\documentclass{amsart}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{mathabx}
\usepackage{stmaryrd}
\usepackage{mathbbol}
\theoremstyle{plain}
\newtheorem{thm}{Theorem}[section]
\newtheorem{cor}[thm]{Corollary}
\newtheorem{lem}[thm]{Lemma}
\theoremstyle{definition}
\newtheorem{defn}[thm]{Definition}
\newtheorem{exer}{Exercise}
\newtheorem{exam}{Example}
\theoremstyle{remark}
\newtheorem{rem}[thm]{Remark}
\newtheorem{fact}[thm]{Fact}
\newtheorem{hyp}[thm]{Hypothesis}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\bP}{\mathbb{P}}
\newcommand{\bQ}{\mathbb{Q}}
\newcommand{\bone}{\mathbb{1}}
\newcommand{\res}{\restriction}
\newcommand{\dom}{\operatorname{dom}}
\newcommand{\ran}{\operatorname{ran}}
\newcommand{\cof}{\operatorname{cof}}
\newcommand{\ot}{\text{o.t.}}
\newcommand{\on}{\text{ON}}
\newcommand{\card}{\text{CARD}}
\newcommand{\ac}{\text{AC}}
\newcommand{\zf}{\text{ZF}}
\newcommand{\zfc}{\text{ZFC}}
\newcommand{\sP}{\mathcal{P}}
\newcommand{\sL}{\mathcal{L}}
\newcommand{\ch}{\text{CH}}
\newcommand{\gch}{\text{GCH}}
\newcommand{\vb}{\text{var}}
\newcommand{\pa}{\text{PA}}
\newcommand{\fa}{\text{F}}
\newcommand{\con}{\text{CON}}
\newcommand{\bz}{\boldsymbol{0}}
\newcommand{\lD}{{\Delta}}
\newcommand{\lS}{{\Sigma}}
\newcommand{\lP}{{\Pi}}
%\newcommand{\models}{\vDash}
\newcommand{\proves}{\vdash}
\newcommand{\sg}{\text{sg}}
\newcommand{\seq}{\text{Seq}}
\newcommand{\lh}{\text{lh}}
\newcommand{\pred}{\text{pred}}
\newcommand{\hod}{\text{HOD}}
\newcommand{\od}{\text{OD}}
\newcommand{\trcl}{\text{tr\,cl}}
\newcommand{\cl}{\text{cl}}
\newcommand{\rank}[1]{|#1|}
\newcommand{\llex}{<_{\text{lex}}}
\newcommand{\comp}{\parallel}
\newcommand{\forces}{\Vdash}
\newcommand{\fn}{\text{FN}}
\newcommand{\ccc}{\text{c.c.c.}}
\newcommand{\cc}{\text{c.c.}}
\newcommand{\sat}{\text{sat}}
\newcommand{\coll}{\text{coll}}
\begin{document}
\begin{center}
\bf Products and Easton's Theorem
\end{center}
\bigskip
\section{Product Forcing}
Let $\bP= \langle P, \leq_P \rangle$, $\bQ=\langle Q, \leq_Q \rangle$
be partial orders. We define their product by $\bP \times \bQ = \{
\langle p, q \rangle \colon p \in P \wedge q \in Q \}$. This is
ordered by $(p',q') \leq_{\bP \times \bQ} (p,q)$ iff
$p' \leq_P p$ and $q' \leq_Q q$
(note: we will frequently use $(p,q)$ instead of the more formal
$\langle p, q \rangle$ when details of the pair coding are
irrelevant).
For example, the forcing for adding two real, $\fn( \omega \times 2, 2)$
is isomorphic to the product $\fn (\omega, 2) \times \fn (\omega, 2)$ (which
in this case is isomorphic to $\fn(\omega,2)$ itself).
If $G \subseteq P$ and $H \subseteq Q$ are filters, then $G \times H
\subseteq P \times Q$ is also easily a filter. Conversely, if $F
\subseteq P \times Q$ is a filter, let $G=\{ p \in P \colon \exists q
\in Q\ (p,q) \in F\}$ and likewise $H=\{ q \in Q \colon \exists p \in
P\ (p,q) \in F\}$. Easily $G$ and $H$ are filters. If $(p,q) \in F$
then by definition $p \in G$ and $q \in H$, so $F \subseteq G \times
H$. For the other direction, suppose $p \in G$ and $q \in H$. then
$(p,q') \in F$ and $(p',q) \in F$ for some $p'$, $q'$. Let $(r,s) \in
F$ with $r \leq p,p'$, $s \leq q, q'$ (note: I've switched to the
other definition of filter now). Since $(p,q) \leq (r,s) \in F$,
$(p,q) \in F$. Thus, filters $F$ in $P \times Q$ are precisely of the
form $F=G \times H$ where $F$, $G$ are filters in $P$, $Q$
respectively. The relation between generics for $\bP \times \bQ$ and
generics for $\bP$, $\bQ$ is clarified in the following lemma.
\begin{lem}
A filter $F=G \times H$ is $M$ generic for $P \times Q$ iff
$G$ is $M$ generic for $P$ and $H$ is $M[G]$ generic for $Q$.
\end{lem}
\begin{rem}
Of course, the situation is symmetrical with respect to $P$ and $Q$,
so we equally well say iff $H$ is $M$ generic for $Q$ and
$G$ is $M[H]$ generic for $P$.
\end{rem}
\begin{proof}
First suppose $G \times H$ is $M$ generic for $P \times Q$.
