text:
\chapter{The Generalized Mitchell Order}\label{GMOChapter}
\section{Introduction}
\subsection{The linearity of the generalized Mitchell order}
The topic of this section is the generalized Mitchell order, which is defined by
extending the definition of the Mitchell order to a broader class of objects:
\begin{defn}\index{Generalized Mitchell order}\index{Mitchell order!generalized}
The {\it generalized Mitchell order} is defined on countably complete
ultrafilters \(U\) and \(W\) by setting \(U\mo W\) if \(U\in M_{W}\).
\end{defn}
The main question we investigate here is to what extent this generalized order
is linear assuming the Ultrapower Axiom. Recall that UA implies the linearity of
the Mitchell order on normal ultrafilters (\cref{UAMO}). On the other hand the
generalized Mitchell order is obviously not a linear order on arbitrary
countably complete ultrafilters (\cref{NonlinearitySection}). The main theorem
of this chapter is the generalization of the linearity of the Mitchell order on
normal ultrafilters to normal fine ultrafilters:

\begin{defn}\index{\(P_\textnormal{bd}(\lambda)\) (bounded powerset)}
	For any ordinal \(\lambda\), the {\it bounded powerset of \(\lambda\)} is
	the set \(P_\text{bd}(\lambda) = \bigcup_{\alpha < \lambda} P(\alpha)\).
\end{defn}

\begin{repthm}{GCHLinear}[UA] Suppose \(\lambda\) is a cardinal such that
	\(2^{<\lambda} = \lambda\). Then the Mitchell order is linear on normal fine
	ultrafilters on \(P_\textnormal{bd}(\lambda)\).
\end{repthm}
This amounts to the most general form of the linearity of the Mitchell order on
normal fine ultrafilters that one could hope for (\cref{NIso}), except for the
cardinal arithmetic assumption on \(\lambda\) (which we dispense with much later
in \cref{ULinearity}).
\subsection{Outline of \cref{GMOChapter}}
We now outline the rest of this chapter.\\

\noindent {\sc\cref{FolkloreSection}.} This contains various folklore facts about large
cardinals and the generalized Mitchell order. None of the results here are due
to the author. We give a brief exposition of the theory of strong embeddings
(\cref{StrengthSection}) and supercompact embeddings (\cref{SCSection}) centered
around the relationship between these concepts and the generalized Mitchell
order. We also exposit the Kunen Inconsistency Theorem, which is closely related
to the wellfoundedness properties of the Mitchell order. Finally we establish
the basic order theoretic properties of the generalized Mitchell order,
especially its transitivity, wellfoundedness (\cref{MOWF}), and nonlinearity
(\cref{NonlinearitySection}).\\

\noindent {\sc\cref{DoddSection}.} This section introduces the notion of Dodd soundness.
This concept first arose in inner model theory, and our exposition is the first
to put it into a general context. We begin by giving a very simple definition of
Dodd soundness that will hopefully help the reader view it as a natural
refinement of supercompactness. We then prove the equivalence of this notion
with the definition of Dodd soundness from fine structure theory
(\cref{DoddEquiv}).

A theorem of Schlutzenberg \cite{Schlutzenberg} (stated as \cref{Schlutz} below)
shows that the Mitchell order is linear on Dodd sound ultrafilters in the
canonical inner models. We prove this theorem (\cref{DoddMO}) here under the
much weaker assumption of UA and by a completely different and much simpler
argument directly generalizing the proof of the linearity of the Mitchell order
on normal ultrafilters.\\

\noindent {\sc\cref{GeneralizedNormalSection}.} We finally turn to the Mitchell order on
normal fine ultrafilters, the natural generalization of normal ultrafilters
associated with supercompact cardinals. Our analysis proceeds by showing that
normal fine ultrafilters are isomorphic to Dodd sound ultrafilters, and then
citing the linearity of the Mitchell order on Dodd sound ultrafilters. To do
this, we introduce the notion of an {\it isonormal ultrafilter} and prove that
every normal fine ultrafilter is isomorphic to an isonormal ultrafilter
(\cref{IsoNormalThm}). The main difficulty is the ``singular case"
(\cref{SingularSolovaySection}) which amounts to generalizing Solovay's Lemma
\cite{Solovay} (proved as \cref{SolovayLemma}) to singular cardinals.
\cref{IsoSound} states that if \(2^{<\lambda} = \lambda\), then isonormal
ultrafilters on \(\lambda\) are Dodd sound. Putting these theorems together, we
obtain that under the Generalized Continuum Hypothesis, normal fine ultrafilters
are isomorphic to Dodd sound ultrafilters, yielding the main theorem of the
chapter (\cref{GCHLinear}), the linearity of the Mitchell order on normal fine
ultrafilters.
\section{Folklore of the generalized Mitchell order}\label{FolkloreSection}

\subsection{Strength and the Mitchell order}\label{StrengthSection}
The generalized Mitchell order is often viewed as a more finely calibrated
generalization of the concept of the strength of an elementary embedding. In
this subsection, we set down the basic theory of strength and discuss its
relationship with the Mitchell order.

\begin{defn}
Suppose \(M\) is a transitive class and \(\lambda\) is a cardinal.
\begin{itemize}
\item An elementary embedding \(j: V\to M\) is {\it \(\lambda\)-strong} if \(P(\lambda)\subseteq M\).\index{Strength!\(\lambda\)-strong embedding}
\item An elementary embedding is {\it \({<}\lambda\)-strong} if
\(P_\text{bd}(\lambda)\subseteq M\).
\end{itemize}
\end{defn}

Notice that the property of being \(\lambda\)-strong depends only on \(M\). The
basic lemmas we prove about \(\lambda\)-strong embeddings almost all apply to
arbitrary inner models containing \(P(\lambda)\). (The embedding \(j\) just
comes along for the ride.)

Most authors define \(j\) to be \(\alpha\)-strong if \(V_\alpha\subseteq M\).
The definition used here is arguably preferable (if one is assuming the Axiom of
Choice and not assuming the Generalized Continuum Hypothesis). This is because
it is more expressive:

\begin{lma} Suppose \(j : V\to M\) is an elementary embedding.
	\begin{itemize}
		\item If \(\alpha\) is an ordinal, then \(V_{\alpha + 1}\subseteq M\) if
		and only if \(j\) is \(\beth_\alpha\)-strong.
		\item If \(\gamma\) is a limit ordinal, then \(V_{\gamma}\subseteq M\)
		if and only if \(j\) is \({<}\beth_\gamma\)-strong.\qed
	\end{itemize}
\end{lma}

It would be strange to define strong embeddings without defining strong
cardinals, so let us include the definition even though we will have little to
say about the concept:
\begin{defn}
	Suppose \(\kappa \leq\lambda\) are cardinals. Then \(\kappa\) is
	{\it\(\lambda\)-strong} if there is an inner model \(M\) and a
	\(\lambda\)-strong elementary embedding \(j : V\to M\) such that
	\(\textsc{crt}(j) = \kappa\) and \(j(\kappa) > \lambda\). \(\kappa\) is {\it
	strong} if \(\kappa\) is \(\lambda\)-strong for all \(\lambda\).
\end{defn}

The requirement that \(j(\kappa) > \lambda\) above is not actually necessary as
a consequence of \cref{KunenInconsistency}. We use standard notation for
hereditary cardinality:
\begin{defn}
	If \(x\) is a set, \(\text{tc}(x)\) denotes the smallest transitive set
	\(y\) with \(x\subseteq y\). The {\it hereditary cardinality} of \(x\) is
	the cardinality of \(\text{tc}(x)\). For any cardinal \(\lambda\),
	\(H(\lambda)\) denotes the collection of sets of hereditary cardinality less
	than \(\lambda\).\index{\(H(\lambda)\)}
\end{defn}

\begin{lma}
	For any infinite cardinal \(\lambda\),
	\begin{itemize}
		\item \(H(\lambda^+)\) is a transitive set.
		\item \(H(\lambda^+)\) is bi-interpretable with \(P(\lambda)\).
		\item \(H(\lambda)\) is bi-interpretable with
		\(P_\textnormal{bd}(\lambda)\).\qed
	\end{itemize}
\end{lma}
The bi-interpretability of \(H(\lambda^+)\) and \(P(\lambda)\) yields the
following lemma:
\begin{lma}
An embedding \(j: V\to M\) is \({<}\lambda\)-strong if and only if
\(H(\lambda)\subseteq M\) and \(\lambda\)-strong if and only if
\(H(\lambda^+)\subseteq M\).\qed
\end{lma}

\begin{defn}
The {\it strength} of an elementary embedding \(j : V\to M\), denoted
\(\textsc{str}(j)\), is the largest cardinal \(\lambda\) such that \(j\) is
\({<}\lambda\)-strong.\index{Strength!of an embedding}
\end{defn}

The following fact specifies exactly which powersets are contained in the target
model of an elementary embedding in terms of its strength:
\begin{lma}\label{StrengthInvariance}
Suppose \(j : V\to M\) is an elementary embedding and \(\lambda\) is a cardinal.
Then the following are equivalent:
\begin{enumerate}[(1)]
\item \(\textsc{str}(j) = \lambda\).
\item For all \(X\in M\), \(P(X)\subseteq M\) if and only if \(|X|^M <
\lambda\).\qed
\end{enumerate}
\end{lma}

The main limitation on the strength of an elementary embedding is known as the
{\it Kunen Inconsistency Theorem} \cite{Kunen}:
\begin{thm}[Kunen] Suppose \(j : V\to M\) is a nontrivial elementary embedding
and \(\lambda\) is the first fixed point of \(j\) above \(\textsc{crt}(j)\).
Then \(\textsc{str}(j)\leq\lambda\).\qed
\end{thm}
We prove this and other related facts in \cref{KunenSection}.

The basic relationship between strength and the Mitchell order is given by the
following two lemmas:
\begin{lma}\label{MOStrength}
Suppose \(U\) and \(W\) are countably complete ultrafilters and \(U\mo W\). Then
\(M_W\) is \(\lambda\)-strong where \(\lambda\) is the cardinality of the
underlying set \(X\) of \(U\). In fact, \(P(X)\subseteq M_W\).
\begin{proof}
Clearly \(X\in M_W\) since \(X\in U\in M_W\). It suffices to show that
\(P(X)\subseteq M_W\). Fix \(A\subseteq X\), and we will show \(A\in M_W\).
Since \(U\) is an ultrafilter, either \(A\in U\) or \(X\setminus A\in U\). If
\(A\in U\), then \(A\in U\in M_W\), so \(A\in M_W\). If \(X\setminus A\in U\),
then similarly \(X\setminus A\in M_W\), and since \(X\in M_W\), it follows that
\(A = X\setminus (X\setminus A)\in M_W\). Therefore in either case, \(A\in
M_W\).
\end{proof}
\end{lma}

\begin{lma}
Suppose \(W\) is a countably complete ultrafilter and \(j_W\) is
\(2^\lambda\)-strong. Then for any countably complete ultrafilter \(U\) on
\(\lambda\),  \(U\mo W\).
\begin{proof}
Since \(U\subseteq P(\lambda)\), \(U\in H((2^\lambda)^+)\subseteq M_W\).
\end{proof}
\end{lma}
This strength requirement implicit in the definition of the generalized Mitchell
order may seem somewhat unnatural. What if one modified the Mitchell order,
considering for example the {\it amenability relation} defined on countably
complete ultrafilters by setting \(U\A W\) if and only if \(U\) concentrates on
\(M_W\) and \(U\cap M_W\in M_W\)? Such modified Mitchell orders are the subject
of \cref{InternalSection}.

For the time being, we must point out some irritating properties of the
generalized Mitchell order that suggest that in some sense it may be a little
bit {\it too} general. The issue is that the definition of \(U\mo W\) above has
a strong dependence on the choice of the underlying set of \(U\). For example,
if \(W\) is nonprincipal, then the following hold:
\begin{itemize}
\item There is a principal ultrafilter \(D\) on an ordinal such that \(D\not\mo
W\).
\item There is a set \(x\) such that the principal ultrafilter \(\{\{ x
\}\}\not\mo W\).
\end{itemize}
For the first bullet point, let \(\lambda\) be the strength of \(j_W\), and let
\(D\) be any principal ultrafilter on \(\lambda\). For the second bullet point,
let \(x\) be any set that does not belong to \(M_W\).

These silly counterexamples suggest that the generalized Mitchell order is only
a well-behaved relation on a restricted class of ultrafilters. Recall that for
any ultrafilter \(U\) on a set \(X\), \(\lambda_U\) is defined to be the least
cardinality of a set in \(U\), and \(U\) is said to be uniform if \(|X| =
\lambda_U\). Hereditary uniformity is a strengthening of uniformity:
\begin{defn}
An ultrafilter \(U\) on a set \(X\) is {\it hereditarily uniform} if
\(\lambda_U\) is the hereditary cardinality of \(X\).\index{Hereditarily uniform
ultrafilter}
\end{defn}
Any ultrafilter \(U\) is isomorphic to a hereditarily uniform ultrafilter since
in fact \(U\) is isomorphic to an ultrafilter on \(\lambda_U\)
(\cref{UniformIso}). The following lemma argues that the generalized Mitchell
order is a reasonable relation on the class of hereditarily uniform
ultrafilters:

\begin{lma}\label{HeredLemma0}
Suppose \(U'\RK U\mo W\) are countably complete ultrafilters. Let \(X\) and
\(X'\) be the underlying sets of \(U\) and \(U'\), and assume \(X'\in M_W\) and
\(M_W\) satisfies \(|X'|\leq|X|\). Then \(U'\mo W\) and \(M_W\) satisfies
\(U'\RK U\). If \(U'\cong U\), then \(M_W\) satisfies \(U'\cong U\).
\end{lma}

\begin{lma}\label{HeredLemma}
Suppose \(U'\RK U\mo W\) are countably complete ultrafilters and \(U'\) is
hereditarily uniform. Then \(U'\mo W\) and \(M_W\) satisfies \(U'\RK U\).  If
\(U'\cong U\), then \(M_W\) satisfies \(U'\cong U\). In particular, the
restriction of the generalized Mitchell order to hereditarily uniform
ultrafilters is isomorphism invariant.
\end{lma}

\cref{HeredLemma0} and \cref{HeredLemma} follow from a fact that is both more
general and easier to prove:
\begin{lma}\label{ZFCStrength}
	Suppose \(M\) is an inner model of \textnormal{ZFC}, \(\lambda\) is a
	cardinal, and \(X\in M\) is a set of cardinality \(\lambda\) such that
	\(P(X)\subseteq M\).
	\begin{itemize}
		\item For any set \(Y\in M\) such that \(M\vDash |Y| \leq |X|\),
		\(P(Y)\subseteq M\).
		\item For any set \(Y\in M\) such that \(M\vDash  |Y| \leq |X|\),
		\(P(X\times Y)\subseteq M\).
		\item For any set \(Y\in M\) such that \(M\vDash  |Y| \leq |X|\), every
		function from \(X\) to \(P(Y)\) belongs to \(M\).
		\item \(P(\lambda)\subseteq M\).
		\item Every set of hereditary cardinality at most \(\lambda\) belongs to
		\(M\) and has hereditary cardinality at most \(\lambda\) in \(M\).\qed
	\end{itemize}
\end{lma}
The bullet points are arranged in such a way that the reader should have no
trouble proving each one in turn.\footnote{It is likely, however, that the
second bullet-point cannot be established if \(M\) is not assumed to satisfy the
Axiom of Choice.}

\begin{proof}[Proof of \cref{HeredLemma0}] Fix \(f : X\to X'\) such that
	\(f_*(U) = U'\). By \cref{ZFCStrength}, \(f\in M_W\), and hence \(U'=
	f_*(U)\in M_W\). Moreover \(f\) witnesses \(U'\RK U\) in \(M_W\). Finally if
	\(U'\cong U\), then this is also witnessed by some \(g\in M_W\).
\end{proof}
\begin{proof}[Proof of \cref{HeredLemma}] By \cref{ZFCStrength}, the underlying
	set of \(U'\) belongs to \(M_W\) and has hereditary cardinality at most
	\(\lambda_{U'}\leq \lambda_U \leq |X|\) in \(M_W\), so the lemma follows
	from \cref{HeredLemma0}.
\end{proof}

\subsection{Supercompactness and the Mitchell order}\label{SCSection}
We now turn to a concept that is more pertinent to this dissertation than
strength: supercompactness.

\begin{defn}\index{Supercompactness!\(X\)-supercompact embedding}
Suppose \(M\) is a transitive class and \(X\) is a set. An elementary embedding
\(j: V\to M\) is {\it \(X\)-supercompact} if \(j[X]\in M\).
\end{defn}

The following lemma allows us to focus solely on the case of
\(\lambda\)-supercompact embeddings for \(\lambda\) a cardinal:
\begin{lma}\label{XSC}
Suppose \(X\) and \(Y\) are sets such that \(|X| = |Y|\). Then an elementary
embedding \(j : V\to M\) is \(X\)-supercompact if and only if it is
\(Y\)-supercompact. In particular, \(j\) is \(X\)-supercompact if and only if
\(j\) is \(|X|\)-supercompact.
\begin{proof}
Suppose \(j\) is \(X\)-supercompact and \(f : X\to Y\) is a surjection. Then
\[j(f)[j[X]] = j[Y]\] so \(j\) is \(Y\)-supercompact.
\end{proof}
\end{lma}

\begin{defn}\index{Supercompactness!\(\lambda\)-supercompact cardinal}\index{Supercompact cardinal}
	Suppose \(\kappa\leq \lambda\) are cardinals. Then \(\kappa\) is {\it
	\(\lambda\)-supercompact} if there is a \(\lambda\)-supercompact embedding
	\(j : V\to M\) such that \(\textsc{crt}(j) = \kappa\) and \(j(\kappa) >
	\lambda\); \(\kappa\) is supercompact if \(\kappa\) is
	\(\lambda\)-supercompact for all cardinals \(\lambda \geq \kappa\).
\end{defn}

The results of this dissertation (\cref{PathologicalSection}) single out a class
of ultrapower embeddings that are just shy of \(\lambda\)-supercompact, so the
following is an important definition:

\begin{defn}\index{Supercompactness!\({<}\lambda\)-supercompact embedding}
Suppose \(\lambda\) is a cardinal. An elementary embedding \(j: V\to M\) is {\it
\({<}\lambda\)-supercompact} if \(j\) is \(\delta\)-supercompact for all
cardinals \(\delta < \lambda\).
\end{defn}

The definition of supercompactness is motivated by its relationship with the
closure of \(M\) under \(\lambda\)-sequences:

\begin{lma}\label{SupercompactClosure}
Suppose \(j : V\to M\) is an elementary embedding and \(\lambda\) is a cardinal.
\begin{enumerate}[(1)]
\item \(j\) is \(\lambda\)-supercompact if and only if \(j\restriction
\lambda\in M\).
\item If \(j\) is \(\lambda\)-supercompact, then \(j\) is \(\lambda\)-strong.
\item If \(j\) is \(\lambda\)-supercompact, then \(j[X]\in M\) for all \(X\) of
cardinality \(\lambda\).
\item If \(j\) is \(\lambda\)-supercompact and \(M = H^M(j[V]\cup S)\) for some
\(S\subseteq M\) such that \(S^\lambda\subseteq M\), then \(M^\lambda\subseteq
M\).
\end{enumerate}
\begin{proof}
For (1), note that \(j\restriction \lambda\) is the inverse of the transitive
collapse of \(j[\lambda]\).

For (2), suppose \(A\subseteq \lambda\). Then \(A = (j\restriction
\lambda)^{-1}[j(A)]\), so since \(j\restriction \lambda\) and \(j(A)\) both
belong to \(M\), so does \(A\).

(3) is immediate from \cref{XSC}.

For (4), fix \(\langle x_\alpha : \alpha < \lambda\rangle\in M^\lambda\). Fix
\(\langle f_\alpha : \alpha < \lambda\rangle\) and \(\langle a_\alpha : \alpha <
\lambda\rangle\in S^\lambda\) such that \(x_\alpha = j(f_\alpha)(a_\alpha)\) for
all \(\alpha < \lambda\). The function \(G : j[\lambda]\to M\) defined by
\(G(j(\alpha)) = j(f_\alpha)\) belongs to \(M\) by (3), since \[G =
j[\{(\alpha,f_\alpha) : \alpha < \lambda\}]\] Therefore the sequence \(\langle
j(f_\alpha) : \alpha < \lambda\rangle\) can be computed from \(G\) and
\(j\restriction \lambda\): \[j(f_\alpha) = G\circ(j\restriction
\lambda)(\alpha)\] Since both \(G\) and \(j\restriction \lambda\) belong to
\(M\) by (1), \(\langle j(f_\alpha) : \alpha < \lambda\rangle\in M\). Finally,
\[\langle x_\alpha : \alpha < \lambda\rangle = \langle j(f_\alpha)(a_\alpha) :
\alpha < \lambda\rangle\] can be computed from \(\langle j(f_\alpha) : \alpha <
\lambda\rangle\) and \(\langle a_\alpha : \alpha < \lambda\rangle\). Both these
sequences belong to \(M\), since \(\langle a_\alpha : \alpha < \lambda\rangle\in
S^\lambda\subseteq M\), so \(\langle x_\alpha : \alpha < \lambda\rangle\in M\),
as desired.
\end{proof}
\end{lma}

For the purposes of this dissertation, the most relevant corollary of
\cref{SupercompactClosure} is its application to ultrapower embeddings:

\begin{cor}\label{UltrapowerSC}
An ultrapower embedding \(j : V\to M\) is \(\lambda\)-supercompact if and only
if \(M^\lambda\subseteq M\).
\begin{proof}
Fix \(a\in M\) such that \(M = H^M(j[V]\cup \{a\})\). The corollary follows from
applying \cref{SupercompactClosure} (4) in the case \(S = \{a\}\).
\end{proof}
\end{cor}

We can make good use of \cref{UltrapowerSC} since it is always possible to
derive a \(\lambda\)-supercompact ultrapower embeddings from a
\(\lambda\)-supercompact embedding:
\begin{lma}\label{DerivedSC}
	Suppose \(j : V\to M\) is an \(X\)-supercompact embedding,
	\(V\stackrel{i}\longrightarrow N\stackrel{k}\longrightarrow M\) are
	elementary embeddings, \(k\circ i = j\), and \(j[X]\in k[N]\). Then \(i\) is
	\(X\)-supercompact and \(k(i[X]) = j[X]\). In particular, letting \(\lambda
	= |X|\), \(k\restriction \lambda + 1\) is the identity.
	\begin{proof}
		Fix \(S\in M\) such that \(k(S) = j[X]\). Then \[S = k^{-1}[k(S)] =
		k^{-1}[j[X]] = k^{-1}\circ j[X] = i[X]\] Thus \(i[X]= S \in M\), so
		\(i\) is \(X\)-supercompact, and moreover, \(k(i[X])= k(S) = j[X]\).

