# LST Numbers for $Q^{\text{e.c.}}$ and $I$ style quantifiers

Christopher Turner  School of Mathematics, University of Bristol, Fry
Building. Woodland Road, Bristol, BS8 1UG, UK
<EMAIL_ADDRESS>
###### Abstract.

We introduce two schemes of quantifiers analogous to $I$ and
$Q^{\text{e.c.}}$, which tell us about regular cardinals of small Cantor-
Bendixson rank. We examine how the Löwenheim-Skolem-Tarski numbers of these
quantifiers interact with one another, and with those of $I$ and
$Q^{\text{e.c.}}$. We then find the exact lower bound for each of the
$\operatorname{LST}$ numbers, assuming the consistency of supercompacts.

###### Key words and phrases:

Löwenheim-Skolem-Tarski numbers, Härtig quantifier, equal cofinality
quantifier

This paper is adapted from part of the author’s Ph.D. thesis, and he is
grateful to his supervisor Philip Welch for his invaluable guidance. He would
also like to thank Andrew Brooke-Taylor and Jonathan Osinski for pointing out
various glitches in a previous version of the paper. The author was supported
in his research by EPSRC grant EP/R513179/1 and a scholarship from the
Heilbronn Institute, and by an internal grant from the University of Bristol.
For the purpose of open access, the author has applied a ‘Creative Commons
Attribution’ (CC BY) public copyright licence to any Author Accepted
Manuscript (AAM) version arising from this submission.

##  1\. Introduction

The Löwenheim-Skolem theorem famously says that for any first-order language
$\mathcal{L}$, any first order $\mathcal{L}$ structure contains an elementary
substructure of size less than $\max(\lvert\mathcal{L}\rvert,\omega_{1})$. The
concept of the Löwenheim-Skolem-Tarski number generalises this to simple
second order logics. The $\operatorname{LST}$ number of a second-order logic
is the smallest cardinal $\kappa$ such that every structure contains a
substructure of size less than $\kappa$.

In [4], Magidor and Väänänen investigate $\operatorname{LST}$ numbers for two
second order quantifiers: the Härtig quantifier $I$ and the equal cofinality
quantifier $Q^{\text{e.c.}}$. Roughly speaking, we can think of $I$ as telling
us about the class $\operatorname{Card}$ of all cardinals (or, equivalently,
about the class of all successor cardinals), while $Q^{\text{e.c.}}$ tells us
about the class $\operatorname{Reg}$ of all (infinite) regular cardinals. In
both cases, the lower bound turns out to be precisely the smallest sort of
large cardinal which is not identified by the quantifier:

######  Theorem 1.1 (Magidor,Väänänen).

[4, 7, 20 & 21] If it exists, then $\operatorname{LST}(I)$ is at least the
first inaccessible cardinal. Similarly,
$\operatorname{LST}(I,Q^{\text{e.c.}})$ is at least the first Mahlo cardinal,
if it exists. Moreover, if it is consistent that a supercompact cardinal
exists then it is also consistent that either one of these
$\operatorname{LST}$ numbers is exactly equal to the bound given.

In this paper, we introduce a suite of new intermediate results, establishing
that this rule – that the lower bound of $\operatorname{LST}(Q)$ is the least
cardinal not identified by a quantifier $Q$ – is true more generally in this
region. Specifically, we shall look at quantifiers which tell us about simple
fragments of $\operatorname{Reg}$ in ways similar to how $I$ tells us about
$\operatorname{Card}$ and $Q^{\text{e.c.}}$ tells us about
$\operatorname{Reg}$.

The fragments of $\operatorname{Reg}$ we are interested in are those of low
Cantor-Bendixson rank:

######  Definition 1.2.

For $\alpha\in\operatorname{On}$, we recursively define the classes
$\operatorname{Reg}_{\alpha}$, $\operatorname{Reg}_{\geq\alpha}$ and
$\operatorname{Reg}_{<\alpha}$ as follows:

  * •

$\operatorname{Reg}_{<\alpha}:=\bigcup_{\beta<\alpha}\operatorname{Reg}_{\beta}$

  * •

$\operatorname{Reg}_{\geq\alpha}=\operatorname{Reg}\setminus\operatorname{Reg}_{<\alpha}$

  * •

$\operatorname{Reg}_{\alpha}$ is the class of successors of the club generated
by $\operatorname{Reg}_{\geq\alpha}$

So $\operatorname{Reg}_{0}$ is the class of successor cardinals,
$\operatorname{Reg}_{1}$ is the class of simple inaccessibles, and so on.

The Cantor-Bendixson hierarchy is the natural partition to use if we’re
looking for something intermediate between $\operatorname{Card}$ and
$\operatorname{Reg}$. The class $\operatorname{Card}$ is simply
$\operatorname{Reg}_{0}$ together with its class of limits, and thus
$\operatorname{Card}$ and $\operatorname{Reg}_{0}$ are mutually definable.
Meanwhile the class
${\operatorname{Reg}_{<\infty}:=\bigcup_{\alpha\in\operatorname{On}}\operatorname{Reg}_{<\alpha}}$
is mutually definable with $\operatorname{Reg}$. (It’s not quite the same as
$\operatorname{Reg}$, because $\omega$ is not in any
$\operatorname{Reg}_{\alpha}$. But of course, we can easily handle this by
defining $\omega$ directly.) And each individual class
$\operatorname{Reg}_{\alpha}$ looks a lot like $\operatorname{Reg}_{0}$, the
class defining $\operatorname{Card}$ – for any $\alpha$,
$\operatorname{Reg}_{\alpha}$ is a discrete sequence of $V$-cardinals.

We will introduce two new natural classes of quantifiers, $Q^{\alpha}$ and
$R^{\alpha}$ (for arbitrary $\alpha$), which each tell us about
$\operatorname{Reg}_{<\alpha}$. We will give an account of how the
$\operatorname{LST}$ numbers of these quantifiers relate to one another,
showing they are closely linked with each other and with $I$ and
$Q^{\text{e.c.}}$. We will then show that $\min(\operatorname{Reg}_{\alpha})$
is a lower bound for both $\operatorname{LST}(I,Q^{\alpha})$ and
$\operatorname{LST}(I,R^{\alpha})$, and show that – assuming the consistency
of supercompacts – there exist universes in which the $\operatorname{LST}$
numbers are both equal to this lower bound.

##  2\. Preliminaries

We shall begin with some brief clarificatory definitions of the standard
concepts we referred to in the introduction.

######  Definition 2.1.

A quantifier is a formal symbol $Q$, which is equipped with an arity
$n\in\omega$ and a number $m\in\omega$, called the number of variables it
quantifies over.

######  Definition 2.2.

Let $\mathcal{L}$ be a first order language, and let $Q$ be an $n$-ary
quantifier over $m$ variables. The language $\mathcal{L}\cup\\{Q\\}$ consists
of the following formulas:

  * •

Atomic formulas of $\mathcal{L}$

  * •

$\neg\varphi$ and $\varphi\implies\psi$, whenever $\varphi$ and $\psi$ are
formulas

  * •

$\forall v(\varphi)$ where $v$ is any variable and $\varphi$ is a formula

  * •

$Qv_{1},\ldots,v_{m}(\varphi_{1},\ldots,\varphi_{n})$ where
$v_{1},\ldots,v_{m}$ are any variables (not necessarily distinct) and
$\varphi_{1},\ldots,\varphi_{n}$ are formulas

If $Q_{1},\ldots,Q_{k}$ are quantifiers then
$\mathcal{L}\cup\\{Q_{1},\ldots,Q_{n}\\}$ is defined similarly.

We are specifically interested in quantifiers like $\forall$, which have a
single canonical interpretation.

######  Definition 2.3.

A predefined quantifier is an $n$-ary quantifier symbol $Q$ in $m$ variables
(for some $n$ and $m$), together with a fixed rule on how to interpret
$Qv_{1},\ldots,v_{m}(\varphi_{1},\ldots,\varphi_{n})$, for any second-order
formulas $\varphi_{1},\ldots,\varphi_{n}$ in any language $\mathcal{L}$
including $Q$, over any $\mathcal{L}$ structure $\mathcal{A}$ and assignment
of variable symbols to elements of $\mathcal{A}$. Usually, this guideline will
be expressed in a recursive manner, with the interpretation of
$Qv_{1},\ldots,v_{m}(\varphi_{1},\ldots,\varphi_{n})$ dependent on the
interpretations of $\varphi_{1},\ldots,\varphi_{n}$.

The details of how we formalise this concept are left to the reader: there are
several different options.

A trivial example might make the idea here clearer.

######  Example 1.

$\forall$ is a predefined $1$-ary quantifier, which quantifies over one
variable. $\forall v_{1}\varphi$ is true (for a given
$\varphi=\varphi(v_{1})$, $\mathcal{A}$ and assignment) if and only if for all
$x\in\mathcal{A}$, $\varphi(x)$ is interpreted as true. Similarly, $\exists$
is also a predefined $1$-ary quantifier, which again quantifies over one
variable.

The two well-known quantifiers mentioned in the introduction also have this
predefined property.

######  Definition 2.4.

The Härtig quantifier $I$ is $2$-ary, and quantifies over two variables.
$Iv_{1},v_{2}(\varphi,\psi)$ is interpreted as true (in a structure
$\mathcal{A}$ over which we can interpret $\varphi$ and $\psi$, and have
defined an assignment for variable symbols) if and only if the two sets

$X:=\\{x\in A:\mathcal{A}\vDash\varphi[x/v_{1}]\\}$

and

$Y:=\\{x\in A:\mathcal{A}\vDash\psi[x/v_{2}]\\}$

have the same cardinality in $V$.

######  Definition 2.5.

The Equal Cofinality quantifier $Q^{\text{e.c.}}$ is $2$-ary, and quantifies
over $4$ variables. $Q^{\text{e.c.}}v_{1},\ldots,v_{4}(\varphi,\psi)$ is
interpreted as true (over $\mathcal{A}$, etc.) if the two sets

$X:=\\{(x,y):\mathcal{A}\vDash\varphi[x/v_{1},y/v_{2}]\\}$

and

$Y:=\\{(x,y):\mathcal{A}\vDash\psi[x/v_{3},y/v_{4}]\\}$

are both linear orders, and have the same cofinality.

Throughout this paper, we will simplify notation by committing a slight abuse:
if the free variables of a formula
$Qv_{1},\ldots,v_{k}\varphi_{1},\ldots,\varphi_{n}$ are, say, some subset of
$v_{m+1},\ldots,v_{m+k}$ and $y_{1},\ldots,y_{k}$ are a tuple of elements of
some structure $\mathcal{A}$, we will write

$\mathcal{A}\vDash Q(\varphi_{1},\ldots,\varphi_{n},y_{1},\ldots,y_{k})$

to denote the statement

> $Qv_{1},\ldots,v_{m}(\varphi_{1},\ldots,\varphi_{n})$ is interpreted as true
> over $\mathcal{A}$ with the assignment of $y_{i}$ to $v_{m+i}$ for $0<i\leq
> k$.

We will also use the same notation if instead of $v_{1},\ldots,v_{m}$ and
$v_{m+1},\ldots,v_{m+k}$, we actually want to quantify over some (otherwise
unused) variables $v_{i_{1}},\ldots,v_{i_{m}}$ and assign the variables
$v_{j_{1}},\ldots,v_{j_{k}}$.

The final standard concept to define is the $\operatorname{LST}$ number.

######  Definition 2.6.

Let $Q_{0},\ldots,Q_{n}$ be predefined quantifier symbols of second order
logic. The Löwenheim-Skolem-Tarski number
$\operatorname{LST}(Q_{0},\ldots,Q_{n})$ of $Q_{0},\ldots,Q_{n}$ is the least
infinite cardinal $\kappa$ such that the following holds: For every first
order language $\mathcal{L}$ of cardinality less than $\kappa$, for every
$\mathcal{L}$ structure $\mathcal{A}$, there is an
$\mathcal{L}\cup\\{Q_{0},\ldots,Q_{n}\\}$ elementary substructure
$\mathcal{B}\leq\mathcal{A}$ of size less than $\kappa$.

If no such $\kappa$ exists, then we say that
$\operatorname{LST}(Q_{0},\ldots,Q_{n})=\infty$ or that it does not exist.

##  3\. The new quantifiers

As we discussed in the introduction, in this paper we shall look at
$\operatorname{LST}$ numbers for quantifiers analogous to $I$ and
$Q^{\text{e.c.}}$, which tell us about $\operatorname{Reg}_{<\alpha}$. It is
now time to define the quantifiers we’re going to be looking at. The first is
similar to $Q^{\text{e.c.}}$. However, in order to restrict it to
$\operatorname{Reg}_{<\alpha}$ in a natural way, we add an extra requirement:
as well as requiring that the two linear orders have the same cofinality, we
ask for an auxiliary set deciding the Cantor-Bendixson rank of that
cofinality.

######  Definition 3.1.

Let $\alpha\leq\operatorname{On}$. The quantifier $Q^{\alpha}$ is $3$-ary, and
quantifies over six variables.
$Q^{\alpha}v_{1},\ldots,v_{6}(\varphi,\psi,\chi)$ is interpreted as true (over
$\mathcal{A}$, etc.) if the three sets

$X:=\\{(x,y):\mathcal{A}\vDash\varphi[x/v_{1},y/v_{2}]\\}$
$Y:=\\{(x,y):\mathcal{A}\vDash\psi[x/v_{3},y/v_{4}]\\}$
$Z:=\\{(x,y):\mathcal{A}\vDash\chi[x/v_{5},y/v_{6}]\\}$

satisfy the following four conditions.

  1. (1)

$X$ and $Y$ are both linear orders with the same $V$ cofinality

  2. (2)

$Z$ is a well order

  3. (3)

The order type of $Z$ is less than $\alpha$

  4. (4)

The equal cofinality of $X$ and $Y$ is an element of
$\operatorname{Reg}_{\text{o.t.}(Z)}$

These extra requirements on $Z$ mean that for
$\beta\leq\alpha\leq\operatorname{On}$, we can naturally define $Q^{\beta}$
from $Q^{\alpha}$ and $\beta$. We simply have to restrict $Q^{\alpha}$ to
those cases where the third argument of the quantifier defines a set with
order type less than $\beta$. This is something which can be defined
universally over any $\mathcal{L}$ structure $\mathcal{A}$ which knows even
basic set theory and contains all the ordinals below $\beta$. By converse, if
we had attempted to define a quantifier which simply restricted the domain of
$Q^{\text{e.c.}}$ without using the auxiliary set, then we would have to
produce a complicated definition – inside $\mathcal{A}$ – of the Cantor-
Bendixson rank of a definable class.

Note that $Q^{\operatorname{On}}$ is effectively $Q^{\text{e.c.}}$, but with
this auxiliary set requirement.111There’s one small difference:
$Q^{\operatorname{On}}$ never tells us that two linear orders have the same
cofinality if they happen to both have cofinality $\omega$. But in a
moderately rich $\mathcal{L}$ structure, that’s not a problem: we can discover
whether the linear order has cofinality $\omega$ by comparing it to itself
using $Q^{\operatorname{On}}$ with different auxiliary sets $Z$. If
$Q^{\operatorname{On}}$ is never true, whatever auxiliary set we use, then we
know that the order has cofinality $\omega$. And of course any two orders of
cofinality $\omega$ have the same cofinality, so we can define the “missing”
bit of $Q^{\text{e.c.}}$. The details are left to the reader. Fortunately, it
turns out that the extra requirement is unimportant when studying
$\operatorname{LST}$ numbers:

######  Theorem 3.2.

$\operatorname{LST}(I,Q^{\operatorname{On}})=\operatorname{LST}(I,Q^{\text{e.c.}})$

Thus, we can reasonably say that (for the purposes of $\operatorname{LST}$
numbers) $Q^{\operatorname{On}}$ is just $Q^{\text{e.c.}}$ expressed in a more
convenient form. And therefore, that for $\alpha<\operatorname{On}$, the
quantifier $Q^{\alpha}$ that we have introduced is a fragment of
$Q^{\text{e.c.}}$.

Before we get into proving the theorem, we finish off our definitions by
introducing the second class of quantifiers we’re going to be looking at.
These tell us only about cardinalities (not cofinalities), and specifically
whether a given cardinality is regular. It uses the same auxiliary set
technique as $Q^{\alpha}$.

######  Definition 3.3.

Let $\alpha\leq\operatorname{On}$. The quantifier $R^{\alpha}$ is $2$-ary, and
quantifies over three variables.
$R^{\alpha}v_{1},\ldots,v_{3}(\varphi,\psi,\chi)$ is interpreted as true (over
$\mathcal{A}$, etc.) if the two sets

$X:=\\{x:\mathcal{A}\vDash\varphi[x/v_{1}]\\}$
$Z:=\\{(x,y):\mathcal{A}\vDash\chi[x/v_{2},y/v_{3}]\\}$

satisfy the following three conditions.

  1. (1)

$Z$ is a well order

  2. (2)

The order type of $Z$ is less than $\alpha$

  3. (3)

$\lvert X\rvert\in\operatorname{Reg}_{\text{o.t.}(Z)}$

Morally, $R^{\alpha}$ is true of $X$ if $\lvert
X\rvert\in\operatorname{Reg}_{<\alpha}$. We are interested in looking at the
$\operatorname{LST}$ numbers of these predicates together with $I$. (Examining
$\operatorname{LST}$ numbers of this kind of symbol in contexts where $I$ is
not available is notoriously difficult: little is known even about
$\operatorname{LST}(Q^{\text{e.c.}})$.)

##  4\. Inequalities

The theorem we stated above is only one of several inequalities we can prove
in $\operatorname{\mathsf{ZFC}}$.

######  Theorem 4.1.

In the following statements, if an $\operatorname{LST}$ number does not exist
we consider it to be equal to $\infty$. For
$\alpha,\beta\in\operatorname{On}\cup\\{\operatorname{On}\\}$ and
$\beta<\alpha$:

  1. (1)

$\operatorname{LST}(I,Q^{\text{e.c.}})=\operatorname{LST}(I,Q^{\operatorname{On}})$

  2. (2)

$\operatorname{LST}(I,Q^{\alpha})\geq\operatorname{LST}(I,R^{\alpha})$

  3. (3)

Either $\operatorname{LST}(I,Q^{\alpha})\geq\operatorname{LST}(I,Q^{\beta})$,
or both $\beta\geq\operatorname{LST}(I,Q^{\alpha})$ and
$\operatorname{LST}(I,Q^{\beta})>\min(\operatorname{Reg}_{\beta})$.

  4. (4)

Either $\operatorname{LST}(I,R^{\alpha})\geq\operatorname{LST}(I,R^{\beta})$,
or both $\beta\geq\operatorname{LST}(I,R^{\alpha})$ and
$\operatorname{LST}(I,R^{\beta})>\min(\operatorname{Reg}_{\beta})$.

  5. (5)

$\operatorname{LST}(I,R^{1})=\operatorname{LST}(I,R^{0})=\operatorname{LST}(I)$.

The second case in 3 and 4 is always false unless $\beta$ is above certain
large cardinals. To see this, we will need certain other results, so that
result is deferred until 4.5.

###### Proof.

The idea of all these proofs is the same. Although the predicates are not
definable from one another in arbitrary structures, we show that they can be
defined from each other in a suitable extension of any $\mathcal{L}$
structure.

Throughout these proofs, we shall use the following two formulas (in any
language containing the symbols $\in$ and $=$), for use with $I,Q$ and $R$:

$\varphi_{\in}(v_{0},v_{1}):=v_{0}\in v_{1}$
$\varphi_{\in}^{\prime}(v_{0},v_{1},v_{2}):=v_{0}\in v_{2}\wedge v_{1}\in
v_{2}\wedge v_{0}\in x_{1}$ $\varphi_{\forall}(v_{0}):=v_{0}=v_{0}$
$\varphi_{\operatorname{On}}(v_{0}):=``v_{0}\text{ is transitive and well-
ordered by}\in"$

These are very simple formulas, of course, but having names for them will make
things clearer when we want to use them in formulas with quantifiers. We will
always be quantifying over $v_{0}$, or $v_{0}$ and $v_{1}$ in the case of
$\varphi_{\in}^{\prime}$.

