappa}$, $\cal{F}_{0}$ is unbounded in $(\prod
A,<)$. Let

$\cal{F}_{\leq\kappa}:=\left\\{f\upharpoonright(\kappa+1):f\in\cal{F}\right\\}$

and

$\cal{F}_{>\kappa}:=\left\\{f\upharpoonright(\kappa+1,\mu]:f\in\cal{F}\right\\}.$

Note that since $\kappa<\mu=\sup(A)$, $A\setminus(\kappa+1)$ is a non-empty
set of regular cardinals. Also, note that $\cal{F}_{>\kappa}$ is bounded in
$\prod(A\setminus(\kappa+1),<)$, since $\cal{F}_{>\kappa}$ consists of at most
$\kappa$-many functions, and we can take a sup of $\kappa$-many elements on
each coordinate in $A\setminus(\kappa+1)$.

We next argue that $\cal{F}_{\leq\kappa}$ has size $\kappa$ and that every
$\kappa$-sized subset is unbounded in $(\prod A\cap(\kappa+1),<)$; this will
show the desired result. First suppose for a contradiction that
$\cal{F}_{\leq\kappa}$ has size $<\kappa$. Then there is a single
$\bar{f}\in\prod A\cap(\kappa+1)$ so that for $\kappa$-many $g\in\cal{F}$,
$g\upharpoonright(\kappa+1)=\bar{f}$. Let $\cal{F}_{0}$ be this set of
$g\in\cal{F}$. But then $\cal{F}_{0}$ is bounded in the entire product $(\prod
A,<)$, using the observation from the previous paragraph to bound the elements
of $\cal{F}_{0}$ on coordinates in $A$ above $\kappa$. A nearly identical
argument shows that $\cal{F}_{\leq\kappa}$ must satisfy that every
$\kappa$-sized subset is unbounded in $(\prod A\cap(\kappa+1),<)$. ∎

The next lemma will also be used in the proof of Theorem 3.5:

######  Lemma 3.4.

Suppose that $\kappa$ is a regular limit cardinal. Let $A\subseteq\kappa$ be a
non-stationary set of regular cardinals which is unbounded in $\kappa$, and
let $D$ be an ultrafilter on $A$ which extends the tail filter on $A$. Then
$\operatorname{cf}(\prod A/D)\geq\kappa^{+}$.

###### Proof.

To begin, let $\left<\mu_{\nu}:\nu<\kappa\right>$ be the increasing
enumeration of $A$, and let $\left<\zeta_{i}:i<\kappa\right>$ enumerate a club
$C\subseteq\kappa$ with $C\cap A=\emptyset$. By relabeling if necessary, we
take $\zeta_{0}=0$.

Suppose for a contradiction that $\operatorname{cf}(\prod A/D)\leq\kappa$;
then the cofinality must equal exactly $\kappa$ since $D$ extends the tail
filter on $A$. Let $\vec{f}=\left<f_{\alpha}:\alpha<\kappa\right>$ be a
sequence of functions in $\prod A$ which is increasing and cofinal modulo $D$.
We obtain our contradiction by showing that $\vec{f}$ is bounded in $\prod A$,
modulo the tail filter on $A$, and hence modulo $D$.

Given $\nu<\kappa$ (corresponding to $\mu_{\nu}$), let $i<\kappa$ so that
$\nu\in[\zeta_{i},\zeta_{i+1})$ (such an $i$ exists since $C$ is club and
$\zeta_{0}=0$). The following observation, though simple, is crucial: for each
$i<\kappa$, $\mu_{\zeta_{i}}>\zeta_{i}$: indeed,
$\mu_{\zeta_{i}}\geq\zeta_{i}$ since the enumeration of $A$ is increasing. But
$\zeta_{i}\in C$ and $\mu_{\zeta_{i}}\in A\subseteq\kappa\setminus C$, so
$\mu_{\zeta_{i}}>\zeta_{i}$. Therefore $\mu_{\nu}$ is also strictly above
$\zeta_{i}$, and so

$\sup\left\\{f_{\alpha}(\mu_{\nu}):\alpha\leq\zeta_{i}\right\\}$

is below $\mu_{\nu}$. We let $g(\mu_{\nu})<\mu_{\nu}$ be above this sup. $g$
is then a member of $\prod A$.

We finish the proof by showing that $g$ bounds each of the $f_{\xi}$ on a
tail. To this end, fix $\alpha<\kappa$, and let $i<\kappa$ so that
$\alpha\leq\zeta_{i}$. We claim that for all $\nu\geq\zeta_{i}$,
$g(\mu_{\nu})>f_{\alpha}(\mu_{\nu})$. Fix $\nu\geq\zeta_{i}$, and let $j\geq
i$ so that $\nu\in[\zeta_{j},\zeta_{j+1})$. Then

$g(\mu_{\nu})>\sup\left\\{f_{\beta}(\mu_{\nu}):\beta\leq\zeta_{j}\right\\}\geq
f_{\alpha}(\mu_{\nu}),$

where the last inequality follows since $\alpha\leq\zeta_{i}\leq\zeta_{j}$. ∎

######  Theorem 3.5.

Let $V$ be a model of $\mathsf{ZFC}$ in which $2^{\mu}=\mu^{+}$ for all limit
cardinals $\mu$ and in which there are no Mahlo cardinals. Then $V$ satisfies
that for any set $A$ of regular cardinals,
$\operatorname{spec}(A)\subseteq\operatorname{pcf}(A)\cup\lim(A)$.

