\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}