Note that $\varphi_{\operatorname{On}}$ is saying that $v_{0}$ is an ordinal
of the structure it is interpreted over, not an ordinal of $V$. We will only
use it in situations where we already know these two notions to agree before
using it.

###### Proof (1).

First, we shall show that
$\operatorname{LST}(I,Q^{\text{e.c.}})\geq\operatorname{LST}(I,Q^{\operatorname{On}})$.
Obviously, this is trivial if the former is $\infty$. So suppose that
$\operatorname{LST}(I,Q^{\text{e.c.}})=\kappa<\infty$. Let $\mathcal{L}$ be a
first order language of cardinality less than $\kappa$, and let $\mathcal{A}$
be an $\mathcal{L}$ structure of cardinality $\lambda\geq\kappa$. Without loss
of generality, assume that $\mathcal{A}$ does not contain any ordinals, but is
an element of $H_{\lambda^{+}}$, and that $\mathcal{L}$ does not include the
symbol $\in$.

Let $P$ be a new 1-ary first order predicate symbol, and let
$\mathcal{L}^{\prime}=\mathcal{L}\sqcup\\{\in,P\\}$. Let
$\mathcal{A}^{\prime}$ be the $\mathcal{L}^{\prime}$ structure with domain
$H_{\lambda^{+}}$, with $\in$ interpreted in the usual way, the predicate $P$
interpreted as true of elements of $\mathcal{A}$, and all the symbols of
$\mathcal{L}$ interpreted in line with $\mathcal{A}$ on the elements of
$\mathcal{A}$, and in some trivial way everywhere else.

Let $\mathcal{B}^{\prime}$ be an
$\mathcal{L}^{\prime}\cup\\{I,Q^{\text{e.c.}}\\}$ elementary substructure of
$\mathcal{A}^{\prime}$ of cardinality less than $\kappa$ (which we are
assuming exists), and let $\mathcal{C}^{\prime}$ be its transitive collapse.
Let $j:\mathcal{C}^{\prime}\rightarrow\mathcal{A}^{\prime}$ be the
$\mathcal{L}^{\prime}\cup\\{I,Q^{\text{e.c.}}\\}$ elementary embedding. Let

$\mathcal{B}=\\{x\in\mathcal{B}^{\prime}:P(x)\\}=\mathcal{B}^{\prime}\cap\mathcal{A}$

viewed as an $\mathcal{L}$ structure, and let $\mathcal{C}$ be the analogous
substructure of $\mathcal{C}^{\prime}$.

We shall show that $j$ is an
$\mathcal{L}^{\prime}\cup\\{I,Q^{\operatorname{On}}\\}$ elementary embedding
of $\mathcal{C}^{\prime}$ into $\mathcal{A}^{\prime}$. From this it
immediately follows that $\mathcal{B}^{\prime}=\operatorname{ran}(j)$ is an
$\mathcal{L}^{\prime}\cup\\{I,Q^{\operatorname{On}}\\}$ elementary
substructure of $\mathcal{A}^{\prime}$, and it hence that
$\mathcal{B}=\mathcal{B}^{\prime}\cap P$ is an
$\mathcal{L}\cup\\{I,Q^{\operatorname{On}}\\}$ elementary substructure of
$\mathcal{A}=\mathcal{A}^{\prime}\cap P$.

We show $\mathcal{L}^{\prime}\cup\\{I,Q^{\operatorname{On}}\\}$ elementarity
of $j$ in several stages.

######  Claim 4.2.

  1. (1)

Both $\mathcal{A}^{\prime}$ and $\mathcal{C}^{\prime}$ believe a set is an
ordinal if and only if it is an ordinal of $V$.

  2. (2)

Both $\mathcal{A}^{\prime}$ and $\mathcal{C}^{\prime}$ believe an ordinal is a
cardinal if and only if it is a cardinal of $V$.

  3. (3)

Both $\mathcal{A}^{\prime}$ and $\mathcal{C}^{\prime}$ believe a cardinal is
regular if and only if it is a regular cardinal of $V$.

  4. (4)

There is an $\mathcal{L}^{\prime}\cup\\{I,Q^{\text{e.c.}}\\}$ formula
$\Phi_{R}(x,y)$ which both $\mathcal{A}^{\prime}$ and $\mathcal{C}^{\prime}$
interpret as true of $(x,y)$ if and only if $y$ is an ordinal and
$x\in\operatorname{Reg}_{y}$.

  5. (5)

Suppose that $\varphi,\psi,\chi$ are $\mathcal{L}\cup\\{I,Q^{\text{e.c.}}\\}$
formulas, which are elementary between $\mathcal{A}^{\prime}$ and
$\mathcal{C}^{\prime}$. Suppose further that
$\varphi(v_{1},\ldots,v_{k})\implies P(v_{1})\wedge P(v_{2})$, and similarly
for $\psi$ and $\chi$. So $\varphi$, $\psi$ and $\chi$, together with any
assignment $\vec{y}$, define subsets of $\mathcal{A}^{2}$ and
$\mathcal{C}^{2}$ in $\mathcal{A}^{\prime}$ and $\mathcal{C}^{\prime}$
respectively.

Then there is a formula $\Psi(\varphi,\psi,\chi,\vec{v})$ such that for any
assignment $\vec{y}$ of variables to elements of $\mathcal{A}$ (resp. of
$\mathcal{C}$), $\mathcal{A}^{\prime}$ (resp. $\mathcal{C}^{\prime}$)
interprets $\Psi(\varphi,\psi,\chi,\vec{y})$ as true if and only if
$\mathcal{A}$ (resp. $\mathcal{C}$) believes
$Q^{\operatorname{On}}(\varphi,\psi,\chi,\vec{y})$.

###### Proof (Claim, 1).

Trivial, since $\mathcal{A}^{\prime}$ and $\mathcal{C}^{\prime}$ are
transitive. ∎

###### Proof (Claim, 2).

First, notice that since $\mathcal{A}^{\prime}$ has domain $H_{\lambda^{+}}$,
it is true that the cardinals of $\mathcal{A}^{\prime}$ are precisely the
cardinals of $V$ which are in $\mathcal{A}^{\prime}$.

In both $\mathcal{A}^{\prime}$ and $\mathcal{C}^{\prime}$, the
$\mathcal{L}^{\prime}\cup{I}$ formula

$\Phi_{C}(x):=x\in\operatorname{On}\wedge\forall y\in x\neg
I(\varphi_{\in},\varphi_{\in},x,y)$

is true of a set $x$ if and only if $x$ is a cardinal of $V$. So we have a way
to “test” whether an ordinal is a cardinal. But then as we saw a moment ago,

$\mathcal{A}^{\prime}\vDash\forall x\,x\in\operatorname{Card}\iff\Phi_{C}(x)$

As usual, “$x\in\operatorname{Card}$” is shorthand for the first order formula
in the language of set theory which says that $x$ is a cardinal.

By elementarity, then,

$\mathcal{C}^{\prime}\vDash\forall x\,x\in\operatorname{Card}\iff\Phi_{C}(x)$

Hence, the cardinals of $\mathcal{C}^{\prime}$ are precisely the cardinals of
$V$ that are in $\mathcal{C}^{\prime}$. ∎

###### Proof (Claim, 3).

Similar to the previous case, but instead of $\Phi_{C}$ we use the formula:

$\Phi_{R}^{\prime}(x):=\Phi_{C}(x)\wedge\forall y\in x\neg
Q^{\text{e.c.}}(\varphi_{\in}^{\prime},\varphi_{\in}^{\prime},x,y)$

which is true in both $\mathcal{A}^{\prime}$ and $\mathcal{C}^{\prime}$ if and
only if $x$ is a $V$-regular cardinal. We know that $\mathcal{A}^{\prime}$ is
correct about the regular cardinals because its domain is $H_{\lambda^{+}}$,
so by elementarity $\mathcal{C}^{\prime}$ is as well. ∎

###### Proof (Claim, 4).

Note that $\mathcal{A}^{\prime}$ believes in recursion. Since
$\operatorname{Reg}^{\mathcal{A}^{\prime}}$ is a definable subclass of
$\mathcal{A}^{\prime}$, we know that $\mathcal{A}^{\prime}$ can recursively
calculate the Cantor-Bendixson ranks of elements of
$\operatorname{Reg}^{\mathcal{A}^{\prime}}$ and therefore can calculate
$\operatorname{Reg}_{\alpha}^{\mathcal{A}^{\prime}}$ for all
$\alpha\in\mathcal{A}^{\prime}$. Let $\Phi_{R}(x,y)$ be the (first order)
formula which does this. That is, for $x,y\in\mathcal{A}^{\prime}$,
$\Phi_{R}(x,y)$ is true if $y\in\operatorname{On}$ and
$x\in\operatorname{Reg}_{y}^{\mathcal{A}^{\prime}}$.

By elementarity, this property of $\Phi_{R}$ is also true in
$\mathcal{C}^{\prime}$. That is, $\mathcal{C}^{\prime}$ believes that
$\Phi_{R}(x,y)$ is true if and only if
$y\in\operatorname{On}^{\mathcal{C}^{\prime}}$ and
$x\in\operatorname{Reg}_{y}^{\mathcal{C}^{\prime}}$.

But we know that both $\operatorname{Reg}^{\mathcal{A}^{\prime}}$ and
$\operatorname{Reg}^{\mathcal{C}^{\prime}}$ are simply
$\operatorname{Reg}^{V}$ intersected with their respective models. So in fact
$\operatorname{Reg}_{y}^{\mathcal{A}^{\prime}}=\operatorname{Reg}_{y}^{\mathcal{C}^{\prime}}=\operatorname{Reg}_{y}^{V}$.
∎

###### Proof (Claim, 5).

We begin by defining a formula $\Psi_{0}(\chi,\vec{v},v_{k})$ (where $v_{k}$
is an otherwise unused variable).

$\Psi_{0}(\chi,\vec{y},z):=z\in\operatorname{On}\wedge\exists
s\,\big{(}\forall x_{0},x_{1}\,(x_{0},x_{1})\in
s\iff\chi(x_{0},x_{1},\vec{y})\big{)}\wedge s\text{ is a well
order}\wedge\text{o.t.}(s)=z$

So $\Psi_{0}$ holds if and only if $\chi$ and $\vec{y}$ define a well order,
and that well order is an element of the structure, and it has order type
$z\in\operatorname{On}$.

Since $\mathcal{A}^{\prime}$ has domain $H_{\lambda^{+}}$, we know that if
$\chi$ and $\vec{y}$ define a well order of size less than $\lambda^{+}$, then
that well order will be an element of $\mathcal{A}^{\prime}$. So in
particular, if $\chi$ and $\vec{y}$ define a linear order over $\mathcal{A}$
(which, remember, has cardinality $\lambda$) then that linear order will be an
element of $\mathcal{A}^{\prime}$, and if it is a well order then its order
type will then also be an ordinal in $\mathcal{A}^{\prime}$.

We claim the same is true of $\mathcal{C}$ and $\mathcal{C}^{\prime}$. If
$\chi$ and $\vec{y}$ define a linear order of elements of $\mathcal{C}$ over
$\mathcal{C}^{\prime}$, then by elementarity they define a linear order of
elements of $\mathcal{A}$ over $\mathcal{A}^{\prime}$. That linear order is in
$\mathcal{A}^{\prime}$. By elementarity then the original linear order over
$\mathcal{C}$ is in $\mathcal{C}^{\prime}$. For the order type part, note that
$\mathcal{A}^{\prime}$ and hence $\mathcal{C}^{\prime}$ believe that for all
$\gamma\in\operatorname{On}$, $\neg
I(\varphi_{\operatorname{On}},\varphi_{\in},\gamma)$, meaning that
$\operatorname{On}\cap\mathcal{C}^{\prime}$ is a cardinal of $V$. And letting
$\varphi_{P}(x):=P(x)$, we also know (by elementarity) that
$\mathcal{C}^{\prime}$ believes there is some cardinal $\gamma$ such that
$I(\varphi_{P},\varphi_{\in},\gamma)$. So the cardinality of $\mathcal{C}$ is
below $\operatorname{On}\cap\mathcal{C}^{\prime}$. Hence all the ordinals of
cardinality $\leq\lvert\mathcal{C}\rvert$ are in $\mathcal{C}^{\prime}$, and
so any well order of elements of $\mathcal{C}$ will have order type in
$\mathcal{C}^{\prime}$.

So if $\chi$ and $\vec{y}$ define a subset of $\mathcal{C}^{2}$ or
$\mathcal{A}^{2}$, over $\mathcal{C}^{\prime}$ or $\mathcal{A}^{\prime}$ then
they define a well order if and only if there is some ordinal $z$ of
$\mathcal{C}^{\prime}$ or $\mathcal{A}^{\prime}$ such that
$\mathcal{C}^{\prime}$ or $\mathcal{A}^{\prime}$, respectively, believes
$\Psi_{0}(\chi,\vec{y},z)$; and if so, then the well order has order type $z$.

Now let

$\displaystyle\Psi(\varphi,\psi,\chi,\vec{y}):=$ $\displaystyle
Q^{\text{e.c.}}(\varphi,\psi,\vec{y})\wedge$
$\displaystyle\exists\alpha\in\operatorname{On}\Psi_{0}(\chi,\vec{y},\alpha)\wedge$
$\displaystyle\exists\beta\in\operatorname{On}\Phi_{R}(\beta,\alpha)\wedge
Q^{\text{e.c.}}(\varphi,\varphi_{\in}^{\prime},\vec{y},\beta)$

An easy definition chase shows that if $\varphi,\psi,\chi$ together with
$\vec{y}$ define subsets of $\mathcal{A}^{2}$ or $\mathcal{C}^{2}$ over
$\mathcal{A}^{\prime}$ or $\mathcal{C}^{\prime}$ (respectively) then
$\mathcal{A}^{\prime}$ or $\mathcal{C}^{\prime}$ believe
$\Psi(\varphi,\psi,\chi,\vec{y})$ if and only if $\mathcal{A}$ believes
$Q^{\operatorname{On}}(\varphi,\psi,\chi,\vec{y})$. Since we already know how
$\Psi_{0}$ is interpreted, the only “trick” is showing that the equal
cofinality of the orders defined by $\varphi$ and $\psi$ must be an element of
$\mathcal{A}^{\prime}$ or $\mathcal{C}^{\prime}$. This is trivial for
$\mathcal{A}^{\prime}$, and for $\mathcal{C}^{\prime}$ it follows from the
fact that (as we saw above) $\mathcal{C}^{\prime}$ contains all the cardinals
$\leq\lvert\mathcal{C}\rvert$. ∎

The existence of $\Psi$ shows that $\mathcal{C}$ is
$\mathcal{L}\cup\\{I,Q^{\operatorname{On}}\\}$ elementarily equivalent to
$\mathcal{A}$ by standard arguments, and therefore that $\mathcal{B}$ is an
$\mathcal{L}\cup\\{I,Q^{\operatorname{On}}\\}$ elementary substructure of
$\mathcal{A}$. So
$\operatorname{LST}(I,Q^{\operatorname{On}})\leq\operatorname{LST}(I,Q^{\text{e.c.}})$.

To show the converse, we use a similar trick, but the technique is much
simpler: given an $\mathcal{L}$ structure $\mathcal{A}$ of cardinality
$\lambda>\kappa=\operatorname{LST}(I,Q^{\operatorname{On}})$, let
$\mathcal{A}^{\prime}=(\lambda+1,\in)$ and let
$\mathcal{A}^{\prime\prime}=\mathcal{A}\sqcup\mathcal{A}^{\prime}$. Let
$\mathcal{B}^{\prime\prime}$ be an
$\mathcal{L}\cup\\{I,Q^{\operatorname{On}}\\}$ elementary substructure of size
less than $\kappa$, and let $\mathcal{B}$ be the part corresponding to
$\mathcal{A}$.

Now, the cardinality of $\mathcal{A}$ is equal to the cardinality of
$\mathcal{A}^{\prime}$, and $\mathcal{A}^{\prime\prime}$ can easily express
this using $I$. Hence, by elementarity $\mathcal{B}$ and
$\mathcal{B}^{\prime}$ also have the same cardinality as each other. Since
$\mathcal{B}^{\prime}$ has a largest element, if $X$ is any linear order of
elements of $\mathcal{B}^{\prime}$ then its $V$ cofinality, and the Cantor-
Bendixson rank of its cofinality in $\operatorname{Reg}^{V}$, will be in the
transitive collapse of $\mathcal{B}^{\prime}$.

Hence, in both $\mathcal{A}^{\prime\prime}$ and $\mathcal{B}^{\prime\prime}$,
$Q^{\text{e.c.}}(\varphi,\psi,\vec{y})$ is equivalent to
“$\exists\alpha\in\operatorname{On}Q^{\operatorname{On}}(\varphi,\psi,\varphi_{\in}^{\prime},\vec{y},\alpha)$”.
So $\mathcal{B}$ is an $\mathcal{L}\cup\\{I,Q^{\text{e.c.}}\\}$ elementary
substructure of $\mathcal{A}$. Hence
$\operatorname{LST}(I,Q^{\text{e.c.}})\leq\operatorname{LST}(I,Q^{\operatorname{On}})$.
∎

###### Proof (2).

Similar to the previous part. Let $\mathcal{A}$ be an $\mathcal{L}$ structure
with cardinality $\lambda\geq\kappa=\operatorname{LST}(I,Q^{\alpha})$. Without
loss of generality assume $\mathcal{A}\in H_{\lambda^{+}}$. Let
$\mathcal{L}^{\prime}=\mathcal{L}\sqcup\\{\in,P\\}$ and let
$\mathcal{A}^{\prime}$ be a $\mathcal{L}^{\prime}$ structure which extends
$\mathcal{A}$, with $P^{\mathcal{A}^{\prime}}=\mathcal{A}$ and all the symbols
of $\mathcal{L}$ interpreted trivially outside $\mathcal{A}$.

Let $\mathcal{B}^{\prime}$ be an
$\mathcal{L}^{\prime}\sqcup\\{I,Q^{\alpha}\\}$ substructure of cardinality
$<\kappa$, and let $\mathcal{B}=\mathcal{B}^{\prime}\cap P\subset\mathcal{A}$.
Now, $R^{\alpha}$ can be defined in terms of $Q^{\alpha}$ in both
$\mathcal{A}^{\prime}$ and $\mathcal{B}^{\prime}$. For
$\mathcal{L}^{\prime}\cup\\{I,Q^{\operatorname{On}}\\}$ formulas $\varphi$ and
$\chi$, let

$\displaystyle\Phi(\varphi,\chi,\vec{y}):=\exists\gamma\in\operatorname{On}($
$\displaystyle I(\varphi,\varphi_{\in},\vec{y},\gamma)\wedge$
$\displaystyle\forall\delta<\gamma\neg
I(\phi_{\in},\phi_{\in},\gamma,\delta)\wedge$
$\displaystyle\exists\epsilon\in\operatorname{On}Q^{\alpha}(\varphi_{\in}^{\prime},\varphi_{\in}^{\prime},\varphi_{\in}^{\prime},\gamma,\gamma,\epsilon)\wedge$
$\displaystyle\forall\delta<\gamma\forall\epsilon\in\operatorname{On}\neg
Q^{\alpha}(\varphi_{\in}^{\prime},\varphi_{\in}^{\prime},\varphi_{\in}^{\prime},\gamma,\delta,\epsilon))$

Then in both $\mathcal{A}^{\prime}$ and $\mathcal{B}^{\prime}$,
$\varphi(x,\vec{y})\implies P(x)$ implies that $\Phi(\varphi,\chi,\vec{y})$
holds if and only if $R^{\alpha}(\varphi,\chi,\vec{y})$ is true.