Consequently, if $2^{\mu}=\mu^{+}$ for all singular $\mu$ and if there are no
regular limit cardinals, then for all $A$,
$\operatorname{spec}(A)=\operatorname{pcf}(A)$.

###### Proof.

Fix such a $V$ (for example, work in $L$ up to the first $\lambda$ which is
Mahlo in $L$, if such a $\lambda$ exists, and otherwise work in all of $L$).
Let $A$ be a set of regular cardinals, and we will prove the result by
induction on the order type of $A$.

We first dispense with the case when $\max(A)$ exists. Let $\lambda:=\max(A)$.
Then, applying Lemma 2.9 and Fact 2.11, as well as our inductive assumption,
we get

$\displaystyle\operatorname{spec}(A)$
$\displaystyle=\operatorname{spec}((A\cap\lambda)\cup\left\\{\lambda\right\\})$
$\displaystyle=\operatorname{spec}(A\cap\lambda)\cup\underbrace{\operatorname{spec}(\left\\{\lambda\right\\})}_{=\left\\{\lambda\right\\}}$
$\displaystyle\subseteq\operatorname{pcf}(A\cap\lambda)\cup\lim(A\cap\lambda)\cup\left\\{\lambda\right\\}$
$\displaystyle=\operatorname{pcf}(A\cap\lambda)\cup\operatorname{pcf}(\left\\{\lambda\right\\})\cup\lim(A\cap\lambda)$
$\displaystyle=\operatorname{pcf}(A)\cup\lim(A\cap\lambda)$
$\displaystyle\subseteq\operatorname{pcf}(A)\cup\lim(A).$

Now we suppose that $A$ does not have a max. Let $\mu:=\sup(A)$, and fix
$\kappa\in\operatorname{spec}(A)$. We have a few cases on $\kappa$.

First suppose that $\kappa<\mu$. Then by Lemma 3.3,
$\kappa\in\operatorname{spec}(A\cap(\kappa+1))$. Since the order type of
$\bar{A}:=A\cap(\kappa+1)$ is less than the order type of $A$, we apply the
induction hypothesis to conclude that

$\kappa\in\operatorname{spec}(\bar{A})\subseteq\operatorname{pcf}(\bar{A})\cup\lim(\bar{A})\subseteq\operatorname{pcf}(A)\cup\lim(A).$

Now suppose that $\kappa\geq\mu$. If $\kappa=\mu$ (and in particular, $\mu$ is
a regular limit cardinal), then because $A$ is unbounded in $\mu$,
$\kappa=\mu\in\lim(A)$.

The final case is that $\kappa>\mu$ (here $\mu$ may be either regular or
singular). Since $\mu$ is a limit cardinal, our cardinal arithmetic assumption
implies that $\prod A$ has size $\mu^{\mu}=\mu^{+}$. Hence no cardinal greater
than $\mu^{+}$ is in $\operatorname{spec}(A)$ or in $\operatorname{pcf}(A)$.

To finish the proof in this final case, we will argue that
$\mu^{+}\in\operatorname{pcf}(A)$. Towards this end, let $D$ be an ultrafilter
on $A$ which extends the tail filter on $A$. Then $\operatorname{cf}(\prod
A/D)\geq\mu=\sup(A)$. If $\mu$ is singular, then the regular cardinal
$\operatorname{cf}(\prod A/D)$ is greater than $\mu$. On the other hand, if
$\mu$ is regular, then $\mu$ is not a Mahlo cardinal by assumption. This in
turn, with the help of Lemma 3.4, implies that $\operatorname{cf}(\prod
A/D)>\mu$. Thus in either case on $\mu$, $\operatorname{cf}(\prod A/D)>\mu$.
This cofinality must be exactly $\mu^{+}$, however, since $\prod A$ has size
$\mu^{+}$.

For the “consequently” part of the theorem, recall that
$\operatorname{spec}(A)$ consists, by definition, of regular cardinals. Thus
if there are no regular limit cardinals, $\lim(A)$ contains no regular
cardinals, and this implies that
$\operatorname{spec}(A)\subseteq\operatorname{pcf}(A)$. By Lemma 2.12, we
conclude that $\operatorname{pcf}(A)=\operatorname{spec}(A)$. ∎

Thus Question 1.1 has a consistent positive answer.

##  4\. Small Large Cardinals and the Tukey Spectrum

In this section, we prove some results showing the relationship between
certain small large cardinals (Mahlo and weakly compact) and the Tukey
spectrum. The first of this gives a sufficient condition for including a
regular limit cardinal in the Tukey spectrum. After this, we prove Theorem 4.2
which gives a sufficient condition for excluding a cardinal from
$\operatorname{spec}(A)$. After the proof of Theorem 4.2, we comment on
applications.