Let $D \subseteq P$, $D \in M$, be dense.
Then $Q \times Q$ is dense in $P \times
Q$, so let $(p,q) \in (G \times H) \cap (D \times Q)$. Thus,
$p \in G \cap D$. This shows $G$ is $M$ generic for $P$.
Let now $E \subseteq Q$, $E \in M[G]$, be dense in $Q$.
Let $E=\tau_G$, where $\tau \in M^\bP$. Let $p_0 \in P \cap G$
with $p \forces (\tau$ is dense in $\check{Q})$. Let
$$
D=\{ (p,q) \in P \times Q \colon (p \perp p_0) \vee (p \leq p_0
\wedge (p \forces \check{q} \in \tau)) \}.
$$
$D$ is easily dense in $P \times Q$ [Let $(r,s) \in P \times Q$.
If $r \perp p_0$, then $(r,s) \in D$. Otherwise, let
$(r',s) \leq (r,s)$ with $r' \leq p_0$. Since
$r' \forces (\tau$ is dense $)$, $r' \forces
\exists q \leq \check{s} \ ( q \in \tau)$. Then there is a $p \leq r'$
and a $q \leq s$ with $p \forces (\check{q} \in \tau)$.
Thus, $(p,q) \leq (r,s)$ and $(p,q) \in D$.]
Let $(p,q) \in (G \times H) \cap D$. We must have $p \leq p_0$,
and so $p \forces (\check{q} \in \tau)$. Since $p \in G$,
$q \in E$, so $q \in H \cap E$. Thus, $H$ is $M[G]$ generic for $Q$.
Conversely, suppose $G$ is $M$ generic for $P$ and $H$ is $M[G]$
generic for $Q$. Let $D \subseteq P \times Q$ be dense.
Let $E =\{ q \in Q \colon \exists p \in G\ (p,q) \in D\}$.
Clearly $E \in M[G]$. It is enough to show that $E$ is dense in $Q$
for then we would have $q \in H \cap E$. By definition of $E$ there
would then be a $p \in G$ with $(p,q) \in D$. Hence,
$(p,q) \in (G \times H) \cap D$. To see $E$ is dense, let
$s \in Q$. Let $A = \{ p \in P \colon \exists q\ (q \leq s \wedge
(p,q) \in D\}$. Clearly $A$ is dense in $P$, and $A \in M$.
So, let $p \in G \cap A$. Let $q \leq s$ with $(p,q) \in D$.
Then $q \in E$ and $q \leq s$.
\end{proof}
Thus, if $\bP$, $\bQ$ are partial orders in $M$, forcing with the
product $\bP \times \bQ$ is equivalent to doing a two-step
forcing where we first force over $M$ with $\bP$ to get
$M[G]$, and then force over $M[G]$ with $\bQ$ to get
$M[G][H]$. Note that $M[G][H]=M[G \times H]$, as $G$, $H$ are
definable from $G \times H$ and conversely.
The following technical lemma combines lemmas~\ref{lemclosed} and
\ref{cccov}.
\begin{lem} \label{lemsplit}
Let $\kappa$ be a cardinal and $\bP$ be $\kappa^+$-c.c.\ and
$\bQ$ be $\leq \kappa$ closed in a transitive model $M$ of $\zfc$.
Let $G \times H$ be $M$ generic for $\bP \times \bQ$. Then
any $f \colon \kappa \to M$ in $M[G][H]$ lies in $M[G]$.
\end{lem}
\begin{proof}
Let $f \colon \kappa \to M$ lie in $M[G][H]$. Let $\tau \in
M^{\bP \times \bQ}$ with $f= \tau_{G \times H}$. For each $\alpha < \kappa$
let $D_\alpha=\{ q \in Q \colon \forall p \in P\
\exists p' \leq p \ \exists x \in M \ ( (p',q) \forces
\tau(\check{\alpha})= \check{x}) \}$. We claim that $D_\alpha$
is dense in $Q$. To see this, let $q \in Q$. Let $p_0 \in P$, $q_0 \leq q$,
and $x_0 \in M$ with $(p_0,q_0) \forces \tau(\check{0})=\check{x}$.
We construct $(p_0,q_0) \leq \dots \leq (p_\beta,q_\beta)\leq $
as follows. Assume $(p_\gamma,q_\gamma)$ is defined for $\gamma < \beta$,
and $\beta < \kappa^+$. If $\{ p_\gamma \}_{\gamma < \beta}$
is a maximal antichain on $P$, then we let stop the
construction and let $q$ extend all of the $q_\gamma$, which we
can do as $\bQ$ is $\leq \kappa$ closed. Otherwise, let
$p_\beta$ be incompatible with all of the $p_\gamma$, $\gamma < \beta$.
Let $q_\beta$ extend all of the $q_\gamma$ for $\gamma < \beta$
and such that for some $x_\beta \in M$, $(p_\beta,q_\beta)
\forces \tau(\check{\beta})=\check{x_\beta}$.
As $\bP$ is $\kappa^+$-c.c.\ this construction cannot go on
$\kappa^+$ times. Thus, for some $\beta < \kappa^+$,
$\{ p_\gamma \}_{\gamma < \beta}$ is a maximal antichain of $P$.
The corresponding $q$ (which extends all of the $q_\gamma$)
lies in $D_\alpha$.
Using again that $\bQ$ is $\leq \kappa$ closed, we get that
$D=\bigcap_{\alpha < \kappa} D_\alpha$ is dense in $Q$.