To see that this works in $\mathcal{B}^{\prime}$, it suffices to know that the
transitive collapse of $\mathcal{B}^{\prime}$ contains ordinals of every order
type below $\lvert\mathcal{B}\rvert$. This follows as before: in
$\mathcal{B}^{\prime}$, $I(\varphi_{\operatorname{On}},\varphi_{\in},\gamma)$
fails for all $\gamma<\operatorname{On}\cap\mathcal{B}^{\prime}$, so
$\operatorname{On}\cap\mathcal{B}^{\prime}$ is a cardinal of $V$. But there is
some $\gamma\in\operatorname{On}\cap\mathcal{B}^{\prime}$ such that
$I(\varphi_{P},\varphi_{\in},\gamma)$ holds, where $\varphi_{P}(x)=P(x)$. So
$\operatorname{On}\cap\mathcal{B}^{\prime}$ is a cardinal of $V$ which is
larger than $\lvert\mathcal{B}\rvert$.

Now $\mathcal{B}^{\prime}$ is an $\mathcal{L}^{\prime}\cup\\{I,\Phi\\}$
elementary substructure of $\mathcal{A}^{\prime}$. But $\Phi$ agrees with
$R^{\alpha}$ on $\mathcal{A}$ and $\mathcal{B}$, so $\mathcal{B}$ is an
$\mathcal{L}\cup\\{I,R^{\alpha}\\}$ elementary substructure of $\mathcal{A}$
of cardinality less than $\kappa$. ∎

###### Proof (3).

First suppose that $\beta<\operatorname{LST}(I,Q^{\alpha})$. We shall show
that $\operatorname{LST}(I,Q^{\alpha})\geq\operatorname{LST}(I,Q^{\beta})$.
The technique is (once again) similar. However, we need some way to name
$\beta$.

Let $\mathcal{A}$ be an $\mathcal{L}$ structure, of cardinality
$\lambda\geq\operatorname{LST}(I,Q^{\alpha})$. Assume that $\mathcal{L}$ has
cardinality less than $\operatorname{LST}(I,Q^{\alpha})$. Expand $\mathcal{L}$
to a language $\mathcal{L}^{\prime}$ which contains $\in$, a predicate $P$ and
$\beta$ many constant symbols $c_{i}:i<\beta+1$. Note that
$\mathcal{L}^{\prime}$ has cardinality less than
$\operatorname{LST}(I,Q^{\alpha})$ still. Let
$\mathcal{A}^{\prime}=\mathcal{A}\sqcup(\lambda+1)$ be an expansion of
$\mathcal{A}$ which adds the first $\lambda+1$ ordinals, and assigns constant
symbols to the first $\beta+1$ many of those ordinals, and interprets $P$ as
true on $\mathcal{A}$. Let $\mathcal{B}^{\prime}\leq\mathcal{A}^{\prime}$ be
an $\mathcal{L}^{\prime}\cup\\{I,Q^{\alpha}\\}$ elementary substructure of
cardinality $<\operatorname{LST}(I,Q^{\alpha})$. We know that
$\mathcal{B}^{\prime}$ contains all the ordinals up to and including
$\beta=c_{\beta}$, so the set defined over $\mathcal{B}^{\prime}$ by
$\varphi_{\in}$ and $c_{\beta}$ is just $\beta$ itself.

It follows that we can define $Q^{\beta}$, in both $\mathcal{A}^{\prime}$ and
$\mathcal{B}^{\prime}$, via:

$Q^{\beta}(\varphi,\psi,\chi,\vec{y})\iff
Q^{\alpha}(\varphi,\psi,\chi,\vec{y})\wedge\exists\gamma\in
c_{\beta}Q^{\alpha}(\varphi,\psi,\varphi_{\in},\vec{y},\gamma)$

So $\mathcal{B}=\mathcal{B^{\prime}}\cap\mathcal{A}$ is a
$\mathcal{L}\cup\\{I,Q^{\beta}\\}$ elementary substructure of $\mathcal{A}$,
and $\operatorname{LST}(I,Q^{\beta})\leq\operatorname{LST}(I,Q^{\alpha})$.

Now suppose instead that
$\operatorname{LST}(I,Q^{\beta})\leq\min(\operatorname{Reg}_{\beta})$, and
suppose (seeking a contradiction) that
$\operatorname{LST}(I,Q^{\alpha})<\operatorname{LST}(I,Q^{\beta})$. Let
$\mathcal{A}$ be an $\mathcal{L}$ structure (where the cardinality of
$\mathcal{L}$ is less than $\operatorname{LST}(I,Q^{\alpha})$). Then
$\mathcal{A}$ contains an $\mathcal{L}\cup\\{I,Q^{\beta}\\}$ elementary
substructure $\mathcal{B}$ of cardinality less than
$\operatorname{LST}(I,Q^{\beta})$. And $\mathcal{B}$ has an
$\mathcal{L}\cup\\{I,Q^{\alpha}\\}$ substructure $\mathcal{C}$ of cardinality
less than $\operatorname{LST}(I,Q^{\alpha})$. But $\mathcal{B}$ is small
enough that every linear order we can construct from its elements has
cofinality in $\operatorname{Reg}_{<\beta}$. Hence, in $\mathcal{B}$ and all
its substructures, $Q^{\alpha}$ and $Q^{\beta}$ agree with each other. So
$\mathcal{C}$ is an $\mathcal{L}\cup\\{I,Q^{\beta}\\}$ elementary substructure
of $\mathcal{B}$ and hence of $\mathcal{A}$. So any $\mathcal{L}$ structure
contains an $\mathcal{L}\cup\\{I,Q^{\beta}\\}$ elementary substructure of
cardinality
$<\operatorname{LST}(I,Q^{\alpha})<\operatorname{LST}(I,Q^{\beta})$.
Contradiction. ∎

###### Proof (4).

Just like the previous case. ∎

###### Proof (5).

First note that $R^{0}$ is simply always false, since there are no ordinals
below $0$. So the claim $\operatorname{LST}(I,R^{0})=\operatorname{LST}(I)$ is
trivial.

$\operatorname{LST}(I,R^{1})\geq\operatorname{LST}(I)$ is also trivial. The
only thing to show is $\operatorname{LST}(I,R^{1})\leq\operatorname{LST}(I)$.
We prove this in the same way as the previous cases. Let $\mathcal{A}$ be an
$\mathcal{L}$ structure of cardinality $\lambda\geq\operatorname{LST}(I)$. Let
$\mathcal{L}^{\prime}=\mathcal{L}\cup\\{\in,P\\}$ and let
$\mathcal{A}^{\prime}=\mathcal{L}\sqcup\lambda+1$ with $P$ interpreted as true
on $\mathcal{A}^{\prime}$. Now, $R^{1}(\varphi,\chi,\vec{y}_{0},\vec{y}_{1})$
holds in $\mathcal{A}^{\prime}$ if and only if $\chi$ and $\vec{y}_{1}$ define
the empty set, and $\varphi$ defines a set whose cardinality is an (infinite)
successor cardinal. Both of these properties can be easily defined in
$\mathcal{A}^{\prime}$:

$\displaystyle\mathcal{A}^{\prime}\vDash
R^{1}(\varphi,\chi,\vec{y})\iff\mathcal{A}^{\prime}\vDash$
$\displaystyle\forall x\neg\chi(x,\vec{y})\wedge$
$\displaystyle\exists\kappa\in\operatorname{On}I(\varphi,\varphi_{\in},\vec{y}_{0},\kappa)\wedge$
$\displaystyle\forall\gamma\in\kappa\neg
I(\varphi_{\in},\varphi_{\in},\kappa,\gamma)\wedge$
$\displaystyle\exists\mu\in\kappa\forall\gamma\in(\mu,\kappa)I(\varphi_{\in},\varphi_{\in}\mu,\gamma)$

So $\mathcal{B}:=P(\mathcal{B}^{\prime})$ is an $\mathcal{L}\cup\\{I,R^{1}\\}$
elementary substructure of $\mathcal{A}=P(\mathcal{A}^{\prime})$. ∎

∎

This theorem shows that if we exclude very large values of $\alpha$,
$\operatorname{LST}(I,Q^{\alpha})$ and $\operatorname{LST}(I,R^{\alpha})$ move
downward as we decrease $\alpha$ or move from $Q$ to $R$. At the bottom of
this hierarchy is $\operatorname{LST}(I)$, and we saw earlier (from [4]) that
the minimum possible value of this is precisely the first inaccessible. At the
top is
$\operatorname{LST}(I,Q^{\operatorname{On}})=\operatorname{LST}(I,Q^{\text{e.c.}})$,
and we saw that the minimum possible value of this is precisely the first
Mahlo cardinal. The remainder of the section is devoted to finding a similar
result for the other members of the hierarchy. The tagline for this section
is:

> For $\alpha>0$, the minimum consistent values for
> $\operatorname{LST}(I,Q^{\alpha})$ and $\operatorname{LST}(I,R^{\alpha})$
> are both precisely the first element of $\operatorname{Reg}_{\alpha}$,
> provided $\alpha$ is not too large.

So for example, this says the minimum value of
$\operatorname{LST}(I,R^{1})=\operatorname{LST}(I)$ is the first element of
$\operatorname{Reg}_{1}$, which is the least inaccessible as we had expected.
We will, of course, explain what we mean by “too large” shortly; it depends on
whether we’re dealing with $Q$ or $R$.

First, we shall prove that these cardinals are lower bounds for the
$\operatorname{LST}$ numbers.

######  Theorem 4.3.

Let $\alpha>0$ be such that there are no hyperinaccessibles below $\alpha$.
Then if it exists, $\operatorname{LST}(I,R^{\alpha})$ is at least the first
element of $\operatorname{Reg}_{\alpha}$.

###### Proof.

Suppose $\kappa=\operatorname{LST}(I,R^{\alpha})$ is less than the first
element of $\operatorname{Reg}_{\alpha}$. Let
$\mathcal{A}=(H_{\kappa^{+}},\in)$. The cardinals of $\mathcal{A}$ are
precisely the cardinals of $V$ up to (and including) $\kappa$.

By definition of the $\operatorname{LST}$ number, we can find an elementary
substructure $\mathcal{B}\subset\mathcal{A}$ of cardinality less than
$\kappa$. Since $\mathcal{A}$ is well founded, so is $\mathcal{B}$, so we can
take its transitive collapse $\mathcal{C}=(C,\in)$ and the elementary
embedding

$j:\mathcal{C}\rightarrow\mathcal{A}$

Now, letting $\varphi_{\in}$ be as in the previous theorem,

$\displaystyle\epsilon\in\operatorname{Card}^{\mathcal{C}}$ $\displaystyle\iff
j(\epsilon)\in\operatorname{Card}^{\mathcal{A}}$ $\displaystyle\iff
j(\epsilon)\in\operatorname{Card}^{V}$
$\displaystyle\iff\mathcal{A}\vDash\forall\gamma\in j(\epsilon)\,\neg
I(\varphi_{\in},\varphi_{\in},\gamma,j(\epsilon))$
$\displaystyle\iff\mathcal{C}\vDash\forall\gamma\in\epsilon\,\neg
I(\varphi_{\in},\varphi_{\in},\gamma,\epsilon)$
$\displaystyle\iff\epsilon\in\operatorname{Card}^{V}$

The last $\iff$ follows because $\mathcal{C}$ is transitive. Similarly,

$\displaystyle\epsilon\in\operatorname{Reg}^{\mathcal{C}}$ $\displaystyle\iff
j(\epsilon)\in\operatorname{Reg}^{\mathcal{A}}$ $\displaystyle\iff
j(\epsilon)\in(\operatorname{Reg}_{<\alpha})^{V}$ $\displaystyle\text{because
}\kappa<\min(\operatorname{Reg}_{\alpha})$
$\displaystyle\iff\mathcal{A}\vDash\exists\delta\in\operatorname{On}R^{\alpha}(\varphi_{\in},\varphi_{\in}^{\prime},j(\epsilon),\delta)$
$\displaystyle\iff\mathcal{C}\vDash\exists\delta\in\operatorname{On}R^{\alpha}(\varphi_{\in},\varphi_{\in}^{\prime},\epsilon,\delta)$
$\displaystyle\iff\epsilon\in(\operatorname{Reg}_{<\alpha})^{V}$
$\displaystyle\iff\epsilon\in\operatorname{Reg}^{V}$

$\mathcal{A}$ and $\mathcal{C}$ can then (using the same recursion) calculate
which cardinals are in $\operatorname{Reg}_{\beta}^{V}$ for each
$\beta<\alpha$.

Now, let $\epsilon\in\mathcal{C}$ be the least ordinal such that
$j(\epsilon)>\epsilon$. (Such a $\epsilon$ must exist: $\mathcal{C}$ has a
largest cardinal $\delta$ which must be smaller than $\kappa$, and
$j(\delta)=\kappa>\delta$.)

Clearly $\epsilon$ is a cardinal of $\mathcal{C}$. Otherwise there would be
some bijection $f:\gamma\rightarrow\epsilon$ for some $\gamma<\epsilon$; and
then $j(f)=f$ would be a bijection from $j(\gamma)=\gamma$ to
$j(\epsilon)>\epsilon$, which is clearly nonsense. So $\epsilon$ is a cardinal
of $\mathcal{C}$, and hence of $V$. Moreover, $\epsilon$ cannot be a successor
cardinal: if $\epsilon=(\delta^{+})^{\mathcal{C}}$ then

$j(\epsilon)=(j(\delta)^{+})^{\mathcal{A}}=(j(\delta)^{+})^{V}=(\delta^{+})^{V}=(\delta^{+})^{\mathcal{C}}=\epsilon$

Suppose that $\epsilon$ is singular (in $\mathcal{C}$ or equivalently in $V$).
Then we can take some sequence
$\mu=(\mu_{\beta})_{\beta<\delta}\in\mathcal{C}$ which is cofinal in
$\epsilon$, with $\delta<\epsilon$. Since
$j{\upharpoonleft}\epsilon=\operatorname{id}$, $j(\mu_{\beta})=\mu_{\beta}$
for all $\beta$, and the length of $j(\mu)$ is $j(\delta)=\delta$. Hence
$j(\mu)=\mu$. But $j(\mu)$ is cofinal in $j(\epsilon)$, so then
$j(\epsilon)=\epsilon$. Contradiction.

So $\epsilon$ is a regular limit cardinal, and hence is in
$\operatorname{Reg}_{\gamma}^{V}$ for some $0<\gamma$. We know that
$\gamma<\alpha$ since
$\lvert\mathcal{C}\rvert<\min(\operatorname{Reg}^{V}_{\alpha})$. As usual,
since $\epsilon\in\operatorname{Reg}_{\gamma}$ we know $\gamma\leq\epsilon$.
Moreover, if $\gamma=\epsilon$ then it is a hyperinaccessible, and we are
assuming none exist below $\alpha$. So $\gamma<\epsilon$ and
$j(\gamma)=\gamma$.

Say $\epsilon$ is the $\delta$’th element of $\operatorname{Reg}_{\gamma}$.
Since $\operatorname{Reg}_{\gamma}\cap\epsilon$ must be bounded below
$\epsilon$, we know $\delta<\epsilon$ and hence $j(\delta)=\delta$. But then

$\mathcal{C}\vDash``\epsilon\text{ is the }\delta\text{'th element
of}\operatorname{Reg}_{\gamma}^{\mathcal{C}}"$

and hence by elementarity

$\mathcal{A}\vDash``j(\epsilon)\text{ is the }\delta\text{'th element
of}\operatorname{Reg}_{\gamma}^{\mathcal{A}}"$

But this is saying that $\epsilon$ and $j(\epsilon)$ are both the same element
of $\operatorname{Reg}_{\gamma}^{V}$. Contradiction. ∎

By the above and Theorem 4.1 it also follows, under the above conditions, that
$\operatorname{LST}(I,Q^{\alpha})\geq\min(\operatorname{Reg}_{\alpha})$. In
fact, if $\min(\operatorname{Reg}_{\alpha})$ is strongly inaccessible (e.g.
because we are assuming GCH) then this will hold under weaker conditions:

######  Theorem 4.4.

Let $\alpha>0$ and suppose that the first element of
$\operatorname{Reg}_{\alpha}$ is strongly inaccessible. Suppose there are no
Mahlo cardinals below $\alpha$. Then if it exists,
$\operatorname{LST}(I,Q^{\alpha})$ is at least the first element of
$\operatorname{Reg}_{\alpha}$.

###### Proof.

Suppose that
$\operatorname{LST}(I,Q^{\alpha})<\min(\operatorname{Reg}_{\alpha})=:\kappa$.
Since $\kappa$ is strongly inaccessible, $M:=H_{\kappa}$ is a model of
$\operatorname{\mathsf{ZFC}}$. It is easy to see that
$\operatorname{LST}^{M}(I,Q^{\alpha})\leq\operatorname{LST}^{V}(I,Q^{\alpha})\in
M$: if $\mathcal{A}\in M$ is an $\mathcal{L}$ structure, then it contains some
$\mathcal{L}\cup\\{I,Q^{\alpha}\\}$ elementary substructure $\mathcal{B}\in V$
of cardinality $<\operatorname{LST}(I,Q^{\alpha})^{V}$. Then $\mathcal{B}\in
H_{\kappa}=M$.

Moreover, $M$ contains no Mahlo cardinals. This is because (by assumption) it
has none below $\alpha$, and since any Mahlo cardinal is hyperinaccessible,
there cannot be any in the interval
$[\alpha,\min(\operatorname{Reg}_{\alpha}))=[\alpha,\kappa)$. So we know that
$\operatorname{LST}^{M}(I,Q^{\text{e.c.}})=\infty$.

But $M$ doesn’t contain any cardinals which are in
$\operatorname{Reg}_{\alpha},\operatorname{Reg}_{\alpha+1},\ldots$. So as far
as $M$ is concerned $Q^{\alpha}$ is evaluated in exactly the same way as
$Q^{\operatorname{On}}$. Hence,
$\operatorname{LST}^{M}(I,Q^{\alpha})=\operatorname{LST}^{M}(I,Q^{\operatorname{On}})=\infty$.
Contradiction. ∎

A similar argument shows that if there are no Mahlo cardinals below $\alpha$
and $\operatorname{Reg}_{\alpha}=\emptyset$ then
$\operatorname{LST}(I,Q^{\alpha})=\infty$.

This gives us the promised largeness criterion, necessary for the bad option
of 3 and 4 of Theorem 4.1 to hold:

######  Corollary 4.5.

If $\beta<\alpha$ and there are no hyperinaccessibles below $\alpha$ then
$\operatorname{LST}(I,R^{\beta})\leq\operatorname{LST}(I,R^{\alpha})$ and
$\operatorname{LST}(I,Q^{\beta})\leq\operatorname{LST}(I,Q^{\alpha})$. If
instead there are no Mahlo cardinals below $\alpha$ and
$\min(\operatorname{Reg}_{\alpha})$ is strongly inaccessible, then
$\operatorname{LST}(I,Q^{\beta})\leq\operatorname{LST}(I,Q^{\alpha})$.

###### Proof.

If $\beta=0$ then this is trivial, as both $Q^{0}$ and $R^{0}$ are always
false. So assume $\beta>0$.

We address the $Q$ case first. We are assuming there are no Mahlo cardinals
below $\alpha$, and we have just seen that this implies
$\operatorname{LST}(I,Q^{\alpha})\geq\min(\operatorname{Reg}_{\alpha})$. But
$\min(\operatorname{Reg}_{\alpha})\geq\alpha>\beta$, so
$\beta<\operatorname{LST}(I,Q^{\alpha})$. We saw in Theorem 4.1 that this
implies $\operatorname{LST}(I,Q^{\beta})\leq\operatorname{LST}(I,Q^{\alpha})$.

The $R$ case follows in exactly the same way. ∎

So we have seen that if $\alpha$ is not too large, the first element of
$\operatorname{Reg}_{\alpha}$ is a lower bound for
$\operatorname{LST}(I,Q^{\alpha})$ and $\operatorname{LST}(I,R^{\alpha})$. We
now want to show these lower bounds are optimal. This our main result, and the
rest of this paper is devoted to its proof.

Just before we begin, we shall note in passing that there is also an upper
bound for $\operatorname{LST}(I,Q^{\alpha})$ and
$\operatorname{LST}(I,R^{\alpha})$: they can never be above a supercompact.
The proof is fairly simple, but introduces several of the techniques we will
use in proving the main theorem.