		Since \(k(i[X]) = j[X]\), the argument of \cref{XSC} shows
		\(k(i[\lambda]) = j[\lambda]\). But then if \(\alpha \leq \lambda\),
		\(k(\alpha) = k(\text{ot}(i[\lambda]\cap i(\alpha)))
		=\text{ot}(k(i[\lambda])\cap k(i(\alpha))) = \text{ot}(j[\lambda]\cap
		j(\alpha)) = \alpha\).
	\end{proof}
\end{lma}

\begin{defn}
	The {\it supercompactness} of an elementary embedding \(j : V\to M\) is the
	least cardinal \(\lambda\) such that \(j\) is not \(\lambda\)-supercompact.
\end{defn}
Which cardinals are the supercompactness of an elementary embedding? Which are
the supercompactness of an ultrapower embedding? This turns out to be a major
distinction:
\begin{prp}
	Suppose \(\lambda\) is a singular cardinal and \(j : V\to M\) is an
	elementary embedding such that \(M^{<\lambda}\subseteq M\). Then
	\(M^\lambda\subseteq M\).\qed
\end{prp}
Thus the supercompactness of an ultrapower embedding is always regular, while it
is easy to see this must fail for arbitrary embeddings if there is a
\(\kappa^+\omega\)-supercompact cardinal. An important point is that if the
cofinality of \(\lambda\) is small, \(\lambda\)-supercompactness is equivalent
to \(\lambda^+\)-supercompactness:
\begin{lma}\label{SmallCfCompact}\label{SmallCfSC}
	Suppose \(\lambda\) is a cardinal, \(j : V\to M\) is elementary embedding,
	and \(\kappa = \textsc{crt}(j)\). If \(j\) is \(\lambda\)-supercompact, then
	\(j\) is \(\lambda^{<\kappa}\)-supercompact.
	\begin{proof}
		Assume \(j[\lambda]\in M\), and we will show that
		\(j[P_\kappa(\lambda)]\in M\).  Note that for \(\sigma\in
		P_\kappa(\lambda)\), \(j(\sigma) = j[\sigma]\). Thus
		\[j[P_\kappa(\lambda)] = \{ j[\sigma] : \sigma\in P_\kappa(\lambda)\} =
		P_\kappa(j[\lambda])\] One consequence of this is that
		\(P_\kappa(j[\lambda])\subseteq M\), since
		\(j[P_\kappa(\lambda)]\subseteq M\), and therefore
		\(P_\kappa(j[\lambda]) = (P_\kappa(j[\lambda]))^M\in M\). It follows
		that \(j[P_\kappa(\lambda)] \in M\), as desired.
	\end{proof}
\end{lma}
It follows for example that the supercompactness of an elementary embedding is
never the successor of a singular cardinal \(\gamma\) of countable cofinality,
since \(\gamma^\omega = \gamma^+\). This is an important component in the proof
of Kunen's Inconsistency Theorem (\cref{KunenInconsistency}).

We now begin to examine the relationship between supercompactness and the
Mitchell order, which turns out to be central to the rest of this dissertation.
The key point is that if \(U\mo W\), then the supercompactness of \(M_W\)
determines the extent to which the ultrapower of \(M_W\) by \(U\) is correctly
computed by \(M_W\).
\begin{lma}\label{MOFactor}
	Suppose \(U\mo W\) are countably complete ultrafilters.  Then there is a
	unique elementary embedding \(k : (M_U)^{M_W}\to j_U(M_W)\) such that
	\(k\circ (j_U)^{M_W} = j_U\restriction M_W\) and \(k(\id_U^{M_W}) = \id_U\).
	Let \(X\) be the underlying set of \(U\). Then \(k \restriction
	j_U((2^\lambda)^{M_W}) + 1\) where \(\lambda = |X|\).
	\begin{proof}
		Since \(P(X)\subseteq M_W\), \(U\) is the ultrafilter derived from
		\(j_U\restriction M_W\) using \(\id_U\). Thus there is a unique factor
		embedding \(k : (M_U)^{M_W}\to j_U(M_W)\) such that \(k\circ (j_U)^{M_W}
		= j_U\restriction M_W\) and \(k(\id_U^{M_W}) = \id_U\). This establishes
		the first part of the lemma.

		As for the second part, since \(U\mo W\), we have \(P(X)\subseteq M_W\)
		and hence by \cref{ZFCStrength}, \(P(\lambda)\subseteq M_W\) and every
		function from \(X\) to \(P(\lambda)\) belongs to \(M_W\). It follows
		that \(j_U(P(\lambda))\subseteq \text{ran}(k)\): if \(A\in
		j_U(P(\lambda))\), then \(A = j_U(f)(\id_U)\) for some \(f : X\to
		P(\lambda)\), and therefore \[A =
		k(j_U^{M_W}(f)(\id_U^{M_W}))\in\text{ran}(k)\] Since there is a
		surjection \(g: P(\lambda)\to (2^\lambda)^{M_W}\) in \(M_W\),
		\[j_U(g)[j_U(P(\lambda))] =  j_U((2^\lambda)^{M_W})\subseteq
		\text{ran}(k)\] Moreover \(j_U((2^\lambda)^{M_W})\in j_U[M_W]\subseteq
		\text{ran}(k)\). Thus \(j_U((2^\lambda)^{M_W}) + 1 \subseteq
		\text{ran}(k)\), or in other words, \(k \restriction
		j_U((2^\lambda)^{M_W}) + 1 = \text{id}\).
	\end{proof}
\end{lma}
We will refer to the embedding of \cref{MOFactor} as a {\it factor
embedding}.\index{Factor embedding}

\begin{lma}\label{MOSuper2}
	Suppose \(U\) and \(W\) are countably complete ultrafilters with \(U\mo W\).
	Let \(X\) be the underlying set of \(U\), let \(\lambda = |X|\) and let
	\(\delta = ((2^{\lambda})^+)^{M_W}\). Then \[j^{M_W}_U\restriction
	H^{M_W}(\delta) = j_U\restriction H^{M_W}(\delta)\]
	\begin{proof}
		Let \(k : (M_U)^{M_W}\to j_U(M_W)\) be the factor embedding with
		\(k\circ (j_U)^{M_W} = j_U\restriction M_W\) and \(k(\id_U^{M_W}) =
		\id_U\). Then \cref{MOFactor} implies \(k\restriction
		j_U^{M_W}(\delta)\) is the identity, and therefore \(k\restriction
		j_U^{M_W}(H^{M_W}(\delta))\) is the identity. Now
		\[j^{M_W}_U\restriction H^{M_W}(\delta) = (k\restriction
		j_U^{M_W}(H^{M_W}(\delta)))\circ (j^{M_W}_U\restriction H^{M_W}(\delta))
		= j_U\restriction H^{M_W}(\delta)\qedhere\]
	\end{proof}
\end{lma}

Our next proposition, \cref{MOSuper}, suggests that the Mitchell order on
ultrafilters be seen as a generalization of supercompactness that asks for one
ultrapower \(M_W\) how much it can see of another embedding \(j_U\). (On this
view supercompactness is the special case in which we ask how much of \(j_U\) is
seen by \(M_U\) itself.)

\begin{prp}\label{MOSuper}
Suppose \(U\) and \(W\) are countably complete ultrafilters. Let \(X\) be the
underlying set of \(U\), let \(\lambda = |X|\) and let \(\delta =
((2^{\lambda})^+)^{M_W}\). Then the following are equivalent:
\begin{enumerate}[(1)]
	\item \(U\mo W\).
	\item \(j_U\restriction H^{M_W}(\delta)\in M_W\).
	\item \(j_U\restriction P(\lambda)\in M_W\).
	\item  \(j_U\restriction P(X) \in M_W\).
\end{enumerate}
\begin{proof}
{\it (1) implies (2).} Immediate from \cref{MOSuper2}.

{\it (2) implies (3).} Immediate since \(P(\lambda)\subseteq H^{M_W}(\delta)\).

{\it (2) implies (3).} This is probably clear enough (and in any case, (1)
implies (4) is easy), but let us just make sure. By \cref{ZFCStrength}, \(|X|^M
= \lambda\). Let \(\rho : \lambda\to X\) be a surjection in \(M_W\). For \(A\in
P(X)\), \[j_U(A) = j_U(\rho)[j_U(\rho^{-1}[A])]\]

{\it (3) implies (1).} If \(j_U\restriction P(X)\) belongs to \(M_W\), then \(U
= \{A\subseteq X : \id_U\in j_U(A)\}\) belongs to \(M_W\) as well.
\end{proof}
\end{prp}

Given \cref{MOSuper2}, it is reasonable to wonder whether the whole embedding
\(j_U\restriction M_W\) might be correctly computed by \(M_W\) as well; that is,
perhaps the factor embedding \(k\) is always trivial. We provide a
counterexample in \cref{CummingsExample}.\footnote{This counterexample also
shows that in the context of  \cref{MOFactor}, the lower bound given there on
\(\textsc{crt}(k)\) can be tight in the sense that (consistently) one can have
\[\textsc{crt}(k) = j_U\left((2^\lambda)^{M_W}\right)^{+(M_U)^{M_W}}\]} This is
equivalent to the supercompactness of \(j_W\), a phenomenon we exploit later:

\begin{prp}\label{USupercompact}
	Suppose \(U\mo W\) are countably complete ultrafilters. Then the following
	are equivalent:
	\begin{enumerate}[(1)]
		\item \((j_U)^{M_W} = j_U\restriction M_W\).
		\item \(j_W\) is \(\lambda_U\)-supercompact.
	\end{enumerate}
	\begin{proof}
		{\it (1) implies (2):} Let \(k : (M_U)^{M_W}\to j_U(M_W)\) be the factor
		embedding of \cref{MOFactor}, with \(k\circ j_U^{M_W} = j_U\restriction
		M_W\) and \(k(\id_U^{M_W}) = \id_U\). Since \((j_U)^{M_W} =
		j_U\restriction M_W\), we have that \(k : j_U(M_W)\to j_U(M_W)\) and
		\(k\circ j_U\circ j_W = j_U\circ j_W\). Hence by the basic theory of the
		Rudin-Keisler order (\cref{RKRigid}), \(k\) is the identity.

		It follows in particular that \(j_U(j_W)(\id_U)\in \text{ran}(k)\). Fix
		\(f:X\to M_W\) in \(M_W\) such that \[k(j_U^{M_W}(f)(\id^{M_W}_U)) =
		j_U(j_W)(\id_U)\] Thus \(j_U(f)(\id_U) = j_U(j_W)(\id_U)\), so by \L
		o\'s's Theorem, there is a set \(A\in U\) such that \(f\restriction A =
		j_W\restriction A\). Since \(P(X)\subseteq M_W\), \(A\in M_W\), and
		hence \(j_W\restriction A = f\restriction A\in M_W\). In particular,
		\(j_W[A]\in M_W\), so \(j_W\) is \(A\)-supercompact. By \cref{XSC},
		\(j_W\) is \(|A|\)-supercompact, and since \(\lambda_U\leq |A|\), it
		follows that \(j_W\) is \(\lambda_U\)-supercompact.

		{\it (2) implies (1):} Obvious.
	\end{proof}
\end{prp}

Down the line (\cref{UltrapowerCorrectness}) we will show that under UA,
whenever \(U\mo W\), in fact \(j_W\) is \(\lambda_U\)-supercompact (and in fact
it suffices that \(P(\lambda_U)\subseteq M_W\)), and thus \(j_W^{M_U} =
j_W\restriction M_U\). For now, let us mention a generalization of
\cref{USupercompact}, whose proof we omit:
\begin{prp}
	Suppose \(U\) and \(W\) are countably complete ultrafilters such that \(U\)
	concentrates on a set in \(M_W\). The following are equivalent:
	\begin{enumerate}[(1)]
		\item \(j_{U\cap M_W}^{M_W} = j_U\restriction M_W\)
		\item There is a function \(f\in M_W\) such that \(f \restriction A =
		j_W\restriction A\) for some \(A\in U\).
		\item For all \(f: I\to M_W\) where \(I\in U\), there is some \(g\in
		M_W\) such that \(g\restriction A = f\restriction A\) for some \(A\in
		U\).\qed
	\end{enumerate}
\end{prp}

\begin{comment}
\begin{lma}
Suppose \(U\) and \(W\) are countably complete ultrafilters such that \(U\)
concentrates on a set in \(M_W\). Let \(\bar U = U\cap M_W\). The following are
equivalent:
\begin{enumerate}[(1)]
\item \(j_{U\cap M_W}^{M_W} = j_U\restriction M_W\)
\item There is a function \(f\in M_W\) such that \(f \restriction A =
j_W\restriction A\) for some \(A\in U\).
\item For all \(f: I\to M_W\) where \(I\in U\), there is some \(g\in M_W\) such
that \(g\restriction A = f\restriction A\) for some \(A\in U\).
\end{enumerate}
\begin{proof}
Let \(k : M^{M_W}_{U\cap M_W}\to j_U(M_W)\) be the factor embedding. defined by
\(k([f]^{M_W}_{U\cap M_W}) = [f]_U\). Therefore \(k\) is the unique elementary
embedding such that \begin{equation}\label{kcommute} k\circ  j_{U\cap M_W}^{M_W}
= j_U\restriction M_W\end{equation}.

{\it (1) implies (2)}: Assume \(j_{U\cap M_W}^{M_W} = j_U\restriction M_W\).
Then \[k\circ j_{U\times W} = k\circ j_U\circ j_W = k\circ  j_{U\cap M_W}^{M_W}
\circ j_W = j_U\circ j_W = j_{U\times W}\] Therefore by \cref{RKRigid}, \(k\) is
the identity. In particular, \([j_W\restriction M_W]_U \in \text{ran}(k)\). Fix
\(f \in M_W\) such that \(k(f) = [j_W\restriction M_W]_U\). Then \([f]_U =
[j_W\restriction M_W]_U\) so for some \(A\in U\), \(f\restriction A =
j_W\restriction A\).

{\it (2) implies (3)}: This follows from the proof of \cref{SupercompactClosure}
(4).

{\it (3) implies (1)}: (3) simply states that the embedding \(k\) is surjective.
Hence \(k\) is the identity so by \cref{kcommute}, \(j_{U\cap M_W}^{M_W} =
j_U\restriction M_W\).
\end{proof}
\end{lma}

One can parse \cref{USupercompact} (2) as the assertion that \(j_W\) is ``almost
supercompact" with respect to \(U\). The fact that correctly computation of
ultrapower embeddings is related to supercompactness will become very important
in \cref{SupercompactChapter} when we use the Ultrapower Axiom to deduce forms
of supercompactness from forms of strong compactness.
\end{comment}

We finish this section with a restriction on the supercompactness of an
ultrafilter:
\begin{prp}\label{UFSuperBound}
Suppose \(U\) is an ultrafilter and \(j_U\) is \(\lambda_U^+\)-supercompact.
Then \(U\) is principal.
\end{prp}

We use the following lemma:

\begin{lma}\label{Ineqs}
Suppose \(j : V\to M\) is an elementary embedding that is discontinuous at the
infinite cardinal \(\lambda\). Let \(\lambda_* = \sup j[\lambda]\). Then
\[\lambda^+ \leq \lambda_*^{+{M}} < j(\lambda)^{+M} = j(\lambda^+)\] If \(j\) is
continuous at \(\lambda^+\), then \(j(\lambda^+)\) is a singular ordinal of
cofinality \(\lambda^+\), so \(j(\lambda^+) < j(\lambda)^+\).
\begin{proof}
We first show that \(\lambda^+ \leq \lambda_*^{+M}\). Suppose \(\alpha <
\lambda^+\). Let \(\prec\) be a wellorder of \(\lambda\) such that
\(\text{ot}(\prec) = \alpha\). Then \({\prec}_* = j(\prec)\restriction
\lambda_*\) is a wellorder of \(\lambda_*\) and \(j\) restricts to an
order-preserving embedding from \((\lambda,\prec)\) into
\((\lambda_*,{\prec}_*)\). Therefore \[\alpha \leq
\text{ot}(\lambda_*,{\prec}_*) < \lambda_*^{+{M}}\] The final inequality follows
from the fact that \((\lambda_*,{\prec}_*)\) belongs to \(M\). Since \(\alpha <
\lambda^+\) was arbitrary, it follows that \(\lambda^+ \leq \lambda_*^{+{M}}\).

To prove \(\lambda_*^{+{M}} < j(\lambda)^{+M}\), it is of course enough to show
\(\lambda_*^{+{M}}\leq j(\lambda)\). But \(j(\lambda)\) is a cardinal of \(M\)
that is greater than \(\lambda_*\), and hence \(\lambda_*^{+{M}}\leq
j(\lambda)\).

Finally, assume that \(j\) is continuous at \(\lambda^+\). Obviously
\(j(\lambda^+)\) has cofinality \(\lambda^+\), but the point is that this
implies \(j(\lambda^+)\) is {\it singular}, since the inequalities above show
\(\lambda^+ < j(\lambda^+)\). We can therefore conclude \(j(\lambda)^{+M} <
j(\lambda)^+\): obviously \(j(\lambda)^{+M} \leq j(\lambda)^+\), but the point
is that equality cannot hold since \(j(\lambda^+)\) is singular and
\(j(\lambda)^+\) is regular.
\end{proof}
\end{lma}

\begin{proof}[Proof of \cref{UFSuperBound}] Let \(\lambda = \lambda_U\). Without
loss of generality, we may assume that \(U\) is a uniform ultrafilter on
\(\lambda\) and \(\lambda\) is infinite. Thus \(j_U\) is discontinuous at
\(\lambda\). Assume towards a contradiction that \(j_U[\lambda^+]\in M_U\). By
\cref{Ineqs}, \(j_U(\lambda^+) > \lambda^+\). But by \cref{UFContinuity},
\(j_U\) is continuous at \(\lambda^+\), and therefore \(j_U[\lambda^+]\in M_U\)
is a cofinal subset of \(j_U(\lambda^+)\) of ordertype \(\lambda^+\). Hence
\(\textnormal{cf}^{M_U}(j_U(\lambda)^{+M_U}) = \lambda^+< j_U(\lambda^+)\), and
this contradicts that \(j_U(\lambda^+)\) is regular in \(M_U\).
\end{proof}

\subsection{The Kunen Inconsistency}\label{KunenSection}
The story of the Kunen Inconsistency Theorem is often cast as a cautionary tale
with the moral that a large cardinal hypothesis may turn out to be false for
nontrivial combinatorial reasons:
\begin{thm}[Kunen]\index{Kunen Inconsistency Theorem}
	There is no nontrivial elementary embedding from the universe to itself.\qed
\end{thm}
A more pragmatic perspective is to view the Kunen Inconsistency as a proof
technique, providing at least some constraint on the elementary embeddings a
large cardinal theorist is bound to analyze. Examples pervade this work, but to
pick the closest one, the Kunen Inconsistency will form a key component of the
proof of the wellfoundedness of the Mitchell order in \cref{MOWFSection}.  Since
our applications of Kunen's theorem will require the basic concepts from the
proof (especially the notion of a critical sequence), we devote this subsection
for a brief exposition of this topic.

We first give a proof of a version of Kunen's inconsistency Theorem that is due
to Harada. (Another writeup of this proof appears in Kanamori's textbook
\cite{Kanamori2}.) The methods are purely ultrafilter-theoretic and very
much in the spirit of this dissertation:
\begin{prp}[Kunen]\label{KunenPrp}
	Suppose \(j : V\to M\) is an elementary embedding and \(\eta\) is a strong
	limit cardinal such that  \(j\) is \(\eta\)-supercompact and \(j[\eta]
	\subseteq \eta\). Then \(j \restriction \eta = \textnormal{id}\).
	\begin{proof}
		Assume the proposition holds for all \(\bar \eta <\eta\).

		If \(\eta\) is has uncountable cofinality, then there is an
		\(\omega\)-closed unbounded set of \(\bar \eta < \eta\) such that
		\(j[\bar \eta]\subseteq \bar \eta\). Therefore \(j \restriction \bar
		\eta=\text{id}\) for unboundedly many \(\bar \eta < \eta\), so
		\(j\restriction \eta = \text{id}\).