Fix $q \in H \cap D$. Working in $M$ we may define dense sets
$D_\alpha$, $\alpha < \kappa$, such that for all $\alpha$ and all
$p \in D_\alpha$, there is an $x \in M$ such that
$(p,q) \forces \tau(\check{\alpha})=\check{x}$. In $M[G]$ we may
then compute $f$, namely $f(\alpha)=x$ iff there is a
$p \in D_\alpha \cap G$ such that $(p,q) \forces
\tau(\check{\alpha})=\check{x}$ (we are using the replacement
and comprehension
axioms in $M$ to get that $f$ is a set in $M$).
\end{proof}
As a warm-up for Easton's theorem, let us give another, perhaps
more direct, proof of lemma~\ref{lemfinreg}. So, let $M$
satisfy $\gch$ and
$\kappa_1 < \dots < \kappa_n$ be regular, and
$\lambda_1 \leq \dots \leq \lambda_n$ with $\cof(\lambda_i)
> \kappa_i$. Let $\bP_i= \fn( \lambda_i, 2, \kappa_i)$
be the partial order for adding $\lambda_i$ many subsets of $\kappa_i$.
Let $\bP=\bP_1 \times \dots \times \bP_n$ be the product. Let
$G=G_0 \times \dots \times G_n$ be $M$ generic for $\bP$.
First we show that $\bP$ preserves all cardinalities and cofinalities.
Let $\delta$ be a regular cardinal of $M$, but suppose
$f \colon \rho \to \delta$ is cofinal where $\rho < \delta$
and $f \in M[G]$. Let $\bP^-= \bP_1 \times \dots \times \bP_i$
where $i$ is maximal so that $\kappa_i \leq \rho$. Let
$\bP^+= \bP_{i+1} \times \dots \times \bP_n$. Clearly $\bP^+$
is $\leq \rho$ closed in $M$. Also, $\bP^-$ is $(2^{< \rho})^+=
\rho^+$-c.c.\ in $M$ (note that $P$ can be viewed as a subset
of $\fn(\lambda_{i} ,2, \kappa_{i})$ since
$\sum_{j \leq i} (\kappa_j \times \lambda_j) \cong \lambda_i$; use
then lemma~\ref{lemgcc}). From lemma~\ref{lemsplit} we have
$f \in M[\bP^-]$. From lemma~\ref{lemcccov} we then have that there is an
$F \colon \rho \to \sP(\delta)$, $F \in M$ with $|F(\alpha)| \leq \rho$
for all $\alpha < \delta$. This is a contradiction as $\delta$
is regular in $M$. Som $\bP$ preserves all cofinalities and hence
cardinalities (which the same argument also shows directly; thus
the preservation of cardinals only requires $M$ to satisfy $\zf$).
Clearly $(2^{\kappa_i}) \geq \lambda_i)^{M[\bP]}$. To get the upper
bound for $(2^{\kappa_i})^{M[\bP]}$, write $\bP= \bP^- \times \bP^+$
where $\bP^-= \bP_1 \times \dots \times \bP_i$
and $\bP^+= \bP_{i+1} \times \dots \times \bP_n$. From lemma~\ref{lemsplit}
we have $\sP(\kappa_i)^{M[\bP]}=\sP(\kappa_i)^{M[\bP^-]}$.
Since $\bP^-$ is $\kappa_i^+$-c.c.\ in $M$, there are at most
$(\lambda_i^{\kappa_i})^{\kappa_i}= \lambda_i^{\kappa_i}= \lambda_i$
many nice names for a subset of $\kappa_i$ in $M$ (these computations are
done in $M$; the last equality uses $\cof(\lambda_i)> \kappa_i$ and the
$\gch$ in $M$). Thus, $(2^{\kappa_i} \leq \lambda_i)^{M[\bP]}$.
We have thus shown that $(2^{\kappa_i} = \lambda_i)^{M[\bP]}$ for all $i\leq n$.
\section{Remarks on Class Forcing}
In most applications we will be in the situation where $\bP \in M$,
that is, $\bP$ is a set in $M$ (this is what we have been
considering up to this point). For some purposes, including Easton's
theorem, we would like to generalize this to allow $\bP$ being a class
in $M$. Note that we are still assuming that $M$ is a transitive set
in $V$, thus there is no problem in quantifying over the classes
of $M$ (as statements in $V$). For this section, when we say
a class of $M$, we mean a formula with set parameters from $M$.
Let $M$ be a set which is a transitive model of $\zf$ (or $\zfc$).
Let $\bP= \langle P, \leq \rangle$ where $\bP$, $\leq \subseteq M$
are classes of $M$ (i.e., definable in $M$ from parameters in $M$), and
such that $\bP$ is a partial order. Note that $M$ also satisfies that
$\bP$ is a partial order in the sense that, for example,
$(\forall x, y, z\ [((x,y) \in \, \leq) \wedge ((y,z) \in \, \leq) \rightarrow
((x,z) \in \leq)])^M$. If $D \subseteq P$ is a class of $M$, we say
$D$ is dense just as before; if $\forall p \in P \ \exists
q \in D (q \leq p)$. For a given $D$, this is expressible in $M$.
We say $G \subseteq P$ is $M$ generic for $\bP$ exactly as before;
if $G \cap D \neq \emptyset$ for all dense classes $D \subseteq P$ of $M$.