######  Theorem 4.6.

Let $\kappa$ be a supercompact, and let $\alpha<\kappa$. Then
$\operatorname{LST}(I,Q^{\alpha})$ exists, and is no larger than $\kappa$.
Hence, the same is true of $\operatorname{LST}(I,R^{\alpha})$.

###### Proof.

Let $\mathcal{L}$ be a first order language of cardinality less than $\kappa$,
and let $\mathcal{A}\in V$ be an $\mathcal{L}$ structure, of cardinality
$\lambda\geq\kappa$. Without loss of generality, let us say that the domain of
$\mathcal{A}$ is $\lambda$. Without loss of generality, let us further say
that $\mathcal{L}\in H_{\kappa}$. We want to show that there is a substructure
of cardinality less than $\kappa$.

Let $j:V\rightarrow M$ be an elementary embedding with critical point
$\kappa$, such that $j(\kappa)>\lambda$ and $M^{\lambda}\subset M$. (Such an
embedding exists by definition of a supercompact). Note that since
$\mathcal{L}\in H_{\kappa}$, $j$ acts as the identity on $\mathcal{L}$.
Moreover, $\mathcal{A}$ can be coded as a $\lambda$ sequence. Since $M$
contains all its $\lambda$ sequences, we know that $\mathcal{A}\in M$. Also
notice that since $\alpha<\kappa\leq\lambda$ we know that
$j((Q^{\alpha})^{V})=(Q^{\alpha})^{M}$. And of course, $j(I^{V})=I^{M}$.

Another consequence of $M$ containing all its $\lambda$ sequences is that it
correctly calculates the cardinalities and cofinalities of all subsets of
$\lambda$ and $\lambda\times\lambda$. In particular, both $I$ and $Q^{\alpha}$
are evaluated the same way over $\mathcal{A}$ in both $M$ and $V$. It follows
that if $\Phi$ is an $\mathcal{L}\cup\\{I,Q^{\alpha}\\}$ formula (with
parameters from $\mathcal{A}$ and $\mathcal{L}$) then

$(\mathcal{A}\vDash\Phi)^{M}\iff(\mathcal{A}\vDash\Phi)^{V}$

By elementarity, we know that

$(\mathcal{A}\vDash\Phi)^{V}\iff(j(\mathcal{A})\vDash\Phi)^{M}$

There is a small technicity which is disguised by this notation. $\Phi$ can
have parameters from $\mathcal{A}$ and $\mathcal{L}$, which we have
suppressed. For elementarity to hold, the $\Phi$ on the right must take the
images of those parameters under $j$. But in fact, since $j$ acts as the
identity on $\mathcal{L}$ and on the domain $\lambda$ of $\mathcal{A}$, doing
$j$ to those parameters just gives us back the original parameters.

This argument implies that in $M$,

$\mathcal{A}\vDash\Phi\iff j(\mathcal{A})\vDash\Phi$

Hence, from the perspective of $M$, $\mathcal{A}$ is an
$\mathcal{L}\cup\\{I,Q^{\alpha}\\}$ elementary substructure of
$j(\mathcal{A})$ of cardinality $\lambda<j(\kappa)$. So $M$ believes:
“$j(\mathcal{A})$ contains an $\mathcal{L}\cup\\{I,Q^{\alpha}\\}$ elementary
substructure of cardinality less than $j(\kappa)$.’ So by elementarity, $V$
believes: “$\mathcal{A}$ contains an $\mathcal{L}\cup\\{I,Q^{\alpha}\\}$
elementary substructure of cardinality less than $\kappa$.” Which is what we
wanted to show. ∎

##  5\. The main theorem

######  Theorem 5.1.

Let $0<\alpha\in\operatorname{On}$. Suppose it is consistent that there is a
supercompact cardinal larger than $\alpha$ and GCH holds. Then it is also
consistent that $\alpha$ is countable (and so has no hyperinaccessibles below
it) and $\operatorname{LST}(I,Q^{\alpha})$ and
$\operatorname{LST}(I,R^{\alpha})$ are both no larger than the first element
of $\operatorname{Reg}_{\alpha}$.

###### Proof.

We adapt a technique from [4]. The first step is to show that without loss of
generality, we may assume that $\epsilon$ is countable. Take a universe
$V_{0}$ which believes GCH and has a supercompact $\kappa>\epsilon$. Let $V$
be a generic extension of $V_{0}$ by the collapsing forcing
$\operatorname{Col}(\omega,\lvert\epsilon\rvert)$. We show that $V$ inherits
the properties of $V_{0}$: $\kappa$ is supercompact, and GCH holds. GCH
follows from Lemma LABEL:Lemma_CH_preservation_for_forcing but the
supercompactness of $\kappa$ requires a little work. We need the following
lemmas, which is known as the Silver Lifting Criterion. We will prove it for
an arbitrary universe $\tilde{V}$ and forcing $\mathbb{P}$, because we’ll need
it again in a different context later on in the proof.

######  Lemma 5.2.

Let $j:\tilde{V}\rightarrow M$ be an elementary embedding from some universe
$\tilde{V}$ to a model $M$. Let $\mathbb{P}\in\tilde{V}$ be a forcing. Let $G$
be $\mathbb{P}$ generic, and let $H$ be $j(\mathbb{P})$ generic. Suppose that
for all $p\in G$, $j(p)\in H$. Then there is an elementary embedding
$j^{*}:\tilde{V}[G]\rightarrow M[H]$, which extends $j$ and is such that the
following diagram commutes.

$M$$M[H]$$\tilde{V}$$\tilde{V}[G]$$j$$j^{*}$

###### Proof.

We define $j^{*}$ in the only way we can, and verify that the definition
works. Let $S\in\tilde{V}[G]$ be a set. Let $\sigma$ be a $\mathbb{P}$ name
with $\sigma^{G}=S$. We fix

$j^{*}(S)=(j(\sigma))^{H}$

We must verify that $j^{*}$ is well defined. Suppose that $\sigma$ and $\tau$
are two $\mathbb{P}$ names, and $\sigma^{G}=\tau^{G}$. Then there is some
$p\in G$ such that $p\mathrel{\Vdash}\sigma=\tau$. By elementarity
$j(p)\mathrel{\Vdash}j(\sigma)=j(\tau)$, and by assumption $j(p)\in H$. Hence
$(j(\sigma))^{H}=(j(\tau))^{H}$.

It is trivial to verify that $j^{*}$ extends $j$, and that the diagram
commutes. The last thing to check is that $j^{*}$ is elementary, and this is
done in a similar way to the “well-defined” proof. Let $\varphi$ be a formula
with parameters $\vec{S}=\vec{\sigma}^{G}$ and suppose that
$\tilde{V}[G]\vDash\varphi(\vec{S})$. Then let $p\in G$ force
$\varphi(\vec{\sigma})$; by elementarity
$j(p)\mathrel{\Vdash}\varphi(j(\vec{\sigma})$ and by assumption $j(p)\in H$.
So $M[H]\vDash\varphi(j^{*}(\vec{S}))$. ∎

If we add some extra assumptions, we can also prove that $j^{*}$ preserves
$\lambda$ sequences like a supercompact embedding.

######  Lemma 5.3.

Suppose the conditions of the above lemma hold, that
$j(\mathbb{P})=\mathbb{P}$, $H=G$ and that $\mathbb{P}$ satisfies the
$\lambda^{+}$ chain condition. Then $\tilde{V}[G]$ believes that
$M[G]^{\lambda}\subset M[G]$.

In fact, this can be proved in much more general circumstances: rather than
assuming that $j(\mathbb{P})=\mathbb{P}$ and $H=G$ we only need that $G\in
M[H]$ and $H\in\tilde{V}[G]$. But that’s overkill for this proof.

###### Proof.

First, note that since $M$ is definable in $\tilde{V}$, $M[G]$ is definable in
$V[G]$. Let $S=(S_{\gamma})_{\gamma<\lambda}\in\tilde{V}[G]$ be a $\lambda$
sequence of elements of $M[G]$. Since $M$ and $G$ are definable in
$\tilde{V}[G]$, we can also define a sequence
$\dot{S}=(\dot{S}_{\gamma})_{\gamma<\lambda}\in\tilde{V}[G]$ of names for the
terms of $S$. (Note that $S$ itself is not a name, it’s just a sequence of
names.) Let $\dot{\Sigma}\in\tilde{V}$ be a $\mathbb{P}$ name for $\dot{S}$.
Let $p\in G$ force $\dot{\Sigma}$ to be a $\lambda$ sequence of $\mathbb{P}$
names, each of which is an element of $M$.

For $\gamma<\lambda$, let $A_{\gamma}$ be a maximal antichain of conditions
below $p$ which decide which name $\dot{\Sigma}(\gamma)$ is going to be
interpreted as. So for any $q\in A_{\gamma}$ there is some name
$\sigma_{q,\gamma}\in M$ such that
$q\mathrel{\Vdash}\dot{\Sigma}(\gamma)=\sigma_{q,\gamma}$. Then $\dot{\Sigma}$
is forced by $p$ to be equal to

$\tau:=\\{\langle(\sigma_{q,\gamma},\check{\gamma}),q\rangle:\gamma<\lambda,q\in
A_{\gamma}\\}$

By the chain condition, $\lvert A_{\gamma}\rvert\leq\lambda$ for all
$\gamma<\lambda$, and hence $\tau$ has cardinality $\leq\lambda$. All the
elements of $\tau$ are in $M$, $\tau\in\tilde{V}$ and $M$ is closed under
$\lambda$ sequences, so $\tau\in M$. So $\dot{S}=\tau^{G}\in M[G]$. The
conclusion that $S\in M[G]$ is now immediate. ∎

######  Corollary 5.4.

$\kappa$ is supercompact in $V$.

###### Proof.

Let $\lambda>\kappa$. Let $j:V_{0}\rightarrow M$ be a $\lambda$ embedding:
i.e. elementary, with critical point $\kappa$, $j(\kappa)>\lambda$ and
$M^{\lambda}\subset M$. Let $G$ be the
$\mathbb{P}:=\operatorname{Col}(\omega,\lvert\alpha\rvert)$ generic filter
used in constructing $V$ (so $V=V_{0}[G]$). Note that $\mathbb{P}$ is small
compared to $\kappa$, so $j(\mathbb{P})=\mathbb{P}$ and $G$ is also generic
over $M$. Clearly for all $p\in G$, $j(p)=p\in G$. So $j$ extends to an
elementary embedding $j^{*}:V_{0}[G]\rightarrow M[G]$ by the first lemma.
Since $j^{*}$ extends $j$ it has critical point $\kappa$ and sends $\kappa$ up
above $\lambda$. Since $\mathbb{P}$ is small compared to $\kappa$ it certainly
satisfies the $\lambda^{+}$ chain condition, and therefore by the second lemma
$V=V_{0}[G]$ believes that $M[G]^{\lambda}\subset M[G]$. ∎

From now on, we will forget about $V_{0}$ and the collapsing forcing, and just
work in $V$, a universe where $\alpha$ is countable, GCH holds and $\kappa$ is
supercompact, and where $\operatorname{Reg}_{<\alpha}$ is unbounded. Note that
since $\kappa$ is supercompact it is hyperinaccessible, and therefore
$\operatorname{Reg}_{<\alpha}$ is also unbounded below $\kappa$. We will also
assume (by cutting off the top of the model if necessary) that
$\operatorname{Reg}_{\alpha}\setminus\kappa^{+}=\emptyset$. Note that $\kappa$
will be an inaccessible limit of elements of $\operatorname{Reg}_{<\alpha}$
(so it is in $\operatorname{Reg}_{\beta}$ for some $\beta>\alpha$), and by the
assumption we just made it is the largest element of this
$\operatorname{Reg}_{\beta}$.

###  5.1. The Structure of the Proof

We will construct a forcing which will make $\kappa$ the first element of
$\operatorname{Reg}_{\alpha}$ and ensure
$\operatorname{LST}(I,Q^{\alpha}),\operatorname{LST}(I,R^{\alpha})\leq\kappa$.
The actual forcing we want to use is complex to define – it’s a delicate
combination of several other forcings, some of which are rather complicated in
their own right. We’ll start by giving an informal sketch of how the proof is
going to work, before we dive into the technicalities.

The forcing $\mathbb{P}$ we use has three components. The first is a non-Mahlo
forcing $\operatorname{NM}_{\lambda}$ which adds a club $C$ of singular
cardinals below some cardinal $\lambda$. The second is a collapsing forcing
$\operatorname{Col}_{\lambda}$ which collapses all the cardinals below that
$\lambda$, except for those which are a “short distance” above some element of
$C$. This will leave $\lambda$ as the first element of
$\operatorname{Reg}_{\alpha}$.

The third component is a Prikry-style forcing $\mathbb{Q}_{\lambda}$, for a
cardinal $\lambda$. It adds an $\operatorname{NM}_{\lambda}$ generic club,
adds a new condition to the $\operatorname{Col}_{\lambda}$ forcing, and
singularises $\lambda$.

$\mathbb{P}$ will be an iteration of $\mathbb{Q}_{\lambda}$ forcings (for
strategically chosen $\lambda<\kappa$), followed by
$\operatorname{NM}_{\kappa}$ and then $\operatorname{Col}_{\lambda}$.

This gives us a universe $V^{\mathbb{P}}$ in which $\kappa$ is the first
element of $\operatorname{Reg}_{\alpha}$. How do we show that
$\operatorname{LST}(I,Q^{\alpha}),\operatorname{LST}(I,R^{\alpha})=\kappa$? We
already know that
$\kappa\leq\operatorname{LST}(I,R^{\alpha})\leq\operatorname{LST}(I,Q^{\alpha})$
from the previous lemmas; we need to show that given any language
$\mathcal{L}\in V^{\mathbb{P}}$ (of cardinality $<\kappa$) and any
$\mathcal{L}$ structure $\mathcal{A}\in V^{\mathbb{P}}$, we can find a
substructure of cardinality less than $\kappa$ which is
$\mathcal{L}\cup(I,Q^{\alpha})$-elementarily equivalent to $\mathcal{A}$. We
use a similar approach to Theorem 4.6.

Say that $\mathcal{A}$ has cardinality $\mu\geq\kappa$. Because $\kappa$ is
supercompact in $V$, we can find an embedding $j:V\rightarrow M$ such that
$j(\kappa)>\mu$ and such that $\mu$ is only a “short distance” above $\kappa$
(in the sense we were discussing a moment ago). With a good deal of effort, we
can show that $\mathbb{P}$ embeds nicely into $j(\mathbb{P})$, and that the
conditions of Lemma 5.2 and Lemma
LABEL:lemma_M[H]_closed_under_lambda_sequences hold for some suitable
$j(\mathbb{P})$ generic extension of $M$. For this to happen, we need the club
of singulars added by $\operatorname{NM}_{\kappa}$ in $\mathbb{P}$ to be an
initial segment of the club of singulars added by
$\operatorname{NM}_{j(\kappa)}$ in $j(\mathbb{P})$. So we need $\kappa$ to be
singular. This is one of the reasons for introducing the Prikry-style forcing
$\mathbb{Q}_{\lambda}$: we will show that $j(\mathbb{P})$ contains
$\mathbb{Q}_{\kappa}$ and thus singularises $\kappa$. We will also need to
ensure that the collapsing forcings in $\mathbb{P}$ and $j(\mathbb{P})$ play
nicely with each other.

Now, in $M^{j(\mathbb{P})}$, we know that no cardinals between $\kappa$ and
$\mu$ are collapsed or singularised, because $\mu$ is only a “short distance”
above $\kappa$ and $\kappa$ is in the club of singulars added by
$\operatorname{NM}_{j(\kappa)}$. And below $\kappa$ we can also show that
$V^{\mathbb{P}}$ and $M^{j(\mathbb{P})}$ agree about the cardinals and the
regular cardinals. Hence, $\mathcal{L}\cup\\{I,Q^{\alpha}\\}$ interprets all
statements about $\mathcal{A}$ in the same way in $V^{\mathbb{P}}$ and
$M^{j(\mathbb{P})}$. (There are some technicalities involved with dealing with
$\kappa$ itself here, but they turn out not to be a problem.)

From there, the proof follows that of Theorem 4.6.

So in summary, the proof will consist of the following steps:

  1. (1)

Define a function $f$ that formalises the concept of “a short distance”

  2. (2)

Define the non-Mahlo and collapsing forcings $\operatorname{NM}_{\lambda}$ and
$\operatorname{Col}_{\lambda}$ we discussed above

  3. (3)

Define the forcing $\mathbb{Q}_{\lambda}$, which adds an
$\operatorname{NM}_{\lambda}$ generic filter, adds a condition to
$\operatorname{Col}_{\lambda}$ and singularises $\lambda$

  4. (4)

Show that $\operatorname{NM}_{\lambda}*\operatorname{Col}_{\lambda}$ embeds
nicely in $\mathbb{Q}_{\lambda}*\operatorname{Col}_{\lambda}$

  5. (5)

Put these forcings together in some way to get the forcing $\mathbb{P}$ we
will be using

  6. (6)

Show that in a $\mathbb{P}$ generic extension, the $\operatorname{LST}$ is
number at most $\kappa$

###  5.2. Defining $f$

Recall the Laver function, for any supercompact cardinal $\kappa$:

######  Lemma.

[3] There is a function $h:\kappa\rightarrow V_{\kappa}$ such that given any
$x\in V$ and any $\mu\geq\kappa$, there is an $M$ with $M^{\mu}\subset M$ and
an embedding $j:V\rightarrow M$ with critical point $\kappa$, such that
$j(\kappa)>\mu$ and $j(h)(\kappa)=x$.

We use this $h$ to define a related function $f$ specifically for the
ordinals.

######  Lemma 5.5.

There is a strictly increasing function $f:\kappa\rightarrow\kappa$, whose
range is a subset of $\operatorname{Reg}_{\alpha}$, such that:

  1. (1)

For all $\gamma<\kappa$, the interval $(\gamma,f(\gamma))$ contains
unboundedly many elements of $\operatorname{Reg}_{\beta}$ for every
$\beta<\alpha$, but no elements of $\operatorname{Reg}_{\alpha}$

  2. (2)

For any $\mu>\kappa$, there is a model $M$ with $M^{\mu}\subset M$ and an
elementary embedding $j:V\rightarrow M$ with critical point $\kappa$ such that
$j(\kappa)>\mu$ and $j(f)(\kappa)>\mu$.

###### Proof.

For $\gamma<\kappa$, let $g(\gamma):=h(\gamma)$ if the latter is an ordinal
greater than $\gamma$ but smaller than $\kappa$, and
$(\gamma,g(\gamma))\cap\operatorname{Reg}_{\alpha}=\emptyset$, and otherwise
let $g(\gamma):=\gamma$. Let $f(\gamma)$ be the least element of
$\operatorname{Reg}_{\alpha}$ which is (strictly) above $g(\gamma)$. It is
trivial to see that $f$ is strictly increasing and satisfies the first
requirement.

Fix $\mu>\kappa$. Using the previous result, let $M$ and $J:V\rightarrow M$ be
such that $j(h)(\kappa)=\mu$. We know that $M$ is closed under $\mu$
sequences, so it correctly calculates the cardinalities and cofinalities of
all cardinals below $\mu$. In particular, $\operatorname{Reg}_{\alpha}^{M}$
agrees with $\operatorname{Reg}_{\alpha}^{V}$ up to $\mu$, and hence
$(\kappa,\mu)\cap\operatorname{Reg}_{\alpha}=\emptyset$ from the perspective
of $M$. Hence $j(g)(\kappa)=j(h)(\kappa)=\mu$. So $j(f)(\kappa)>\mu$.

∎

In the language we used in the preamble to this proof, $\delta$ is “a short
distance” above $\gamma$ if $\delta\in(\gamma,f(\gamma))$.