######  Proposition 4.1.

Suppose that $\kappa$ is a Mahlo cardinal and that $A\subseteq\kappa$ is any
stationary set of regular cardinals. Then $\kappa\in\operatorname{spec}(A)$.

###### Proof.

We show that functions which are constant on a tail witness the result. For
each $\alpha<\kappa$, let $f_{\alpha}$ be the function in $\prod A$ which
takes value $0$ on all $a\in A$ with $a\leq\alpha$, and which takes value
$\alpha$ on all $a\in A$ with $\alpha<a$. We claim that the sequence
$\left<f_{\alpha}:\alpha<\kappa\right>$ enumerates a set which witnesses that
$\kappa\in\operatorname{spec}(A)$.

Towards this end, let $X\in[\kappa]^{\kappa}$, and we will show that
$\left<f_{\alpha}:\alpha\in X\right>$ is unbounded in $(\prod A,<)$. Since $A$
is stationary, we may find some $a\in A\cap\lim(X)$. Then for all $\alpha\in
X\cap a$, $f_{\alpha}(a)=\alpha$. Since $a$ is a limit point of $X$, we have
that $\left\\{f_{\alpha}(a):\alpha\in X\cap a\right\\}$ is cofinal in $a$.
This completes the proof. ∎

The next result shows that we can use weak compactness to exclude a regular
limit $\kappa$ from $\operatorname{spec}(A)$, for certain $A$.

######  Theorem 4.2.

Suppose that $\kappa$ is weakly compact and that $A\subseteq\kappa$ is a non-
stationary set of regular cardinals which is unbounded in $\kappa$. Then
$\kappa\notin\operatorname{spec}(A)$.

Therefore, if $\kappa$ is weakly compact and $A\subseteq\kappa$ is an
unbounded set of regular cardinals, $\kappa\in\operatorname{spec}(A)$ iff $A$
is stationary.

###### Proof.

Let $C\subseteq\kappa$ be a club with $C\cap A=\emptyset$. Enumerate $C$ in
increasing order as $\left<\zeta_{i}:i<\kappa\right>$, where we assume
$\zeta_{0}=0$. Also, enumerate $A$ in increasing order as
$\left<\mu_{i}:i<\kappa\right>$. As in the proof of Lemma 3.4, we have that
for each $i<\kappa$, $\mu_{\zeta_{i}}>\zeta_{i}$.

Next, let $\left<f_{\alpha}:\alpha<\kappa\right>$ be an enumeration of a set
$\cal{F}$ of $\kappa$-many functions in $\prod A$, and we will show that
$\cal{F}$ does not witness that $\kappa\in\operatorname{spec}(A)$. Fix a
$\kappa$-model $M$ (see Definition 2.20) which contains $A$, $C$, and
$\left<f_{\alpha}:\alpha<\kappa\right>$ as elements. By the weak compactness
of $\kappa$, let $\mathcal{U}$ be an $M$-normal ultrafilter on
$\cal{P}(\kappa)\cap M$.

Our strategy is to use $\cal{U}$ to successively freeze out longer and longer
initial segments of many functions on the sequence
$\left<f_{\alpha}:\alpha<\kappa\right>$. We will then bound their tails using
the non-stationarity of $A$.

For each $i<\kappa$, the product

$\prod_{\nu\in[\zeta_{i},\zeta_{i+1})}\mu_{\nu}$

is a member of $M$, and hence a subset of $M$. Since $\kappa$ is strongly
inaccessible, this product has size $<\kappa$. Applying the fact that
$\cal{U}$ is a $\kappa$-complete ultrafilter on $\cal{P}(\kappa)\cap M$, we
may find a function $\varphi_{i}$ in
$\prod_{\nu\in[\zeta_{i},\zeta_{i+1})}\mu_{\nu}$ so that

$Z_{i}:=\left\\{\beta<\kappa:f_{\beta}\upharpoonright\left\\{\mu_{\nu}:\nu\in[\zeta_{i},\zeta_{i+1})\right\\}=\varphi_{i}\right\\}\in\mathcal{U}.$

Next, define an increasing sequence $\left<\beta_{j}:j<\kappa\right>$ below
$\kappa$ so that for all $j<\kappa$,

$\beta_{j}\in\bigcap_{i<j}Z_{i}.$

This also uses the completeness of $\cal{U}$ to see that for each $j<\kappa$,
$\bigcap_{i<j}Z_{i}$ is in $\cal{U}$ and has size $\kappa$.