		Assume instead that \(\eta\) has countable cofinality. Then \(j\) is
		continuous at \(\eta\), so since \(j[\eta]\subseteq \eta\), we have
		\(j(\eta) = \eta\). We essentially reduce to the case that \(j\) is the
		ultrapower of the universe by an ultrafilter \(U\) on \(P(\eta)\). Let
		\(U\) be the ultrafilter on \(P(\eta)\) derived from \(j\) using
		\(j[\eta]\). Let \(k : M_U\to M\) be the factor embedding. Then
		\(j[\eta]\in k[M_U]\), so by our analysis of the supercompactness of
		derived embeddings (\cref{DerivedSC}), \(j_U\) is \(\eta\)-supercompact
		and \(k\restriction \eta + 1 = \text{id}\). If follows that \(j_U(\eta)
		= \eta\). Moreover, if we show \(j_U\restriction \eta = \text{id}\) then
		we can conclude \(j\restriction \eta = \text{id}\). In fact, we will
		show that \(U\) is principal.

		By \cref{SmallCfCompact}, \(j_U\) is actually
		\(\eta^\omega\)-supercompact. Since \(\eta\) is a strong limit cardinal
		of countable cofinality, \(\eta^\omega = 2^\eta\). Thus \(j_U\) is
		\(2^\eta\)-supercompact. Recall that \cref{UFSuperBound} states that if
		\(j_U\) is \(\lambda_U^+\)-supercompact, then \(U\) is principal. Thus
		to show \(U\) is principal, it suffices to show that \(\lambda_U <
		2^\eta\).

		Since \(U\) is an ultrafilter on \(P(\eta)\), \(\lambda_U \leq |P(\eta)|
		\leq 2^\eta\), so in fact, we need only show \(\lambda_U \neq 2^\eta\).
		Since \(U\) is isomorphic to a uniform ultrafilter on \(\lambda_U\),
		\(j_U\) is discontinuous at \(\lambda_U\), and in particular
		\(j_U(\lambda_U) \neq \lambda_U\). On the other hand, since \(M_U\) is
		closed under \(2^\eta\)-sequences, \((2^\eta)^{M_U} = 2^\eta\), and
		hence \[j_U(2^\eta) =  (2^{j_U(\eta)})^{M_U} = (2^\eta)^{M_U} = 2^\eta\]
		Since \(\lambda_U\) is moved by \(j_U\) while \(2^\eta\) is fixed,
		\(\lambda_U \neq 2^\eta\), and hence \(\lambda_U < 2^\eta\), as desired.
	\end{proof}
\end{prp}

Suppose \(j : V\to M\) is an elementary embedding with critical point
\(\kappa\). Let \(\lambda\) be the first ordinal above \(\kappa\) such that
\(j[\lambda]\subseteq \lambda\), the first ordinal at which one might be able to
apply the Kunen argument. \cref{KunenPrp} tells us that \(j[\lambda]\notin M\)
if \(\lambda\) is a strong limit; we would like to see that in fact
\(j[\lambda]\) never belongs to \(M\). This follows from the critical sequence
analysis of \(\lambda\):

\begin{defn}\label{CriticalSequence}\index{Critical sequence}
	Suppose \(N\) and \(P\) are transitive models of ZFC with the same ordinals
	and \(j : N\to P\) is a nontrivial elementary embedding. The {\it critical
	sequence} of \(j\) is the sequence \(\langle \kappa_n : n < \omega\rangle\)
	defined by recursion: \(\kappa_0 = \textsc{crt}(j)\) and for all \(n <
	\omega\), \(\kappa_{n+1} = j(\kappa_n)\).
\end{defn}

In the context of \cref{CriticalSequence}, let \(\lambda = \sup_{n < \omega}
\kappa_n\). Clearly \(\lambda\) is the least ordinal such that
\(j[\lambda]\subseteq \lambda\). If \(\text{cf}^M(\lambda) = \omega\), \(j\) is
continuous at \(\lambda\), so \(j(\lambda) = \lambda\). In particular, if \(N =
V\), which is the case of interest in this section, then \(\lambda\) is the
first fixed point of \(j\) above \(\textsc{crt}(j)\).

In the case \(n > 1\), the conclusion of the following lemma is a considerable
understatement:
\begin{lma}\label{CrtSeqMeas}
	Suppose \(j : V\to M\) is a nontrivial elementary embedding and \(\langle
	\kappa_n :  n < \omega\rangle\) is its critical sequence. For any \(n <
	\omega\), if \(j\) is \(\kappa_n\)-strong then \(\kappa_n\) is measurable.
	\begin{proof}
		The proof is by induction on \(n\). Certainly \(\kappa_0 =
		\textsc{crt}(j)\) is measurable. Assume the lemma is true for \(n = m\),
		and we will show it is true for \(n = m+1\). Therefore assume \(j\) is
		\(\kappa_{m+1}\)-strong. In particular, \(j\) is \(\kappa_m\)-strong, so
		by our induction hypothesis, \(\kappa_m\) is measurable. By
		elementarity, \(\kappa_{m+1} = j(\kappa_m)\) is measurable in \(M\).
		Since \(j\) is \(\kappa_{m+1}\)-strong, \(P(\kappa_{m+1})\subseteq M\).
		Thus the measurability of \(\kappa_{m+1}\) in \(M\) is upwards absolute
		to \(V\), so \(\kappa_{m+1}\) is measurable.
	\end{proof}
\end{lma}

\begin{thm}[Kunen]\label{KunenInconsistency}
	Suppose \(j : V\to M\) is a nontrivial elementary embedding and \(\lambda\)
	is the least ordinal above \(\textsc{crt}(j)\) with \(j[\lambda]\subseteq
	\lambda\). Then \(j[\lambda]\notin M\).
	\begin{proof}
		Assume towards a contradiction that \(j[\lambda]\in M\). By
		\cref{SupercompactClosure}, \(j\) is \(\lambda\)-strong. Therefore
		\(\kappa_n\) is measurable for all \(n < \omega\) by \cref{CrtSeqMeas},
		and so \(\lambda\) is a strong limit cardinal. Since \(j[\lambda]
		\subseteq \lambda\) and \(P(\lambda)\subseteq M\), \(j[\lambda]\in M\).
		Thus \(\lambda\) is a strong limit cardinal, \(j[\lambda]\subseteq
		\lambda\), and \(j\) is \(\lambda\)-supercompact. From \cref{KunenPrp}
		we can therefore conclude that \(\textsc{crt}(j) \geq \lambda\),
		contradicting that \(\textsc{crt}(j) < \lambda\).
	\end{proof}
\end{thm}

A useful structural consequence of Kunen's Inconsistency Theorem is the
following lemma:

\begin{lma}\label{Kunen}
	Suppose \(\gamma\) is a cardinal, \(j : V\to M\) is a nontrivial elementary
	embedding, \( \textsc{crt}(j)\leq \gamma\), and \(P(\gamma)\subseteq M\).
	Then there is a measurable cardinal \(\kappa \leq\gamma\) such that
	\(j(\kappa) > \gamma\).
	\begin{proof}
		Let \(\langle \kappa_n : n < \omega\rangle\) be the critical sequence of
		\(j\) and  \(\lambda = \sup_{n <\omega} \kappa_n\). Thus \(\lambda\) is
		the least ordinal with \(j[\lambda]\subseteq \lambda\). By
		\cref{KunenInconsistency}, \(P(\lambda)\not\subseteq M\), so since
		\(P(\gamma)\subseteq M\), we have \(\gamma < \lambda\). Let \(n <
		\omega\) be least such that \(\kappa_n \leq \gamma < \kappa_{n+1}\).
		\cref{CrtSeqMeas} implies \(\kappa_n\) is measurable, and \(j(\kappa_n)
		= \kappa_{n+1} > \gamma\). Thus taking \(\kappa= \kappa_n\) proves the
		lemma.
	\end{proof}
\end{lma}

In one instance (\cref{GeneralizedSolovay}), we will need a strengthening of
\cref{Kunen} which has essentially the same proof:
\begin{lma}\label{Kunen2}
	Suppose \(\gamma\leq \lambda\) are cardinals and \(j : V\to M\) is a
	nontrivial \(\lambda\)-supercompact elementary embedding with
	\(\textsc{crt}(j)\leq \gamma\). Then there is a \(\lambda\)-supercompact
	cardinal \(\kappa \leq\gamma\) such that \(j(\kappa) > \gamma\).\qed
\end{lma}

\subsection{The wellfoundedness of the generalized Mitchell order}\label{MOWFSection}
The main theorem of this subsection states that the generalized Mitchell order
is a wellfounded partial order when restricted to a reasonable class of
countably complete ultrafilters. In fact, the wellfoundedness of the generalized
Mitchell order on countably complete ultrafilters is a special case of Steel's
wellfoundedness theorem for the Mitchell order on extenders
\cite{SteelWellfounded}, since countably complete ultrafilters are amenable
extenders in the sense of \cite{SteelWellfounded}, but we will give a much
simpler proof here.

We start with the fundamental fact that the Mitchell order is irreflexive:
\begin{lma}\label{MOStrict}
Suppose \(U\) is a countably complete nonprincipal ultrafilter. Then \(U\not\mo
U\).
\begin{proof}
Suppose towards a contradiction that \(U\mo U\). By \cref{HeredLemma}, if \(U'
\cong U\) is a uniform ultrafilter on a cardinal (as given by \cref{UniformIso})
then  \(U'\mo U'\) as well. We can therefore assume without loss of generality
that \(U\) is a uniform ultrafilter on a cardinal \(\lambda\).  By
\cref{MOSuper}, \(j_U\restriction P(\lambda)\in M_U\). In particular,
\(j_U\restriction \lambda\in M_U\), so \(M_U^\lambda\subseteq M_U\) by
\cref{SupercompactClosure}. Therefore \(j^{M_U}_U = j_U\restriction M_U\), for
example as a consequence of \cref{USupercompact}. Thus \(j_U\) is
\(\delta\)-supercompact for all cardinals \(\delta\). This contradicts
\cref{UFSuperBound}.
\end{proof}
\end{lma}

We now turn to the transitivity and wellfoundedness of the generalized Mitchell
order. The following lemma (which in the language of \cite{SteelWellfounded}
states that countably complete ultrafilters are {\it amenable}), is the key to
the proof.
\begin{lma}\label{AmenabilityLemma}
Suppose \(U\) is a nonprincipal countably complete ultrafilter on a set \(X\).
Suppose \(\lambda\) is a cardinal such that \(P(\lambda)\subseteq M_U\). Then
\(M_U\vDash 2^{\lambda} < j_U(|X|)\).
\begin{proof}
	The proof proceeds by finding a measurable cardinal \(\kappa\leq |X|\) such
	that \(2^\lambda < j_U(\kappa)\).

	If \(\lambda < \textsc{crt}(j_U)\), then \(\kappa = \textsc{crt}(j_U)\)
	works. Therefore assume \(\textsc{crt}(j_U)\leq\lambda\). By \cref{Kunen},
	there is an measurable cardinal \(\kappa \leq \lambda\) such that
	\(j_U(\kappa) > \lambda\). We claim that \(\kappa \leq |X|\), which
	completes the proof. Assume not. Then \(|X|\) is smaller than the
	inaccessible cardinal \(\kappa\), and hence \(j_U(\kappa) = \kappa \leq
	\lambda\) (\cref{UFCardinality}), a contradiction.
\end{proof}
\end{lma}

We really only use the following consequence of \cref{AmenabilityLemma}:

\begin{cor}\label{Hered0}
Suppose \(U_0\mo U_1\) are countably complete nonprincipal hereditarily uniform
ultrafilters. Then \(M_{U_1}\vDash 2^{\lambda_{U_0}} < j_{U_1}(\lambda_{U_1})\).
\begin{proof}
This is immediate from \cref{AmenabilityLemma}, using the fact
(\cref{MOStrength}) that if \(U_0\mo U_1\) then \(P(\lambda_{U_0})\subseteq
M_{U_1}\).
\end{proof}
\end{cor}

\begin{cor}
Suppose \(U_0\mo U_1\) are countably complete nonprincipal hereditarily uniform
ultrafilters. Let \(\lambda = \lambda_{U_1}\). Then \(U_0\in
j_{U_1}(H(\lambda))\).
\begin{proof}
Since \(U_0\) is hereditarily uniform, \(M_{U_1}\vDash |\text{tc}(U_0)| =
2^{\lambda_{U_0}}\) By \cref{Hered0}, \(M_{U_1}\vDash 2^{\lambda_{U_0}} <
j_{U_1}(\lambda)\). Therefore \(U_0\in H^{M_{U_1}}(j_{U_1}(\lambda)) =
j_{U_1}(H(\lambda))\).
\end{proof}
\end{cor}

\begin{prp}\label{StrongTransitivity}
Suppose \(U_0\mo U_1\mo U_2\) are countably complete nonprincipal hereditarily
uniform  ultrafilters. Then \(U_0\mo U_2\) and \(M_{U_2}\vDash U_0\mo U_1\).
\begin{proof}
Let \(\lambda =\lambda_{U_1}\). Then \(U_0\in j_{U_1}(H(\lambda))\). By
\cref{MOSuper2}, \(M_{U_2}\) contains \(j_{U_1}(H(\lambda))\), so \(U_0\in
M_{U_2}\), which yields \(U_0\mo U_2\). In fact, by \cref{MOSuper2},
\(j_{U_1}(H(\lambda))= j_{U_1}^{M_{U_2}}(H(\lambda))\), and so \(U_0\in
j_{U_1}^{M_{U_2}}(H(\lambda))\subseteq M^{M_{U_2}}_{U_1}\). Thus \(U_0\in
M^{M_{U_2}}_{U_1}\), or other words, \(M_{U_2}\vDash U_0\mo U_1\).
\end{proof}
\end{prp}

\begin{cor}
The generalized Mitchell order is transitive on countably complete nonprincipal
hereditarily uniform ultrafilters.\qed
\end{cor}

It is worth pointing out that the generalized Mitchell order on extenders is
{\it not} transitive if there is a cardinal that is \(\kappa\)-strong where
\(\kappa\) is a measurable cardinal. The counterexample is described in
\cite{SteelWellfounded}. (The generalized Mitchell order is not  transitive on
arbitrary countably complete ultrafilters either as a consequence of the silly
counterexamples in \cref{StrengthSection}.)

\begin{prp}\label{MOUniformWF}
The generalized Mitchell order is wellfounded on countably complete nonprincipal
hereditarily uniform ultrafilters.
\begin{proof}
Suppose not, and let \(\lambda\) be the least cardinal such that there is a
descending sequence \[U_0\gmo U_1\gmo U_2\gmo \cdots\] of  countably complete
hereditarily uniform ultrafilters with \(\lambda_{U_0} = \lambda\).

 By \cref{StrongTransitivity} and the closure of \(M_{U_0}\) under countable
sequences, the sequence \(\langle U_n : {1\leq n < \omega}\rangle\) belongs to
\(M_{U_0}\) and \[M_{U_0}\vDash U_1\gmo U_2\gmo \cdots\] Note that in
\(M_{U_0}\), \(U_1\) is a countably complete nonprincipal hereditarily uniform
ultrafilter, and by \cref{Hered0}, \(\lambda_{U_1} < j_{U_0}(\lambda)\).

On the other hand, by the elementarity of \(j_{U_0}\), from the perspective of
\(M_{U_0}\), \(j_{U_0}(\lambda)\) is the least cardinal \(\lambda'\) such that
there is a descending sequence \[W_0\gmo W_1\gmo W_2\gmo \cdots\] of countably
complete hereditarily uniform ultrafilters such that \(\lambda_{W_0} =
\lambda'\). This is a contradiction.
\end{proof}
\end{prp}
One can prove a slightly more general result than \cref{MOUniformWF} although
this generality is never useful.
\begin{thm}\label{MOWF}
The generalized Mitchell order is wellfounded on nonprincipal countably complete
ultrafilters.\index{Generalized Mitchell order!wellfoundedness}
\begin{proof}
Assume towards a contradiction that \(U_0\gmo U_1\gmo \cdots\) are nonprincipal
countably complete ultrafilters. For each \(n < \omega\), let \(U'_n\) be a
hereditarily uniform ultrafilter such that \(U_n'\cong U_n\). Then by
\cref{HeredLemma}, \(U_0'\gmo U_1'\gmo\cdots\). This contradicts
\cref{MOUniformWF}.
\end{proof}
\end{thm}

\subsection{The nonlinearity of the generalized Mitchell order}\label{NonlinearitySection}
Before we discuss the extent to which the generalized Mitchell order is linear
under UA, it is worth pointing out:
\begin{itemize}
\item the obvious counterexamples to linearity
\item the maximal amount of linearity one can reasonably hope for.
\end{itemize}

The fact is that if there is a measurable cardinal, then the generalized
Mitchell order is not linear, even restricting to uniform countably complete
ultrafilters on cardinals. The known counterexamples to the linearity of the
generalized Mitchell order are closely related to the {\it Rudin-Frol\'ik order}
(the subject of \cref{RFChapter}):
\begin{defn}\index{Rudin-Frol\'ik order}
The {\it Rudin-Frol\'ik order} is defined on countably complete ultrafilters
\(U\) and \(W\) by setting \(U\D W\) if there is an internal ultrapower
embedding \(i : M_D\to M_W\) such that \(i\circ j_D = j_W\).
\end{defn}
By \cref{RKChar}, the Rudin-Keisler order can be defined in exactly the same way
except omitting the requirement that \(i\) be internal.
\begin{prp}
If \(U\D W\) are nonprincipal countably complete ultrafilters, then \(U\) and
\(W\) are incomparable in the generalized Mitchell order.
\begin{proof}
We first show \(U\not \mo W\). Since \(U\D W\), \(M_W\subseteq M_U\). Therefore
the fact that \(U\notin M_U\) implies that \(U\notin M_W\), and hence \(U\not
\mo W\).

We now show \(W\not \mo U\). Assume towards a contradiction that \(W\mo U\).
Assume without loss of generality that \(U\) is a uniform ultrafilter on a
cardinal \(\lambda\). (Since the Mitchell order is isomorphism invariant in its
second argument, this does not change our situation.) Since \(U\D W\), we have
\(U\RK W\) by \cref{RKChar}. Since \(U\) is hereditarily uniform and \(U\RK W
\mo U\), our lemma on the invariance of the Mitchell order (\cref{HeredLemma})
yields that \(U\mo U\). This contradicts \cref{MOStrict}.
\end{proof}
\end{prp}
A similar argument shows the following:
\begin{prp}
Suppose \(U\) and \(W\) are countably complete ultrafilters and there is a
nonprincipal \(D\D U,W\). Then \(U\) and \(W\) are incomparable in the
generalized Mitchell order.\qed
\end{prp}
Even this does not exhaust the known counterexamples to the linearity of the
generalized Mitchell order:

\begin{prp}
Suppose \(U_0\mo U_1\mo U_2\). Suppose \(U_0,U_2\D W\). Then \(U_1\) and \(W\)
are incomparable in the Mitchell order.\qed
\end{prp}
We omit the proof. The hypotheses of the proposition are satisfied if
\(U_0,U_1,U_2\) are normal ultrafilters on measurable cardinals
\(\kappa_0<\kappa_1<\kappa_2\) respectively and \(W = U_0\times U_2\).

All known examples of nonlinearity in the generalized Mitchell order are
accompanied by nontrivial relations in the Rudin-Frol\'ik order. A driving
question in this work is whether assuming UA, these are the only
counterexamples.
\begin{defn}\index{Irreducible ultrafilter}
A nonprincipal countably complete ultrafilter \(W\) is {\it irreducible} if for
all \(U\D W\), either \(U\) is principal or \(U\) is isomorphic to \(W\).
\end{defn}
The Irreducible Ultrafilter Hypothesis (IUH) essentially states that the sort of
counterexamples to the linearity of the Mitchell order that we have described
are the only ones.
\begin{iuh}
Suppose \(U\) and \(W\) are hereditarily uniform irreducible ultrafilters.
Either \(U\cong W\), \(U\mo W\), or \(W\mo U\). \index{Irreducible Ultrafilter
Hypothesis}
\end{iuh}
We can now make precise the question of the extent of the linearity of the
Mitchell order under UA:
\begin{qst}
Does UA imply IUH?
\end{qst}
With this in mind, let us turn to the positive results on linearity.
\section{Dodd soundness}\label{DoddSection}
\subsection{Introduction}
{\it Dodd soundness} is a fine-structural generalization of supercompactness,
introduced by Steel \cite{Schimmerling} in the context of inner model theory as
a strengthening of the initial segment condition. The following remarkable
theorem is due to Schlutzenberg \cite{Schlutzenberg}:
\begin{thm}[Schlutzenberg]\label{Schlutz}
Suppose \(L[\mathbb E]\) is an iterable Mitchell-Steel model and \(U\) is a
countably complete ultrafilter of \(L[\mathbb E]\). Then the following are
equivalent:\index{Schlutzenberg}
\begin{enumerate}[(1)]
\item \(U\) is irreducible.
\item \(U\) is isomorphic to a Dodd sound ultrafilter.
\item \(U\) is isomorphic to an extender on the sequence \(\mathbb E\).\qed
\end{enumerate}
\end{thm}
Since the total extenders on \(\mathbb E\) are linearly ordered by the Mitchell
order, this has the following consequence:
\begin{thm}[Schlutzenberg] Suppose \(L[\mathbb E]\) is an iterable
Mitchell-Steel model. Then \(L[\mathbb E]\) satisfies the Irreducible
Ultrafilter Hypothesis.\qed\index{Irreducible Ultrafilter Hypothesis}
\end{thm}
It is open whether this theorem can be extended to the Woodin models at the
finite levels of supercompactness. The main result of this section
(\cref{DoddMO}) states that UA alone suffices to prove the linearity of the
generalized Mitchell order on Dodd sound ultrafilters.