###  5.3. NM and Col

Now we have $f$, we can define the simpler forcings used in this construction,
$\operatorname{NM}_{\lambda}$ and $\operatorname{Col}_{\lambda}$. Throughout
this section, let $\tilde{V}$ be any universe which contains the $f$ we found
in the previous section. We will also assume that $f$ has the same relation to
$\operatorname{Reg}$ in $\tilde{V}$ as it does in $V$ (although we do not
assume that
$\operatorname{Reg}_{\beta}^{\tilde{V}}=\operatorname{Reg}_{\beta}^{V}$ for
any particular $\beta\leq\alpha$), and (in preparation for the future) we
assume that GCH holds in $\tilde{V}$ except that there are perhaps a few
singular cardinals $\lambda$ such that $2^{\lambda}=\lambda^{++}$. Let us fix
$\lambda\leq\kappa$ to be a $\lambda^{+}$ supercompact cardinal in
$\tilde{V}$.

A Mahlo cardinal is one for which the class of regulars below it is
stationary, so to stop it being Mahlo we need to add a club which consists
only of singular cardinals. We use a variant of the usual club shooting
forcing which also chooses some elements of $\operatorname{Reg}_{\alpha}$.

######  Definition 5.6.

We define the non-Mahlo forcing $\operatorname{NM}_{\lambda}$ as follows. Its
elements are closed bounded subsets of $\lambda$, whose minimum element is
$\omega$, whose successors are all taken from $\operatorname{Reg}_{\alpha}$,
and whose limits are all singular. We order $\operatorname{NM}_{\lambda}$ by
end inclusion.

Note that this is not quite the usual non-Mahlo forcing: it gives us a club
whose limits are singular, but whose successors are elements of
$\operatorname{Reg}_{\alpha}$. (Since $\operatorname{Reg}_{\alpha}$ is
unbounded below $\lambda$, we do get an unbounded subset of $\lambda$ despite
the odd requirement for successors.) By contrast, a standard non-Mahlo forcing
would just add a club of singular cardinals. Of course, we can obtain a fully
singular club here simply by deleting all the successors, so
$\operatorname{NM}_{\lambda}$ does indeed force that $\lambda$ is no longer
Mahlo.

Although $\operatorname{NM}_{\lambda}$ is not strictly $<\lambda$ closed
(since the limit of a sequence of conditions could be inaccessible) it is
almost $<\lambda$ closed, and we can prove the usual results of $<\lambda$
closed-ness.

######  Lemma 5.7.

Let $\mu<\lambda$. Then there are densely many conditions $p$ such that
$\operatorname{NM}_{\lambda}{\upharpoonleft}p$ is $\mu$ closed.

###### Proof.

Let $p\in\operatorname{NM}_{\lambda}$ be any condition whose final element is
larger than $\mu$. Let $p\geq p_{0}\geq p_{1}\geq\ldots$ be a descending
sequence of conditions of length $\mu$. Then $p_{0}\subset p_{1}\subset\ldots$
are all end extensions of one another, so $\bigcup_{i<\mu}p_{i}$ is a set
whose successors are all in $\operatorname{Reg}_{\alpha}$ and whose limit
elements are all singular. It is closed, except that it might not contain its
supremum $\gamma$. But $\gamma\geq\sup p>\mu$ has cofinality
$\mathrm{Cof}(\mu)$, so it is singular. This implies that $\gamma<\lambda$,
and that $\bigcup_{i<\mu}p_{i}\cup\\{\gamma\\}\in\operatorname{NM}_{\lambda}$.
∎

######  Corollary 5.8.

$\operatorname{NM}_{\lambda}$ is $<\lambda$ distributive, does not collapse or
singularise any cardinals, and preserves GCH, except perhaps that
$2^{\lambda}$ becomes $\lambda^{++}$.

###### Proof.

Let $\mu<\lambda$. Let $D_{i}:i<\mu$ be a collection of dense sets. Fix some
$p$ such that $\operatorname{NM}_{\lambda}{\upharpoonleft}p$ is $\mu$ closed.
For $i<\mu$ let
$D_{i}^{\prime}=D_{i}\cap(\operatorname{NM}_{\lambda}{\upharpoonleft}p)$. Then
$D_{i}^{\prime}$ is a dense subset of
$\operatorname{NM}_{\lambda}{\upharpoonleft}p$. We know that
$\operatorname{NM}_{\lambda}{\upharpoonleft}p$ is $\mu$ distributive since it
is $\mu$ closed, and hence $\emptyset\neq\bigcap D_{i}^{\prime}\subset\bigcap
D_{i}$.

It follows immediately from distributivity that $\operatorname{NM}_{\lambda}$
preserves all cardinals $<\lambda$ and does not singularise any of them, and
that it preserves GCH below $\lambda$.

Showing that $\lambda$ is not collapsed or singularised requires a slightly
more technical argument, imitating the standard proof we’d use for a $\lambda$
closed forcing. Suppose that $\tilde{V}[G]$ collapses $\lambda$. Let
$g:\mu\rightarrow\lambda$ be a bijection for some $\mu<\lambda$, and let
$\dot{g}$ be a name for $g$. Let $p\in G$ force “$\dot{g}$ is a bijection from
$\mu$ to $\lambda$”. Let $q\leq p$ be such that
$\operatorname{NM}_{\lambda}{\upharpoonleft}q$ is $\mu$ closed. We construct a
descending sequence of conditions $q=q_{0}\geq q_{1}\geq\ldots$ (not
necessarily elements of $G$) of length $\mu+1$ as follows. If $i=j+1$ then we
choose $q_{i}\leq q_{j}$ which decides the value of $\dot{g}(j)$. If $i$ is a
limit, then by $\mu$ closed-ness, we can choose some $q_{i}$ below every
earlier $q_{j}$. At the end, $q_{\mu}$ has decided $\dot{g}(i)$ for every $i$,
and hence has defined a bijection from $\mu$ to $\lambda$ in $\tilde{V}$.
Contradiction. A similar proof shows that $\lambda$ is not singularised.

Finally, note that $\operatorname{NM}_{\lambda}\subset\lambda^{<\lambda}$, so

$\lvert\operatorname{NM}_{\lambda}\rvert\leq\sum_{\gamma<\lambda}\lambda^{\gamma}=\sum_{\gamma<\lambda}\lambda=\lambda$

Hence $\operatorname{NM}_{\lambda}$ has the $\lambda^{+}$ chain condition, and
so does not collapse or singularise any cardinals $\geq\lambda^{+}$, and
preserves GCH for cardinals above $\lambda$.

Finally, then, in the generic extension $2^{\lambda}\leq
2^{\lambda^{+}}=\lambda^{++}$. ∎

Throughout this proof, we shall write $C$, and variants thereof, to refer to
the generic club added by $\operatorname{NM}_{\lambda}$. It should (hopefully)
always be clear from context to which $\lambda$ we are referring. Notice that
for any $\gamma\in C$, we know $\operatorname{Succ}(\gamma)\geq f(\gamma)$,
because there are no $\operatorname{Reg}_{\alpha}$’s between $\gamma$ and
$f(\gamma)$.

Next, we define the collapsing forcing. Recall that if $\gamma$ is regular and
$\delta$ is strongly inaccessible, the forcing
$\operatorname{Col}(\gamma,<\delta)$ collapses all the cardinals between
$\gamma$ and $\delta$ down to $\gamma$.

For the rest of this section, let us fix a club $C$ which is generic for
$\operatorname{NM}_{\lambda}$. We will work in some generic extension
$\tilde{V}[G]$ of $\tilde{V}$ containing $C$, but we do not assume that
$\tilde{V}[G]=\tilde{V}[C]$. We will, however, assume that $G$ does not
collapse or singularise any cardinals below $\lambda$, and preserves GCH below
$\lambda$. In particular, this means that all the successors of $C$ will be
strongly inaccessible in $\tilde{V}[G]$. The reason for this slightly arcane
scenario is that we will eventually need to have definitions for
$\operatorname{Col}_{\lambda}$ in two different universes:
$\tilde{V}^{\operatorname{NM}_{\lambda}}$, and
$\tilde{V}^{\mathbb{Q}_{\lambda}}$. We have already seen that
$\operatorname{NM}_{\lambda}$ preserves GCH; once we define
$\mathbb{Q}_{\lambda}$ we will discover that it also preserves GCH below
$\lambda$.

The forcing we want to work with is a combination of these collapsing
forcings, and is defined in terms of $C$. We want a forcing which will
collapse, for each $\gamma\in C$, every cardinal between $f(\gamma)$ and
$\operatorname{Succ}_{C}(\gamma)$. Once we have done this, it will be easy to
see that it makes $\lambda$ the first element of
$\operatorname{Reg}_{\alpha}$. The forcing we want to use is the Easton
product of all the collapsing forcings:

######  Definition 5.9.

Let $\tilde{V}[G]$ be some generic extension of $\tilde{V}$, which contains an
$\operatorname{NM}_{\lambda}^{\tilde{V}}$ generic club $C$, and preserves GCH
below $\lambda$. We define the forcing $\operatorname{Col}_{\lambda}$ to be
the set of all the elements $h$ of

$\prod_{\gamma\in
C}\operatorname{Col}(f(\gamma),<\operatorname{Succ}_{C}(\gamma))$

such that for all regular cardinals $\mu\leq\lambda$, the set

$\\{\gamma\in C\cap\mu:h(\gamma)\neq\emptyset\\}$

is bounded in $\mu$.

We order $\operatorname{Col}_{\lambda}$ in the obvious way.

Note that $\operatorname{Col}_{\lambda}$ depends critically on our choice of
$\operatorname{NM}_{\lambda}$ generic club $C$. So really, we should include
$C$ as a subscript. But there will never be more than a single generic club
below $\lambda$, so we will leave this implicit.

$\operatorname{Col}_{\lambda}$ also depends on which cardinals are regular. So
even if two universes both contain the same generic club $C$, we may find that
$\operatorname{Col}_{\lambda}$ is different. In particular, once we have
defined $\mathbb{Q}_{\lambda}$ we will find that
$\operatorname{Col}_{\lambda}$ will be different in
$\tilde{V}^{\operatorname{NM}_{\lambda}}$ and
$\tilde{V}^{\mathbb{Q}_{\lambda}}$, even if both generic extensions add the
same club.

If $c$ is a closed subset of $\lambda$, then we extend the above definition by
defining $\operatorname{Col}(c)$ in the obvious way, as the Easton support
product of $\operatorname{Col}(f(\gamma),<\operatorname{Succ}_{c}(\gamma))$
forcings for $\gamma\in c\setminus\max(c)$. If $c\in\tilde{V}$ then this can
be defined in the ground model $\tilde{V}$, before we add an
$\operatorname{NM}_{\lambda}$ generic club $C$. Of course, once we do add $C$,
we know that if $c$ is an initial segment of $C$ then $\operatorname{Col}(c)$
will be a subset of $\operatorname{Col}(C)=\operatorname{Col}_{\lambda}$.
(Technically, this is only true if we allow a minor abuse of notation about
the domains of the functions in $\operatorname{Col}(c)$ and
$\operatorname{Col}_{\lambda}$.)

######  Proposition 5.10.

Let $c$ be a closed set of cardinals (in $\tilde{V}$ or $\tilde{V}[G]$) whose
successors are all inaccessibles below $\kappa$. If $\gamma=\min(c)$ then
$\operatorname{Col}(c)$ is $<f(\gamma)$ closed in any universe where this
makes sense. In particular, this means it is $\gamma^{+}$ closed.

###### Proof.

An easy definition chase we leave to the reader. ∎

######  Lemma 5.11.

In any universe where $\operatorname{Col}_{\lambda}$ is defined (i.e. any
universe containing an $\operatorname{NM}_{\lambda}$ generic club $C$ whose
successors are strongly inaccessible), it does not collapse any cardinals
except those which it is supposed to (i.e. those in the interval
$(f(\gamma),\operatorname{Succ}_{C}(\gamma))$ for some $\gamma\in C$). Nor
does it singularise any other cardinals.

###### Proof.

Let $\mu$ be a cardinal which is not in any of the intervals that are supposed
to be collapsed by $\operatorname{Col}_{\lambda}$. Let
$\mathbb{P}=\operatorname{Col}(C{\upharpoonleft}\mu)$ (in the sense of
$\tilde{V}[C]$, not $\tilde{V}$) and let
$\mathbb{R}=\operatorname{Col}(C\setminus\mu)$. Then
$\operatorname{Col}_{\lambda}=\mathbb{P}\times\mathbb{R}$. Now $\mathbb{P}$
has cardinality less than $\mu$ (by GCH) and therefore does not collapse or
singularise $\mu$. On the other hand, $\mathbb{R}$ is $\mu$ closed, and so
again it does not collapse or singularise $\mu$. ∎

######  Corollary 5.12.

In any universe where $\operatorname{Col}_{\lambda}$ is defined, if
$0<\beta<\alpha$ then $\operatorname{Col}_{\lambda}$ does not modify
$\operatorname{Reg}_{\beta}$, other than removing those elements which are in
an interval $(f(\gamma),\operatorname{Succ}_{C}(\gamma))$.

###### Proof.

Let $\mu\in\operatorname{Card}$ not be in
$(f(\gamma),\operatorname{Succ}_{C}(\gamma))$ for any $\gamma\in C$. We show
that $\mu\in\operatorname{Reg}_{\beta}$ in the $\operatorname{Col}_{\lambda}$
generic extension if and only if $\mu\in\operatorname{Reg}_{\beta}$ in
$\tilde{V}[G]$.

Case 1: $\mu<\min(C)$, $\mu\in(\gamma,f(\gamma)]$ or $\mu>\lambda$. All the
cardinals in these intervals are preserved, and not singularised, by
$\operatorname{Col}_{\lambda}$. So $\operatorname{Reg}_{\beta}$ is preserved
over these intervals too.

Case 2: $\mu\in C$ is a successor of $C$. By definition of
$\operatorname{NM}_{\lambda}$, we know $\mu$ is an element of
$\operatorname{Reg}_{\alpha}$ in $\tilde{V}[G]$, and hence not in
$\operatorname{Reg}_{\beta}$. In the generic extension $\mu$ becomes a
successor cardinal, and hence again not in $\operatorname{Reg}_{\beta}$.

Case 3: $\mu\in C$ is a limit of $C$. By definition of
$\operatorname{NM}_{\lambda}$, we know $\mu$ is a successor cardinal in
$\tilde{V}[G]$ and hence also in the $\operatorname{Col}_{\lambda}$ generic
extension.

Case 4: $\mu=\lambda$. In $\tilde{V}$, we know that $\lambda>\beta$ is
hyperinaccessible, and therefore a limit of elements of
$\operatorname{Reg}_{\beta}$. By assumption, $\tilde{V}[G]$ agrees with
$\tilde{V}$ on $\operatorname{Reg}{\upharpoonleft}\lambda$, so in particular
in $\tilde{V}[G]$ we know that $\lambda$ is still a limit of
$\operatorname{Reg}_{\beta}$. So it cannot be an element of
$\operatorname{Reg}_{\beta}$. Recall that every interval $(\gamma,f(\gamma)]$
contains an element of $\operatorname{Reg}_{\beta}$. So by Case 1, we know
that in the generic extension $\lambda$ is still a limit of elements of
$\operatorname{Reg}_{\beta}$, and hence cannot be in
$\operatorname{Reg}_{\beta}$. ∎

Notice also that in the $\operatorname{Col}_{\lambda}$ generic extension,
there are no elements of $\operatorname{Reg}_{\alpha}$ below $\lambda$, and
that $\lambda$ is a limit of elements of $\operatorname{Reg}_{\beta}$ for all
$\beta<\alpha$. So if $\lambda$ is regular in $\tilde{V}[G]$ (and hence in the
$\operatorname{Col}_{\lambda}$ generic extension) then it will be the first
element of $\operatorname{Reg}_{\alpha}$.

###  5.4. The Prikry-style forcing Q

Again, we fix a $\lambda^{+}$ supercompact $\lambda\leq\kappa$ in a universe
$\tilde{V}$ which knows about $f$.

The forcing $\mathbb{Q}_{\lambda}$ is similar to a Prikry forcing, but with
some extra components. We want any $\mathbb{Q}_{\lambda}$ generic filter to
not only singularise $\lambda$, but also to define a
$\operatorname{NM}_{\lambda}$ generic club $C$. For reasons that will become
clearer later, we also want it to define some kind of “generic element” of
$\operatorname{Col}_{\lambda}$.

Where a standard Prikry forcing would add an $\omega$ sequence of single
ordinals, we will arrange for $\mathbb{Q}_{\lambda}$ to add an $\omega$
sequence whose terms are taken from the following set:

######  Definition 5.13.

$K_{\lambda}$ is the set of all triples $\langle c,x,\epsilon\rangle$ where:

  1. (1)

$c\in\operatorname{NM}_{\lambda}$;

  2. (2)

$h\in\operatorname{Col}(c)$;

  3. (3)

$\epsilon\in\operatorname{On}$ and $\max(c)<\epsilon<\lambda$.

For $\beta<\lambda$, $K_{\lambda}^{\beta}$ is the subset of $K_{\lambda}$
consisting of all the triples $\langle c,h,\epsilon\rangle$ where the least
element of $c$ is greater than $\beta$ (and thus also $\epsilon>\beta$).

The Prikry-style generic sequence we’re aiming for will consist of an element
$\langle c_{0},h_{0},\epsilon_{0}\rangle$ of $K_{\lambda}$, then an element
$\langle c_{1},h_{1},\epsilon_{1}\rangle$ of $K_{\lambda}^{\epsilon_{0}}$, and
so on up through the all $n\in\omega$.

From such a sequence $G$, we will be able to extract a club
$C=\bigcup_{n\in\omega}c_{n}$ in $\lambda$, an element
$H=\bigcup_{n\in\omega}h_{n}$ of $\prod_{\gamma\in
C}\operatorname{Col}(\gamma,<\operatorname{Succ}_{C}(\gamma))$ and an
$\omega$-sequence $(\epsilon_{n})$. If we set up the forcing correctly, we
will later discover that $C$ is $\operatorname{NM}_{\lambda}$ generic, that
$H\in\operatorname{Col}^{\tilde{V}[G]}_{\lambda}$, and that $(\epsilon_{n})$
is cofinal in $\lambda$.

Recall that a condition in a Prikry forcing consists of two components: a
finite stem of the sequence we’re constructing, and a “large” (i.e. measure
$1$) set of places that are allowed to be in later parts of the sequence.

It’s fairly easy to see what the finite stem should look like in this context.
But how can we find an analogue of a measure $1$ set of ordinals? We must
define a measure on $K_{\lambda}$, and in fact on $K_{\lambda}^{\beta}$ for
every $\beta<\lambda$.

To do this, we first define an ordering on $K_{\lambda}$:

######  Definition 5.14.

Let $\langle c,h,\epsilon\rangle$ and
$\langle\tilde{c},\tilde{h},\tilde{\epsilon}\rangle$ be elements of
$K_{\lambda}$. We say that
$\langle\tilde{c},\tilde{h},\tilde{\epsilon}\rangle\leq^{*}\langle
c,h,\epsilon\rangle$ if:

  1. (1)

$\tilde{c}\leq c$, that is, $\tilde{c}$ is an end extension of $c$;

  2. (2)

$\tilde{h}{\upharpoonleft}\max(c)=h$;

  3. (3)

$\tilde{\epsilon}\geq\epsilon$.

Notice the unusual nature of the second clause. It’s not enough that
$\tilde{h}\leq h$ in the $\operatorname{Col}(\tilde{c})$ ordering. It must
actually agree completely with $h$, although it is allowed to add more things
once we’re above the area where $h$ is defined. This is because $H=\bigcup
h_{n}$ is supposed to be a condition in $\operatorname{Col}_{\lambda}$, not a
$\operatorname{Col}_{\lambda}$ generic filter.

Of course, this also defines an ordering on $K_{\lambda}^{\beta}\subset
K_{\lambda}$ for every $\beta<\lambda$.

######  Lemma 5.15.

Let $\beta<\lambda$. Let $F$ be the family of all subsets of
$K_{\lambda}^{\beta}$ which contain a $\leq^{*}$ dense open subset of
$K_{\lambda}^{\beta}$. Then $F$ is a $\lambda$ complete filter in
$K_{\lambda}^{\beta}$ in the usual $\subset$ ordering of
$\mathcal{P}(K_{\lambda}^{\beta})$.