As a result of this freezing out, we have the following: for a fixed
$i<\kappa$, and all $j>i$, $\beta_{j}\in Z_{i}$. Hence, for all
$\nu\in[\zeta_{i},\zeta_{i+1})$,

$f_{\beta_{j}}(\mu_{\nu})=\varphi_{i}(\mu_{\nu}).$

Now consider an $i<\kappa$ and $\nu\in[\zeta_{i},\zeta_{i+1})$. Let

$R(\nu):=\left\\{f_{\beta_{j}}(\mu_{\nu}):j<\kappa\right\\},$

i.e., all values of all of the $f_{\beta_{j}}$ on the column $\mu_{\nu}\in A$.
By applying the argument in the previous paragraph, we conclude that $R(\nu)$
in fact equals

$\left\\{f_{\beta_{j}}(\mu_{\nu}):j\leq
i\right\\}\cup\left\\{f_{\beta_{j}}(\mu_{\nu}):j>i\right\\}=\left\\{f_{\beta_{j}}(\mu_{\nu}):j\leq
i\right\\}\cup\left\\{\varphi_{i}(\mu_{\nu})\right\\}.$

Now fix an arbitrary $\nu<\kappa$. Since $C$ is a club and $\zeta_{0}=0$,
there is an $i$ so that $\nu\in[\zeta_{i},\zeta_{i+1})$. We claim that
$R(\nu)=\left\\{f_{\beta_{j}}(\mu_{\nu}):j\leq
i\right\\}\cup\left\\{\varphi_{i}(\mu_{\nu})\right\\}$ has size less than
$\mu_{\nu}$. If $i$ is finite, then $R(\nu)$ is finite, and hence has size
smaller than $\mu_{\nu}$ (which is, after all, an infinite cardinal). On the
other hand, if $i$ is infinite, then

$|R(\nu)|\leq|i|\leq i\leq\zeta_{i}<\mu_{\zeta_{i}}\leq\mu_{\nu}.$

Consequently, for each $\nu<\kappa$, $R(\nu)$ is bounded in $\mu_{\nu}$, as
$\mu_{\nu}$ is regular. But by definition of $R(\nu)$, this means that for all
$\nu<\kappa$, $\left\\{f_{\beta_{j}}(\mu_{\nu}):j<\kappa\right\\}$ is bounded
in $\mu_{\nu}$. Thus $\left\\{f_{\beta_{j}}:j<\kappa\right\\}$ enumerates a
set of $\kappa$-many functions from $\cal{F}$ which is bounded in $(\prod
A,<)$. Since $\cal{F}$ was arbitrary, this shows that
$\kappa\notin\operatorname{spec}(A)$.

For the final statement of the theorem, note that if $A\subseteq\kappa$ is
non-stationary, then since $\kappa$ is Mahlo, Proposition 4.1 shows that
$\kappa\notin\operatorname{spec}(A)$. ∎

Note that the converse of the above theorem may fail, since a regular limit
cardinal which is not weakly compact may also fail to be in
$\operatorname{spec}(A)$:

######  Corollary 4.3.

Suppose that $\kappa$ is weakly compact and that $A\subseteq\kappa$ is a non-
stationary set of regular cardinals unbounded in $\kappa$. Let
$\mathbb{P}:=\operatorname{Add}(\omega,\kappa)$, the poset to add
$\kappa$-many Cohen subsets of $\omega$. Then $\mathbb{P}$ forces that
$\kappa\notin\operatorname{spec}(A)$.

###### Proof.

Let $\left<\dot{f}_{\alpha}:\alpha<\kappa\right>$ be a sequence of
$\mathbb{P}$-names for elements of $\prod A$. We will find an
$X\in[\kappa]^{\kappa}$ in $V$ so that $\mathbb{P}$ forces that
$\left<\dot{f}_{\alpha}:\alpha\in X\right>$ is bounded.

Indeed, using the c.c.c. of $\mathbb{P}$, for each $\alpha<\kappa$, we may
find a function $\varphi_{\alpha}\in\prod A$ so that $\mathbb{P}\Vdash(\forall
a\in A)\,[\dot{f}_{\alpha}(a)<\varphi_{\alpha}(a)]$. Namely, let
$\varphi_{\alpha}(a)$ be above the sup of the countably-many $\gamma\in a$ so
that $\gamma$ is forced to be the value of $\dot{f}_{\alpha}(a)$ by some
condition in $\mathbb{P}$.

By the Theorem 4.2, let $X\in[\kappa]^{\kappa}$ so that
$\left<\varphi_{\alpha}:\alpha\in X\right>$ is bounded in the product $(\prod
A,<)$, say with $h$ as a bound. Then $\mathbb{P}$ forces that for each
$\alpha\in X$, $\dot{f}_{\alpha}$ is pointwise below $h$. ∎

We conclude this section with a discussion of a promising suggestion of James
Cummings about separating $\operatorname{pcf}(A)$ and
$\operatorname{spec}(A)$. Given that
$\operatorname{pcf}(A)\subseteq\operatorname{spec}(A)$ always holds, we’d like
to create a forcing extension in which, for some $A$, there is a cardinal
$\kappa\in\operatorname{spec}(A)\setminus\operatorname{pcf}(A)$.

The strategy is to start with a cardinal $\kappa$ which is at least Mahlo.
Then let $A=\left\\{\mu^{+}:\mu<\kappa\right\\}$ and attempt to force the
existence of a set $\cal{F}$ of $\kappa$-many functions in $\prod A$ which
witnesses that $\kappa\in\operatorname{spec}(A)$. This strategy appears
promising due to the next observation.