We define $M^\bP$ essentially as before. Thus, $M^\bP= \bigcup_{\alpha \in
\on^M} M^\bP_\alpha$, where we take unions at limit ordinals and
$$M^\bP_{\alpha+1}= \{ \tau \in M \cap V_{\alpha+1} \colon
(\tau \text{ is a relation }) \wedge \dom (\tau) \subseteq
M^\bP_\alpha \wedge \ran(\tau) \subseteq P \}.$$
Again, the transfinite recursion theorem shows that $M^\bP$ is a
well-defined class of $M$. Given a filter $G \subseteq P$,
we define the evaluation map $\tau \to \tau_G$ exactly as before,
and again define $M[G]=\{ \tau_G \colon \tau \in M^\bP \}$.
For $p \in P$, $\phi(x_1,\dots,x_n)$ a formula,
and $\tau_1,\dots, \tau_n \in M^\bP$, we define the forcing relation
$p \forces \phi(\tau_1,\dots,\tau_n)$ exactly as before.
We again have that for all formuals $\phi(x_1,\dots,x_n)$ that
$\{ (p, \tau_1,\dots, \tau_n) \colon p \forces \phi(x_1,\dots,x_n) \}$
is a class of $M$.
Finally, the forcing theorem goes through as before. For example, consider
the atomic case $\phi=(\tau_1 \in \tau_2)$. Suppose $p \in P$
and $p \forces \phi$. Then the class
$$D= \{ q \in P \colon \exists
\langle \sigma, r \rangle \in \tau_2\ (q \leq r \wedge
r \forces (\tau_1 \approx \sigma)) \}$$ is dense below $p$.
$D$ is now a class, not a set in $M$, but $G$ being generic still
implies $G \cap D \neq \emptyset$. If $q \in G \cap D$, let
$\langle \sigma, r \rangle \in \tau_2$ be such that
$q \leq r \wedge r \forces (\tau_1 \approx \sigma)$. So,
$\sigma_G \in (\tau_2)_G$ and by induction $(\tau_1)_G= \sigma_G$.
Hence, $(\tau_1)_G \in (\tau_2)_G$. The other direction is also as before.
Consider now which axioms of $\zf$ hold in $M[G]$. Certainly
founadation, extensionality, pairing and union hold in $M[G]$
(and again only require $G$ to be a filter).
We run into problems, though, when we try to
show power set, replacement, and comprehension. The proofs of these axioms
given previously for set forcing use the fact that $\bP$
is a set in $M$. For example the proof of power set required us
to consider $\{ \sigma \colon \dom(\sigma) \subseteq \dom(\tau)
\wedge \ran(\sigma) \subseteq P \}$, where $\tau \in M^\bP$ is fixed.
If $\bP$ is not a set in $M$, this will clearly be a proper
class of $M$ as well. Similarly, the previous proof of replacement
required the apllication of replacement in $M$ to the set
$\dom(\tau) \times P$ for some $\tau \in M^\bP$; here again this will
be a proper class.
For general class forcing, the power set, replacement, and
comprehension axioms may fail in $M[G]$. For example, power set will
fail if we add
$\on$ many reals, or if we collapse $\on$ to
$\omega$. Thus, some restiction on the forcing is necessary. The
following gives a sufficient condition.
\begin{thm} \label{classforcing}
Let $\bP$ be a class partial order of $M$, where $M$ is a transitive
model of $\zf$ (or $\zfc$). Suppose that for arbitrarily large
cardinals $\kappa$ of $M$ that we can write $\bP= \bP^- \times \bP^+$
where $\bP^-$ is
$\kappa^+$-c.c.\ and $\bP^+$ is $\leq \kappa$
closed. Let $G$ be $M$ generic for $\bP$. Then $M[G]$ satisfies $\zf$
(or $\zfc$).
\end{thm}
\begin{proof}
We show comprehension, power set, and replacement in $M[G]$.
To show comprehension, fix $a_1=(\tau_1)_G, \dots,$, $a_n= (\tau_n)_G$,
$A= \tau_G$, and a formula $\phi(x_1,\dots,x_n,y,z)$.
We must show that $\{ z \in A \colon \phi^{M[G]}(a_1,\dots,a_n,A,z) \}$
exists as a set in $M[G]$. Let $\kappa > |\tau |$ be such that
$\bP= \bP^- \times \bP^+$ with $\bP^-$ $\kappa^+$ c.c., and $\bP^+$
$\leq \kappa$ closed.
As in the proof of
lemma~\ref{lemsplit}, let $Q \subseteq P^+$ be those $q \in P^+$
such that for all $\langle \pi ,p \rangle \in \tau$, there is a
dense below $p$ set of conditions $p' \in P^-$ such that $(p', q)$
decides $\phi(\tau_1,\dots, \tau_n, \tau, \pi)$. More precisely,
\begin{equation*}
\begin{split}
Q= & \{ q \in P^+ \colon \forall \pi \in \dom(\tau)
\ \forall r \in P^- \ \ \exists s \in P^-\ [ ((s,q) \forces
\phi(\tau_1,\dots, \tau_n, \tau, \pi)) \vee
\\ & \quad
((s,q) \forces \neg \phi(\tau_1,\dots, \tau_n, \tau, \pi)) ].
\end{split}
\end{equation*}
As in the proof of lemma~\ref{lemsplit}, it follows from the
$\leq \kappa$ closure of $\bP^+$ that $Q$ is dense in $P^+$.