###### Proof.

Clearly, $F$ is upwards closed and nonempty. The intersection of fewer than
$\lambda$ many $\leq^{*}$ dense open sets can be easily seen to be dense open,
so $F$ is a $\lambda$ complete filter. ∎

The following is a standard consequence of $\lambda^{+}$ supercompactness:

######  Lemma 5.16.

[2, 22.17] Let $S$ be a set of size $\lambda$, and let $F$ be a $\lambda$
complete filter on $\mathcal{P}(S)$. Then $F$ can be extended to a $\lambda$
complete ultrafilter $U$.

###### Proof.

Without loss of generality, let us assume $S=\lambda$. Note that $\lvert
F\rvert\leq 2^{\lambda}=\lambda^{+}$. Let $j:\tilde{V}\rightarrow M$ be an
embedding with critical point $\lambda$ such that $j(\lambda)>\lambda^{+}$ and
$M^{\lambda^{+}}\subset M$. By elementarity $j(F)\in M$ is $j(\lambda)$
complete. Since $\lvert j^{\prime\prime}(F)\rvert\leq\lambda^{+}$ (in
$\tilde{V}$) and $M$ is closed under $\lambda^{+}$ sequences, we know that
$j^{\prime\prime}(F)\in M$. Since also $\lambda^{+}\leq j(\lambda)$ and
$j^{\prime\prime}(F)\subset j(F)$, we know by $j(\lambda)$ completeness of
$j(F)$ that $\cap j^{\prime\prime}(F)\in j(F)$. In particular, there is some
$\gamma\in\cap j^{\prime\prime}(F)$. Now define an ultrafilter $U\in\tilde{V}$
by:

$X\in U\iff X\subset\lambda\wedge\gamma\in j(X)$

It is easy to check that $F\subset U$, that $U$ is an ultrafilter, and that it
is $\lambda$ complete. ∎

######  Corollary 5.17.

Let $\beta<\lambda$. There is an ultrafilter $U_{\beta}$ on
$K_{\lambda}^{\beta}$ which contains all the $\leq^{*}$ dense open subsets of
$K_{\lambda}^{\beta}$.

In fact, of course, there will be many such ultrafilters, but we will fix a
single one for the rest of this section to call $U_{\beta}$.

With $U_{\beta}$ in hand, we can define the analogue of the measure $1$ set of
ordinals in a Prikry forcing.

######  Definition 5.18.

Let $T$ be a tree of height $\omega$, whose nodes are all elements of
$K_{\lambda}$. We abuse notation by allowing the same element of $K_{\lambda}$
to appear multiple times, provided no element appears twice as direct
successors of the same node. We say $T$ is nice if the following hold:

  1. (1)

If $\langle\tilde{c},\tilde{h},\tilde{\epsilon}\rangle<_{T}\langle
c,h,\epsilon\rangle\in T$, then
$\langle\tilde{c},\tilde{h},\tilde{\epsilon}\rangle\in
K_{\lambda}^{\epsilon}$;

  2. (2)

If $\langle c,h,\epsilon\rangle\in T$ then the set of its direct successors
(which is a subset of $K_{\lambda}^{\epsilon}$ by the previous condition) is
in $U_{\epsilon}$.

Recall that a condition in a Prikry forcing contains two components: a finite
sequence of ordinals, and a measure $1$ set. The finite sequence fixes an
initial segment of the $\omega$ sequence we are going to add, and the rest of
the sequence is chosen from elements of the measure $1$ set. Analogously, in
the forcing $\mathbb{Q}_{\lambda}$, our conditions will have two components: a
finite sequence $s$ of terms from $K_{\lambda}$ and a nice tree $T$ which
gives us a map of where the sequence is allowed to go from there. The root of
$T$ will the final term of $s$. Then for $n>0$, the $n$’th level contains all
the elements of $K_{\lambda}$ which we are allowing to appear as the $n$’th
term in the undetermined part of the sequence. A branch through the tree
corresponds to a (not necessarily generic) way to complete the sequence.

######  Definition 5.19.

The forcing $\mathbb{Q}_{\lambda}$ has conditions of the form

$\big{(}(\langle c_{0},h_{0},\epsilon_{0}\rangle,\langle
c_{1},h_{1},\epsilon_{1}\rangle,\ldots,\langle
c_{n-1},h_{n-1},\epsilon_{n-1}\rangle),T\big{)}$

for some $n\in\omega$, where

  1. (1)

$\langle c_{0},h_{0},\epsilon_{0}\rangle\in K_{\lambda}$ if $n>0$;

  2. (2)

For $0<i<n$, $\langle c_{i},h_{i},\epsilon_{i}\rangle\in
K_{\lambda}^{\epsilon_{i-1}}$;

  3. (3)

$T$ is a nice tree

  4. (4)

The root of $T$ is $\langle c_{n-1},h_{n-1},\epsilon_{n-1}\rangle$.

We call $\epsilon_{n-1}$ the height of the condition (writing
$\operatorname{ht}$ in symbols).

The conditions are ordered in the usual way for a Prikry style forcing:
$(s^{\prime},T^{\prime})\leq(s,T)$ if $s^{\prime}$ is an end extension of $s$
and there is a path $B=b_{0},b_{1},\ldots,b_{k}$ through $T$ of length
$k:=\lvert s^{\prime}\setminus s\rvert+1$ such that:

  1. (1)

$b_{0}$ is the root of $T$ (i.e. the last element of $s$)

  2. (2)

For all $0<i\leq k$, $b_{i}$ is the $i$’th element of $s^{\prime}\setminus s$

  3. (3)

$T^{\prime}$ is a subtree of $T$ whose root is $b_{k}$

As usual for Prikry forcings, if $s^{\prime}=s$ we say
$(s^{\prime},T^{\prime})$ is a direct extension of $(s,T)$ and write
$(s^{\prime},T^{\prime})\leq^{*}(s,T)$.

Note: We now have two different definitions of $\leq^{*}$. One talks about
elements of $K_{\lambda}$ and the other about conditions in
$\mathbb{Q}_{\lambda}$, so it should be easy to understand which one we are
talking about.

######  Proposition 5.20.

The forcing $(\mathbb{Q}_{\lambda},\leq^{*})$ is ${<}\lambda$ closed. That is,
any descending sequence $T_{0}\supset T_{1}\supset T_{2}\ldots$ of nice trees
with the same root will have a nice tree as their intersection.

###### Proof.

Follows from the fact that $U_{\epsilon}$ is closed under $<\lambda$
intersections. ∎

The following lemma is very standard for Prikry style forcings.

######  Lemma 5.21.

Let $p\in\mathbb{Q}_{\lambda}$. Let $\varphi$ be first order (perhaps with
parameters). Then there is some $q\leq^{*}p$ deciding $\varphi$.

###### Proof.

We’ll essentially follow the standard proof of this result for Prikry
forcings. However, the argument gets rather technical to state, because we
need to construct a tree analogue of the diagonal intersection of measure $1$
sets. The way we do this is really quite simple and natural, but unfortunately
it’s also rather messy to write out.

Let $p=(s,T)$. We shall recursively construct a condition
$r=(s,T^{\prime})\leq^{*}p$, together with a collection of nice subtrees
$\\{T_{t}\subset T:t\in T^{\prime}\\}$, where $T_{t}$ has root $t$. For $t\in
T^{\prime}$, we define $s_{t}$ to be the sequence of predecessors of $t$ in
$T^{\prime}$, starting at the root of $T^{\prime}$ and ending at $t$ itself.
For each $t\in T^{\prime}$, the intention is that $(s\cup
s_{t},T^{\prime}{\upharpoonleft}t)$ should be a condition of
$\mathbb{Q}_{\lambda}$ below $(s\cup s_{t},T_{t})$; and that if at all
possible both will decide $\varphi$.

First, suppose $t$ is the root of $T$ (which will also have to be the root of
$T^{\prime}$, since $(s,T^{\prime})$ is supposed to be a condition of
$\mathbb{Q}_{\lambda}$). If there is some $r_{t}\leq^{*}p$ which decides
$\varphi$, then let $T_{t}$ be its associated tree. (Of course, if such an
$r_{t}$ exists, then we’re already done!). Otherwise, let $T_{t}=T$.

Now, let $n>0$ and assume that we have already defined levels $1\ldots,n$ of
$T^{\prime}$, as well as $T_{t}$ for all $t$ in these levels of $T^{\prime}$.
Let $t\in T^{\prime}$ be at level $n$. Then we define the direct successors of
$t$ in $T^{\prime}$ to be the level $1$ elements of $T_{t}$ (all of which are,
by inductive hypothesis, direct successors of $t$ in $T$). In doing this, we
have define the whole of level $n+1$ of $T^{\prime}$. It remains to define the
trees $T_{u}$ of elements $u$ of this level.

If $u\in T^{\prime}$ is a direct successor of $t$, then let $T_{u}^{0}$ be the
restriction $T_{t}{\upharpoonleft}u$ of $T_{t}$ to $u$. (So $T_{u}^{0}$ has
root $u$ and consists of all elements of $T_{t}$ which are below $u$.)
Consider the condition $p_{u}:=(s\cup s_{u},T_{u}^{0})$. Since $u\in T$ and
$T_{t}$ is a subtree of $T$, we know that $p_{u}\leq p$. If there is some
direct extension $r_{u}\leq^{*}p_{u}$ which decides $\varphi$, then let
$T_{u}$ be the tree in $r_{u}$. If no such $r_{t}$ exists, then we define
$T_{u}:=T_{u}^{0}$. This completes the recursive definition.

It is easy to verify that $r:=(s,T^{\prime})\leq^{*}p$, and that it has the
following property: if $\tilde{r}=(\tilde{s},\tilde{T})\leq r$ decides
$\varphi$, then so does $(\tilde{s},T^{\prime}{\upharpoonleft}\max\tilde{s})$.

The next step is to recursively construct another condition
$q=(s,T^{\prime\prime})\leq^{*}r$ which actually decides $\varphi$ itself.
This time, the construction is a little simpler: we don’t need the auxiliary
$T_{t}$ trees. The root of $T^{\prime\prime}$ is the same as that of
$T^{\prime}$ and $T$, of course.

Say that we have built the first $n$ levels of $T^{\prime\prime}$. Let $t$ be
in the $n$’th level of $T^{\prime\prime}$, and let $A\subset K_{\lambda}$ be
the set of all its direct successors in $T^{\prime}$. Recall that $A$ is
measure $1$ in the sense of $U_{\beta}$, where $\beta$ is the height of $t$.
We partition $A$ into three parts: we say that $u\in A$ is in $A_{1}$ if
$q_{u}:=(s\cup s_{u},T^{\prime}{\upharpoonleft}u)$ decides that $\varphi$ is
true, $u$ is in $A_{2}$ if $q_{u}$ decides $\varphi$ is false, and $u$ is in
$A_{3}$ if $q_{u}$ doesn’t decide $\varphi$ at all.

One of $A_{1}$, $A_{2}$ and $A_{3}$ will be measure $1$ in the sense of
$U_{\beta}$. The direct successors of $t$ in $T^{\prime\prime}$ are defined to
be the elements of that measure $1$ set. It is again easy to verify that this
defines a nice tree $T^{\prime\prime}$ and that
$q:=(s,T^{\prime\prime})\in\mathbb{Q}_{\lambda}$. It is also easy to check
that $q\leq^{*}r\leq^{*}p$ and that $q$ has the following two properties:

  1. (1)

If $\tilde{q}=(\tilde{s},\tilde{T})\leq q$ decides $\varphi$, then so does
$q^{\prime}:=(\tilde{s},T^{\prime\prime}{\upharpoonleft}\tilde{s})$ (this
follows from $q\leq r$); and

  2. (2)

For any extension
$q^{\prime}=(\tilde{s},T^{\prime\prime}{\upharpoonleft}\tilde{s})$ of $q$,
either all the one step extensions of $q^{\prime}$ fail to decide $\varphi$,
or they all decide it and make the same decision.

Let $\tilde{q}=(\tilde{s},\tilde{T})\leq q$ be some condition which decides
$\varphi$, and let it be such that $\tilde{s}$ has minimal length among
conditions which decide $\varphi$. Then by the first property, $\varphi$ is
decided by
$q^{\prime}:=(\tilde{s},T^{\prime\prime}{\upharpoonleft}\tilde{s})$. Suppose,
seeking a contradiction, that $\tilde{s}$ is a proper extension of $s$, and
let $\tilde{s}^{\prime}$ be $\tilde{s}$ with its final term $t$ omitted. (So
by assumption, $\tilde{s}^{\prime}$ still extends $s$.)

Then $q^{\prime}$ is a one step extension of
$q^{\prime\prime}:=(\tilde{s}^{\prime},T^{\prime\prime}{\upharpoonleft}\tilde{s}^{\prime})$,
and decides $\varphi$. So by the second property, all the one-step extensions
of $q^{\prime\prime}$ decide $\varphi$, and agree on that decision. But then
$q^{\prime\prime}$ decides $\varphi$ as well, and this is a contradiction
since $\tilde{s}^{\prime}$ is shorter than $\tilde{s}$.

So in fact $\tilde{s}=s$ and $\tilde{q}\leq^{*}q$ decides $\varphi$. ∎

######  Corollary 5.22.

Let $\varphi(x)$ be a formula with one free variable, let $\mu\leq\lambda$,
and let $p\mathrel{\Vdash}\exists x<\check{\mu}\,\varphi(x)$. Then there is
some $q\leq^{*}p$ and some $\gamma<\mu$ such that
$q\mathrel{\Vdash}\varphi(\check{\gamma})$.

###### Proof.

If no such $q$ exists, then let $p=q_{0}\geq^{*}q_{1}\geq^{*}q_{2}\ldots$ be a
descending sequence of conditions of length $\mu+1$ defined as follows. For
$i=j+1$, we choose $q_{i}\leq^{*}q_{j}$ deciding $\varphi(\check{j})$; by
assumption $q_{i}\mathrel{\Vdash}\neg\varphi(\check{j})$. At limit $i$ we take
some $q_{i}$ which is $\leq^{*}$ every $q_{j},j<i$. This exists by $<\lambda$
closure of $(\mathbb{Q}_{\lambda},\leq^{*})$. Then
$q_{\mu}\leq^{*}p\mathrel{\Vdash}\forall x<\check{\nu}\neg\varphi(x)$.
Contradiction. ∎

This tells us that $\mathbb{Q}_{\lambda}$ does not do any unexpected
collapsing or singularising of cardinals.

######  Lemma 5.23.

$\mathbb{Q}_{\lambda}$ does not add any new bounded subsets of $\lambda$,
collapse any cardinals, or singularise any cardinals apart from $\lambda$. It
preserves GCH where it holds in $\tilde{V}$, except at $\lambda$.

###### Proof.

First, note that $\mathbb{Q}_{\lambda}$ has the $\lambda^{+}$ chain condition,
because any two conditions of $\mathbb{Q}_{\lambda}$ with the same stem are
compatible. Hence, it does not collapse or singularise any cardinals
$\geq\lambda^{+}$, or change the cardinalities of their power sets.

Let $\mu<\lambda$ be a cardinal of $\tilde{V}$, and suppose that
$\mathbb{Q}_{\lambda}$ collapses it to some cardinal $\nu<\mu$. Let $\dot{g}$
be a name for a bijection $g:\nu\rightarrow\mu$. Let
$p\mathrel{\Vdash}\dot{g}:``\check{\nu}\rightarrow\check{\mu}\text{ is a
bijection}"$. We construct a descending chain of direct extensions
$p=p_{0}\geq^{*}p_{1}\geq^{*}\ldots$ of length $\nu+1$. If $i$ is a successor,
then we use the previous corollary to take $p_{i}$ deciding the value of
$\dot{g}(i)$; at limit $i$ we take some $p_{i}$ which is $\leq^{*}p_{j}$ for
all $j<i$. Then $p_{\nu}$ decides what $g$ is, and hence $g\in\tilde{V}$.
Contradiction. A similar proof shows that $\mathbb{Q}_{\lambda}$ does not
singularise $\mu$ either, and that it adds no bounded subsets of $\lambda$.
(For the latter, we start with some $p\in\mathbb{Q}_{\lambda}$ which decides
what the bound will be, and then take a $\leq^{*}$ descending chain whose
length is that bound, deciding which elements of $\lambda$ will be in the new
subset.)

This also means that $\lambda$ is a limit of cardinals in $\tilde{V}[G]$, and
hence is still a cardinal, and that the power sets of cardinals below
$\lambda$ are preserved. ∎

All our definitions of $\operatorname{NM}_{\lambda}$,
$\operatorname{Col}_{\lambda}$ and $\mathbb{Q}_{\lambda}$ have been given only
for $\lambda\leq\kappa$. The only reason for this is because we use $f$ in
their definition, and $f$ is only defined up to $\kappa$. We will later want
to deal with elementary embeddings $j:\tilde{V}\rightarrow M$ with critical
point $\kappa$. From the perspective of such a model $M$, we can extend the
definition by introducing analogous forcings in terms of $j(f)$, for any
$\lambda\leq j(\kappa)$. We will extend the notation by referring to these
forcings also as $\operatorname{NM}_{\lambda}$, $\operatorname{Col}_{\lambda}$
and $\mathbb{Q}_{\lambda}$. Since $j(f)$ will be an end-extension of $f$, for
$\lambda\leq\kappa$ the forcings $\operatorname{NM}_{\lambda}$,
$\operatorname{Col}_{\lambda}$ and $\mathbb{Q}_{\lambda}$ are defined the same
way in $M$ whether we use $f$ or $j(f)$ in their definitions, so there is no
ambiguity in doing this.

###  5.5. Embedding NM*Col into Q*Col

Once again, fix a $\lambda^{+}$ supercompact $\lambda\leq\kappa$ in a universe
$\tilde{V}$ which knows about $f$. As we discussed earlier, we can extract
elements of $\operatorname{NM}_{\lambda}$ and $\operatorname{Col}_{\lambda}$
from a $\mathbb{Q}_{\lambda}$ condition.

######  Definition 5.24.

We define two schemes of abbreviations.

  1. (1)

Let

$p=((\langle c_{0},h_{0},\epsilon_{0}\rangle,\langle
c_{1},h_{1},\epsilon_{1}\rangle,\ldots,\langle
c_{n-1},h_{n-1},\epsilon_{n-1}\rangle),T)$

be a condition in $\mathbb{Q}_{\lambda}$. We define $c_{p}=\bigcup_{i<n}c_{i}$
and $h_{p}=\bigcup_{i<n}h_{i}$.

  2. (2)

Let $G$ be $\mathbb{Q}_{\lambda}$ generic. We define $C_{G}=\bigcup_{p\in
G}c_{p}$ and $H_{G}=\bigcup_{p\in G}h_{p}$.

An equivalent, but less friendly, definition is that

$C_{G}=\bigcup\\{c:\exists h,\epsilon\,\,\langle c,h,\epsilon\rangle\text{ is
a term in the first part of some }p\in G\\}$

and that

$H_{G}=\bigcup\\{h:\exists c,\epsilon\,\,\langle c,h,\epsilon\rangle\text{ is
a term in the first part of some }p\in G\\}$

We shall now establish what these four objects actually are, in terms of
$\operatorname{NM}_{\lambda}$ and $\operatorname{Col}_{\lambda}$. The first
two objects, $c_{p}$ and $h_{p}$, are simply conditions of the relevant
forcings in $\tilde{V}$:

######  Proposition 5.25.

Let $p\in\mathbb{Q}_{\lambda}$. Then $c_{p}\in\operatorname{NM}_{\lambda}$ and
$h_{p}\in\operatorname{Col}(c_{p})$ in $\tilde{V}$.

###### Proof.

$c_{p}$ is a union of finitely many conditions in
$\operatorname{NM}_{\lambda}$, so it is certainly closed, bounded, and its
successors and limits have the right properties. So
$c_{p}\in\operatorname{NM}_{\lambda}$.