######  Lemma 4.4.

Suppose that $\kappa$ is strongly inaccessible and that $A\subseteq\kappa$ is
a nonstationary set of regular cardinals unbounded in $\kappa$. Then
$\kappa\notin\operatorname{pcf}(A)$.

###### Proof.

If $D$ is an ultrafilter on $A$ that concentrates on a bounded subset of $A$,
then the strong inaccessibility of $\kappa$ implies that
$\operatorname{cf}(\prod A/D)<\kappa$. On the other hand, if $D$ extends the
tail filter on $\kappa$, then by Lemma 3.4, $\operatorname{cf}(\prod
A/D)>\kappa$. ∎

While the above strategy is natural, problems remain. First, natural Easton-
style forcings to add a witness to $\kappa\in\operatorname{spec}(A)$ seem
either to fail to add such a witness, or seem to change the Mahlo $\kappa$
into a weakly, non-strongly inaccessible cardinal, i.e., they increase the
continuum function below $\kappa$ to take values at or above $\kappa$. Thus
the crucial assumption of Lemma 4.4 fails. Or phrased differently,
ultrafilters which concentrate on bounded subsets may give rise to reduced
products with very high cofinality.

Moreover, Theorem 4.2 provides another obstacle: if a forcing $\mathbb{P}$
places $\kappa$ inside $\operatorname{spec}(A)\setminus\operatorname{pcf}(A)$
for some $A$ which is non-stationary and unbounded in $\kappa$, then one of
two things needs to happen. Either $\kappa$ starts off as non-weakly compact
(and this assumption plays a role in the argument) or $\mathbb{P}$ must ensure
that $\kappa$ loses its weak compactness.

##  5\. The Strong Part of the Tukey Spectrum

In this section we introduce the idea of the strong part of the Tukey
spectrum, and then we will show how this idea can be used in place of scales
to lift the property of not being a Jónsson cardinal. Recall the notation
$\operatorname{ub}(\cal{F})$ from Definition 2.8.

First we make a simple observation about having infinitely-many unbounded
coordinates.

######  Lemma 5.1.

Let $A$ be a set of regular cardinals without a max, and let
$\kappa\in\operatorname{spec}(A)$ with $\kappa\geq\sup(A)$. Let $\cal{F}$ be
any witness that $\kappa\in\operatorname{spec}(A)$. Then for all
$\cal{F}_{0}\in[\cal{F}]^{\kappa}$, $\operatorname{ub}(\cal{F}_{0})$ is
infinite.

###### Proof.

Suppose otherwise, with $\cal{F}_{0}$ as a counterexample. Then since
$\operatorname{ub}(\cal{F}_{0})$ is finite and has a max below $\kappa$,
$\prod\operatorname{ub}(\cal{F}_{0})$ has size below $\kappa$. Let
$\cal{F}_{1}\in[\cal{F}_{0}]^{\kappa}$ so that the function
$f\in\cal{F}_{1}\mapsto f\upharpoonright\prod\operatorname{ub}(\cal{F}_{0})$
is constant, say with value $\bar{f}$. Then we can bound all of $\cal{F}_{1}$
in the entire product $\prod A$ using $\bar{f}$ on the coordinates in
$\operatorname{ub}(\cal{F}_{0})\supseteq\operatorname{ub}(\cal{F}_{1})$. ∎

Of course, $\operatorname{ub}(\cal{F})$ can very well be finite, or even a
singleton, for instance, if $\kappa$ is a member of $A$.

We want to isolate cases in which there are plenty of unbounded coordinates.
This leads to the next definition.

######  Definition 5.2.

Suppose that $A$ is a set of regular cardinals. The _strong part_ of the Tukey
spectrum of $A$, denoted $\operatorname{spec}^{*}(A)$, consists of all regular
$\lambda$ satisfying the following: there is a set $\cal{F}\subseteq\prod A$
of size $\lambda$, so that for every $\cal{F}_{0}\in[\cal{F}]^{\lambda}$,
$\operatorname{ub}(\cal{F}_{0})$ is unbounded in $\sup(A)$.

Thus $\lambda\in\operatorname{spec}^{*}(A)$ iff there is a witness $\cal{F}$
to $\lambda\in\operatorname{spec}(A)$ with the additional property that every
$\lambda$-sized subset has unboundedly-many unbounded coordinates.

Observe that if $A$ is a set of regular cardinals without a max, then
$\operatorname{spec}^{*}(A)\cap\sup(A)=\emptyset$. Indeed, if
$\lambda<\sup(A)$ and $\cal{F}\subseteq\prod A$ has size $\lambda$, then we
can bound $\cal{F}$ on all coordinates in $A\setminus\lambda^{+}$.

Under cardinal arithmetic assumptions, it is easy to see that every
$\lambda\in\operatorname{spec}(A)\setminus\sup(A)$ is in
$\operatorname{spec}^{*}(A)$:

######  Lemma 5.3.

Suppose that $A$ is a set of regular cardinals with no max and that $\sup(A)$
is a strong limit cardinal (regular or singular). Then
$\operatorname{spec}(A)\setminus\sup(A)\subseteq\operatorname{spec}^{*}(A)$.