\subsection{Dodd sound embeddings, extenders, and ultrafilters}
In this subsection, we present a definition of Dodd soundness  due to the author
that is much simpler than the one given in \cite{Schimmerling} and
\cite{Schlutzenberg}, and that is easier to use in certain contexts. (The other
definition is also useful.) We then show that the two definitions are
equivalent.

\begin{defn}\label{EmbeddingSolid}\index{Soundness of an embedding}
Suppose \(M\) is a transitive class, \(j : V\to M\) is an elementary embedding,
and \(\alpha\) is an ordinal. Let \(\delta\) be the least ordinal such that
\(j(\delta)\geq \alpha\). Then \[j^\alpha: P(\delta)\to M\] is the function
defined by \(j^\alpha(X) = j(X)\cap \alpha\). The embedding \(j\) is said to be
{\it \(\alpha\)-sound} if \(j^\alpha\) belongs to \(M\).
\end{defn}
Recall that the bounded powerset of an ordinal \(\delta\) is defined by
\(P_\text{bd}(\delta) = \bigcup_{\xi < \delta} P(\xi)\). In the context of
\cref{EmbeddingSolid}, if \(\alpha = \sup j[\delta]\) it have been natural to
define \(j^\alpha = j\restriction P_{\text{bd}}(\delta)\). With this alternate
definition, \(j^\alpha\in M\) is an a priori weaker requirement. The next lemma
shows that this does not actually make a difference:

\begin{lma}\label{SpaceIrrel}
Suppose \(M\) is a transitive class, \(j : V\to M\) is an elementary embedding,
and \(\delta\) is an ordinal. Let \(\delta_* = \sup j[\delta]\). Then the
following are equivalent:
\begin{enumerate}[(1)]
\item \(j\) is \(\delta_*\)-sound.
\item \(j[P_\textnormal{bd}(\delta)]\in M\) or equivalently \(j\) is
\(2^{<\delta}\)-supercompact.
\item \(j\restriction P_\textnormal{bd}(\delta)\in M\).
\item \(j\restriction P^M_\textnormal{bd}(\delta)\in M\).
\end{enumerate}
\begin{proof}
{\it (1) implies (2):} Trivial. (The equivalence of
\(j[P_\textnormal{bd}(\delta)]\in M\) with \(2^{<\delta}\)-supercompactness is
immediate from \cref{XSC}.)

{\it (2) implies (3):} \(j\restriction P_\textnormal{bd}(\delta)\) is the
inverse of the transitive collapse of \(j[P_\textnormal{bd}(\delta)]\).

{\it (3) implies (4):} Trivial.

{\it (4) implies (1):} Assume \(j\restriction P^M_\textnormal{bd}(\delta)\in
M\). Since \(\delta\subseteq P^M_\textnormal{bd}(\delta)\), \[j\restriction
\delta = (j\restriction P^M_\textnormal{bd}(\delta))\restriction \delta\in M\]
Therefore \(j\) is \(\delta\)-supercompact. Since supercompactness implies
strength (\cref{SupercompactClosure}), \(P(\delta)\subseteq M\). In particular
\(j\restriction P^M_\textnormal{bd}(\delta) = j\restriction
P_\text{bd}(\delta)\). Finally for \(X\subseteq \delta\), \(j^{\delta_*}(X) =
\bigcup_{\xi < \delta} j(X\cap \xi)\), so \(j^{\delta_*}\) is definable from
\(j\restriction   P_\text{bd}(\delta)\) and hence \(j^{\delta_*}\in M\), which
shows (1).
\end{proof}
\end{lma}

\begin{lma}\label{SetDodd}
Suppose \(M\) is a transitive class, \(j : V\to M\) is an elementary embedding,
and \(\alpha\) is an ordinal. Then \(j\) is \(\alpha\)-sound if and only if
\(\{j(X)\cap \alpha : X\in V\}\in M\).
\begin{proof}
The forward direction is immediate since \(\{j(X)\cap \alpha : X\in V\} =
\text{ran}(j^\alpha)\). The reverse direction follows from the fact that
\(j^\alpha\) is the inverse of the transitive collapse of \(\{j(X)\cap \alpha :
X\in V\}\).
\end{proof}
\end{lma}

Our next lemma states that the fragments \(j^\alpha\) ``pull back" under
elementary embeddings.

\begin{lma}\label{DoddPullback}
Suppose \(V\stackrel{i}{\longrightarrow} N\stackrel{k}{\longrightarrow} M\) are
elementary embeddings and \(j = k\circ i\). Suppose \(j^\alpha\in
\textnormal{ran}(k)\). Then \(k^{-1}(j^\alpha) = i^{k^{-1}(\alpha)}\).
\begin{proof}
Let \(\delta\) be the least ordinal such that \(j(\delta) \geq \alpha\). Note
that \(j^\alpha[\text{Ord}] = j[\delta]\in \text{ran}(k)\), so by our analysis
of derived embeddings (\cref{DerivedSC}), \(k\restriction \delta + 1\) is the
identity and \(i\) is \(\delta\)-supercompact. In particular,
\(P_\text{bd}(\delta)\subseteq M\) and \(k(P_\text{bd}(\delta)) =
P_\text{bd}(\delta)\).

Let \(h = k^{-1}(j^\alpha)\). Then \(\text{dom}(h) = k^{-1}(P_\text{bd}(\delta))
= P_\text{bd}(\delta)\). Thus for \(X\in \text{dom}(h)\), \(k(X) = X\), and
hence \[k(h(X)) = k(h)(k(X)) = k(h)(X) = j^\alpha(X) = j(X)\cap \alpha =
k(i(X))\cap \alpha\] By the elementarity of \(k\), this implies that \(h(X) =
i(X)\cap k^{-1}(\alpha)\), or in other words \(k^{-1}(j^\alpha) = h =
i^{k^{-1}(\alpha)}\), as desired.
\end{proof}
\end{lma}

We now turn to Dodd soundness.

\begin{defn}
If \(j : V\to M\) is an extender embedding, the {\it Dodd length of \(j\)},
denoted \(\alpha(j)\), is the least ordinal \(\alpha\) such that every element
of \(M\) is of the form \(j(f)(\xi)\) for some \(\xi < \alpha\).\index{Dodd
length}
\end{defn}

On first glance, one might believe that the Dodd length of an elementary
embedding \(j\) is the same as its {\it natural length}, denoted \(\nu(j)\), the
least \(\nu\) such that \(M = H^M(j[V]\cup \nu)\). In fact, equality may fail:
the issue is that \(\nu(j)\) is the least ordinal such that every element of
\(M\) is of the form \(j(f)(p)\) for a finite set \(p\subseteq \nu\), whereas in
the definition of \(\alpha(j)\), one must write every element of \(M\) in the
form \(j(f)(\xi)\) where \(\xi\) is not a finite set but a single ordinal below
\(\nu\).

Our main focus, of course, is on ultrafilters, and in this case the Dodd length
has an obvious characterization:\footnote{This gives us a counterexample to the
equality of Dodd length and natural length. Suppose \(U\) is a normal
ultrafilter on \(\kappa\). Let \(W = U^2\). Then \(\nu(j_W) = j_U(\kappa) + 1\)
but \(\alpha(j_W) = j_U(\kappa) + \kappa + 1\).}

\begin{lma}\label{UltraDodd} If \(j: V\to M\) is an ultrapower embedding, then \(\alpha(j) = \xi + 1\) where \(\xi\) is the least ordinal such that \(M = H^{M}(j[V]\cup \{\xi\})\). Therefore \(U\) is incompressible if and only if \(U\) is tail uniform and \(\alpha(j_U) = \id_U + 1\).\qed\end{lma}

Our next lemma establishes a limit on the solidity of an extender embedding. (It
is equivalent to the statement that no extender belongs to its own ultrapower.)

\begin{lma}\label{SoundnessLimit}
Suppose \(j : V\to M\) is an extender embedding and \(\alpha = \alpha(j)\). Then
\(j\) is not \(\alpha\)-sound.
\begin{proof}
Let us first show that if \(U\) is a countably complete tail uniform ultrafilter
on an ordinal \(\delta\), then \(j_U\) is not \(\id_U + 1\)-sound. Note that \[U
= \{A\subseteq \delta : \id_U\in j(A)\} = \{A\subseteq \delta : \id_U\in
j^{\id_U + 1}_U(A)\}\] so since \(U\notin M_U\), \(j^{\id_U + 1}_U\notin M\).
Thus \(j_U\) is not \(\id_U + 1\)-sound, as claimed.

We now handle the case where \(j\) is an arbitrary extender embedding. By the
definition of Dodd length, there is some \(\xi < \alpha\) and some function
\(f\in V\) such that \(j^\alpha = j(f)(\xi)\). Let \(U\) be the tail uniform
ultrafilter derived from \(j\) using \(\xi\), and let \(k : M_U\to M\) be the
factor embedding. Then \(\xi \in \text{ran}(k)\) and so \(j^\alpha\in
\text{ran}(k)\). Applying our lemma on pullbacks of the fragments \(j^\alpha\)
(\cref{DoddPullback}), \(k^{-1}(j^\alpha) = j_U^{k^{-1}(\alpha)}\). Therefore
\(j_U\) is \(k^{-1}(\alpha)\)-sound. But note that \(\id_U = k^{-1}(\xi) <
k^{-1}(\alpha)\). Hence \(j_U\) is \(\id_U + 1\)-sound, and this contradicts the
first paragraph.
\end{proof}
\end{lma}

An embedding is {\it Dodd sound} if it is as sound as it can possibly be:

\begin{defn}\label{EmbeddingSoundDef}
Suppose \(M\) is a transitive class and \(j : V\to M\) is an elementary
embedding. Then \(j\) is said to be {\it Dodd sound} if \(j\) is \(\beta\)-sound
for all \(\beta < \alpha(j)\).\index{Dodd soundness}
\end{defn}

We now prove the equivalence between the Dodd soundness of an extender \(E\) as
it is defined in \cite{Schimmerling} and the Dodd soundness of its associated
embedding \(j_E\) as it is defined in \cref{EmbeddingSolid}.

\begin{defn}
\begin{itemize}\index{Parameter}
	\item A {\it parameter} is a finite set of ordinals.
	\item The {\it parameter order}\index{Parameter!parameter order} is defined
	on parameters \(p\) and \(q\) by \[p < q\iff \max(p\mathrel{\triangle} q)\in
	q\]
	\item If \(p\) is a parameter, then \(\langle p_i : i < |p|\rangle\) denotes
	the {\it descending} enumeration of \(p\).
	\item For any \(k \leq |p|\), \(p\restriction k\) denotes the parameter
	\(\{p_i : i < k\}\).
\end{itemize}
\end{defn}

The point of enumerating parameters in descending order is that the parameter
order is then transformed into the lexicographic order:

\begin{lma}
Suppose \(p\) and \(q\) are parameters of length \(n\) and \(m\) respectively.
Then \(p < q\) if and only if \(\langle p_0,\dots,p_{n - 1}\rangle
<_\textnormal{lex} \langle q_0,\dots,q_{m - 1}\rangle\).\qed
\end{lma}

\begin{figure}
	\center
	\includegraphics[scale=.6]{Parameter.pdf}
	\caption{The parameter order}\label{ParamterFig}
\end{figure}

\begin{lma}
The parameter order is a set-like wellorder.\qed
\end{lma}

\begin{defn}
If \(j : V\to M\) is an elementary embedding and \(p\) is a parameter, then
\(\mu_j(p)\) is the least ordinal \(\mu\) such that \(p\subseteq j(\mu)\).
\end{defn}

\begin{defn}\index{Extender}
Suppose \(j : V\to M\) is an elementary embedding, \(p\) is a parameter, and
\(\nu < \min(p)\) is an ordinal. Let \(\delta = \mu_j(p)\). Then the {\it
extender of \(j\) below \((p,\nu)\)} is the set \[E^j\restriction p\cup \nu =
\{(q,A) : q\in [\nu]^{<\omega}\text{, }A\subseteq [\delta]^{<\omega}\text{, and
}p\cup q\in j(A)\}\]
\end{defn}

The restriction \(E^j\restriction p\cup \nu\) can be thought of as an extender
{\it relativized} to the parameter \(p\). It is possible to axiomatize
relativized extenders as directed systems of ultrafilters and associate to them
ultrapower embeddings, namely the direct limit of these systems. Instead we make
the following definition:

\begin{defn}\index{Extender!relativized}
A {\it relativized extender} is a set of the form \(E^j\restriction p\cup \nu\)
for some elementary embedding \(j\). The {\it extender embedding} associated to
a relativized extender \(E\), denoted \[j_E : V\to M_E\] is the unique \(j :
V\to M\) such that \(E = E^j\restriction p\cup \nu\) for some \(p,\nu\) and \(M
= H^M(j[V]\cup p\cup \nu)\).

If \(E\) is a relativized extender, \(\nu\) is an ordinal, and \(p\) is a
parameter, then \[E\restriction p\cup \nu = E^{j}\restriction p\cup \nu\] where
\(j = j_E\).
\end{defn}

The {\it Dodd parameter} of an extender is the key to the fine-structural proofs
of Dodd soundness, which are motivated by the fundamental solidity proofs from
fine structure theory.

\begin{defn}\label{DoddParamDef1}
Suppose \(j : V\to M\) is an extender embedding. Then \(\eta(j)\) is the least
ordinal \(\eta\) such that for some parameter \(p\), \[M = H^M(j[V]\cup p\cup
\eta)\] The {\it Dodd parameter}\index{Dodd parameter}\index{\(p(j)\) (Dodd
parameter)} of \(j\), denoted \(p(j)\), is the least parameter \(p\) such that
\[M = H^M(j[V]\cup p\cup \eta(j))\]
\end{defn}
Thus if \(j\) is an ultrapower embedding, as it always will be in our
applications, then \(\eta = 0\). More generally, \(\eta\) is obviously always a
limit ordinal.

The Dodd parameter can also be defined recursively using the concept of an
\(x\)-generator of an elementary embedding:
\begin{defn}\index{Generator}
Suppose \(M\) and \(N\) are transitive models of ZFC, \(j : M\to N\) is an
elementary embedding, and \(x\in N\). Then an ordinal \(\xi\in N\) is an {\it
\(x\)-generator of \(j\)} if \(\xi\notin H^{N}(j[M]\cup \xi\cup \{x\})\).
\end{defn}

\begin{lma}
Suppose \(j : V\to M\) is an extender embedding. Let \(q\) be the
\(\subseteq\)-maximum parameter with the property that \(q_k\) is the largest
\(q\restriction k\)-generator of \(j\) for all \(k < |q|\). Then \(p(j) = q\)
and \(\eta(j)\) is the strict supremum of the \(q\)-generators of \(j\).
\begin{proof}
	 Let \(p = p(j)\), \(n = |p|\), and \(\eta = \eta(j)\). Fix \(k < n\). We
	 will show \(p_k\) is the largest \(p\restriction k\)-generator.

	 Since \(M = H^M(j[V]\cup p\cup \eta)\subseteq H^M(j[V]\cup p\restriction k
	 \cup (p_k+1))\), there are no \(p\restriction k\)-generators strictly above
	 \(p_k\). It therefore suffices to show that \(p_k\) is a \(p\restriction
	 k\)-generator. Assume not. Then \(p_k \in H^M(j[V]\cup p\restriction k \cup
	 p_k)\). Fix \(u\subseteq p_k\) such that \(p_k = j(f)(p\cup r)\) for some
	 function \(f\in V\). Let \(r = p\setminus \{p_k\} \cup u\). Then \(r < p\)
	 in the parameter order, but \(p\subseteq H^M(j[V]\cup r)\), and hence \(M =
	 H^M(j[V]\cup r \cup \eta)\), contrary to the minimality of the Dodd
	 parameter \(p\).

	 By the maximality of \(q\), this shows that \(p = q\restriction n\). We now
	 show that \(\eta\) is the strict supremum of the \(p\)-generators of \(j\).
	 Since \(M = H^M(j[V]\cup p\cup \eta)\), there are no \(p\)-generators
	 greater than or equal to \(\eta\). It therefore suffices to show that for
	 any \(\alpha < \eta\), there is a \(p\)-generator of \(j\) above
	 \(\alpha\). Suppose \(\alpha < \eta\). By the minimality of \(\eta\), \(M
	 \neq H^M(j[V]\cup p\cup \alpha)\), and so there is a \(p\)-generator of
	 \(j\) above \(\alpha\), as desired.

	 Since \(\eta\) is a limit ordinal, there is no largest \(p\)-generator, and
	 hence \(p = q\).
\end{proof}
\end{lma}

\begin{cor}
Suppose \(j : V\to M\) is an extender embedding and \(p = p(j)\). Then for all
\(i < |p|\), \(p_i\) is a \(\{p_0,\dots,p_{i-1}\}\)-generator.\qed
\end{cor}

The following is Steel's definition of the Dodd soundness of an extender:

\begin{defn}\label{SteelSoundDef}\index{Dodd soundness!Dodd sound extender}\index{Dodd soundness!Dodd solid extender}
Suppose \(E\) is an extender, \(p = p(j_E)\), and \(\eta = \eta(j_E)\).
\begin{itemize}
\item  \(E\) is {\it Dodd solid} if \[E\restriction \{p_0,\dots,p_{i-1}\}\cup
p_i\in M_E\] for all \(i < |p|\).
\item \(E\) is {\it Dodd sound} if \(E\) is Dodd solid and \[E\restriction p\cup
\nu\in M_E\] for all \(\nu < \eta\).
\end{itemize}
\end{defn}

If \(E\) is an extender such that \(j_E\) is an ultrapower embedding, then \(E\)
is Dodd solid if and only if \(E\) is Dodd sound, simply because \(\eta(j_E) =
0\) (so the extra requirement for Dodd soundness holds vacuously).

The following fact is essentially a matter of rearranging definitions:

\begin{thm}\label{DoddEquiv}
Suppose \(E\) is an extender. Then \(E\) is Dodd sound in the sense of
\cref{SteelSoundDef} if and only if \(j_E\) is Dodd sound in the sense of
\cref{EmbeddingSoundDef}.
\begin{proof}
Before we prove the equivalence, we prove three preliminary claims.

Let \(j = j_E\) and \(M = M_E\). Let \(\eta = \eta(j)\) and let \(p = p(j)\) be
the Dodd parameter of \(j\).

\begin{clm}\label{LeastParamClm}
\(p\cup \{\eta\}\) is the least parameter \(s\) such that every element of \(M\)
is of the form \(j(f)(q)\) for some \(q < s\).
\begin{proof}
Suppose not. Then fix \(s < p\cup \{\eta\}\) such that every element of \(M\) is
of the form \(j(f)(q)\) for some \(q < s\). Fix \(q < s\) such that \(p =
j(f)(q)\) for some \(f\). Then \(M = H^M(j[V]\cup q\cup \eta)\). Since \(p\) is
the least parameter with this property (by the definition of the Dodd
parameter), it follows that \(p\leq q\). In particular \(p < s\). Since \(p < s
< p\cup \{\eta\}\), \(s = p \cup r\) for some \(r \in [\eta]^{<\omega}\). Now
let \(\xi < \eta\) be a \(p\)-generator such that \(r \subseteq \xi\). Then
\(p\cup \{\xi\} = j(f)(u)\) for some \(u < s\). Since \(u\) generates \(p\), we
must have \(p\leq u\). Since \(p \leq u \leq p \cup r\), \(u = p \cup t\) for
some \(t < r\). In particular, since \(r\subseteq \xi\), \(t\subseteq \xi\). Now
\(\xi = j(f)(p\cup r)\) where \(r \in [\xi]^{<\omega}\), contradicting that
\(\xi\) is not a \(p\)-generator.
\end{proof}
\end{clm}

Let \(\varphi\) be the function that sends a parameter to its rank in the
parameter order.
\begin{clm}\label{ConstClm} Suppose \(x\in M\) and \(q\) is a parameter. Then \(x = j(f)(q)\) for some function \(f\in V\) if and only if \(x = j(g)(\varphi(q))\) for some function \(g\in V\).\end{clm}
\begin{proof}
For the forwards direction, let \(g = f\circ \varphi^{-1}\), and for the reverse
direction, let \(f = g\circ \varphi\).
\end{proof}
From \cref{LeastParamClm} and \cref{ConstClm}, we obtain the following key
identity:
\begin{equation}\varphi(p\cup \{\eta\}) = \alpha(j)\label{lengthparam}\end{equation}
(Recall that \(\alpha(j)\) denotes the Dodd length of \(j\), the least ordinal \(\alpha\) such that every element of \(M\) is of the form \(j(f)(\xi)\) for some \(\xi < \alpha\).)

\begin{clm}\label{unionclm}
Suppose \(q\) is a parameter and \(m = |q|\). For \(i < m\), let \[F_i = E\restriction \{q_0,\dots,q_{i-1}\}\cup q_i\] Then for any transitive model \(N\) of \textnormal{ZFC}, the following are equivalent:
\begin{enumerate}[(1)]
\item \(F_0,\dots,F_{m-1}\in N\).
\item \(j^{\varphi(q)}\in N\).
\end{enumerate}
\begin{proof}[Sketch]
{\it (1) implies (2):} Let \(\mu = \mu_j(q) = \mu_j(\{q_0\})\). If \(F_0,\dots,F_{m-1}\in N\), then so is the function \(e : P([\mu]^{<\omega})\to M\) defined by \(e(X) = \{r < q : r\in j(X)\}\). (\(e\) is the parameter version of \(j^q\).) This is because \(r\in e(X)\) if and only if \((r,X)\in F_i\) where \(i\) is such that \(\max(q\triangle r) = q_i\).

