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\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}