In the notation used in the definition, $h_{p}$ is a union of one condition
from each $\operatorname{Col}(c_{i})$, $i<n$. Thus it’s certainly an element
of

$\prod_{\gamma\in
c\setminus\\{\max(c)\\}}\operatorname{Col}(f(\gamma),<\operatorname{Succ}_{C}(\gamma))$

Moreover, $\max(c_{i-1})<\epsilon_{i-1}<\min(c_{i})$ for all $0<i<n$, so the
domain of $h_{p}$ is trivially bounded below $\min(c_{i})$ for $i<n$. It is
also bounded below all other regular ordinals, by definition of
$\operatorname{Col}(c_{i})$. Hence, $h_{p}\in\operatorname{Col}(c_{p})$. ∎

$C_{G}$ will be generic for $\operatorname{NM}_{\lambda}$, and there is a
useful correspondence between $\operatorname{NM}_{\lambda}$ and
$\mathbb{Q}_{\lambda}$:

######  Lemma 5.26.

If $G$ is $\mathbb{Q}_{\lambda}$ generic, then $C_{G}$ is a club which is
$\operatorname{NM}_{\lambda}$ generic (over $\tilde{V}$). Moreover, given any
$\operatorname{NM}_{\lambda}$ name $\sigma$, there is a $\mathbb{Q}_{\lambda}$
name $\varphi(\sigma)$ such that for any $\mathbb{Q}_{\lambda}$ filter $G$
(not necessarily generic), $\sigma^{C_{G}}=\varphi(\sigma)^{G}$.

(As usual, in the statement of the lemma we are muddling the definition of a
generic filter over $\operatorname{NM}_{\lambda}$, and the club corresponding
to that generic filter.)

###### Proof.

It is easy to see that $C_{G}$ is a club in $\lambda$, and that all its closed
initial segments are elements of $\operatorname{NM}_{\lambda}$. We must show
that it is generic.

Let $D\subset\operatorname{NM}_{\lambda}$ be open dense. We will show that
$X_{D}:=\\{p\in\mathbb{Q}_{\lambda}:c_{p}\in D\\}$ is dense in
$\mathbb{Q}_{\lambda}$. This implies $G\cap X_{D}\neq\emptyset$ and so $C_{G}$
contains an element of $D$ as required.

Fix a condition $p=(s,T)\in\mathbb{Q}_{\lambda}$.

$D$ is dense in $\operatorname{NM}_{\lambda}$, so the set

$S_{D,p}:=\\{\langle c,h,\epsilon\rangle\in
K_{\lambda}^{\operatorname{ht}(p)}:c_{p}\cup c\in D\\}$

is $\leq^{*}$ dense in $K_{\lambda}^{\operatorname{ht}(p)}$. Clearly, it is
also open.

So $S_{D,p}\in U_{\operatorname{ht}(p)}$ because $U_{\operatorname{ht}(p)}$
contains all the open dense subsets of $K_{\lambda}^{\operatorname{ht}(p)}$.
Since the set of all successors of the root of $T$ is also in
$U_{\operatorname{ht}(p)}$, there must be a level $1$ element of the tree $T$
which is in $S_{D,p}$. But then that gives us a $1$ step extension $q\leq p$
in $\mathbb{Q}_{\lambda}$, such that the extra term $\langle
c,h,\epsilon\rangle$ in $q$ is an element of $S_{D,p}$. It follows that
$c_{q}=c_{p}\cup c\in D$, and hence $q\in X_{D}$ as required.

The second part of the lemma is an easy recursive definition: we take

$\varphi(\sigma):=\\{\langle\varphi(\tau),p\rangle:p\in\mathbb{Q}_{\lambda}\wedge\langle\tau,c_{p}\rangle\in\sigma\\}$

and verify that the required equality holds for any $\mathbb{Q}_{\lambda}$
filter $G$. ∎

So $\mathbb{Q}_{\lambda}$ behaves well regarding
$\operatorname{NM}_{\lambda}$. What about $\operatorname{Col}_{\lambda}$? We
want $\mathbb{Q}_{\lambda}*\operatorname{Col}^{\tilde{V}[G]}_{\lambda}$ to
play nicely with
$\operatorname{NM}_{\lambda}*\operatorname{Col}^{\tilde{V}[C_{G}]}_{\lambda}$.
Recall that in both cases, $\operatorname{Col}_{\lambda}$ is defined using the
same club $C_{G}$. However, $\operatorname{Col}^{\tilde{V}[G]}(C_{G})$ and
$\operatorname{Col}^{\tilde{V}[C_{G}]}(C_{G})$ are not the same. Remember that
the Easton product we used to define $\operatorname{Col}$ only asks for its
conditions to be bounded below regular ordinals. $\lambda$ is regular in
$\tilde{V}[C_{G}]$ but is singular in $\tilde{V}[G]$. So there are sets in
$\tilde{V}[C_{G}]$ which are in $\operatorname{Col}^{\tilde{V}[G]}(C_{G})$ but
not in $\operatorname{Col}^{\tilde{V}[C_{G}]}(C_{G})$. On the other hand, we
do at least know that
$\operatorname{Col}^{\tilde{V}[C_{G}]}(C_{G})\subset\operatorname{Col}^{\tilde{V}[G]}(C_{G})\cap\tilde{V}[C_{G}]$.

This is where the condition $H_{G}$ named by $G$ comes in. It is a sort of
“generic element” of $\operatorname{Col}^{\tilde{V}[G]}(C_{G})$ which forces
any generic extension containing it to cooperate in the way we want despite
this difficulty.

First, we must verify that $H_{G}$ really is a condition.

######  Lemma 5.27.

Let $G$ be $\mathbb{Q}_{\lambda}$ generic. Then
$H_{G}\in\operatorname{Col}^{\tilde{V}[G]}_{\lambda}$.

###### Proof.

It is easy to see that $H_{G}\in\prod_{\gamma\in
C_{G}}\operatorname{Col}(\gamma,<\operatorname{Succ}_{C_{G}}(\gamma))$. We
must verify that its support is bounded below every $\tilde{V}[G]$ regular
cardinal $\mu\in[0,\lambda]$. If $\mu<\lambda$, this follows from Proposition
5.25: take $p\in G$ such that $\sup c_{p}>\mu$, and then
$H_{p}|\mu=h_{p}\in\operatorname{Col}(c_{p})$ and hence (since $\mu$ is
regular in $\tilde{V}$) the support of $H_{p}$ is bounded below $\mu$.

On the other hand, we know that $\mathbb{Q}_{\lambda}$ singularises $\lambda$,
so the case $\mu=\lambda$ is vacuous. ∎

The value of $H_{G}$ is shown in the following rather technical lemma.

######  Lemma 5.28.

Let $G$ be $\mathbb{Q}_{\lambda}$ generic. Let $G^{*}$ be
$\operatorname{Col}^{\tilde{V}[G]}_{\lambda}$ generic over $\tilde{V}[G]$, and
contain $H_{G}$. Then the filter
$G^{**}:=G^{*}\cap\operatorname{Col}^{\tilde{V}[C_{G}]}_{\lambda}$ is
$\operatorname{Col}^{\tilde{V}[C_{G}]}_{\lambda}$ generic over
$\tilde{V}[C_{G}]$.

###### Proof.

It is easy to check that $G^{**}$ is indeed a filter; the challenge is showing
that it’s generic. So let $\dot{D}$ be an $\operatorname{NM}_{\lambda}$ name
for a dense open subset of
$\operatorname{Col}^{\tilde{V}[C_{G}]}(C_{G})=\operatorname{Col}^{\tilde{V}[C_{G}]}_{\lambda}$.
(Formally, this means $\mathbbold{1}_{\operatorname{NM}_{\lambda}}$ should
force that $\dot{D}$ is a dense subset of $\operatorname{Col}(C)$, where $C$
is the generic club added by $\operatorname{NM}_{\lambda}$.)

For any $\mathbb{Q}_{\lambda}$ generic filter $G$, and for any
$\operatorname{Col}^{\tilde{V}[G]}(C_{G})$ generic filter $G^{*}$ containing
$H_{G}$, the set $G^{*}\cap D\neq\emptyset$, where $D=\dot{D}^{C_{G}}$.

To begin with, we shall work in $\tilde{V}[C_{G}]$ for a fixed filter $G$. Let
$D=\dot{D}^{C_{G}}$ as above. For $\delta\in\operatorname{Succ}(C_{G})$,
consider the two forcings
$\mathbb{P}^{\delta}:=\operatorname{Col}^{\tilde{V}[C_{G}]}(C_{G}\setminus\delta+1)$
and
$\mathbb{P}_{\delta}:=\operatorname{Col}^{\tilde{V}[C_{G}]}(C_{G}\cap(\delta+1))$.

For $h\in\mathbb{P}^{\delta}$ we define the set

$D_{h}:=\\{h^{\prime}\in\mathbb{P}_{\delta}:h^{\prime}\cup h\in D\\}$

Then we define

$D^{\delta}:=\\{h\in\mathbb{P}^{\delta}:D_{h}\text{ is open dense}\\}$

######  Claim 5.29.

$D^{\delta}$ is an open dense subset of $\mathbb{P}^{\delta}$.

###### Proof.

$D$ is open, so if $\tilde{h}\leq h\in\mathbb{P}^{\delta}$ then
$D_{\tilde{h}}\supset D_{h}$. Hence $D_{\tilde{h}}$ is dense. Moreover,
$D_{\tilde{h}}$ is also open, since $D$ is open. Hence $D^{\delta}$ is open.

To show $D^{\delta}$ is also dense, let us fix $h\in\mathbb{P}^{\delta}$.

Now $\mathbb{P}_{\delta}$ has cardinality $\delta$, so we can enumerate its
elements $\\{h_{\gamma}:\gamma<\delta\\}$. We construct a decreasing sequence
$(h_{\gamma}^{\prime})_{\gamma<\delta+1}$ of length $\delta$ of conditions in
$\mathbb{P}^{\delta}$. Let $h_{0}^{\prime}=h$. For $\epsilon<\delta$, we
choose some $h_{\epsilon+1}^{\prime}\leq h_{\epsilon}^{\prime}$ such that for
some $h^{*}\leq h_{\epsilon}$, the condition $h^{*}\cup
h_{\epsilon+1}^{\prime}\in D$. We can do this easily, since $D$ is dense: just
take some element of $D$ below $h_{\epsilon}\cup h_{\epsilon}^{\prime}$ and
let $h_{\epsilon+1}^{\prime}$ be the part of it which is above $\delta$.

For limit $\gamma\leq\delta$, we take $h_{\gamma}$ to be below every earlier
term of the sequence, which we can do since $\mathbb{P}^{\delta}$ is
$\delta^{+}$ closed by Proposition 5.10.

Now, $h_{\delta}^{\prime}\leq h$ is such that for all
$h_{\gamma}\in\mathbb{P}_{\delta}$, there exists $h^{*}\leq h_{\gamma}$ such
that $h^{*}\cup h_{\delta}^{\prime}\in D$, since $h_{\delta}^{\prime}\leq
h_{\gamma+1}^{\prime}$ and $D$ is open. It follows immediately that
$D_{h_{\delta}^{\prime}}$ is dense. Since $D$ is open, it’s also immediate
that $D_{h_{\delta}^{\prime}}$ is open. Hence $h_{\delta}^{\prime}\in
D^{\delta}$. But $h^{\prime}_{\delta}\leq h$ so $D^{\delta}$ is dense. ∎

We now work in $\tilde{V}$. We shall show that
$\mathbbold{1}_{\mathbb{Q}_{\lambda}}$ forces the following statement:

> “There are two ordinals $\delta<\beta$ in $C_{G}$, with $\delta$ a successor
> element of $C_{G}$, such that $H_{G}{\upharpoonleft}[\delta,\beta]\in
> D^{\delta}$.”

This statement makes sense, since $\tilde{V}[C_{G}]$ is a definable subclass
of $\tilde{V}[G]$ and so $\tilde{V}[G]$ knows what $D^{\delta}$ looks like for
any $\delta$. Note that it implicitly assumes that
$H_{G}{\upharpoonleft}[\delta,\beta]\in\tilde{V}[C_{G}]$, but that’s
automatically true since it’s an element of
$\operatorname{Col}(C_{G}\cap\beta+1)\in\tilde{V}$.

So fix $p\in\mathbb{Q}_{\lambda}$. Let $\operatorname{ht}(p)=\epsilon$. We
shall show there is a one step extension of $p$ which forces the above
statement.

Let us say an element $\langle c,h,\gamma\rangle$ of $K_{\lambda}^{\epsilon}$
is cooperative if there is some element $\delta\in\operatorname{Succ}(c)$ such
that ${c_{p}\cup
c\Vdash_{\operatorname{NM}_{\lambda}}h{\upharpoonleft}(c\setminus\delta)\in
D^{\delta}}$. (Recall that by definition of $K_{\lambda}$, we know
$h\in\operatorname{Col}^{\tilde{V}}(c)$, so
$h\in\operatorname{Col}^{\tilde{V}[C_{G}]}(C_{G})$. So
$h{\upharpoonleft}(c\setminus\delta)\in\mathbb{P}^{\delta}$ in
$\tilde{V}[C_{G}]$, and thus it makes sense ask whether it is in
$D^{\delta}$.)

######  Claim 5.30.

The set of all cooperative elements of $K_{\lambda}^{\epsilon}$ is $\leq^{*}$
dense.

###### Proof.

Let $\langle c,h,\gamma\rangle\in K_{\lambda}^{\epsilon}$. Without loss of
generality, we can assume $c$ has a largest element. (If it doesn’t, then we
can simply extend $c$ arbitrarily by one step to get a
$(c^{\prime},h,\gamma^{\prime})<^{*}(c,h,\gamma)$ and work with that instead.)
Let $\delta$ be that largest element.

$c_{p}\cup c$ forces that $\delta$ is a successor element of the club that
$\operatorname{NM}_{\lambda}$ adds, so it forces that $D^{\delta}$ is an open
dense subset of $\mathbb{P}^{\delta}$. Hence, there is some end extension
$c^{\prime}\leq c$ and some name $\dot{h}^{\prime}$ for an element of
$\mathbb{P}^{\delta}$ such that $c_{p}\cup
c^{\prime}\Vdash_{\operatorname{NM}_{\lambda}}\dot{h}^{\prime}\in D^{\delta}$.

Now, in $\tilde{V}[C_{G}]$ an element of $\mathbb{P}^{\delta}$ is a sequence
of conditions in collapsing forcings whose support is bounded below $\lambda$.
Each of these collapsing forcings also has cardinality less than $\lambda$. So
any element of $\mathbb{P}^{\delta}$ in $\tilde{V}[C_{G}]$ can be coded as a
subset of $\lambda$. $\operatorname{NM}_{\lambda}$ is $\lambda$ distributive,
so by Theorem LABEL:theorem_distributivity_consequences all the elements of
$\mathbb{P}^{\delta}$ in $\tilde{V}[C_{G}]$ actually already existed in
$\tilde{V}$. (Of course, $\mathbb{P}^{\delta}$ itself does not exist in
$\tilde{V}$, though!)

So without loss of generality, we can choose $\dot{h}^{\prime}$ to be a check
name for some $h^{\prime}\in\tilde{V}$. Now $h^{\prime}$ is certainly bounded
below $\lambda$, so without loss of generality we may assume that $c^{\prime}$
is longer than the support of $h^{\prime}$, that is, that
$h^{\prime}\in\operatorname{Col}^{\tilde{V}}(c^{\prime})=\operatorname{Col}^{\tilde{V}[C_{G}]}(c^{\prime})$.
In fact, since $h^{\prime}\in P^{\delta}$, we know
$h^{\prime}\in\operatorname{Col}(c^{\prime}\setminus\delta)$. In particular,
since $h\in\operatorname{Col}(c)$ and $\sup(c)=\delta$, we know $h\cup
h^{\prime}$ is a well defined element of $\operatorname{Col}(c^{\prime})$.
Take $\gamma^{\prime}\geq\gamma$ to be some ordinal which is larger than
$\sup(c^{\prime})$.

Then $\langle c^{\prime},h\cup h^{\prime},\gamma^{\prime}\rangle\in
K_{\lambda}^{\epsilon}$ and $\langle c^{\prime},h\cup
h^{\prime},\gamma^{\prime}\rangle\leq^{*}\langle c,h,\gamma\rangle$. Using the
same $\delta$ as above, we can see by construction that $c_{p}\cup
c^{\prime}\Vdash_{\operatorname{NM}_{\lambda}}(h\cup
h^{\prime}){\upharpoonleft}(c^{\prime}\setminus\delta)=h^{\prime}\in
D^{\delta}$. Hence $(c^{\prime},h\cup h^{\prime},\gamma^{\prime})$ is
cooperative. ∎

So the set $\tilde{K}$ of all cooperative elements of $K_{\lambda}^{\epsilon}$
is in $U_{\epsilon}$. Since the set of all valid one step extensions of $p$ is
also in $U_{\epsilon}$, there is a cooperative $\langle c,h,\gamma\rangle$
which is a valid way to extend $p$ by one step. Let $q$ be this one step
extension of $p$. Since $\langle c,h,\gamma\rangle$ is cooperative, we can
find $\delta\in\operatorname{Succ}(c)$ such that $c_{p}\cup
c\Vdash_{\operatorname{NM}_{\lambda}}h{\upharpoonleft}(c\setminus\delta)\in
D^{\delta}$. Let $\beta=\sup(c)$.

Now, let $G$ be a $\mathbb{Q}_{\lambda}$ generic filter with $q\in G$. Then
$H_{G}{\upharpoonleft}[\delta,\beta]=h{\upharpoonleft}(c\setminus\delta)$, and
$c_{p}\cup c\in C_{G}$. So $H_{G}{\upharpoonleft}[\delta,\beta]\in
D^{\delta}$. Hence $q$ forces “There are two ordinals $\delta<\beta$ in
$C_{G}$, with $\delta$ a successor element of $C_{G}$, such that
$H_{G}{\upharpoonleft}[\delta,\beta]\in D^{\delta}$”. The condition $p$ was
arbitrary, so $\mathbbold{1}_{Q_{\lambda}}$ forces the statement, which is
what we wanted to show.

Now let $G$ be an arbitrary $\mathbb{Q}_{\lambda}$ generic filter $G$, and
work over $\tilde{V}[G]$. The statement is true of $\tilde{V}[G]$, so find
$\delta<\beta$ that fit it. Let $G^{*}$ be a
$\operatorname{Col}^{\tilde{V}[G]}(C_{G})$ generic filter containing $H_{G}$.
Recall that we are aiming to show that $G^{*}\cap
D=G^{*}\cap\dot{D}^{C_{G}}\neq\emptyset$. Let
$h=H_{G}{\upharpoonleft}[\delta,\beta]$. So $h\in D^{\delta}$. Hence, $D_{h}$
is open and dense in
$\mathbb{P}_{\delta}=\operatorname{Col}^{\tilde{V}[C_{G}]}(C_{G}\cap(\delta+1))$.
Also, $h\in G^{*}$ since $H_{G}\leq h$.

Now, $\mathbb{Q}_{\lambda}$ adds no new bounded subsets of $\lambda$ by Lemma
5.23, so in fact
$\operatorname{Col}^{\tilde{V}}(C_{G}\cap(\delta+1))=\operatorname{Col}^{\tilde{V}[C_{G}]}(C_{G}\cap(\delta+1))=\operatorname{Col}^{\tilde{V}[G]}(C_{G}\cap(\delta+1))$.
So $D_{h}$ is open dense over the latter forcing. Since $G^{*}$ is generic
over $\operatorname{Col}^{\tilde{V}[G]}(C_{G})$, the restriction of $G^{*}$ to
$\operatorname{Col}^{\tilde{V}[G]}(C_{G}\cap(\delta+1))$ is generic over that
forcing. Hence $G^{*}\cap D_{h}\neq\emptyset$, so it contains some
$h^{\prime}$. But then $h^{\prime}\cup h\in D$ by definition of $D_{h}$ (and
hence also $h^{\prime}\cup h\in\operatorname{Col}^{\tilde{V}[C_{G}]}(C_{G})$).
Since $h^{\prime},h\in G^{*}$ also $h^{\prime}\cup h\in G^{*}$. So $G^{**}\cap
D=G^{*}\cap\operatorname{Col}^{\tilde{V}[C_{G}]}(C_{G})\cap D$ contains
$h^{\prime}\cup h$ and is therefore nonempty.