###### Proof.

Fix $\lambda\in\operatorname{spec}(A)$ at least as large as $\sup(A)$. It
suffices to show that if $\cal{F}$ is any witness that
$\lambda\in\operatorname{spec}(A)$, then $\operatorname{ub}(\cal{F})$ is
unbounded. Fix $\delta\in A$. Then the product $\prod(A\cap(\delta+1))$ has
size below $\sup(A)$. Thus there is $\cal{F}_{0}\in[\cal{F}]^{\lambda}$ so
that the function taking $f\in\cal{F}_{0}$ to
$f\upharpoonright(A\cap(\delta+1))$ is constant on $\cal{F}_{0}$. Since
$\cal{F}$ witnesses that $\lambda\in\operatorname{spec}(A)$, we must have that
$\operatorname{ub}(\cal{F}_{0})$ is non-empty. Let $\delta^{*}$ be the least
element of $\operatorname{ub}(\cal{F}_{0})$, and note that
$\delta^{*}>\delta$, since we froze out the values of the functions in
$\cal{F}_{0}$ on $A\cap(\delta+1)$. ∎

We’d now like to connect the strong part of the Tukey spectrum with PCF
theory. First we recall a few more definitions from PCF theory, beginning with
the following standard version of the notion of a scale:

######  Definition 5.4.

Let $\mu$ be a singular cardinal,
$\left<\mu_{i}:i<\operatorname{cf}(\mu)\right>$ an increasing sequence of
regular cardinals which is cofinal in $\mu$, and $I$ an ideal on
$\left\\{\mu_{i}:i<\operatorname{cf}(\mu)\right\\}$. Let
$\vec{f}=\left<f_{\nu}:\nu<\rho\right>$ be a sequence of functions in
$\prod_{i<\operatorname{cf}(\mu)}\mu_{i}$. The tuple $(\vec{\mu},\vec{f},I)$
is called a _scale of length $\rho$ modulo $I$_ if $\vec{f}$ is increasing and
cofinal in $\prod_{i<\operatorname{cf}(\mu)}\mu_{i}$ modulo $I$.

If $I$ is just the ideal of bounded subsets of
$\left\\{\mu_{i}:i<\operatorname{cf}(\mu)\right\\}$, then we simply say that
$(\vec{\mu},\vec{f})$ is a _scale of length $\rho$_ for $\mu$.

We now connect scales with the strong part of the Tukey spectrum:

######  Lemma 5.5.

Suppose that $(\vec{\mu},\vec{f})$ is a scale of length $\rho$, where $\rho$
is a regular cardinal. Then $\rho$ is in $\operatorname{spec}^{*}(A)$, where
$A=\left\\{\mu_{i}:i<\operatorname{cf}(\mu)\right\\}$.

###### Proof.

This follows since we are working modulo the ideal of bounded subsets of $A$.
Indeed, let $\cal{F}=\left\\{f_{\alpha}:\alpha<\rho\right\\}$, and we will
show that $\cal{F}$ witnesses the result. Note that if $Z\in[\rho]^{\rho}$,
then $\left<f_{\alpha}:\alpha\in Z\right>$ is also a scale. But then
$\operatorname{ub}(\left\\{f_{\alpha}:\alpha\in Z\right\\})$ must be unbounded
in $\sup(A)$, as otherwise we contradict that $\left<f_{\alpha}:\alpha\in
Z\right>$ is cofinal in $\prod A$ modulo the bounded ideal. ∎

Now we examine one way in which $\operatorname{spec}^{*}(A)$ can play a
traditional PCF-theoretic role. We begin with some background: a remarkable
phenomenon in PCF theory is that scales of length $\mu^{+}$ (where $\mu$ is
singular) can be used to “lift” a property which holds at the $\mu_{i}$ to
hold at $\mu^{+}$. For example, Shelah proved that the failure of being a
Jónsson cardinal lifts in this way; we will discuss this in more detail in a
moment. Other examples include Theorem 3.5 of [4] and a theorem of Todorcevic
about lifting the failure of certain square bracket partition relations
([33]).

Here we include a very short review of the notion of a Jónsson cardinal,
referring the reader to [9] for more details.

######  Definition 5.6.

  1. (1)

An _algebra_ is a structure $\cal{A}=\left<A,f_{n}\right>_{n<\omega}$ so that
each $f_{n}$ is a finitary function mapping $A$ to $A$.

  2. (2)

A _Jónsson algebra_ is an algebra without a proper subalgebra of the same
cardinality.

  3. (3)

A cardinal $\lambda$ is said to be a _Jónsson cardinal_ if there does not
exist a Jónsson algebra of cardinality $\lambda$.

Jónsson cardinals can be characterized in terms of a coloring relation.

######  Fact 5.7.

$\lambda$ is a Jónsson cardinal iff for any $F:[\lambda]^{<\omega}\to\lambda$,
there exists an $H\in[\lambda]^{\lambda}$ so that the range of
$F\upharpoonright[H]^{<\omega}$ is a proper subset of $\lambda$.