Let \(\delta\) be least such that \(j(\delta) \geq \varphi(q)\). Then \(\varphi[\delta]\subseteq\mu\) and for \(A\subseteq\delta\), \(j(A)\cap \varphi(q) = \varphi^{-1}[e(\varphi[A])]\). This shows \(j^{\varphi(q)}\in N\).

{\it (2) implies (1):} Similar.
\end{proof}
\end{clm}

Having proved the three claims, we finally turn to the equivalence of the two notions of Dodd soundness. (We will leave some of the parameter order theoretic details to the reader.)

Assume first that \(E\) is Dodd sound in the sense of \cref{SteelSoundDef}. Suppose \(\beta < \alpha(j)\), and we will show that \(j\) is \(\beta\)-sound. It suffices to show that \(j\) is \(\beta'\)-sound for some \(\beta' \geq \beta\), which allows us to increase \(\beta\) throughout the argument if necessary. By \cref{lengthparam}, by increasing \(\beta\), we may assume \(\varphi(p)\leq \beta\). Thus \(p\leq \varphi^{-1}(\beta) < \varphi^{-1}(\alpha(j)) = p\cup \{\eta\}\), as a consequence of \cref{lengthparam}. Let \(q = \varphi^{-1}(\beta)\). Then \(p\leq q  < p\cup \{\eta\}\), so \(q = p \cup r\) for some \(r \subseteq \eta\). Since \(\eta\) is a limit ordinal, by increasing \(\beta\) if necessary, we may assume \(|r| \leq 1\). By the Dodd soundness of \(E\), for all \(i < |q|\), \[E \restriction \{q_0,\dots,q_{i-1}\}\cup q_i\in M\]
This is because either \(\{q_0,\dots,q_{i-1}\}\cup q_i = \{p_0,\dots,p_{i-1}\}\cup p_i\) or \(\{q_0,\dots,q_{i-1}\}\cup q_i =p\cup \xi\) for some \(\xi < \eta\). Therefore by \cref{unionclm}, \(j^\beta\in M\) so \(j\) is \(\beta\)-sound.

Conversely, assume that \(j\) is Dodd sound as an elementary embedding. Let \(\beta = \varphi(p)\). Since \(p < p\cup \{\eta\}\), by \cref{lengthparam}, \(\beta < \alpha\). Therefore \(j^\beta\in M\) by the Dodd soundness of \(j\). By \cref{unionclm}, it follows that \(E\restriction \{p_0,\dots,p_{i-1}\}\cup p_i\) for all \(i < |p|\), so \(E\) is Dodd solid. If \(\eta = 0\), it follows that \(E\) is Dodd sound. Assume instead that \(\eta > 0\). Fix \(\xi < \eta\), and we will show \(E\restriction p \cup \xi\in M\). Let \(q = p \cup \{\xi\}\). Then \(q < p\cup \{\eta\}\), so \(\varphi(q) < \alpha\). Therefore by the Dodd soundness of \(j\), \(j^{\varphi(q)}\in M\). Applying \cref{unionclm}, it follows that \(E\restriction p \cup \xi\in M\).
\end{proof}
\end{thm}

It is worth remarking that the proof shows that an extender \(E\) is Dodd solid if and only if \(j_E\) is \(\beta\)-solid where \(\beta\) is the rank of \(p(j_E)\) in the canonical wellorder on parameters.

We now define Dodd sound ultrafilters. One could define an ultrafilter to be Dodd sound if its ultrapower embedding is Dodd sound, but then there would be many isomorphic Dodd sound ultrafilters all with the same associated embedding, which complicates the statements of our theorems and adds no real generality. Instead, we ensure that a Dodd sound ultrafilter is the canonical element of its isomorphism class:

\begin{defn}\index{Dodd soundness!Dodd sound ultrafilter}
A countably complete ultrafilter is {\it Dodd sound} if it is incompressible and its ultrapower embedding is Dodd sound.
\end{defn}

The following alternate characterization of Dodd soundness for ultrafilters is immediate from \cref{UltraDodd} and \cref{SoundnessLimit}:

\begin{lma}\label{IdDodd}
	A tail uniform ultrafilter \(U\) on an ordinal \(\delta\) is Dodd sound if and only if \(j_U\) is \(\id_U\)-sound. That is, \(U\) is Dodd sound if and only if the function \(h : P(\delta)\to M_U\) defined by \(h(X) = j_U(X)\cap \id_U\) belongs to \(M_U\).\qed
\end{lma}

We finally provide a combinatorial characterization of Dodd soundness for ultrafilters:

\begin{defn}
Suppose \(U\) is an ultrafilter on an ordinal \(\delta\).
\begin{itemize}
	\item A sequence \(\langle S_\alpha : \alpha < \delta\rangle\) is {\it \(U\)-threadable} if there is a set \(S\subseteq \delta\) such that \(S\cap \alpha = S_\alpha\) for \(U\)-almost all \(\alpha < \delta\).
	\item A {\it soundness sequence} for \(U\) is a sequence \(\langle \mathcal A_\alpha :\alpha < \delta\rangle\) such that for any sequence \(\langle S_\alpha : \alpha <\delta\rangle\), the following are equivalent:
\begin{enumerate}[(1)]
\item \(\langle S_\alpha :\alpha < \delta\rangle\) is \(U\)-threadable.
\item \(S_\alpha\in \mathcal A_\alpha\) for \(U\)-almost all \(\alpha\).
\end{enumerate}
\end{itemize}
\end{defn}

\begin{thm}
A tail uniform ultrafilter \(U\) is {\it Dodd sound} if and only if it has a soundness sequence.
\end{thm}
\begin{proof}
Note that a sequence \(\langle S_\alpha : \alpha < \delta\rangle\) is \(U\)-threadable if and only if \[[\langle S_\alpha : \alpha < \delta\rangle] = j_U(S)\cap a_U\] some \(S\subseteq \delta\). Thus \(\langle \mathcal A_\alpha :\alpha < \delta\rangle\) is a soundness sequence for \(U\) if and only if \[[\langle \mathcal A_\alpha :\alpha < \delta\rangle]_U = \{j_U(S)\cap \id_U : S\subseteq \delta\}\]
By \cref{SetDodd}, it follows that \(U\) has a soundness sequence if and only if \(j_U\) is \(\id_U\)-sound, or in other words (applying \cref{IdDodd}) \(U\) is Dodd sound.
\end{proof}

\subsection{The generalized Mitchell order on Dodd sound ultrafilters}
In this short section, we prove the linearity of the Mitchell order on Dodd sound ultrafilters. We first prove a stronger statement that characterizes \(P(P(\lambda))\cap M_W\) when \(W\) is Dodd solid in terms of the Lipschitz order on subsets of \(P(\lambda)\).
\begin{prp}\label{LipschitzDodd}
Suppose \(W\) is a Dodd sound ultrafilter on a cardinal \(\lambda\). Then
\[P(P(\lambda))\cap M_W = \{X\subseteq P(\lambda) : X \sLi W\}\]
\begin{proof}
Suppose \(X\subseteq P(\lambda)\).

Assume first that \(X\sLi W\). By our characterization of the Lipschitz order where the second argument is an ultrafilter (\cref{LipChar}), this means that there is a set \(Z\in M_W\) such that for all \(A\subseteq \delta\), \(A\in X\) if and only if \(j_W(A)\cap \id_W \in Z\). But then \(X = (j^{\id_W})^{-1}[Z]\), so \(X\in M_W\).

Conversely, suppose \(X\in M_W\). Let \(Z = j^{\id_W}[X]\). Then \(Z\in M_W\) and for all \(A\subseteq \delta\), \(A\in X\) if and only if \(j_W(A)\cap \id_W= j^{\id_W}(A) \in Z\). It follows that \(X\sLi W\).
\end{proof}
\end{prp}

\begin{cor}\label{LipMO}
Suppose \(U\) and \(W\) are countably complete ultrafilters on \(\lambda\) and \(W\) is Dodd sound. Then \(U\sLi W\) if and only if \(U\mo W\). In particular, if \(U\E W\) then \(U\mo W\).\qed.
\end{cor}

In particular, the Lipschitz order is wellfounded on Dodd sound ultrafilters.

\begin{thm}[UA]\label{DoddMO}
The generalized Mitchell order is linear on Dodd sound ultrafilters.\index{Generalized Mitchell order!linearity!on Dodd sound ultrafilters}
\begin{proof}
Suppose \(U\) and \(W\) are Dodd sound ultrafilters. By the linearity of the Lipschitz order on \(\Un\), either \(U\sLi W\), \(U = W\), or \(U\sgLi W\). Therefore by \cref{LipschitzDodd}, either \(U\mo W\), \(U = W\), or \(U\gmo W\), as desired.
\end{proof}
\end{thm}
Notice that the linearity of the Mitchell order on Dodd sound ultrafilters actually follows from the linearity of the Lipschitz order, which perhaps is weaker than UA.

As a consequence of \cref{LipMO}, if \(W\) is Dodd sound and \(U\sE W\), then \(U\mo W\). We now prove a strong converse, which is closely related to \cref{USupercompact}:

\begin{prp}\label{SuperMOKet}\index{Seed order!vs. the generalized Mitchell order}\index{Ketonen order!vs. the generalized Mitchell order}
Suppose \(U\) is a countably complete ultrafilter on a cardinal \(\lambda\) and \(W\) is a nonprincipal uniform ultrafilter on a cardinal \(\delta\) such that \(j_W\) is \(\lambda\)-supercompact. If \(U \mo W\), then \(U\swo W\).
\begin{proof}
Note that \((j_U(j_W),j_U\restriction M_W)\) is a \(0\)-internal comparison of \((j_U,j_W)\) by the standard identity: \[j_U(j_W)\circ j_U = j_U\circ j_W\] Since \(j_W\) is \(\lambda\)-supercompact, \(j_U\restriction M_W = j_U^{M_W}\), which is definable over \(M_W\) since \(U\mo W\).

Since \(j_W\) is \(\lambda\)-supercompact, \(\lambda \leq \delta\) by \cref{UFSuperBound}. Therefore for all \(\alpha < \lambda\), \(j_W(\alpha) < \id_W\). Applying \L o\'s's Theorem,
\[j_U(j_W)(\id_U) = [j_W\restriction \lambda]_U < j_U(\id_W)\]
Thus \((j_U(j_W),j_U\restriction M_W)\) witnesses that \(U\swo W\).
\end{proof}
\end{prp}

This raises the question of whether the Ketonen order extends the generalized Mitchell order. One should restrict attention here to countably complete uniform ultrafilters on cardinals, or else there are silly counterexamples. If this were true, it would complete the picture in which the wellfoundedness of the Ketonen order explains that of all the other known wellfounded orders. It is consistently false, however (\cref{CummingsExample}):
\begin{prp}
Suppose \(\kappa\) is \(2^\kappa\)-supercompact and \(2^\kappa = 2^{\kappa^+}\). Then there are \(\kappa\)-complete uniform ultrafilters \(U\) and \(W\) on \(\kappa\) and \(\kappa^+\) respectively such that \(W\mo U\).\qed
\end{prp}

Thus \(W\mo U\) but \(U\sE W\) simply because \(\delta_U < \delta_W\). (This is a consequence of \cref{SpaceLemma}.) By \cref{SuperMOKet}, if \(U\) and \(W\) are uniform ultrafilters on the same cardinal \(\lambda\) and both \(j_U\) and \(j_W\) are \(\lambda\)-supercompact, then \(U\mo W\) implies \(U\sE W\).

\begin{lma}\label{MOKetExtend}
Suppose \(\lambda\) is a cardinal, \(W\) is a countably complete ultrafilter on \(\lambda\), and \(Z\) is a countably complete ultrafilter such that \(W\mo Z\). Assume that for all \(\alpha < \lambda\), \(\mathscr B(\lambda,\alpha)\subseteq M_Z\) and \(M_Z\vDash \mathscr B(\lambda,\alpha) \leq 2^\lambda\). Then for any \(U\sE W\), \(U\mo Z\).
\begin{proof}
Since \(W\mo Z\), \(P(\lambda)\subseteq M_Z\) and in fact \(P(\lambda)^\lambda\subseteq M_Z\). Moreover \[M_Z\vDash \left|\bigcup_{\alpha < \lambda} \mathscr B(\lambda,\alpha)\right| \leq 2^\lambda = |P(\lambda)|\] Hence \((\bigcup_{\alpha < \lambda} \mathscr B(\lambda,\alpha))^\lambda\subseteq M_Z\), so \(\prod_{\alpha\in I}\mathscr B(\lambda,\alpha)\in M_Z\) for any set \(I\subseteq \lambda\).

Now suppose \(U\sE W\). Fix \(I\in W\) and \(\langle U_\alpha : \alpha\in I\rangle\in \prod_{\alpha\in I}\mathscr B(\lambda,\alpha)\) such that \(U = W\text{-}\lim_{\alpha \in I} U_\alpha\). Then the sequence \(\langle U_\alpha : \alpha\in I\rangle\in M_Z\), so \(U\in M_Z\), so \(U\mo Z\), as desired.
\end{proof}
\end{lma}

In fact, this lemma yields the somewhat stronger result that for any \(I\in W\) and sequence \(\langle U_\alpha : \alpha \in I\rangle\) of ultrafilters with \(\delta_{U_\alpha} < \lambda\), \(W\text{-}\lim_{\alpha\in I} U_\alpha \mo Z\).

\begin{cor}[UA]
Assume \(\lambda\) is a cardinal such that \(2^{<\lambda} = \lambda\). If \(W\) and \(Z\) are countably complete ultrafilters on \(\lambda\) such that \(W\mo Z\), then \(W\sE Z\).
\begin{proof}
Given the assumption that \(2^{<\lambda} = \lambda\) and the fact that \(P(\lambda)\subseteq M_Z\), it is not hard to show that \(\Un_\alpha\in M_Z\) and \(M_Z\vDash |\Un_\alpha| \leq 2^\lambda\) for all \(\alpha < \lambda\). Therefore we are in a position to apply \cref{MOKetExtend} to any ultrafilter \(U\sE W\). Assume towards a contradiction that \(W\not \sE Z\). By the linearity of the Ketonen order, \(Z\sE W\). Now \(Z\sE W\mo Z\), so by \cref{MOKetExtend}, \(Z\mo Z\). This contradicts the strictness of the Mitchell order (\cref{MOStrict}).
\end{proof}
\end{cor}

\begin{cor}[UA + GCH]\index{Ketonen order!vs. the generalized Mitchell order}The Ketonen order extends the generalized Mitchell order on countably complete uniform ultrafilters on infinite cardinals.\qed\end{cor}
\cref{MOExtensions} shows that the same conclusion can be deduced from UA alone. This will be achieved by proving from UA that if \(W\mo Z\), then \(Z\) is \(\lambda_W\)-supercompact. The result then follows from \cref{SuperMOKet}.
\section{Generalizations of normality}\label{GeneralizedNormalSection}
In this section, we develop the theory of normal fine ultrafilters, the natural combinatorial generalization of normal ultrafilters, and a central component of the classical theory of supercompact cardinals. The main result of the section (\cref{GCHLinear}) states roughly that UA + GCH implies that all these ultrafilters are linearly ordered by the Mitchell order.

\begin{defn}\index{\(\mathscr N_\lambda\) (normal ultrafilters on \(P_\text{bd}(\lambda)\))}
	For any infinite cardinal \(\lambda\), let \(\mathscr N_\lambda\) be set of normal fine ultrafilters on \(P_\textnormal{bd}(\lambda)\). Let \(\mathscr N = \bigcup_{\lambda }\mathscr N_\lambda\).
\end{defn}

We provide the definitions of normality and fineness in \cref{NFSection}.

\begin{thm}[UA]\label{GCHLinear}\index{Generalized Mitchell order!linearity!on normal fine ultrafilters}
	Suppose \(\lambda\) is a cardinal such that \(2^{<\lambda} = \lambda\). Then \(\mathscr N_\lambda\) is wellordered by the Mitchell order. Therefore assuming the Generalized Continuum Hypothesis, \(\mathscr N\) is linearly ordered by the Mitchell order.
\end{thm}

We asserted that UA + GCH would roughly imply that the Mitchell order is linear on the class of all normal fine ultrafilters, but our theorem only mentions the subclass \(\mathscr N\). In fact, the class of all normal fine ultrafilters is not literally linearly ordered by the Mitchell order for a number of reasons: one reason is that distinct normal fine ultrafilters can be isomorphic and hence Mitchell incomparable. \cref{NIso} below, however, shows that every normal fine ultrafilter is isomorphic to an element of \(\mathscr N\), so \cref{GCHLinear} essentially covers all the bases.

A key concept in the proof of \cref{GCHLinear}, introduced here for the first time, is that of an isonormal ultrafilter.
\begin{defn}\index{Isonormal ultrafilter}
Suppose \(\lambda\) is a cardinal. An ultrafilter \(U\) on \(\lambda\) is {\it isonormal} if \(U\) is weakly normal and \(j_U\) is a \(\lambda\)-supercompact embedding.
\end{defn}

We define weak normality in \cref{WeaklyNormalSection}. The concept dates back to Solovay and Ketonen \cite{Ketonen2}. The other main theorem of this section explains how isonormal ultrafilters get their name:
\begin{repthm}{IsoNormalThm}
Suppose \(U\) is a nonprincipal ultrafilter. Then \(U\) is isonormal if and only if \(U\) is the incompressible ultrafilter isomorphic to a normal fine ultrafilter. In particular, every normal fine ultrafilter is isomorphic to a unique isonormal ultrafilter.
\end{repthm}

The proof appears in \cref{SingularSolovaySection}. The forwards direction is quite easy, but the reverse implication requires quite a bit of work amounting to a generalization of the theorem of \cite{Solovay} known as {\it Solovay's Lemma} to singular cardinals. This generalization constitutes a fundamental and (apparently) new fact about supercompactness whose proof requires some basic notions from PCF theory.

The investigation of isonormal ultrafilters is related back to the linearity of the Mitchell order by the following proposition:
\begin{repthm}{IsoSound}
Suppose \(2^{<\lambda} = \lambda\). Then every isonormal ultrafilter \(U\) on \(\lambda\) is Dodd sound.
\end{repthm}

This is basically just a matter of defining weakly normal ultrafilters on singular cardinals.  We actually prove our main theorem (\cref{GCHLinear}) right now. But we will need to assume \cref{IsoSound} and \cref{IsoNormalThm}. We also need a lemma that shows \(\mathscr N\) is well-behaved under the Mitchell order assuming GCH:
\begin{lma}\label{NHered}
If \(2^{<\lambda} = \lambda\), then any \(\mathcal U \in \mathscr N_\lambda\) is hereditarily uniform and satisfies \(\lambda_\mathcal U = \lambda\).
\begin{proof}
Since \(P_\text{bd}(\lambda)\) is transitive, \(|\text{tc}(P_\text{bd}(\lambda))| = |P_\text{bd}(\lambda)| = 2^{<\lambda} = \lambda\). On the other hand, since \(j_\mathcal U\) is \(\lambda\)-supercompact, \cref{UFSuperBound} implies \(\lambda_\mathcal U\geq \lambda\). Thus \(|\text{tc}(P_\text{bd}(\lambda))| = \lambda_\mathcal U\), so \(\mathcal U\) is hereditarily uniform.
\end{proof}
\end{lma}

We finally prove \cref{GCHLinear} assuming \cref{IsoSound} and \cref{IsoNormalThm}.

\begin{proof}[Proof of \cref{GCHLinear}]
Suppose \(\mathcal U\) and \(\mathcal W\) are elements of \(\mathscr N_\lambda\). We show that either \(\mathcal U\mo \mathcal W\), \(\mathcal U = \mathcal W\), or \(\mathcal U\gmo\mathcal W\). Applying \cref{IsoNormalThm}, let \(U\) be the isonormal ultrafilter isomorphic to \(\mathcal U\) and let \(W\) be the isonormal ultrafilter isomorphic to \(\mathcal W\). Note that \(U\) and \(W\) are uniform ultrafilters on the cardinal \(\lambda_\mathcal U = \lambda_\mathcal W = \lambda\) (\cref{NHered}). We have \(2^{<\lambda} = \lambda\) by assumption, so \cref{IsoSound} yields that \(U\) and \(W\) are Dodd sound. By the linearity of the Mitchell order on Dodd sound ultrafilters (\cref{DoddMO}), we are in one of the following cases:

\begin{case} \label{EqLin}\(U = W\).\end{case}
\begin{proof}[Proof in \cref{EqLin}]
Since \(\mathcal U \cong U = W\cong \mathcal W\),  \cref{NormalIso} below implies \(\mathcal U=\mathcal W\).
\end{proof}
\begin{case} \label{moCase}\(U \mo W\).\end{case}
\begin{proof}[Proof in \cref{moCase}]
Since \(W\cong \mathcal W\), we have \(U\mo \mathcal W\). Since \(\mathcal U\) is hereditarily uniform (\cref{NHered}) and isomorphic to \(U\), the isomorphism invariance of the generalized Mitchell order on hereditarily uniform ultrafilters (\cref{HeredLemma}) implies \(\mathcal U\mo \mathcal W\).
\end{proof}
\begin{case}\label{gmoCase}\(U \gmo W\).\end{case}
\begin{proof}[Proof in \cref{gmoCase}]
Proceeding as in \cref{moCase}, we obtain \(\mathcal U\gmo \mathcal W\).
\end{proof}
This shows that either \(\mathcal U\mo \mathcal W\), \(\mathcal U = \mathcal W\), or \(\mathcal U\gmo\mathcal W\), as desired.