Let $(p_0,q_0) \in G$ with $q_0 \in Q$. For $\pi \in \dom(\tau)$ let
\begin{equation*}
\begin{split}
D_\pi=
\{ p \in P^- \colon
((p,q_0) \forces \phi(\tau_1,\dots, \tau_n, \tau, \pi)) \vee
((p,q_0) \forces \neg \phi(\tau_1,\dots, \tau_n, \tau, \pi)) \}.
\end{split}
\end{equation*}
Thus, $D_\pi$ is dense in $P^-$. Moreover, as in lemma~\ref{lemsplit}
we may assume each $D_\pi$ is an antichain which is a set of size
$\leq \kappa$. Let
$$
\sigma= \{ \langle \pi , (p,q_0) \rangle \colon (\pi \in \dom(\tau)) \wedge
(p \in D_\pi) \wedge ((p,q_0)
\forces \phi(\tau_1,\dots, \tau_n, \tau, \pi)) \}.
$$
Note that $\sigma$ is a valid name, that is, $\sigma$ is a set in $M$.
If $x \in \sigma_G$, then $x=\pi_G$ where $(p,q_0) \in G$,
and $(p,q_0) \forces \phi(\tau_1,\dots, \tau_n, \tau, \pi)$. Thus,
$\phi^{M[G]}(a_1,\dots,a_n,A,x)$. Suppose next that $x \in M[G]$
and $\phi^{M[G]}(a_1,\dots,a_n,A,x)$. Then $x= \pi_G$ where
$\pi \in \dom(\tau)$. Let $(p_1,q_1) \in G$ with $p_1 \in D_\pi$
and $(p_1,q_1) \forces \phi(\tau_1,\dots, \tau_n, \tau, \pi)$.
Since $p_1 \in D_\pi$, either $(p_1,q_0) \forces
\phi(\tau_1,\dots, \tau_n, \tau, \pi)$ or
$(p_1,q_0) \forces \neg \phi(\tau_1,\dots, \tau_n, \tau, \pi)$.
The latter case is impossible as $(p_1,q_0)$, $(p_1,q_1)$
are compatible. This shows $\langle \pi , (p_1,q_0) \rangle \in
\sigma$, and hence $x=\pi_G \in \sigma_G$ (note that $(p_1,q_0) \in G$,
as $G$ is a filter).
Consider next power set. Let $x = \tau_G \in M[G]$, let
$\kappa > |\tau|$ and again write
$\bP=\bP^- \times \bP^+$ as before. Let $\rho= \{ \langle \sigma, \bone \rangle
\colon \dom(\sigma) \subseteq \dom(\tau) \wedge \ran(\sigma) \subseteq \bP^- \}$.
We show that $\sP(x) \subseteq \rho_G$. Fix $y \subseteq x$, say $y = \mu_G$.
Arguing as in the previous case,
we get a $(p_0,q_0) \in G$ such that for all $\pi \in \dom(\tau)$,
the set $D_\pi \subseteq P^-$ is dense, where now
\begin{equation*}
\begin{split}
D_\pi=
\{ p \in P^- \colon
((p,q_0) \forces \pi \in \mu ) \vee
((p,q_0) \forces \neg (\pi \in \mu)) \}.
\end{split}
\end{equation*}
Let
$$
\sigma= \{ \langle \pi , (p, \bone) \rangle \colon (\pi \in \dom(\tau)) \wedge
(p \in D_\pi) \wedge ((p,q_0)
\forces \pi \in \mu ) \}.
$$
Clearly $\sigma_G \subseteq \mu_G$ (note that if $(p, \bone) \in G$, then
$(p, q_0) \in G$, since $(p_0,q_0) \in G$).
The other direction, $\mu_G
\subseteq \sigma_G$ follows now exactly as in the previous case.
Consider replacement. The proof is again similar to the previous
cases. Let $A= \tau_G$, $a_1= (\tau_1)_G, \dots$, $a_n=(\tau_n)_G$,
and $\phi(x_1,\dots,x_n,y,z,w)$ be a formula. Assume that
$$
\forall y \in A\ \exists z \in M[G]\ \phi^{M[G]}(a_1,\dots,a_n,A,y,z).
$$
Fix $\kappa> |\tau |$, and again write $\bP= \bP^- \times \bP^+$.
As in the previous cases (using the fact that $\bP^+$ is $\leq \kappa$
closed and $\bP^-$ is $\kappa^+$-c.c.) we get a $(p_0,q_0) \in G$
such that for each $\pi \in \dom(\tau)$ the set $D_\pi \subseteq
\bP^-$ is a dense set (which we may assume has size $\leq \kappa$), where
\begin{equation*}
\begin{split}
D_\pi=
\{ p \in P^- \colon \exists z \in M\
((p,q_0) \forces \phi(\tau_1,\dots,\tau_n,\tau, \pi,\check{z}) \}.
\end{split}
\end{equation*}
Using replacement in $M$, let $S$ be a set in $M$ such that
for all $\pi \in \dom(\tau)$ and all $p \in D_\pi$,
there is a $z \in S$ such that $(p,q_0) \forces
\phi(\tau_1,\dots,\tau_n,\tau, \pi,\check{z})$. Let
$$
\sigma= \{ \langle \rho, \bone \rangle \colon \rho \in S \}.
$$
To see this works, let $y= \pi_G \in A= \tau_G$, where $\pi \in
\dom(\tau)$. Let $(p_1,q_1) \in G$ with $p_1 \in
D_\pi$. By definition of $D_\pi$, let $z \in M$ be such that
$(p_1,q_0) \forces \phi(\tau_1,\dots,\tau_n,\tau, \pi,\check{z})$.