We use $[\lambda]\to[\lambda]^{<\omega}_{\lambda}$ to denote the coloring
property from the previous fact. We can also characterize this in terms of
elementary submodels. The next item is almost exactly Lemma 5.6 from [9]; we
have added a parameter to the statement, which does not change the proof. In
the statement of the lemma, $<_{\chi}$ denotes a wellorder of $H(\chi)$.

######  Lemma 5.8.

The following two statements are equivalent:

  1. (1)

$\lambda$ is a Jónsson cardinal.

  2. (2)

For every sufficiently large regular $\chi>\lambda$, every cardinal $\kappa$
so that $\kappa^{+}<\lambda$, and every parameter $P\in H(\chi)$, there is an
$M\prec\left<H(\chi),\in,<_{\chi},P\right>$ so that

    1. (a)

$\left\\{\lambda,\kappa\right\\}\in M$;

    2. (b)

$|M\cap\lambda|=\lambda$;

    3. (c)

$\lambda\not\subseteq M$; and

    4. (d)

$\kappa+1\subseteq M$.

Shelah ([27]) has proven the following remarkable theorem:

######  Theorem 5.9.

(Shelah) Suppose that $\mu$ is singular and that $(\vec{\mu},\vec{f})$ is a
scale (modulo the ideal of bounded sets) of length $\mu^{+}$. Additionally,
suppose that each $\mu_{i}$ carries a Jónsson algebra (i.e., is not a Jónsson
cardinal). Then $\mu^{+}$ carries a Jónsson algebra.

Here we show that it suffices to assume that $\mu^{+}$ is in the strong part
of the Tukey spectrum of $A$, provided the order type of $A$ is not too high.

######  Theorem 5.10.

Suppose that $A$ is a set of regular cardinals with
$\operatorname{ot}(A)<\mu:=\sup(A)$ so that every $a\in A$ carries a Jónsson
algebra, and suppose that $\mu^{+}\in\operatorname{spec}^{*}(A)$. Then
$\mu^{+}$ carries a Jónsson algebra.

###### Proof.

We will show that $\mu^{+}$ carries a Jónsson algebra by showing that (2) of
Lemma 5.8 is false.

Fix a large enough regular cardinal $\chi$. Letting $\mu^{+}$,
$|\operatorname{ot}(A)|$, and $A$ play the respective roles of $\lambda$,
$\kappa$, and $P$ in (the negation of) Lemma 5.8(2), fix an arbitrary
$M\prec\left<H(\chi),\in,<_{\chi},A\right>$ so that $\mu^{+}\in M$,
$|M\cap\mu^{+}|=\mu^{+}$, and $|\operatorname{ot}(A)|+1\subseteq M$. We will
show that $\mu^{+}\subseteq M$. Observe that $A\subseteq M$, since $M$
contains a bijection from $|\operatorname{ot}(A)|$ onto $A$, and since
$|\operatorname{ot}(A)|+1\subseteq M$.

Applying the elementarity of $M$, we may find a set $\cal{F}\subseteq\prod A$
of functions witnessing $\mu^{+}\in\operatorname{spec}^{*}(A)$ with
$\cal{F}\in M$. Using $\cal{F}$, we will show that there are unboundedly-many
$a\in A$ so that $|M\cap a|=a$. The upshot of this is that for each such $a$,
since $a$ is not a Jónsson cardinal, $a\subseteq M$. Since there are
unboundedly-many such $a$, we conclude that $\sup(A)=\mu\subseteq M$. And
finally, since $|M\cap\mu^{+}|=\mu^{+}$, we can conclude that
$\mu^{+}\subseteq M$.

To show the existence of unboundedly-many such $a$, let
$\cal{F}_{M}:=\cal{F}\cap M$. Since $|M\cap\mu^{+}|=\mu^{+}$, we know that
$\cal{F}_{M}\in[\cal{F}]^{\mu^{+}}$. Since
$\mu^{+}\in\operatorname{spec}^{*}(A)$, the set
$\operatorname{ub}(\cal{F}_{M})$ is unbounded in $A$.

Now let $a\in\operatorname{ub}(\cal{F}_{M})$. Then for all $f\in\cal{F}_{M}$,
$f(a)\in M$, since $f$ and $a$ are each members of $M$. Since
$\left\\{f(a):f\in\cal{F}_{M}\right\\}$ is unbounded in $a$ (by definition of
$a$ being an unbounded coordinate) and a subset of $M$, we conclude that
$M\cap a$ has size $a$. This completes the proof. ∎

We close this section by providing a bound on the strong part of the Tukey
spectrum. First note that it follows almost immediately from the definitions
that

$\sup(\operatorname{spec}(A))\leq\operatorname{cf}(\prod A,<).$

Now let $J_{\text{bd}}$ denote the ideal of bounded sets on $A$. Note that
$\prod A/J_{\text{bd}}$ does have a cofinality, but it needn’t have a _true_
cofinality (i.e., a linearly-ordered, cofinal subset).