We finally sketch the proof that \(\mathscr N\) is linearly ordered by the Mitchell order assuming UA + GCH. It suffices to show the following: suppose \(\mathcal U\in \mathscr N_\gamma\), \(\mathcal W\in \mathscr N_\lambda\), and \(2^{<\lambda} = \lambda\). Then \(\mathcal U \mo \mathcal W\). Let \(U\) be the isonormal ultrafilter of \(\mathcal U\), so by the proof of \cref{NHered}, \(U\) is an ultrafilter on \(\gamma\). Since \(2^\gamma \leq 2^{<\lambda} = \lambda\), \(U\in H_{(2^\gamma)^+}\subseteq H_{\lambda^+}\subseteq M_{\mathcal W}\) Since \(P(P_{\text bd}(\gamma))\subseteq M_{\mathcal W}\), this easily implies that \(\mathcal U\mo \mathcal W\).
\end{proof}

\subsection{Normal fine ultrafilters}\label{NFSection}
In this section, we give the general definition of a normal fine ultrafilter, which is the natural combinatorial generalization of the notion of a normal ultrafilter on a cardinal. This begins with the generalized diagonal intersection operation:
\begin{defn}\index{Diagonal intersection}\index{\(\triangle_{\alpha < \delta}\) (diagonal intersection)}
Suppose \(X\) is a set and \(\langle A_x : x\in X\rangle\) is a sequence with \(A_x\subseteq P(X)\) for all \(x\in X\). The {\it diagonal intersection} of \(\langle A_x : x\in X\rangle\) is the set \[\triangle_{x\in X} A_x = \left\{\sigma\in P(X) : \sigma\in\textstyle \bigcap_{x\in \sigma} A_x\right\}\]
\end{defn}

\begin{defn}\index{Family over a set}
	If \(X\) is a set, a {\it family over \(X\)} is a family \(Y\) of subsets of \(X\) such that every element of \(X\) belongs to some element of \(Y\).
\end{defn}

Thus any set \(Y\) is a family on a unique set (namely \(X = \bigcup Y\)).

\begin{defn}\label{NormalFineDef}\index{Normal fine ultrafilter}\index{Fine ultrafilter}
Suppose \(Y\) is a family over \(X\). A filter \(\mathcal F\) on \(Y\) is:
\begin{itemize}
\item {\it fine} if for any \(x\in X\), \(\mathcal F\) concentrates on \(\{\sigma : x\in \sigma\}\).
\item {\it normal} if for any \(\{A_x : x\in X\}\subseteq \mathcal F\), \(\triangle_{x\in X} A_x\in \mathcal F\).
\end{itemize}
\end{defn}

\begin{rmk}
Let us make some remarks regarding this definition.
\begin{enumerate}[(1)]
	\item It makes sense to discuss normal fine filters on \(Y\) without mention of \(X\), since \(X = \bigcup Y\) is determined from \(Y\).
	\item The structure of the underlying set \(Y\) is usually not that important since a normal fine ultrafilter \(\mathcal U\) on \(Y\) can always be lifted to a normal fine ultrafilter on \(P(X)\) where \(X = \bigcup_{\sigma\in Y} \sigma\). Therefore it is tempting to restrict consideration to normal fine ultrafilters on \(P(X)\) for some \(X\). It is often important for technical reasons, however, that the underlying set \(Y\) be small; usually we want \(|Y| = |\bigcup Y|\).
	\item The structure of the set \(X\) is also usually irrelevant, but sometimes it is useful that \(X\) be transitive or that \(X\) be a cardinal. Suppose \(X\) and \(X'\) are sets and \(f : X\to X'\) is a surjection. If \(Y\) is a family over \(X\), then \(Y' = \{f[\sigma] : \sigma \in Y\}\) is a family over \(X'\) and \(g(\sigma) = f[\sigma]\) defines a surjection from \(Y\) to \(Y'\). If \(\mathcal U\) is an ultrafilter on \(Y\), then \(g_*(\mathcal U)\) is an isomorphic ultrafilter on \(Y'\) and moreover \(\mathcal U'\) is normal (fine) if and only if \(\mathcal U\) is normal (fine). (This is the ultrafilter theoretic analog of \cref{XSC}.)
	\item An ultrafilter on an ordinal is fine if and only if it is tail uniform. Thus a normal fine ultrafilter on \(\kappa\) is the same thing as a normal ultrafilter on \(\kappa\).
\end{enumerate}
\end{rmk}

The connection between normality and supercompactness is clear from the following lemma:
\begin{lma}\label{NormalFineChar}
Suppose \(Y\) is a family over \(X\) and \(\mathcal U\) is an ultrafilter on \(Y\).
\begin{enumerate}[(1)]
\item \(\mathcal U\) is fine if and only if \(j_\mathcal U[X]\subseteq \id_\mathcal U\).
\item \(\mathcal U\) is normal if and only if \(\id_\mathcal U\subseteq j_\mathcal U[X]\).\qed
\end{enumerate}
Thus \(\mathcal U\) is a normal fine ultrafilter on \(Y\) over \(X\) if and only if \(\id_\mathcal U = j_\mathcal U[X]\), or in other words, \(\id_\mathcal U\) witnesses that \(j_\mathcal U\) is \(X\)-supercompact.
\end{lma}
 \cref{NormalFineChar} yields the main source of normal fine ultrafilters.

\begin{lma}\label{DerivedNF}
Suppose \(j :V\to M\) is an \(X\)-supercompact elementary embedding and \(Y\subseteq P(X)\) is such that \(j[X]\in j(Y)\).
\begin{itemize}
	\item \(Y\) is a family over \(X\).
	\item The ultrafilter \(\mathcal U\) on \(Y\) derived from \(j\) using \(j[X]\) is a normal fine ultrafilter on \(Y\).
	\item Let \(k : M_\mathcal U\to M\) be the factor embedding. Then \(k(\alpha) = \alpha\) for all \(\alpha \leq |X|\).
\end{itemize}
\begin{proof}
Immediate from \cref{DerivedSC} and \cref{NormalFineChar}.
\end{proof}
\end{lma}

Another consequence of \cref{NormalFineChar} is the following fact, which does not seem to have a simple combinatorial proof:
\begin{lma}\label{NormalIso}
Suppose \(\mathcal U\) and \(\mathcal W\) are normal fine ultrafilters on \(Y\). If \(\mathcal U\cong \mathcal W\) then \(\mathcal U = \mathcal W\).
\begin{proof}
Let \(X = \bigcup Y\). Since \(\mathcal U\cong\mathcal W\), \(j_\mathcal U = j_\mathcal W\). By \cref{NormalFineChar}, \(\id_\mathcal U = j_\mathcal U[X] =  j_\mathcal W[X]= \id_\mathcal W\). Thus \(\mathcal U = \{A\subseteq Y : \id_\mathcal U \in j_\mathcal U(A)\} =  \{A\subseteq Y : \id_\mathcal W \in j_\mathcal W(A)\}  = \mathcal W\).
\end{proof}
\end{lma}

It also follows that any normal fine ultrafilter is countably complete. This is because the proof that an \(\omega\)-supercompact ultrapower embedding \(j :V\to M\) has the property that \(M^\omega\subseteq M\) does not really require that \(M\) is wellfounded. (The reader will lose nothing by simply appending countable completeness to the definition of normality, rather than proving it from the definition we have given.)

Recall the class \(\mathscr N\) defined in the previous section. We finish this section by proving that every normal fine ultrafilter is isomorphic to a unique element of \(\mathscr N\).
\begin{prp}\label{NIso}
Any nonprincipal normal fine ultrafilter \(\mathcal D\) is isomorphic to a unique ultrafilter \(\mathcal U\in \mathscr N\).
\end{prp}
For this we will use a basic lemma about supercompactness:
\begin{lma}\label{SmallCfCompact2}
Suppose \(j : V\to M\) is \(\lambda\)-supercompact and \(\sup j[\lambda] = j(\lambda)\). Then \(j\) is \(\lambda^{\iota}\)-supercompact where \(\iota = \textnormal{cf}(\lambda)\). In particular, \(j\) is \(\lambda^+\)-supercompact.
\begin{proof}
Let \(\kappa = \textsc{crt}(j)\). \cref{SmallCfCompact} states that \(j\) is \(\lambda^{<\kappa}\)-supercompact. It suffices to show that \(\iota < \kappa\): then since \(j\) is \(\lambda^{<\kappa}\)-supercompact, \(j\) is \(\lambda^{\iota}\)-supercompact, and so since \(\lambda^\iota>\lambda\), \(j\) is \(\lambda^+\)-supercompact.

We now show \(\iota < \kappa\). Since \(\sup j[\lambda] = j(\lambda)\) and \(j[\lambda]\in M\), \(\text{cf}^M(j(\lambda)) = \text{cf}(\lambda) = \iota\). On the other hand, by elementarity \(\text{cf}^M(j(\lambda)) = j(\text{cf}(\lambda)) = j(\iota)\). It follows that \(j(\iota) = \iota\). Since \(j\) is \(\iota\)-supercompact, the Kunen Inconsistency Theorem (\cref{KunenInconsistency}) implies \(\iota < \kappa\) where \(\kappa = \textsc{crt}(j)\).
\end{proof}
\end{lma}
Actually, we always have \(\lambda^+ = \lambda^{<\kappa}\) in the context of \cref{SmallCfCompact2}, and this is how SCH above a supercompact is proved.

\begin{proof}[Proof of \cref{NIso}]
Obviously, any normal fine ultrafilter is isomorphic to a normal fine ultrafilter on \(P(\lambda)\) for some cardinal \(\lambda\). Therefore assume \(\mathcal D\) is a normal ultrafilter on \(P(\lambda)\), and we will show that \(\mathcal D\) is isomorphic to a normal fine ultrafilter on \(P_\text{bd}(\lambda')\) for some cardinal \(\lambda'\).

If \(\mathcal D\) concentrates on \(P_\text{bd}(\lambda)\), we are done, since \(\mathcal D\) is then isomorphic to \(\mathcal D\mid P_\text{bd}(\lambda)\). So assume \(\mathcal D\) does not concentrate on \(P_\text{bd}(\lambda)\). By \L o\'s's Theorem, \(\id_\mathcal D = j_\mathcal D[\lambda]\) is unbounded in \(j_\mathcal D(\lambda)\). In other words, \(j_\mathcal D\) is continuous at \(\lambda\). Therefore by \cref{SmallCfCompact2}, \(j_\mathcal D\) is \(\lambda^+\)-supercompact. Note that \(j_\mathcal D[\lambda^+]\) is not cofinal in \(j_\mathcal D(\lambda^+)\): otherwise \(j_\mathcal D(\lambda^+) =\text{cf}^{M_\mathcal D}(j_\mathcal D(\lambda^+) ) = \lambda^+\), so \(\textsc{crt}(j_\mathcal D) > \lambda^+\) by \cref{KunenInconsistency0}, which implies that \(\mathcal D\) is principal. Therefore let \(\mathcal U\) be the normal fine ultrafilter on \(P_\text{bd}(\lambda^+)\) derived from \(j_\mathcal D\) using \(j_\mathcal D[\lambda^+]\). Then \(\mathcal U\) is isomorphic to \(\mathcal D\): by construction \(\mathcal U\RK \mathcal D\), and on the other hand, the map \(f : P_\text{bd}(\lambda^+)\to Y\) defined by \(f(\sigma) = \sigma\cap \lambda\) pushes \(\mathcal U\) forward to \(\mathcal D\) so \(\mathcal D\RK \mathcal U\).
\end{proof}
\subsection{Weakly normal ultrafilters}\label{WeaklyNormalSection}
Another combinatorial generalization of the notion of a normal ultrafilter, due to Solovay and Ketonen \cite{Ketonen2}, is the notion of a weakly normal ultrafilter.
\begin{defn}\index{Weakly normal ultrafilter}
A uniform ultrafilter \(U\) on a cardinal \(\lambda\) is {\it weakly normal} if for any set \(A\in U\), if \(f : A\to\lambda\) is regressive, then there is some \(B\subseteq A\) such that \(B\in U\) and \(f[B]\) has cardinality less than \(\lambda\).
\end{defn}
Solovay's definition of a weakly normal ultrafilter applied only to regular cardinals \(\lambda\), asserting that every regressive function on \(\lambda\) is {\it bounded} on a set of full measure. The generalization of the concept of weak normality to singular cardinals is due to Ketonen.
\begin{lma}
Suppose \(U\) is a uniform ultrafilter on a cardinal \(\lambda\). Then the following are equivalent:
\begin{enumerate}[(1)]
\item \(U\) is weakly normal.
\item Suppose \(\langle A_\alpha : \alpha < \lambda\rangle\) is a sequence of subsets of \(\lambda\) such that \(\bigcap_{\alpha\in \sigma} A_\alpha\in U\) for all nonempty \(\sigma\in P_\lambda(\lambda)\). Then \(\triangle_{\alpha <\lambda} A_\alpha\in U\).\qed
\end{enumerate}
\end{lma}
\begin{cor}\index{Weakly normal ultrafilter!on a regular cardinal}A uniform ultrafilter on a regular cardinal is weakly normal if and only if it is closed under decreasing diagonal intersections.\qed\end{cor}

Weakly normal ultrafilters on regular cardinals have a simple characterization in terms of their ultrapowers:
\begin{lma}\label{RegWeaklyNormal}
Suppose \(\lambda\) is a regular cardinal. An ultrafilter \(U\) on \(\lambda\) is weakly normal if and only if \(\id_U = \sup j_U[\lambda]\).
\begin{proof}
Suppose \(U\) is weakly normal. Since \(U\) is a tail uniform ultrafilter on \(\lambda\), \(\id_U > j_U(\alpha)\) for all \(\alpha < \lambda\). We will show that \(j_U[\lambda]\) is cofinal in \(\id_U\), which proves \(\id_U = \sup j_U[\lambda]\). Suppose \(\xi < \id_U\). Then \(\xi = [f]_U\) for some \(f : \lambda\to \lambda\) that is regressive on a set in \(U\). Since \(U\) is weakly normal, there is a set \(A\in U\) such that \(|f[A]| < \lambda\). Since \(\lambda\) is regular, \(f[A]\) is bounded below \(\lambda\). Fix \(\alpha < \lambda\) such that \(f(\xi) < \alpha\) for all \(\xi \in A\). Then \([f]_U < j_U(\alpha)\).

Conversely suppose \(\id_U = \sup j_U[\lambda]\). Since \(\id_U > j_U(\alpha)\) for all \(\alpha <\lambda\), \(\delta_U \geq \lambda\), and hence \(U\) is tail uniform. Since \(\lambda\) is regular, it follows that \(\lambda\) is uniform. Next, suppose \(A\in U\) and \(f :A\to \lambda\) is regressive. Then \([f]_U <\id_U\). Since \(j_U[\lambda]\) is cofinal in \(\id_U\), fix \(\alpha < \lambda\) with \([f]_U < j_U(\alpha)\). Then for a set \(B\in U\) with \(B\subseteq A\), \(f(\beta) < \alpha\) for all \(\beta\in B\). In particular, \(f\) takes fewer than \(\lambda\) values on \(B\).
\end{proof}
\end{lma}

\cref{RegWeaklyNormal} yields the main source of weakly normal ultrafilters on regular cardinals:

\begin{cor}\label{DerivedWN}
Suppose \(j : V\to M\) is an elementary embedding and \(\lambda\) is a regular cardinal such that \(\sup j[\lambda] < j(\lambda)\). Then the ultrafilter on \(\lambda\) derived from \(j\) using \(\sup j[\lambda]\) weakly normal.
\end{cor}

To help motivate the concept of weak normality on singular cardinals, let us explain its relationship to an isomorphism invariant notion:
\begin{defn}\index{Minimal ultrafilter}
Suppose \(\lambda\) is an infinite cardinal. An ultrafilter \(U\) is {\it \(\lambda\)-minimal} if \(\lambda_U = \lambda\) and for any \(W\sRK U\), \(\lambda_W < \lambda\).
\end{defn}

If \(2^\lambda = \lambda^+\), there is a \(\lambda\)-minimal (countably incomplete) ultrafilter on \(\lambda\), according to a result of Comfort-Negrepontis \cite{Negrepontis}. On the other hand, the existence of a weakly normal ultrafilter (with no completeness assumptions) implies the existence of an inner model with a measurable cardinal \cite{Donder}. Weakly normal ultrafilters, however, are the revised Rudin-Keisler analog (\cref{rRKDef}) of \(\lambda\)-minimal ones:

\begin{lma}\label{WeaklyNormalrRK}
An ultrafilter \(U\) on a cardinal \(\lambda\) is weakly normal if and only if \(\lambda_U = \lambda\) and for all \(W\rRK U\), \(\lambda_W < \lambda\).\qed
\end{lma}

\cref{WeaklyNormalrRK} yields a generalization of Scott's theorem that every countably complete ultrafilter has a derived normal ultrafilter:
\begin{cor}
If \(Z\) is a countably complete uniform ultrafilter on \(\lambda\), there is a weakly normal ultrafilter \(U\) on \(\lambda\) such that \(U\RK Z\).
\begin{proof}
Since \(\rRK\) is wellfounded on countably complete ultrafilters, there is a countably complete ultrafilter \(U\) that is \(\rRK\)-minimal with the property that \(\lambda_U = \lambda\) and \(U\leq_{\textnormal{RK}} Z\). Then \(U\) satisfies the conditions of \cref{WeaklyNormalrRK}: if \(W\rRK U\), then \(W \leq_{\textnormal{RK}} Z\), so by the \(\rRK\)-minimality of \(U\), it must be the case that \(\lambda_W < \lambda\)
\end{proof}
\end{cor}

The following theorem shows that every countably complete \(\lambda\)-minimal ultrafilter is isomorphic to a weakly normal ultrafilter.
\begin{prp}\label{MinimalIncompressible}
A countably complete uniform ultrafilter \(U\) on a cardinal \(\lambda\) is weakly normal if and only if it is \(\lambda\)-minimal and incompressible.
\begin{proof}
Suppose \(U\) is weakly normal. To see \(U\) is incompressible, note that any function that is regressive on a set in \(U\) takes less than \(\lambda\)-many values on a set in \(U\), and hence is not one-to-one. To see \(U\) is \(\lambda\)-minimal, suppose \(W\sRK U\) and we will show that \(\lambda_W < \lambda\). Since \(W\RK U\), \(W\) is countably complete, and hence \(W\) is isomorphic to an incompressible ultrafilter. We can therefore assume without loss of generality that \(W\) is incompressible. Then by the key lemma about the strict Rudin-Keisler order on incompressible ultrafilters (\cref{IncomRK}) the fact that \(W\sRK U\) implies \(W \rRK U\). Now by \cref{WeaklyNormalrRK}, \(\lambda_W < \lambda\).

Conversely suppose \(U\) is \(\lambda\)-minimal and incompressible. Suppose \(W\rRK U\), and we will show \(\lambda_W < \lambda\). We can then conclude that \(U\) is weakly normal using \cref{WeaklyNormalrRK}. Since \(U\) is incompressible, \(W\rRK U\) implies \(W\sRK U\) (\cref{IncomRK0}, essentially the definition of incompressibility). Therefore by the definition of \(\lambda\)-minimality, \(\lambda_W < \lambda\), as desired.
\end{proof}
\end{prp}

It is not clear to us whether \cref{MinimalIncompressible} can be proved without the assumption of countable completeness, though of course countable completeness is not required if \(\lambda\) is regular.

The following characterization of weak normality is the one that is most relevant to our investigations of supercompactness.

\begin{prp}\label{WeaklyNormalGenerator}
Suppose \(\lambda\) is an infinite cardinal. A countably complete ultrafilter \(U\) on \(\lambda\) is weakly normal if and only if \(\id_U\) is the unique generator of \(j_U\) that lies above \(j(\delta)\) for all \(\delta < \lambda\).
\end{prp}

For the proof, we will need an obvious lemma:
\begin{lma}\label{UniformGenerator}
Suppose \(\lambda\) is an infinite cardinal.  An ultrafilter \(U\) on \(\lambda\) is uniform if and only if \(\id_U \notin H^{M_U}(j_U[V]\cup j_U(\delta))\) for any \(\delta < \lambda\). \qed
\end{lma}

\begin{proof}[Proof of \cref{WeaklyNormalGenerator}]
We begin with the forwards direction. Suppose \(U\) is weakly normal.

We first show that for any ordinal \(\xi\) such that \(\xi < \id_U\), \(\xi\in H^{M_U}(j_U[V]\cup j_U(\delta))\) for some \(\delta < \lambda\). Assume not, towards a contradiction. Let \(W\) be the tail uniform ultrafilter derived from \(j_U\) using \(\xi\). Then \(W\rRK U\), as witnessed by the factor embedding \(k :M_W\to M_U\). By \cref{WeaklyNormalrRK}, it follows that \(W\) is not a uniform ultrafilter on \(\lambda\), and so by \cref{UniformGenerator}, there is some \(\delta < \lambda\) such that \(\xi\in H^{M_W}(j_W[V]\cup j_W(\delta))\). It follows that \(\xi\in H^{M_U}(j_U[V]\cup j_U(\delta))\).