So fixing an arbitrary $\mathbb{Q}_{\lambda}$ filter $G$, we know that if
$G^{*}$ is any $\operatorname{Col}^{\tilde{V}[G]}(C_{G})$ generic filter
containing $H_{G}$ and $D$ is any
$\operatorname{Col}^{\tilde{V}[C_{G}]}(C_{G})$ dense set in
$\tilde{V}[C_{G}]$, then
$G^{**}:=G^{*}\cap\operatorname{Col}^{\tilde{V}[C_{G}]}(C_{G})$ meets $D$.
Hence this $G^{**}$ is generic over $\tilde{V}[C_{G}]$, as required. ∎

As a consequence of our earlier results about $\operatorname{NM}$,
$\mathbb{Q}$ and $\operatorname{Col}$, we can also conclude that the generic
extensions agree about $\operatorname{Card}$ and $\operatorname{Reg}_{\beta}$
for all $\beta<\alpha$

######  Lemma 5.31.

Let $G$ be $\mathbb{Q}_{\lambda}$ generic, let $G^{*}$ be
$\operatorname{Col}^{\tilde{V}[G]}_{\lambda}$ generic and contain $H_{G}$, and
let $G^{**}$ be obtained from $G^{*}$ as in lemma 5.28. The cardinals of
$\tilde{V}[G][G^{*}]$ and $\tilde{V}[C_{G}][G^{**}]$ are both precisely the
cardinals of $\tilde{V}$ which are not in the interval
$(f(\gamma),\operatorname{Succ}_{C}(\gamma))$ for any $\gamma\in C$. Moreover,
in both these generic extensions, for all $0<\beta<\alpha$, the class
$\operatorname{Reg}_{\beta}$ in the generic extension is precisely
$\operatorname{Reg}_{\beta}^{\tilde{V}}$ but with the intervals
$(f(\gamma),\operatorname{Succ}_{C}(\gamma))$ omitted. In particular,
$\operatorname{Reg}_{\beta}^{\tilde{V}}$ and $\operatorname{Card}$ agree up in
$\tilde{V}[G][G^{*}]$ and $\tilde{V}[C_{G}][G^{**}]$, for all $\beta<\alpha$.

###### Proof.

Corollary 5.8 tells us that $\tilde{V}[C_{G}]$ has the same cardinals, and
regular cardinals, as $\tilde{V}$. Likewise, Lemma 5.11 tells us that
$\tilde{V}[G]$ has the same cardinals and regular cardinals as $\tilde{V}$,
except that $\lambda$ has been singularised (which does not modify
$\operatorname{Reg}_{\beta}$ for $\beta<\alpha$). Then Lemma 5.23 tells us
that the generic extensions of $\tilde{V}[G]$ and $\tilde{V}[C_{G}]$ by
$G^{*}$ and $G^{**}$ respectively both remove all the cardinals in the
intervals $(f(\gamma),\operatorname{Succ}_{C}(\gamma))$ but do not otherwise
change $\operatorname{Card}$ or $\operatorname{Reg}_{\beta}$ for
$0<\beta<\alpha$. For the final sentence, we’ve covered the case $\beta>0$;
and showing that the two models agree on $\operatorname{Reg}_{0}$ just
requires us to recall that $\operatorname{Reg}_{0}$ can be trivially
calculated from $\operatorname{Card}$. ∎

######  Corollary 5.32.

Let $G,G^{*},G^{**}$ be as above. Then $\tilde{V}[C_{G}][G^{**}]$ and
$\tilde{V}[G][G^{*}]$ have the same cardinals, and their
$\operatorname{Reg}_{\beta}$’s are the same for all $\beta<\alpha$.

###  5.6. Putting the forcings together

We are finally ready to put together these forcings, and define the overall
forcing we’re going to be using. We use a Prikry style iteration of
$\mathbb{Q}_{\lambda}$ forcings, followed by an $\operatorname{NM}_{\kappa}$
forcing and the corresponding $\operatorname{Col}_{\kappa}$ forcing. This will
give us a universe in which $\kappa$ is the first element of
$\operatorname{Reg}^{\alpha}$, and is also $\operatorname{LST}(I,Q^{\alpha})$.
We now drop our discussions of $\tilde{V}$, and just work in the universe $V$
we fixed near the start of the proof. Recall that $V$ believes GCH.

######  Definition 5.33.

Recursively, we define forcings $P_{\gamma}$ ($\gamma\leq\kappa)$ with two
orders $\leq$ and $\leq^{*}$, and (for $\gamma<\kappa$) $P_{\gamma}$ names
$\dot{Q}_{\gamma}$ for forcings, also with two orders $\leq$ and $\leq^{*}$,
as follows.

  * •

For $\gamma\leq\kappa$, the elements of $P_{\gamma}$ are sequences
$\langle\tau_{\epsilon}\rangle_{\epsilon<\gamma}$ of length $\gamma$ such
that:

    1. (1)

For $\delta<\gamma$, the sequence
$\langle\tau_{\epsilon}\rangle_{\epsilon<\delta}\in P_{\delta}$ and forces
$\tau_{\delta}\in\dot{Q}_{\delta}$.

    2. (2)

The sequence has Easton support. That is, for every $V$ regular
$\lambda\leq\gamma$, the set

$\\{\delta<\lambda:\langle\tau_{\epsilon}\rangle_{\epsilon<\delta}\not\Vdash\tau_{\delta}=\mathbbold{1}_{\dot{Q_{\delta}}}\\}$

is bounded below $\lambda$.

  * •

The $\leq$ order of $P_{\gamma}$ is defined as follows:
$\langle\tau_{\epsilon}^{\prime}\rangle_{\epsilon<\gamma}\leq\langle\tau_{\epsilon}\rangle_{\epsilon<\gamma}$
if:

    1. (1)

For all $\delta<\gamma$,
$\langle\tau_{\epsilon}^{\prime}\rangle_{\epsilon<\delta}\Vdash\tau_{\delta}^{\prime}\leq\tau_{\delta}$.

    2. (2)

For all but finitely many $\delta$, either $\tau_{\delta}$ is forced to be
$\mathbbold{1}_{\dot{Q}_{\delta}}$ by
$\langle\tau_{\epsilon}\rangle_{\epsilon<\delta}$, or we can replace $\leq$
with $\leq^{*}$ (in the sense of $\dot{Q}_{\delta}$) on the previous line.

  * •

$\langle\tau_{\epsilon}^{\prime}\rangle_{\epsilon<\gamma}\leq^{*}\langle\tau_{\epsilon}\rangle_{\epsilon<\gamma}$
if in the above, 2 holds for every $\delta$.

  * •

For any $\gamma<\kappa$, $\dot{Q}_{\gamma}$ is a $P_{\gamma}$ name for a
forcing:

    1. (1)

If $\gamma$ is $\gamma^{+}$ supercompact in $V$ and $P_{\gamma}$ forces:
“$\gamma$ is a cardinal which is $\gamma^{+}$ supercompact”, then
$\dot{Q}_{\gamma}$ is a name for the forcing $\mathbb{Q}_{\gamma}$ we defined
earlier.

    2. (2)

Otherwise, $\dot{Q}_{\gamma}$ is the canonical name for the trivial forcing.

This gives us a well defined forcing $P_{\kappa}$. But we don’t immediately
know very much about it. For the rest of the proof to work, we need to verify
that it does in fact singularise every $\lambda^{+}$ supercompact
$\lambda<\kappa$ with a $\mathbb{Q}_{\lambda}$ style forcing, but that
$\kappa$ remains Mahlo. There is also a further complication: we need to make
sure that when we do our
$\operatorname{NM}_{\kappa}*\operatorname{Col}_{\kappa}$ portion of the
forcing, we will end up with $\kappa=\min(\operatorname{Reg}_{\alpha})$.
Showing that there are no elements of $\operatorname{Reg}_{\alpha}$ below
$\kappa$ is pretty much immediate. But making sure that unboundedly many
elements of $\operatorname{Reg}_{\beta}$ survive below $\kappa$, for every
$\beta<\alpha$, is harder.

Of course, once we know that $\kappa$ is a Mahlo cardinal in the $P_{\kappa}$
generic extension, it follows that $\operatorname{Reg}_{\beta}$ will certainly
be unbounded in $\kappa$ at that point. But remember that
$\operatorname{Col}_{\kappa}$ collapses nearly all the cardinals below
$\kappa$. So if we’re not careful, we might get unlucky and find that we’ve
collapsed all the elements of $\operatorname{Reg}_{\beta}$ in the final stage
of hte forcing. To avoid this, we need to make sure that there are elements of
$\operatorname{Reg}_{\beta}$ in the intervals $(\gamma,f(\gamma)),\gamma\in
C$.

Now because of how we defined $f$, we know that there are such regulars in
that interval in $V$. But when we choose $\gamma\in C$, we need to be sure
that these regulars have survived in the $P_{\kappa}$ generic extension. This
is non-trivial, since we cannot (easily) prove $P_{\kappa}$ preserves all the
cardinals, let alone that it doesn’t singularise anything we didn’t ask it to.

We start with some technical lemmas. The first is a special case of[1, 1.3]
and we do not re-prove it here.

######  Lemma 5.34.

Let $\gamma\leq\kappa$ be a Mahlo cardinal (of $V$). Then $P_{\gamma}$ has the
$\gamma$ chain condition, and has cardinality $\leq\gamma$.

######  Corollary 5.35.

Let $\gamma\leq\kappa$. Then $\lvert P_{\gamma}\rvert\leq\gamma^{++}$.

###### Proof.

Case 1: $\gamma$ is a Mahlo cardinal. Then $\lvert P_{\gamma}\rvert\leq\gamma$
by the previous lemma. Case 2: $\gamma$ is a limit of Mahlo cardinals. Let
$(\epsilon_{i})_{i<\operatorname{cf}(\gamma)}\subset\gamma$ be a sequence of
Mahlo cardinals which is cofinal below $\mu$. Again by Lemma 5.34, for all
$i$, $\lvert P_{\epsilon_{i}}\rvert=\epsilon_{i}$. A condition of $P_{\gamma}$
can be expressed as a collection of conditions, one from each
$P_{\epsilon_{i}}$, which all agree with each other. So

$\displaystyle\lvert P_{\gamma}\rvert$
$\displaystyle\leq\prod_{i<\operatorname{cf}(\gamma)}\lvert
P_{\epsilon_{i}}\rvert$
$\displaystyle=\prod_{i<\operatorname{cf}(\gamma)}\epsilon_{i}$
$\displaystyle\leq\prod_{i<\operatorname{cf}(\gamma)}\gamma$
$\displaystyle\leq\gamma^{\gamma}$ $\displaystyle=\gamma^{+}$

Case 3: $\gamma$ is neither a Mahlo cardinal nor a limit of Mahlo cardinals.
Let $\delta<\gamma=\sup\\{\text{Mahlo cardinals}\\}\cap\gamma$. Any
$\lambda^{+}$ supercompact cardinal $\lambda$ is Mahlo, so we know that
$\dot{Q}_{\epsilon}$ is trivial for all $\epsilon\in(\delta,\gamma)$. So

$\displaystyle\lvert P_{\gamma}\rvert$ $\displaystyle\leq\lvert
P_{\delta}*\dot{Q}_{\delta}\rvert$ $\displaystyle\leq\delta^{++}$
$\displaystyle<\gamma^{++}$

∎

Similarly, this next lemma is a special case of [1, 1.4], and again we don’t
prove it here:

######  Lemma 5.36.

Let $\varphi$ be a statement (with parameters) and $p\in P_{\gamma}$. There is
some $q\leq^{*}p$ which decides $\varphi$. The same is true of the forcing
$P_{\gamma}/P_{\delta}$ if $\delta<\gamma$ is a Mahlo cardinal.

Here $P_{\gamma}/P_{\delta}$ is the ($P_{\delta}$ name for the) forcing
defined in the usual way in a $P_{\delta}$ generic extension $V[G]$: we simply
take $P_{\gamma}$ and delete all those conditions which are incompatible with
an element of $G$.

######  Lemma 5.37.

For any $\delta<\gamma\leq\kappa$, the forcing $P_{\gamma}/P_{\delta}$ is
closed under $\leq^{*}$ sequences of length less than $\delta$.

###### Proof.

Let $p_{0}\geq^{*}p_{1}\geq\ldots$ be a $\geq^{*}$ decreasing sequence of
$P_{\gamma}/P_{\delta}$ length $\lambda<\delta$. We want to find a
$p=p_{\lambda}={\tau_{\epsilon}}_{\delta\leq\epsilon<\gamma}$ which is
$\leq^{*}$ below every $p_{i}$. We define it recursively as follows. Let
$\delta\leq\epsilon<\gamma$.

If for all $i$, the $\epsilon$ component of $p_{i}$ is trivial (either because
$\dot{Q}_{\epsilon}$ is trivial, or because $\epsilon$ is not in the support
of $p_{i}$) then we define $\tau_{\epsilon}$ to be trivial as well.

Otherwise, if say $p_{j}(\epsilon)$ is nontrivial, then we know that
$p_{j}(\epsilon)\geq^{*}p_{j+1}(\epsilon)\geq^{*}\ldots$ are names for an
$\leq^{*}$ decreasing sequence of conditions in $\mathbb{Q}_{\epsilon}$ of
length $\lambda<\epsilon$. We know by Proposition 5.20 that there is a
condition in $\mathbb{Q}_{\epsilon}$ below them; let $\tau_{\epsilon}$ be a
name which is forced to be that condition by $p{\upharpoonleft}\epsilon$.

If this construction works, it will obviously give a condition which is
$\leq^{*}p_{i}$ for all $i<\lambda$. We must check that
$p{\upharpoonleft}\epsilon$ is a condition of $P_{\gamma}/P_{\delta}$, for all
$\epsilon$. It suffices to check that $p$ has Easton support, i.e. its support
is bounded below every regular cardinal of $V[G]$. But we can easily see that

$\operatorname{supp}(p)=\bigcup_{i<\lambda}\operatorname{supp}(p_{i})$

For all $i$, we know that $\operatorname{supp}(p_{i})$ is bounded below every
regular cardinal of $V[G]$, and
$\operatorname{supp}(p_{i})\cap\delta=\emptyset$. So if $\mu$ is regular in
$V[G]$, then either $\mu\leq\delta$ (and
$\operatorname{supp}(p)\cap\mu=\emptyset$) or $\operatorname{supp}(p)\cap\mu$
is a union of $\lambda<\mu$ many bounded sets, and therefore is bounded below
$\mu$. ∎

######  Lemma 5.38.

If $\delta<\gamma$ is a Mahlo cardinal, then the forcing
$P_{\gamma}/P_{\delta}$ does not collapse or singularise any cardinals below
$\delta$, or add any new bounded subsets of $\delta$.

###### Proof.

This is similar to Lemma 5.23. Let us write $P$ for $P_{\gamma}/P_{\delta}$.
First, let $p\in P$, and let $\varphi(x)$ be a statement such that

$p\mathrel{\Vdash}\exists x\in\check{\lambda}\,\varphi(x)$

We claim there is some condition $q\leq^{*}p$ which forces, for some
$\epsilon\in\lambda$, that $\varphi(\check{\epsilon})$ holds. By Lemmas 5.36
and 5.37 we can construct a descending chain of conditions
$p=p_{0}\geq^{*}p_{1}\geq^{*}\ldots$ in $P$ of length $\lambda+1$ such that
$p_{i}$ decides $\varphi(\check{i})$. But then $q:=p_{\lambda}$ decides
$\varphi(\check{\epsilon})$ for every $\gamma\in\lambda$, so it must decide at
least one of them positively.

Now let $\lambda<\delta$ be a cardinal of $V[G]$, a generic extension of $V$
by $P_{\delta}$. Suppose that $\lambda$ is collapsed by $P$. Let $\dot{h}$ be
a name for a bijection $h:\nu\rightarrow\lambda$ for some $\nu<\lambda$. Let
$p\in P$ be a condition forcing this. Using the above and Lemma 5.37, we
construct a descending sequence $p=p_{0}\geq^{*}p_{1}\geq^{*}\ldots$ of length
$\nu+1$, such that for all $i$, $p_{i}$ decides the value of
$\dot{h}(\check{i})$. Then $p_{\nu}$ defines bijection between $\nu$ and
$\lambda$ in $V[G]$. Contradiction.

Showing that $\lambda$ is not singularised by $P$, and that $P$ adds no
bounded subsets of $\delta$, are both similar. ∎

######  Corollary 5.39.

If $\gamma$ is a Mahlo cardinal, and $G$ is a $P_{\gamma}$ generic filter,
then GCH holds in $V[G]$ except perhaps at the cardinals $\lambda$ for which
$\dot{Q}_{\lambda}$ is nontrivial. (In other words, if $\lambda$ is not
$\lambda^{+}$ supercompact in $V$, or $\lambda\geq\gamma$, then
$2^{\lambda}=\lambda^{+}$ in $V[G]$.)

###### Proof.

Let $\lambda$ be a cardinal of $V$ which is not $\lambda^{+}$ supercompact. We
first examine the case $\lambda\geq\gamma$. By Lemma
LABEL:Lemma_CH_preservation_for_forcing and Lemma 5.34 we know that if
$\lambda>\gamma^{+}$ then $2^{\lambda}=\lambda^{+}$ in any $P_{\gamma}$
generic extension. Similarly, if $\lambda=\gamma$ or $\lambda=\gamma^{+}$,
then by regularity we know that for all
$\delta<\lambda,\lambda^{\delta}=\lambda$. So again by the same two lemmas, we
know $P_{\gamma}$ preserves $2^{\lambda}$.

Now suppose that $\lambda<\gamma$. Let
$\mu\leq\lambda=\sup\\{\delta\leq\lambda:\delta\text{ is }\delta^{+}\text{
supercompact}\\}$ and let $\nu$ be the least $\delta>\lambda$ such that
$\delta$ is $\delta^{+}$ supercompact. Then up to some trivial notation
changes,

$P_{\gamma}=P_{\mu}*\dot{Q}_{\mu}*P_{\gamma}/P_{\nu}$

We have just seen that $P_{\mu}$ preserves $2^{\lambda}$. If $\dot{Q}_{\mu}$
is nontrivial, then $\mu$ is $\mu^{+}$ supercompact in $V$, and hence
$\mu<\lambda$. If so, then $\dot{Q}_{\mu}$ is $\mathbb{Q}_{\mu}$, which we saw
in Lemma 5.23 preserves $2^{\lambda}$. And finally, $P_{\gamma}/P_{\nu}$ adds
no new bounded subsets of $\nu>\lambda$, so it doesn’t change
$\mathcal{P}(\lambda)$ at all. Hence $P_{\gamma}$ preserves $2^{\lambda}$. ∎

######  Lemma 5.40.

If $\gamma\leq\kappa$ is $\gamma^{+}$ supercompact in $V$ then it is still
$\gamma^{+}$ supercompact in the $P_{\gamma}$ generic extension.

###### Proof.

Let $V[G]$ be a $P_{\gamma}$ generic extension of $V$. Let $j:V\rightarrow M$
be an $\gamma^{+}$ supercompact embedding with critical point
$\gamma\geq\kappa$, such that $\gamma$ is no longer $\gamma^{+}$ supercompact
in $M$. Let $U$ be the ultrafilter on $\mathcal{P}_{\gamma}(\gamma^{+})$
generated by $j$. Let

$(P_{\delta}^{*}:\delta\leq j(\gamma)):=j((P_{\delta}:\delta\leq\gamma))$

and

$(\dot{Q}_{\delta}^{*}:\delta<j(\gamma)):=j((\dot{Q}_{\delta}:\delta<\gamma))$

Now $\gamma$ is strongly inaccessible in $V$, so by Corollary 5.35 we know
that for $\delta<\gamma$, $\lvert P_{\delta}\rvert\leq\delta^{++}<\gamma$.
Also, $\lvert\mathbb{Q}_{\delta}\rvert<\lambda$ for $\delta<\gamma$. Hence,