######  Proposition 5.11.

Let $A$ be a set of regular cardinals, and let
$\lambda\in\operatorname{spec}(A)$. Then either
$\lambda\leq\operatorname{cf}(\prod A/J_{\text{bd}})$ or
$\lambda\in\operatorname{spec}(\bar{A})$ for some proper initial segment
$\bar{A}$ of $A$.

In particular (see Definition 5.2), if $\lambda$ is in
$\operatorname{spec}^{*}(A)$, then $\lambda\leq\operatorname{cf}(\prod
A/J_{\text{bd}})$.

###### Proof.

Let $\mu:=\operatorname{cf}(\prod A/J_{\text{bd}})$, and let
$\left<f_{\alpha}:\alpha<\mu\right>$ enumerate a set of functions which is
cofinal (but not necessarily increasing) in $\prod A/J_{\text{bd}}$. Suppose
that $\lambda>\mu$. Let $\left<h_{\xi}:\xi<\lambda\right>$ enumerate a set
$\cal{F}$ of $\lambda$-many functions witnessing that
$\lambda\in\operatorname{spec}(A)$. Since $\lambda$ is regular and
$\lambda>\mu$, fix $\alpha<\mu$ and $X\in[\lambda]^{\lambda}$ so that for all
$\xi\in X$, $h_{\xi}<_{J_{\text{bd}}}f_{\alpha}$. Now freeze out the tail: let
$Y\in[X]^{\lambda}$ and $\bar{a}\in A$ so that for all $\xi\in Y$ and all
$a\in A\setminus\bar{A}$,

$h_{\xi}(a)<f_{\alpha}(a).$

Thus $\cal{F}_{0}:=\left\\{h_{\xi}:\xi\in Y\right\\}$ is bounded on every
coordinate in $A\setminus\bar{a}$. Since $\cal{F}_{0}\in[\cal{F}]^{\lambda}$,
$\cal{F}_{0}$ is unbounded in $(\prod A,<)$. From this, one can argue that
$\left\\{f_{\xi}\upharpoonright(A\cap\bar{a}):\xi\in Y\right\\}$ has size
$\lambda$ and witnesses that $\lambda\in\operatorname{spec}(A\cap\bar{a})$.

For the “in particular” part of the proposition, note that if
$\lambda>\operatorname{cf}(\prod A/J_{\text{bd}})$, then the previous argument
shows that there is an $\cal{F}_{0}\in[\cal{F}]^{\lambda}$ so that
$\operatorname{ub}(\cal{F}_{0})$ is bounded in $A$. ∎

##  6\. Questions

Here we record a few questions which we find interesting. The first question
restates Question 1.1, the main one driving this line of research:

######  Question 6.1.

Does $\mathsf{ZFC}$ prove that for all sets $A$ of regular cardinals,
$\operatorname{pcf}(A)=\operatorname{spec}(A)$?

A restricted version of Question 6.1, to be read in light of Theorem 3.2, is
this:

######  Question 6.2.

Does $\mathsf{ZFC}$ prove that for all _progressive_ sets $A$ of regular
cardinals, $\operatorname{pcf}(A)=\operatorname{spec}(A)$?

One can also ask about the relationship between $\operatorname{spec}^{*}(A)$
and $\operatorname{pcf}(A)$, as in the next two questions:

######  Question 6.3.

Does $\mathsf{ZFC}$ prove that
$\operatorname{spec}^{*}(A)\subseteq\operatorname{pcf}(A)$?

######  Question 6.4.

Does $\mathsf{ZFC}$ prove that if $A$ is a set of regular cardinals without a
max, then
$\operatorname{spec}(A)\setminus\sup(A)\subseteq\operatorname{spec}^{*}(A)$?

The following question should be read in light of Theorem 3.5:

######  Question 6.5.

Does $\mathsf{ZFC}$ prove that for all sets $A$ of regular cardinals,

$\operatorname{spec}(A)\subseteq\operatorname{pcf}(A)\cup\lim(\operatorname{pcf}(A))?$

The next question connects to Theorem 4.2 and Corollary 4.3:

######  Question 6.6.

Is a weakly compact necessary to get a model in which $\kappa$ is a regular
limit, $A\subseteq\kappa$ is unbounded and non-stationary, and
$\kappa\notin\operatorname{spec}(A)$?

Theorem 5.10 showed that the strong part of the Tukey spectrum can be used in
place of PCF-theoretic scales to lift the property of not being a Jónsson
cardinal. Where else, if at all, can $\operatorname{spec}^{*}(A)$ be used in
this way? In particular, we ask whether the strong part of the Tukey spectrum
is enough to generalize a classic result of Todorcevic ([33]; see the
treatments in [3] and [9]).

######  Question 6.7.

Suppose that $A$ is a set of regular cardinals cofinal in a singular $\mu$ so
that every $\kappa\in A$ fails to satisfy $\kappa\to[\kappa]^{2}_{\kappa}$.
Suppose that $\mu^{+}\in\operatorname{spec}^{*}(A)$. Does this imply that
$\mu^{+}$ fails to satisfy $\mu^{+}\to[\mu^{+}]^{2}_{\mu^{+}}$?

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