Next we show that \(\id_U\) is a generator of \(j_U\). Since \(U\) is uniform, \cref{UniformGenerator} implies \(\id_U\notin H^{M_U}(j_U[V]\cup j_U(\delta))\) for any \(\delta < \lambda\). But by the previous paragraph, for all \(\xi < \id_U\), \(\xi \in H^{M_U}(j_U[V]\cup j_U(\delta))\) for some \(\delta < \lambda\). Thus \(\id_U\notin H^{M_U}(j_U[V]\cup\vec \xi)\) for any \(\vec \xi \in [\id_U]^{<\omega}\). In other words, \(\id_U\) is a generator of \(j_U\). By the previous paragraph, \(\id_U\) is clearly the unique generator above \(j_U(\delta)\) for all \(\delta < \lambda\).

We now turn to the converse. Assume \(\id_U\) is the unique generator of \(j_U\) that lies above \(j(\delta)\) for all \(\delta < \lambda\). We will show \(U\) is weakly normal by verifying the conditions of \cref{MinimalIncompressible}. Since \(\id_U\) is a generator, \(U\) is incompressible. Since \(M_U\) is wellfounded, there is a least ordinal that does not belong to \(H^{M_U}(j_U[V]\cup j_U(\delta))\) for any \(\delta\), and clearly this ordinal is a generator of \(j_U\) that lies above \(j_U(\delta)\) for all \(\delta < \lambda\). Thus it must equal \(\id_U\). In other words, for any \(\xi < \id_U\), \(\xi\in H^{M_U}(j_U[V]\cup j_U(\delta))\) for some \(\delta < \lambda\).

Fix an ultrafilter \(W\) on \(\lambda\) such that \(W\rRK U\). We will show \(\lambda_W < \lambda\), verifying the second condition of \cref{MinimalIncompressible}. Let \(k : M_W\to M_U\) be an elementary embedding with \(k\circ j_W = j_U\) and \(k(\id_W) < \id_U\). Then by the previous paragraph, \(k(\id_W)\in H^{M_U}(j_U[V]\cup j_U(\delta))\) for some \(\delta < \lambda\). It follows that \(\id_W\in H^{M_W}(j_W[V]\cup j_W(\delta))\) (by the proof of \cref{WidthLemma}). By \cref{UniformGenerator}, this implies \(W\) is not uniform on \(\lambda\), or in other words, \(\lambda_W < \lambda\).
\end{proof}

Using \cref{WeaklyNormalGenerator}, we can prove the Dodd soundness of isonormal ultrafilters on \(\lambda = 2^{<\lambda}\).\index{Isonormal ultrafilter!Dodd soundness}
\begin{thm}\label{IsoSound}
	Suppose \(2^{<\lambda} = \lambda\). Then every isonormal ultrafilter \(U\) on \(\lambda\) is Dodd sound.
\end{thm}
\begin{proof}
Let \(j : V\to M\) be the ultrapower of the universe by \(U\). Since \(j\) is \(\lambda\)-supercompact, \(j\) is \(2^{<\lambda}\)-supercompact. By \cref{SpaceIrrel}, \(j\) is \(\lambda_*\)-sound where \(\lambda_* = \sup j[\lambda]\).

We now show that \(j\) is \(\xi\)-sound where \(\xi\) is the least generator of \(j\) such that \(\xi\geq \lambda_*\). Since \(\lambda_*\) is closed under pairing, the \(\lambda_*\)-soundness of \(j\) implies that the extender \[E = E_j\restriction \lambda_* = \{(p,X) : p\in [\lambda_*]^{<\omega}\text{, }X\subseteq [\lambda]^{<\omega}\text{, and }p\in j(X)\}\] belongs to \(M_U\). Let \(j_E : V\to M_E\) be the associated extender embedding and let \(k : M_E\to M\) be the factor embedding. Then \[\textsc{crt}(k) = \min \{\alpha :\alpha\notin H^{M}(j[V]\cup \lambda_*)\} = \xi\] by the definition of a generator. Therefore \(j_E^\xi = j^\xi\). Moreover since \(M\) is closed under \(\lambda\)-sequences by \cref{UltrapowerSC}, \(j_E^{M} = j_E\restriction M\). Therefore \(j^\xi = j_E^\xi = (j_E^{M})^\xi\in M\), so \(j\) is \(\xi\)-sound.

By \cref{WeaklyNormalGenerator}, \(\xi = \id_U\). Therefore \(j\) is \(\id_U\)-sound, which implies that \(U\) is Dodd sound.
\end{proof}

We should point out that the assumption \(\lambda = 2^{<\lambda}\) is {\it necessary}:
\begin{lma}
Suppose \(\lambda\) is a cardinal that carries a Dodd sound ultrafilter \(U\). Then \(2^{<\lambda} = \lambda\).
\begin{proof}
Since \(U\) is Dodd sound, \(j_U\) is \(\id_U\)-sound. In particular, \(j_U\) is \(\sup j_U[\lambda]\)-sound since \(\sup j_U[\lambda] \leq \id_U\). Therefore by \cref{SpaceIrrel}, \(j_U\) is \(2^{<\lambda}\)-supercompact. By \cref{UFSuperBound}, \(j_U\) is not \(\lambda^+\)-supercompact. It follows that \(2^{<\lambda} < \lambda^+\), or in other words \(2^{<\lambda} = \lambda\).
\end{proof}
\end{lma}
\subsection{Solovay's Lemma}
A special case of our main theorem, \cref{IsoNormalThm}, was known long before our work.
\begin{thm}[Solovay's Lemma]\label{SolovayLemma}\index{Solovay's Lemma}
Suppose \(\lambda\) is a regular cardinal. Then there is a set \(B\subseteq P(\lambda)\) such that the following hold:
\begin{itemize}
\item For any family \(Y\) over \(\lambda\), any normal fine ultrafilter \(\mathcal U\) on \(Y\) concentrates on \(B\).
\item If \(\sigma\) and \(\tau\) are elements of \(B\) with the same supremum, then \(\sigma = \tau\).
\end{itemize}
\end{thm}
Before proving Solovay's Lemma, let us explain its relevance to isonormal ultrafilters. Essentially, Solovay's Lemma yields the ``regular case" of the key isomorphism theorem for isonormal ultrafilters (\cref{IsoNormalThm}):
\begin{cor}\label{SolovayCor}
Suppose \(\lambda\) is a regular cardinal, \(Y\) is a family over \(\lambda\), and \(\mathcal U\) is a nonprincipal normal fine ultrafilter on \(Y\). Then \(\mathcal U\) is isomorphic to the ultrafilter \[U = \{A\subseteq \lambda : \{\sigma\in Y : \sup \sigma\in A\}\in \mathcal U\}\]
Moreover, \(U\) is an isonormal ultrafilter.
\begin{proof}
To see \(\mathcal U\cong U\), let \(f : P(\lambda)\to \lambda+1\) be the function \(f(\sigma) = \sup \sigma\). Then \(f_*(\mathcal U) = U\) and by \cref{SolovayLemma}, \(f\) is one-to-one on a set in \(\mathcal U\).

To see \(U\) is isonormal, we must verify that \(U\) is weakly normal and \(j_U\) is \(\lambda\)-supercompact. The latter is trivial: \(j_\mathcal U\) is \(\lambda\)-supercompact by \cref{NormalFineChar}, and \(j_\mathcal U = j_U\) since \(\mathcal U\) and \(U\) are isomorphic. As for weak normality, by \cref{PushDerived}, \(U = f_*(\mathcal U)\) is the ultrafilter on \(\lambda\) derived from \(j_\mathcal U\) using \([f]_{\mathcal U}\) so \(U\) is weakly normal by \cref{DerivedWN}.
\end{proof}
\end{cor}

The proof of Solovay's lemma uses the observation that if \(j :V\to M\) is an elementary embedding, \(j[\lambda]\) is definable from the action of \(j\) on a stationary partition:\footnote{Solovay's published proof \cite{Solovay} uses the combinatorics of \(\omega\)-Jonsson algebras instead of stationary sets. Woodin rediscovered the proof using stationary sets, which was already known to Solovay.}

\begin{lma}\label{StationaryPartition}
Suppose \(\lambda\) is a cardinal, \(j : V\to M\) is an elementary embedding, and \(\mathcal P \subseteq P(\lambda)\) is a partition of \(S^\lambda_\omega = \{\alpha < \lambda:\textnormal{cf}(\alpha) = \omega\}\) into stationary sets. Then \[j[\mathcal P] = \{T \in j(\mathcal P) : T\text{ is stationary in }\sup j[\lambda]\}\]
\end{lma}
It is worth noting that \cref{StationaryPartition} is perfectly general; we really do allow \(j\) to be an arbitrary elementary embedding of \(V\).
\begin{proof}
Let \(\lambda_* = \sup j[\lambda]\).
\begin{clm}\(j[\mathcal P]\subseteq \{T \in j(\mathcal P) : T\text{ is stationary in }\lambda_*\}\).\end{clm}
\begin{proof}Fix \(S\in \mathcal P\). We will show that \(j(S)\) intersects every closed cofinal subset of \(\lambda_*\). Suppose \(C\subseteq \lambda_*\) is closed cofinal in \(\lambda_*\). Then \(j^{-1}[C]\) is \(\omega\)-closed cofinal in \(\lambda\). Since \(S\) is a stationary subset of \(S^\lambda_\omega\), \(S\cap j^{-1}[C]\neq\emptyset\). But \(j(S)\cap C =j(S)\cap C \supseteq j[S\cap j^{-1}[C]]\neq \emptyset\). So \(j(S)\cap C\neq \emptyset\), as desired.
\end{proof}
\begin{clm} \(\{T \in j(\mathcal P) : T\text{ is stationary in }\lambda_*\}\subseteq j[\mathcal P]\). \end{clm}
\begin{proof}
Fix \(T\in j(\mathcal P)\) such that \(T\) is stationary in \(\lambda_*\). We will show that there is some \(S\in \mathcal P\) such that \(j(S) = T\). Since \(j[\lambda]\) is \(\omega\)-closed cofinal in \(\lambda_*\), \(T\cap j[\lambda]\neq \emptyset\). Take \(\xi < \lambda\) such that \(j(\xi)\in T\). Since \(j(\xi)\in T\subseteq j(S^\lambda_\omega)\), \(\xi\in S^\lambda_\omega\). Therefore \(\xi\in S\) for some \(S\in \mathcal P\), since \(\bigcup \mathcal P = S^\lambda_\omega\). Now \(j(\xi)\in j(S)\cap T\). Therefore \(j(S)\) and \(T\) are not disjoint, so since \(j(\mathcal P)\) is a partition, \(j(S) = T\), as desired.
\end{proof}
Combining the two claims yields the lemma.
\end{proof}

\cref{StationaryPartition} leads to a characterization of supercompactness that looks surprisingly weak:

\begin{cor}\label{SCStatCor}\index{Supercompactness!vs. stationary correctness}
Suppose \(j : V\to M\) is an elementary embedding and \(\lambda\) is a regular cardinal. The following are equivalent:
\begin{enumerate}[(1)]
\item \(j\) is \(\lambda\)-supercompact.
\item \(M\) is correct about stationary subsets of \(\lambda_* = \sup j[\lambda]\).
\end{enumerate}
\begin{proof}

{\it (1) implies (2):} Assume \(j\) is \(\lambda\)-supercompact. Suppose \(M\) satisfies that \(S\) is stationary in \(\lambda_*\), and we will show that \(S\) is truly stationary in \(\lambda_*\). Fix a closed cofinal set \(C\subseteq \lambda_*\). We will show \(S\cap C\neq \emptyset\). Note that \(C\cap j[\lambda]\in M\) by \cref{SupercompactClosure} (3). Let \(E\) be the closure of \(C\cap j[\lambda]\) in \(\lambda_*\). Then \(E\in M\), \(E\subseteq C\), and \(E\) is closed cofinal in \(\lambda_*\). Since \(E\in M\) and \(S\) is stationary from the perspective of \(M\), \(S\cap E\neq \emptyset\). In particular, \(S\cap C\neq \emptyset\).

{\it (2) implies (1):} Since \(\lambda\) is regular, there is a partition \(\mathcal P\) of \(S^\lambda_\omega\) into stationary sets such that \(|\mathcal P| = \lambda\). By \cref{StationaryPartition}, \(j[\mathcal P] = \{T\in j(\mathcal P) : T\text{ is stationary in }\lambda_*\}\), which is definable over \(M\) since \(M\) is correct about stationary subsets of \(\lambda_*\). Thus \(j\) is \(\mathcal P\)-supercompact, so by \cref{XSC}, \(j\) is \(\lambda\)-supercompact, as desired.
\end{proof}
\end{cor}
Of course the implication from (1) to (2) is not very surprising, but it allows us to restate \cref{StationaryPartition} in a useful way:

\begin{cor}\label{MStationaryPartition}
	Suppose \(\lambda\) is a regular cardinal, \(j : V\to M\) is a \(\lambda\)-supercompact elementary embedding, and \(\langle S_\alpha : \alpha < \lambda\rangle\) is a partition of \(S^\lambda_\omega\) into stationary sets. Let \(\langle T_\beta : \beta < j(\lambda)\rangle = j(\langle S_\alpha : \alpha < \lambda\rangle)\). Then \(j[\lambda] = \{\beta < j(\lambda) : M\vDash  T_\beta\text{ is stationary in }\lambda_*\}\).\qed
\end{cor}

We now prove Solovay's Lemma.
\begin{proof}[Proof of \cref{SolovayLemma}]
Let \(\langle S_\alpha : \alpha < \lambda\rangle\) be a partition of \(S^\lambda_\omega = \{\alpha < \lambda:\text{cf}(\alpha) = \omega\}\) into stationary sets. Let \[B = \{\sigma\subseteq \lambda : \sigma = \{\beta < \lambda : S_\beta\text{ is stationary in sup}(\sigma)\}\}\]
By construction, any two elements of \(B\) with the same supremum are equal.

To finish, suppose \(Y\) is a family over \(\lambda\) and \(\mathcal U\) is a normal fine on \(Y\). We must show that \(\mathcal U\) concentrates on \(B\), or equivalently, that \(\id_\mathcal U\in j_\mathcal U(B)\). Since \(\id_\mathcal U = j_\mathcal U[\lambda]\) (\cref{NormalFineChar}), this amounts to showing \[j_\mathcal U[\lambda] = \{\beta < j_\mathcal U(\lambda) : M_\mathcal U\vDash j_\mathcal U(S)_\beta\text{ is stationary in sup} j_\mathcal U[\lambda] \}\] which is of course a consequence of \cref{MStationaryPartition}.
\end{proof}

Another corollary of Solovay's Lemma is Woodin's proof of the Kunen Inconsistency Theorem:
\begin{thm}\label{KunenInconsistency0}\index{Kunen Inconsistency Theorem!Woodin's proof}
Suppose \(j : V\to M\) is an elementary embedding, \(\iota\) is a regular cardinal, \(j\) is \(\iota\)-supercompact, and \(j(\iota) = \sup j[\iota]\). Then \(j\restriction \iota + 1\) is the identity.
\end{thm}
\begin{proof}
Let \(\langle S_\alpha : \alpha <\iota\rangle\) be a partition of \(S^\iota_\omega\) into stationary sets. By \cref{MStationaryPartition}, and using the fact that \(j(\iota) = \sup j[\iota]\),
\[j[\iota] = \{\beta < j(\iota) : M\vDash j(S)_\beta\text{ is stationary in }j(\iota)\} = j(\iota)\]
But this means \(j\restriction \iota + 1\) is the identity, as desired.
\end{proof}
Applying \cref{KunenInconsistency0} at \(\iota  = \lambda^+\) where \(\lambda\) is the first fixed point of \(j\) above \(\textsc{crt}(j)\) yields a proof of the Kunen Inconsistency  (\cref{KunenInconsistency}).
\subsection{Supercompactness and singular cardinals}\label{SingularSolovaySection}
In this section, we finish the proof of \cref{IsoNormalThm}. We do this by proving an analog of Solovay's Lemma at singular cardinals.\index{Solovay's Lemma!at singular cardinals} One basic issue, however, is that \cref{SolovayLemma} itself cannot generalize: in fact, if \(\lambda\) is a singular cardinal, \(Y\) is a family over \(\lambda\), and \(\mathcal U\) is a normal fine ultrafilter on \(Y\), then the supremum function is {\it not} one-to-one on any set in \(\mathcal U\).
\begin{prp}\label{NonSolovay}
Suppose \(\lambda\) is a cardinal of cofinality \(\iota\), \(Y\) is a family over \(\lambda\), and \(\mathcal U\) is a normal fine ultrafilter on \(Y\). Define \(f : Y\to \lambda + 1\) by
\[f(\sigma) = \sup \sigma\]
Define \(g : Y\to \iota + 1\) by
\[g(\sigma) = \sup (\sigma\cap \iota)\]
Then \(f_*(\mathcal U) \cong g_*(\mathcal U)\).
\end{prp}
It is a bit easier to prove the following equivalent statement first (which in any case turns out to be more useful):
\begin{prp}\label{iota*lemma}
Suppose \(j : V\to M\) is an elementary embedding and \(\lambda\) is a cardinal of cofinality \(\iota\). Then \(\sup j[\lambda]\) and \(\sup j[\iota]\) are interdefinable in \(M\) from parameters in \(j[V]\).
\begin{proof}
Let \(h : \iota\to \lambda\) be an increasing cofinal function. Then \[\sup j[\lambda] = \sup j[h[\iota]] = \sup j(h)\circ j[\iota] = \sup j(h)[\sup j[\iota]]\] Therefore \(\sup j[\lambda]\) is definable in \(M\) from \(j(h)\) and \(\sup j[\iota]\). Moreover, \[\sup j[\iota] = \sup j(h)^{-1}[\sup j[\lambda]]\] so \(\sup j[\iota]\) is definable in \(M\) from \(j(h)\) and \(\sup j[\lambda]\).
\end{proof}
\end{prp}
\begin{proof}[Proof of \cref{NonSolovay}]
Let \(j :V\to M\) be the ultrapower of the universe by \(\mathcal U\). Then (by \cref{PushDerived})
\(f_*(\mathcal U)\) is the ultrafilter on \(\lambda+ 1\) derived from \(j\) using \([f]_\mathcal U = \sup j[\lambda]\) and \(g_*(\mathcal U)\) is the ultrafilter on \(\iota + 1\) derived from \(j\) using \([g]_\mathcal U = \sup j[\iota]\).
By \cref{iota*lemma}, \[H^M(j[V]\cup \{\sup j[\lambda]\}) = H^M(j[V]\cup \{\sup j[\iota]\})\] But \[(M_{f_*(\mathcal U)}, j_{f_*(\mathcal U)})\cong (H^M(j[V]\cup \{\sup j[\lambda]\}),j)\cong (M_{g_*(\mathcal U)}, j_{g_*(\mathcal U)})\]
It follows that \(f_*(\mathcal U) \cong g_*(\mathcal U)\).
\end{proof}

\begin{cor}
Suppose \(\lambda\) is a cardinal of cofinality \(\iota\), \(Y\) is a family over \(\lambda\), and \(\mathcal U\) is a normal fine ultrafilter on \(Y\). Then there is a set \(B\in \mathcal U\) on which the supremum function takes at most \(\iota\)-many values.
\begin{proof}
Let \(f : Y\to \lambda\) be the supremum function. Since \(f_*(\mathcal U)\) is isomorphic to an ultrafilter on \(\iota + 1\), \(f\) takes at most \(\iota\)-many values on a set in \(\mathcal U\).
\end{proof}
\end{cor}

What we show instead is that an analog of \cref{StationaryPartition} holds:
\begin{thm}\label{GeneralizedSolovay}
Suppose \(\lambda\) is a cardinal and \(j : V\to M\) is a \(\lambda\)-supercompact elementary embedding. Let \(\theta\) be the least generator of \(j\) with \(\theta\geq\sup j[\lambda]\). Then \[j[\lambda] \in H^M(j[V]\cup \{\theta\})\] Moreover if \(\sup j[\lambda] < j(\lambda)\), then \(\theta < j(\lambda)\).
\end{thm}
As a corollary, we prove the second of the main theorems of this section:
\begin{thm}\label{IsoNormalThm}
	Suppose \(U\) is a nonprincipal ultrafilter. Then \(U\) is isonormal if and only if \(U\) is the incompressible ultrafilter isomorphic to a normal fine ultrafilter. In particular, every normal fine ultrafilter is isomorphic to a unique isonormal ultrafilter.
\end{thm}
\begin{proof}
We begin with the forward direction, which turns out to follow from \cref{MinimalIncompressible}. Suppose \(U\) is an isonormal ultrafilter on a cardinal \(\lambda\). We will show that \(U\) is incompressible and isomorphic to a normal fine ultrafilter on \(P_\text{bd}(\lambda)\). Since \(U\) is weakly normal, \cref{MinimalIncompressible} implies \(U\) is incompressible.

Since \(U\) is uniform on \(\lambda\), \(\sup j_U[\lambda] < j_U(\lambda)\) and thus \(j_U\restriction \lambda\in j_U(P_\text{bd}(\lambda))\). Let \(\mathcal U\) be the ultrafilter on \(P_\text{bd}(\lambda)\) derived from \(j_U\) using \(j_U\restriction \lambda\). Then \(\mathcal U\RK U\) and \(\mathcal U\) is a normal fine ultrafilter on \(P_\text{bd}(\lambda)\) by \cref{NormalFineChar}. It follows that \(j_\mathcal U\) is \(\lambda\)-supercompact, and therefore \(\lambda_\mathcal U\geq \lambda\) by \cref{UFSuperBound}. Since \(U\) is weakly normal, \cref{MinimalIncompressible} implies \(U\) is \(\lambda\)-minimal and therefore \(\mathcal U\not \sRK U\). Since \(\mathcal U\RK U\) and \(\mathcal U\not \sRK U\), we must have \(\mathcal U\cong U\) (by definition).