From the definition of $S$ it follows that
$(p_1,q_0) \forces
\phi(\tau_1,\dots,\tau_n,\tau, \pi,\check{z'})$ for some
$z' \in S$.
Hence, $\phi^{M[G]}(a_1,\dots,a_n,A,y,z')$ holds for some $z' \in \sigma_G$.
This verifies replacement.
If $M$ satisfies $\ac$, then so does $M[G]$ by exactly the same argument
as for set forcing, since if $A= \tau_G$, then $\tau$ is a set in $M$,
and so there is map $f \in M$ from an ordinal $\alpha$ onto $\tau$.
Although $G$ itself is no longer
in $M[G]$ (as with set forcing), we nevertheless still get a map
from $\alpha$ onto $\tau_G$ definable from $f$ and $G \cap V_\beta$
for some $\beta$, which is a set in $M[G]$. Thus, in $M[G]$ we still define an
$F$ from $\alpha$ onto $\tau_G$.
\end{proof}
If we assume the factoring hypothesis of theorem~\ref{classforcing}
holds for all regular cardinals, then the class forcing $\bP$ also
preserves all cardinals and cofinalities.
\begin{thm} \label{classforcingpres}
Let $\bP$ be a class partial order of $M$, where $M$ is a transitive
model of $\zfc$. Suppose that for all regular
cardinals $\kappa$ of $M$ that we can write $\bP= \bP^- \times \bP^+$
where $\bP^-$ is
$\kappa^+$-c.c.\ and $\bP^+$ is $\leq \kappa$
closed. Then all cardinals and cofinalites are preserved in forcing
with $\bP$.
\end{thm}
\begin{proof}
Let $\delta$ be a regular cardinal of $M$, and suppose
$\rho=(\cof(\delta))^{M[G]} < \delta$. We again use the argument of
lemma~\ref{lemsplit}. Write $\bP= \bP^- \times \bP^+$
where $\bP^-$ is $\rho^+$-c.c.\ and $\bP^+$ is $\leq \rho$ closed.
Let $f \colon \rho \to \delta$ be onto, $f \in M[G]$.
Fronm lemma~\ref{lemsplit}, $f \in M[G^-]$, where $G= G^- \times G^+$.
Since $\bP^-$ is $\rho^+$ c.c., there is an $F \in M$, $F \colon
\rho \to \delta$ with $\ran(f) \subseteq \ran(F)$.
This contradicts $\delta$ being regular in $M$.
\end{proof}
We are now ready to give Easton's theorem.
\begin{defn}
An {\em Easton} function is a class function $F$ of $M$ with
domain a class of regular cardinals of $M$ and range in cardinals
of $M$ satisfying:
\begin{enumerate}
\item
If $\lambda_1 < \lambda _2$ are in $\dom(F)$, then $F(\lambda_1)
\leq F(\lambda_2)$.
\item
$\forall \lambda \in \dom(F)\ (\cof(F(\lambda)> \lambda)$.
\end{enumerate}
If $F$ is an Easton function for $M$, we define the Easton forcing
$\bP_F$ as follows. Condition $p \in \bP_F$ are functions with
domain $\dom(F)$, and for $\lambda \in \dom(F)$,
$p(\lambda) \in \fn( F(\lambda), 2, \lambda)$. Further, we require
$p$ to satisfy the {\em Easton condition}:
for all regular $\kappa$ of $M$, $\{ \lambda < \kappa \colon
p(\lambda) \neq \bone \}$ has size $< \kappa$.
\end{defn}
Thus, the forcing is just the product of the forcings to
make $2^\lambda$ at least $F(\lambda)$, except we add the
Easton condition which restricts the size of the domains of the
conditions. Note that the Easton condition is only non-trivial
when $\kappa$ is a weakly inaccessible cardinal.
\begin{thm} [Easton] \label{thmeaston}
Let $M$ be a transitive model of $\zfc+\gch$, and $F$ a class of $M$ which
is an Easton function. Assume $G$ is $M$ generic for the Easton
forcing $\bP_F$. Then $M[G]$ satisfies $\zfc$, all cardinalities and
cofinalities are preserved from $M$ to $M[G]$, and
for all regular cardinals $\lambda$ of $M[G]$ we have
$(2^\lambda= F(\lambda))^{M[G]}$.
\end{thm}
\begin{proof}
Fix a regular cardinal $\lambda$ of $M$ (equivalently, of $M[G]$).
Write $\bP_F=\bP^{\leq \lambda} \times \bP^{> \lambda}$
where $\bP^{\leq \lambda}$ consists of those
$p \in \bP_F$ with $\dom(p) \subseteq \lambda +1$, and
$\bP^{> \lambda}$ those $p$ with $\dom(p) \subseteq \card -(\lambda+1)$.
Clearly $\bP^{> \lambda}$ is $\leq \lambda$ closed. We show that $\bP^{\leq \lambda}$
is $\lambda^+$ c.c.\ For every regular $\kappa \leq \lambda$,
$\fn(F(\kappa),2, \kappa)$ is $(2^{< \kappa})^+= \kappa^+$ c.c.\ in $M$,
since $M$ satisfies the $\gch$.
Suppose $\{ p_\alpha\}_{\alpha < \lambda^+}$ were an antichain of size
$\lambda^+$ in $\bP^{\leq \lambda}$. Let $d_\alpha= \dom(p_\alpha)$.
Since $| d_\alpha | < \lambda$ (by the Easton condition if
$\lambda$ is limit, otherwise trivially), there are only
$\lambda^{< \lambda}= \lambda$ many choices for $d_\alpha$. So, we
may assume that all of the $p_\alpha$ have the same domain $d$.
By regularity of $\lambda$, each $p_\alpha$ may be viewed as a
function from $d \times F(\lambda) \to \{ 0, 1\}$
with domain of size $< \lambda$. Since $\fn( F(\lambda), 2, \lambda)$
is $(2^{< \lambda})^+= \lambda^+$ c.c.\ in $M$, this is a contradiction.
Thus, $\bP^{\leq \lambda}$ is $\lambda^+$ c.c.
From lemmas~\ref{classforcing}, \ref{classforcingpres} we know that $M[G]$
satisfies $\zfc$ and all cardinals and cofinalities are preserved from
$M$ to $M[G]$. We clearly have for all regular cardinals of $M[G]$
that $(2^\lambda \geq F(\lambda))^{M[G]}$. To see the other direction,
fix a regular cardinal $\lambda$ of $M$ (equivalently, of $M[G]$)
and consider $\bP_F=\bP^{\leq \lambda} \times \bP^{> \lambda}$ as above.
Every subset of $\lambda$ in $M[G]$ is in $M[G^{\leq \lambda}]$, where
$G= G^{\leq \lambda} \times G^{> \lambda}$, from lemma~\ref{lemsplit}.
Since $\bP^{\leq \lambda}$ is $\lambda^+$ c.c., $(2^\lambda)^{M[G^{\leq \lambda}]}
\leq (| \bP^{\leq \lambda}| ^{\lambda \cdot \lambda})^M$. Also, $| \bP^{\leq \lambda}|
\leq F(\lambda)^{< \lambda} 2^{< \lambda}
= F(\lambda)$ since $\cof(F(\lambda))> \lambda$
and the $\gch$ in $M$ (these computations are done in $M$).
Thus, $(2^\lambda)^{M[G^{\leq \lambda}]} \leq
F(\lambda)^\lambda = F(\lambda)$.
\end{proof}
Finally, we use class forcing to get a model of $\gch$.
\begin{thm}
Let $M$ be a transitive model of $\zfc$. Then there is a class
partial order $\bP$ of $M$ such that if $G$ is $M$-generic for
$\bP$ then $M[G]$ satisfies $\zfc+ \gch$.
\end{thm}
\begin{proof}
Let $M$ be a transitive model of $\zfc$. Let $\alpha \to \beth_\alpha$
be the beth function of $M$ (for this proof, $\beth_\alpha$ always
denotes $(\beth_\alpha)^M$). For each ordinal $\alpha$ of $M$, let
$\bP_\alpha= \coll ( \beth_\alpha^+, \beth_{\alpha+1})^M= \fn (
\beth_\alpha^+, \beth_{\alpha+1}, \beth_\alpha^+)^M$. Note that
$\bP_\alpha$ is $\beth_\alpha$ closed and
$(\beth_{\alpha+1}^{\beth_\alpha})^+= \beth_{\alpha+1}^+$ c.c.\
($\beth_{\alpha+1}^{\beth_\alpha} = (2^{\beth_\alpha})^{\beth_\alpha}=
2^{\beth_\alpha}=\beth_{\alpha+1}$).
Let $\bP$ be the Easton product of the $\bP_\alpha$. That is,
$\bP$ consists of functions $p$ with domain a subset of ordinals,
$p(\alpha) \in \bP_\alpha$ for all $\alpha \in \dom(p)$, and
$p$ satisfies the Easton condition: for all inaccessible
$\lambda$, $\{ \alpha < \lambda \colon p(\alpha) \neq \bone \}$
has size $< \lambda$. For $\alpha \in \on^M$, let
$\bP^{< \alpha}$ denote those $p \in \bP$ with
$\dom(p) \subseteq \alpha$. Likewise, $\bP^{\geq \alpha}$
denotes those $p$ with $\dom(p) \cap \alpha = \emptyset$.
Clearly $\bP= \bP^{< \alpha} \times \bP^{\geq \alpha}$ (at least,
up to isomorphism).
First we show that $M[G]$ satisfies $\zfc$. For $\alpha$ a successor
ordinal of $M$, consider $\bP= \bP^{< \alpha} \times
\bP^{ \geq \alpha}$.
Easily $\bP^{\geq \alpha}$ is $\leq \beth_\alpha$ closed. Any
$p \in \bP^{< \alpha}$ can be viewed as a function from $\beth_{\alpha-1}^+$
to $\beth_{\alpha}$ of size $ \leq \beth_{\alpha-1}$. Since
$\fn(\beth_{\alpha-1}^+,\beth_{\alpha}, \beth_{\alpha-1}^+)$ is
$\beth_\alpha^+$ c.c., it
follows that $\bP^{< \alpha}$ is $\beth_\alpha^+$ c.c.\
From lemma~\ref{classforcing} it now follows that
$M[G]$ satisfies $\zfc$.
Clearly $(|\beth_{\alpha+1}| \leq (\beth_\alpha)^+)^{M[G]}$.
Thus, $|\beth_\alpha|^{M[G]} \leq \aleph_\alpha^{M[G]}$. First we show
that $|\beth_\alpha|^{M[G]} = \aleph_\alpha^{M[G]}$, and for this it suffices
to show that $\beth_{\alpha}^+$ is still a cardinal of $M[G]$.