Conversely, suppose \(U\) is incompressible and isomorphic to a normal fine ultrafilter, and we will show that \(U\) is isonormal. Since every normal fine ultrafilter is isomorphic to an element of \(\mathscr N\) (\cref{NIso}), for some cardinal \(\lambda\), \(U\) is isomorphic to a normal fine ultrafilter \(\mathcal U\) on \(P_\text{bd}(\lambda)\). In particular \(j_U = j_\mathcal U\) is \(\lambda\)-supercompact. To show that \(U\) is isonormal, it therefore suffices to show that \(U\) is a weakly normal ultrafilter on \(\lambda\).

Let \(j: V\to M\) be the ultrapower of the universe by \(U\). Let \(\theta\) be the least generator of \(j\) with \(\theta\geq \sup j[\lambda]\). Since \(P_\text{bd}(\lambda)\in \mathcal U\), \(\sup j[\lambda] < j(\lambda)\), and so by \cref{GeneralizedSolovay}, \(\theta < j(\lambda)\). Since \(\theta\) is a generator of \(j = j_U\), \(\theta \leq \id_U\). In fact, we claim \(\id_U = \theta\). On the other hand, by \cref{GeneralizedSolovay}, \[M = H^M(j[V]\cup \{j[\lambda]\}) = H^M(j[V]\cup \{\theta\})\] The ultrapower theoretic characterization of incompressibility (\cref{IncomPower}) implies that \(\id_U\) is the least ordinal \(\alpha\) such that \(M= H^M(j[V]\cup \{\alpha\})\). Thus \(\id_U\leq\theta\). Hence \(\id_U = \theta\), as desired.

Since \(U\) is tail uniform (by the definition of incompressibility) and \(\id_U < j_U(\lambda)\), \(U\) is an ultrafilter on \(\lambda\). Since \(\id_U\) is the least generator of \(j\) above \(\sup j[\lambda]\), the characterization of weakly normal ultrafilters in terms of generators (\cref{WeaklyNormalGenerator}) implies that \(U\) is a weakly normal ultrafilter on \(\lambda\).
\end{proof}

We conclude this chapter by proving \cref{GeneralizedSolovay}. The proof relies on some basic notions from PCF theory.
\begin{defn}
Suppose \(\iota\) is an ordinal. We denote by \(J^\iota_\text{bd}\)\index{\(J_{\textnormal{bd}}\) (bounded ideal)} the ideal of bounded subsets of \(\iota\), omitting the superscript \(\iota\) when it is clear from context. If \(f\) and \(g\) are functions from \(\iota\) to \(\text{Ord}\),
\begin{itemize}
	\item \(f <_\text{bd} g\) if \(\{\alpha < \iota : f(\alpha) \geq g(\alpha)\}\in J_\text{bd}\).\index{\(<_\text{bd}\) (domination mod bounded)}
	\item \(f =_\text{bd} g\) if \(\{\alpha < \iota : f(\alpha) \neq g(\alpha)\}\in J_\text{bd}\).
\end{itemize}
\end{defn}
\begin{defn}
Suppose \(C\) is a set of functions from \(\iota\) to \(\text{Ord}\). A function \(s: \iota\to \text{Ord}\) is an {\it exact upper bound}\index{Exact upper bound} of \(C\) if the following hold:
\begin{itemize}
\item For all \(f\in C\), \(f <_\text{bd} s\).
\item For all \(g <_\text{bd} s\), for some \(f\in C\), \(g <_\text{bd} f\).
\end{itemize}
\end{defn}

The following trivial fact plays a key role in the proof of \cref{GeneralizedSolovay}:

\begin{lma}\label{EUBUnique}
Suppose \(C\) is a set of functions from \(\iota\) to \(\textnormal{Ord}\) and \(s\) and \(t\) are exact upper bounds of \(C\). Then \(s =_\textnormal{bd} t\).
\begin{proof}
Suppose \(s\) and \(t\) are exact upper bounds of \(C\). Suppose towards a contradiction that \(s\neq_\text{bd} t\). Without loss of generality, we can assume that there is an unbounded set \(A\subseteq \iota\) such that \(s(\alpha) < t(\alpha)\) for all \(\alpha\in A\). Define \(g : \iota\to \text{Ord}\) by setting
\[g(\alpha) = \begin{cases}
s(\alpha)&\text{if }\alpha\in A\\
0&\text{otherwise}\end{cases}\]
Then \(g < t\), so since \(t\) is an exact upper bound of \(C\), there is some \(f\in C\) such that \(g <_\text{bd} f\). Since \(s\) is an upper bound of \(C\), \(f <_\text{bd} s\). Therefore \(g <_\text{bd} s\). This contradicts that \(A = \{\alpha <\iota : g(\alpha) = s(\alpha)\}\) is unbounded in \(\iota\).
\end{proof}
\end{lma}
\begin{defn}\index{Scale}
If \(s : \iota\to \text{Ord}\) is a function and \(\delta\) is an ordinal, a {\it scale of length \(\delta\) in \(\prod_{\alpha < \iota} s(\alpha)\)} is a \(<_\text{bd}\)-increasing cofinal sequence \(\langle f_\alpha : \alpha < \delta\rangle\subseteq \prod_{\alpha < \iota} s(\alpha)\).
\end{defn}
Shelah's Representation Theorem \cite{AbrahamMagidor} states that if \(\lambda\) is a singular cardinal of cofinality \(\iota\), then there is a cofinal continuous sequence \(u : \iota\to \lambda\) such that \(\prod_{\alpha < \iota} u(\alpha)^+\) has a scale of length \(\lambda^+\). This is a deep theorem in the context of ZFC, but since we are assuming large cardinals, we will have enough SCH to get away with using only the following trivial version of Shelah's theorem:
\begin{lma}\label{ScaleExistence}
Suppose \(\lambda\) is a singular cardinal of cofinality \(\iota\) such that \(\lambda^\iota = \lambda^+\). Suppose \(\langle \delta_\alpha :\alpha <\iota\rangle\) is a sequence of regular cardinals cofinal in \(\lambda\). Then there is a scale of length \(\lambda^+\) in \(\prod_{\alpha < \iota} \delta_\alpha\).
\begin{proof}
We start by proving the standard fact that \(\mathbb P = (\prod_{\alpha < \iota} \delta_\alpha,<_\text{bd})\) is a \({\leq}\lambda\)-directed partial order. The proof proceeds in two steps.

First, we prove that \(\mathbb P\) is \({<}\lambda\)-directed. Suppose \(\gamma < \lambda\) and \(\{g_i: i < \gamma\}\subseteq \mathbb P\). We will find a \(<_\text{bd}\)-upper bound \(g\) of \(\{g_i: i < \gamma\}\). Fix \(\alpha_0\) such that \(\gamma < \delta_{\alpha_0}\). For \(\alpha  < \iota\), define \[g(\alpha) = \begin{cases}
\sup_{i < \gamma} g_i(\alpha) + 1&\text{if }\alpha_0\leq \alpha\\
0&\text{otherwise}
\end{cases}\]
If \(\alpha_0\leq \alpha\), then \(\delta_\alpha\) is a regular cardinal greater than \(\gamma\), so \(\sup_{i < \gamma} g_i(\alpha) < \delta_\alpha\). Hence \(g\in \prod_{\alpha < \iota} \delta_\alpha\) and \(g\) is a \(<_\text{bd}\)-upper bound of \(\{g_i: i < \gamma\}\).

Second, we prove that \(\mathbb P\) is \(\lambda\)-directed. Fix \(\{g_i: i < \lambda\}\subseteq \mathbb P\). For \(\alpha<\iota\), let \(h_\alpha\in \mathbb P\) be a \(<_\text{bd}\)-upper bound of \(\{g_i : i < \delta_\alpha\}\). Finally let \(g\in \mathbb P\) be a \(<_\text{bd}\)-upper bound of \(\{g_i : i < \iota\}\). Then \(g\) is a \(<_\text{bd}\)-upper bound of \(\{g_i: i < \lambda\}\), as desired.

Enumerate \(\prod_{\alpha < \iota} \delta_\alpha\) as \(\{g_\xi : \xi < \lambda^+\}\). We define \(\langle f_\xi : \xi < \lambda^+\rangle\) recursively. If \(\langle f_\xi : \xi < \theta\rangle\) has been defined, choose a \(<_\text{bd}\)-upper bound \(f_\theta\in \mathbb P\) of \(\{ f_\xi : \xi < \theta\}\cup \{g_\xi\}\). (Such a function exists by the \(\lambda\)-directedness of \(\mathbb P\).) By construction \(\langle f_\xi : \xi < \lambda^+\rangle\) is a scale in \(\prod_{\alpha < \iota} \delta_\alpha\).
\end{proof}
\end{lma}

This concludes our summary of the basic notions from PCF theory used in the proof of \cref{GeneralizedSolovay}, which we now commence.

\begin{proof}[Proof of \cref{GeneralizedSolovay}]\index{Solovay's Lemma!at singular cardinals}
For the purposes of the proof, let us say that \(x\) is {\it weakly definable from \(y\) (in \(M\))} if \(x\) is definable in \(M\) from parameters in \(j[V]\cup \{y\}\), or in other words, \(x\in H^M(j[V]\cup \{y\})\). Note that weak definability is a transitive relation.

By \cref{StationaryPartition}, we may assume \(\lambda\) is a singular cardinal. Let \(\iota\) be the cofinality of \(\lambda\). Let \(\lambda_* = \sup j[\lambda]\).

\begin{clm}\label{JbdClm}
Suppose \(\langle \delta_\alpha :\alpha < \iota\rangle\) is an increasing sequence of regular cardinals cofinal in \(\lambda\). Let \(e\) be the equivalence class of \(\langle\sup j[\delta_\alpha] : \alpha < \iota\rangle\) modulo \(J_\textnormal{bd}\). Then \(j[\lambda]\) is weakly definable from \(e\) and \(j\restriction \iota\) in \(M\).
\begin{proof}[Proof of \cref{JbdClm}]

Fix a sequence \(\langle S^\alpha : \alpha < \iota\rangle\) such that \(S^\alpha = \{S^\alpha_\beta : \beta < \delta_\alpha\}\) is a partition of \(S^{\delta_\alpha}_\omega\) into stationary sets. Note that \(\langle j(S^\alpha) : \alpha < \iota\rangle= j(\langle S^\alpha : \alpha < \iota\rangle)\circ j\restriction \iota\) is weakly definable from \(j\restriction \iota\).

Solovay's Lemma (\cref{MStationaryPartition}) implies that for all \(\alpha < \iota\), \(j[\delta_\alpha]\) is equal to the set \(\{\beta < j(\delta_\alpha) :M\vDash  j(S^\alpha)_\beta\text{ is stationary in }\sup j[\delta_\alpha]\}\).
It follows that
\begin{align*}\beta\in j[\lambda]&\iff \{\alpha < \iota :  M\vDash j(S^\alpha)_\beta\text{ is stationary in }\sup j[\delta_\alpha]\}\notin J_{\text{bd}}\\
&\iff \exists s\in e\ \{\alpha < \iota : M\vDash  j(S^\alpha)_\beta\text{ is stationary in }s(\alpha)\}\notin J_{\text{bd}}
\end{align*}
Thus \(j[\lambda]\) is weakly definable from \(e\) and \(\langle j(S^\alpha) : \alpha < \iota\rangle\). Since \(\langle j(S^\alpha) : \alpha < \iota\rangle\) is weakly definable from  \(j\restriction \iota\), this proves the claim.
\end{proof}
\end{clm}
It is not hard to see that \(j\restriction \iota\) is itself weakly definable from \(e\), but we will not need this. The following observation, however, will be crucial:
\begin{obs}\label{supobs}\(j\restriction \iota\) and \(j[\iota]\) are weakly definable from \(\sup j[\iota]\).\end{obs}
This is an immediate consequence of Solovay's Lemma (\cref{MStationaryPartition}).

Let \(\mathcal D\) be the normal fine ultrafilter on \(P(\iota)\) derived from \(j\) using \(j[\iota]\) and let \(k : M_\mathcal D\to M\) be the factor embedding. Let \(\lambda_\mathcal D = \sup j_\mathcal D[\lambda]\).\footnote{If \(\eta^\iota < \lambda\) for all \(\eta < \lambda\), then \(\lambda_\mathcal D= \lambda\), but we do not assume this.} By \cref{DerivedNF}, \(\textsc{crt}(k) > \iota\) and hence \(k(\lambda_\mathcal D) = \sup k[\lambda_\mathcal D] =\lambda_*\).

\begin{obs}\label{kobs}
	\(k[M_\mathcal D]\) consists of all \(x\in M\) that are weakly definable from \(\sup j[\iota]\).
\end{obs}
\cref{kobs} follows from the fact that \(k[M_\mathcal D] = H^M(j[V]\cup \{j[\iota]\})\) combined with the fact (\cref{supobs}) that \(j[\iota]\) and \(\sup j[\iota]\) are weakly definable from each other.

Let \[\theta = \sup k[\lambda_\mathcal D^{+M_\mathcal D}]\] The ordinal \(\theta\) will turn out to be the least generator of \(j\) above \(\lambda_*\). For now, let us just show that there is no smaller generator:
\begin{clm}\label{ThetaClm}
\(\theta\subseteq H^M(j[V]\cup \lambda_*)\).
\begin{proof}
Suppose \(\alpha < \theta\). The claim amounts to showing that \(\alpha\) is weakly definable from a finite set of ordinals below \(\lambda_*\). By the definition of \(\theta\), \(\alpha < k(\xi)\) for some \(\xi < \lambda_\mathcal D^{+M_\mathcal D}\). Fix a surjection \(p : \lambda_\mathcal D\to \xi\) with \(p\in M_\mathcal D\). \cref{kobs} implies \(k(p)\) is weakly definable from \(\sup j[\iota]\). Since \(k(p)\) is a surjection from \(\lambda_*\) onto \(k(\xi)\), for some \(\nu < \lambda_*\), \(\alpha = k(p)(\nu)\). Thus \(\alpha\) is weakly definable from \(\sup j[\iota]\) and \(\nu\), which both lie below \(\lambda_*\), proving the claim.
\end{proof}
\end{clm}

Fix a sequence \(\langle \delta_\alpha :\alpha < \iota\rangle\) of regular cardinals greater than \(\iota\) that is increasing and cofinal in \(\lambda\).
\begin{clm}
In \(M_\mathcal D\), there is a scale \(\vec f = \langle f_\alpha : \alpha < \lambda_\mathcal D^{+M_\mathcal D}\rangle\) in \(\prod_{\alpha < \iota} j_\mathcal D(\delta_\alpha)\).
\begin{proof}
Applying \cref{ScaleExistence} in \(M_\mathcal D\), it suffices to show that \(M_\mathcal D\) satisfies \(\lambda_\mathcal D^\iota = \lambda_\mathcal D^{+\mathcal D}\). By the critical sequence analysis given by the Kunen Inconsistency Theorem (\cref{Kunen2}), there is a \(\lambda\)-supercompact cardinal \(\kappa \leq \iota\) such that \(j_\mathcal D(\kappa) > \iota\).
Thus \(\iota < j_\mathcal D(\kappa) < \lambda_*\) and \(j_{\mathcal D}(\kappa)\) is \(\lambda_\mathcal D^{+\mathcal D}\)-supercompact in \(M_\mathcal D\).
By the local version of Solovay's theorem \cite{Solovay} (which appears as \cref{SolovaySCH}) applied in \(M_\mathcal D\), it follows that in \(M_\mathcal D\), \(\lambda_\mathcal D^\iota \leq (\lambda_\mathcal D^+)^{<j_\mathcal D(\kappa)} = \lambda_\mathcal D^{+M_\mathcal D}\), as desired.
\end{proof}
\end{clm}
\begin{clm}\label{EUBClm}
\(\langle\sup j[\delta_\alpha] : \alpha < \iota\rangle\) is an exact upper bound of \(k(\vec f)\restriction \theta\).
\end{clm}

Before proving \cref{EUBClm}, let us show how it implies the theorem.

Let \(e\) be the equivalence class of \(\langle\sup j[\delta_\alpha] : \alpha < \iota\rangle\) modulo the bounded ideal on \(\iota\). Then \cref{EUBClm} and \cref{EUBUnique} imply that \(e\) is definable in \(M\) from the parameters \(\theta\) and \(k(\vec f)\). Thus by \cref{JbdClm}, \(j[\lambda]\) is weakly definable from \(\theta\), \(k(\vec f)\), and \(j\restriction \iota\).

Note that \(\lambda_*\) is definable in \(M\) from \(\theta\): \(\lambda_*\) is the largest \(M\)-cardinal below \(\theta\). By \cref{iota*lemma}, \(\sup j[\iota]\) is weakly definable from \(\lambda_*\) and hence from \(\theta\). Thus by \cref{supobs} \(j\restriction \iota\) is weakly definable from \(\theta\), and by by \cref{kobs}, \(k(\vec f)\) is weakly definable from \(\theta\). Combining this with the previous paragraph, \(j[\lambda]\) is weakly definable from \(\theta\) alone. This yields:
\[j[\lambda] \in H^M(j[V]\cup \{\theta\})\]

We now show \(\theta\) is the least generator of \(j\) above \(\lambda_*\). It suffices by \cref{ThetaClm} to show that \(\theta\) is a generator of \(j\). Assume towards a contradiction that this fails. Then \(\theta\in H^M(j[V]\cup \theta) = H^M(j[V]\cup \lambda_*)\) by \cref{ThetaClm}. Thus \(j[\lambda]\in H^M(j[V]\cup \lambda_*)\). Fix \(\xi < \lambda_*\) such that \(j[\lambda]\in H^M(j[V]\cup \{\xi\})\). Let \(W\) be the ultrafilter derived from \(j\) using \(\xi\). Then by \cref{DerivedSC}, \(j_W\) is \(\lambda\)-supercompact, yet \(\lambda_W < \lambda\), and this contradicts \cref{UFSuperBound}. Thus our assumption was false, and in fact \(\theta\) is a generator of \(j\).

Thus \(j[\lambda]\in H^M(j[V]\cup \{\theta\})\) where \(\theta\) is the least generator of \(j\) greater than or equal to \(\lambda_*\). To finish, we must show that if \(\lambda_* < j(\lambda)\) then \(\theta < j(\lambda)\). But \(\theta \leq \lambda_*^{+M}\) while \(j(\lambda)\) is a limit cardinal of \(M\) above \(\lambda_*\). Hence \(\lambda_*^{+M} < j(\lambda)\), as desired.

We now turn to the proof of \cref{EUBClm}. It will be important here that for any \(s:\iota\to \text{Ord}\), \(k(s) = k\circ s\) since \(\textsc{crt}(k) > \iota\).
\begin{proof}[Proof of \cref{EUBClm}]
We first show that for all \(\nu < \theta\), \[k(\vec f)_\nu <_\text{bd} \langle \sup j[\delta_\alpha] : \alpha < \iota\rangle\] For any \(\nu < \theta\), there is some \(\xi < \lambda_\mathcal D^{+M_\mathcal D}\) such that \(\nu < k(\xi)\). Therefore \[k(\vec f)_\nu <_\text{bd} k(\vec f)_{k(\xi)} = k(f_\xi)\] Hence it suffices to show that for any \(\xi < \lambda_\mathcal D^{+M_\mathcal D}\), \(k(f_\xi) < \langle  \sup j[\delta_\alpha] : \alpha < \iota\rangle\). For all \(\alpha < \iota\), we have that \(\delta_\alpha \) is a regular cardinal above \(\iota\). By \cref{SolovayCor}, \(\lambda_\mathcal D = \iota\), so since ultrapower embeddings are continuous at regular cardinals above their size (\cref{UFContinuity}), \[j_\mathcal D(\delta_\alpha) = \sup j_\mathcal D[\delta_\alpha]\]  Since \(f_\xi\in \prod_{\alpha < \iota}j_\mathcal D(\delta_\alpha)\), we therefore have \(f_\xi(\alpha) < \sup j_\mathcal D[\delta_\alpha]\) and hence \(k(f_\xi)(\alpha) = k(f_\xi(\alpha)) < \sup j[\delta_\alpha]\) for all \(\alpha < \iota\), as desired.

We finish by showing that for any \(g :\iota\to \text{Ord}\) such that \(g<_\text{bd}\langle \sup j[\delta_\alpha] : \alpha < \iota\rangle\), there is some \(\xi < \lambda_\mathcal D^{+M_\mathcal D}\) such that \(g <_\text{bd} k(f_\xi)\). For \(\alpha < \iota\), let \(h(\alpha) < \delta_\alpha\) be least such that \(g(\alpha) \leq j(h(\alpha))\). Then \(j_\mathcal D\circ h\in M_\mathcal D\) (since \(M_\mathcal D\) is closed under \(\iota\)-sequences by \cref{UltrapowerSC}). Since \(\langle f_\xi : \xi < \lambda^+\rangle\) is cofinal, in \(\prod_{\alpha < \iota}j_\mathcal D(\delta_\alpha)\), there is some \(\xi < \lambda_\mathcal D^{+M_\mathcal D}\) such that \(j_\mathcal D\circ h <_\text{bd} f_\xi\) . It follows that \[g \leq j\circ h = k\circ j_\mathcal D\circ h = k(j_\mathcal D\circ h) <_\text{bd} k(f_\xi)\] as desired.
\end{proof}
This completes the proof of \cref{GeneralizedSolovay}.
\end{proof}