First assume that $\alpha$ is a successor
and write $\bP= \bP^{< \alpha} \times \bP^{\geq \alpha}$ as above. Thus,
$\bP^{\geq \alpha}$ is $\leq \beth_\alpha$ closed and
$\bP^{< \alpha}$ is $\beth_\alpha^+$
c.c.\ If $\rho < \beth_\alpha^+$ and $f \colon \rho \to \beth_\alpha^+$,
then from lemma~\ref{lemsplit} it follows that $f \in M[G^{< \alpha}]$, where
$G= G^{< \alpha} \times G^{\geq \alpha}$.
Since $\bP^{< \alpha}$ is $\beth_\alpha^+$ c.c.,
this gives an onto $F \colon \beth_\alpha \to \beth_{\alpha}^+$ in $M$, a
contradiction.
Suppose next that $\alpha$ is limit. We consider cases as to
whether $\beth_\alpha$ is regular (i.e., inaccessible). Suppose
first that $\beth_\alpha$ is regular. Again write
$\bP= \bP^{< \alpha} \times \bP^{ \geq \alpha}$. $\bP^{\geq \alpha}$
is $\leq \beth_\alpha$ closed. From the Easton condition,
any $p \in \bP^{< \alpha}$ has domain bounded in $\alpha$.
This gives that $\bP^{< \alpha}$ is $\beth_\alpha^+$ c.c., since if
there were an antichain in $\bP^{< \alpha}$ of size $\beth_\alpha^+$
we could assume the domains of the conditions were constant, and a simple
computation would then show there are $< \beth_\alpha$ many conditions
in the antichain.
If $f \colon \beth_\alpha \to \beth_\alpha^+$ were onto and in $M[G]$,
lemma~\ref{lemsplit} would give a function $F \colon \beth_\alpha \to
\beth_\alpha^+$ in $M$ which was also onto, a contradiction.
Suppose then that $\beth_\alpha$ is singular, say
$\rho= \cof(\beth_\alpha) < \beth_\alpha$. Let $\{ \alpha _i \}_{i < \rho}$
be increasing cofinal in $\beth_{\alpha}$ with $\beth_{\alpha_0}
> \rho$. Let $D$ be those $p \in \bP$ such that there is a sequence
$\{ A_i^\gamma \}$ for $i < \rho$, $\gamma < \beth_{\alpha_i}$,
each $A_i$ a maximal antichain of
$\bP^{< \alpha_i}$, such that for all $i < \rho$,
$\gamma < \beth_{\alpha_i}$, and all
$q \in A_i^\gamma$ there is an ordinal $\beta$ such that
$(q, p^{\geq \alpha_i}) \forces \tau(\check{\gamma})=\check{\beta}$.
Iterating the argument of lemma~\ref{lemsplit} $\rho$ times
shows that $D$ is dense. Fix $p \in \bP \cap D$, and let
$A_i^\gamma$ be the corresponding antichains. From the
$A_i^\gamma$ we may construct in $M$ a set of size
$| \bigcup_{i,\gamma} A_i^\gamma |$ which contains the range of $f$.
Thus in $M$ we have a set of size $\beth_\alpha$ which
contains $\beth_{\alpha}^+$, a contradiction.
We now know that $\beth_\alpha$ has cardinality $\aleph_\alpha^{M[G]}$
in $M[G]$. To show the $\gch$ in $M[G]$ it is thus enough
to show that there are at most $\beth_{\alpha+1}$ many
subsets of $\beth(\alpha)$ in $M[G]$. Suppose first that
$\alpha$ is a successor. Write $\bP= \bP^{< \alpha} \times
\bP^{\geq \alpha}$. Every subset of $\beth_\alpha$ in $M[G]$ lies in
$M[G^{< \alpha}]$, so it is enough to count these. In this case
$\bP^{< \alpha}$ has size $\beth_\alpha$, so there are at most
$\beth_\alpha^{\beth_\alpha}=\beth_{\alpha+1}$
many nice names for subsets of $\beth_\alpha$.
Suppose next that $\beth_\alpha$ is innaccessible.
Again write $\bP= \bP^{< \alpha} \times
\bP^{\geq \alpha}$. Every subset of $\beth_\alpha$ in $M[G]$ again lies in
$M[G^{< \alpha}]$. From the Easton condition, $\bP_{< \alpha}$ has
size $\beth_\alpha$ in $M$. So again there are at most
$\beth_\alpha^{\beth_\alpha}= \beth_{\alpha+1}$ many nice
names for subsets of $\beth_\alpha$. Finally, suppose $\alpha$ is limit
and $\beth_\alpha$ is singular, say $\rho=\cof(\beth_\alpha)< \beth_\alpha$.
Let $\beta < \alpha$ be a successor with $\beth_\beta > \rho$.
Proceeding inductively, we may assume the $\gch$ holds in $M[G]$
below $\beth_\alpha$. Thus it suffices to show that
$((\beth_\alpha) ^\rho \leq \beth_{\alpha+1})^{M[G]}$.
Write $\bP= \bP^{< \beta} \times \bP^{\geq \beta}$. Every $f \in
(\beth_\alpha) ^\rho \cap M[G]$ lies in $M[G^{< \beta}]$. Since
$| \bP^{< \rho}| \leq \beth_\beta$, there are at most
$(\beth_\beta^{\beth_\alpha})^\rho \leq 2^{\beth_\alpha}= \beth_{\alpha+1}$
many nice names for functions $f \in (\beth_\alpha )^\rho$.
\end{proof}
\end{document}