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Use standard notation for forcing: $q\le p$ means that $q$ is stronger. A generic filter is nonempty, upward closed and downward directed, and meets every dense subset of a forcing order in the ground model. Names below are forcing names in $M$, with pairs ordered as $(\text{name},\text{condition})$.

The semantic forcing relation isHere ${\tau }^K=\mathrm{val}(\tau ,K)$ is the evaluation of a forcing name. Countability of $M$ supplies such generic filters through every condition by the Rasiowa–Sikorski lemma.

Define the syntactic forcing relation by mutual well-founded recursion on the forcing names for the atomic cases, then induction on the first-order formula. Its membership clause isFor equality, first abbreviateand set $p{⊩}^{\ast }\sigma =\tau$ exactly when both inclusions hold. Each recursive call lowers the rank of at least one forcing name without raising the other. This makes the mutual recursion well-founded.

For a basis of connectives consisting of logical conjunction, negation and existential quantification, the remaining clauses areThe last line is the existential clause of syntactic forcing; witnesses need only occur densely, rather than be forced by $p$ itself with one preselected name. Other connectives are defined by logical abbreviations. The recursion gives a definable relation inside $M$ for each fixed first-order formula, with quantification over the class of its names. If $\mathbb{P}$ has no greatest element, use canonical forcing names $\check{a}=\{(\check{b},p):b\in a,p\in \mathbb{P}\}$, which still evaluate to $a$. It also proves monotonicity: strengthening a condition preserves what it forces. The forcing theorem identifies the two relations and supplies the truth lemma.

Let $D\in M$ be a dense subset of a forcing order $\mathbb{P}$. Then $D\times \mathbb{Q}\in M$ is dense in the product forcing order: below $(p,q)$, choose $p^{\prime }\le p$ in $D$ and retain $q$. The generic filter $H$ meets it, so its first projection $G_0$ meets $D$. Together with the filter properties proved above, this givesThe same argument proves that $G_1$ is $\mathbb{Q}$-generic over $M$, but genericity over the larger generic extension $M[G_0]$ needs the next argument.

We prove mutual genericity for product forcing. Let $D\in M[G_0]$ be a dense subset of $\mathbb{Q}$, and take a forcing name $\dot{D}\in M$ evaluating to $D$. By the forcing theorem, some $p_0\in G_0$ forces that $\dot{D}$ is a dense subset of the canonical forcing name for $\mathbb{Q}$.

In $M$ defineThis set is dense in the product forcing order. Given $(p,q)$, the incompatible case is immediate. Otherwise first strengthen $p$ below $p_0$. The forced density assertion and the existential clause of syntactic forcing supply a further $r\le p,p_0$ and a ground-model $q^{\prime }\le q$ with $r⊩{\check{q}}^{\prime }\in \dot{D}$. To justify choosing a ground-model $q^{\prime }$, a name forced to lie in $\check{\mathbb{Q}}$ can densely be made equal to some ${\check{q}}^{\prime }$ by the atomic membership clause; strengthen to that equality and use the forced order comparison. Thus $(r,q^{\prime })\in E$ lies below $(p,q)$.

The generic filter $H$ meets $E$. It cannot meet the first part, because its first projection contains $p_0$ and is directed. Hence there is $(p,q)\in H$ with $p\in G_0$ and $p⊩\check{q}\in \dot{D}$. Soundness of the forcing theorem gives $q\in D$, and the projection gives $q\in G_1$. Since every such $D$ is met,This establishes the stronger property, rather than merely meeting dense ground-model subsets.

Suppose, towards a contradiction, that $G\in M$. Then $D=\mathbb{P}\setminus G$ is a ground-model set by axiom schema of separation. It is dense. If $p\notin G$, it already lies in $D$. If $p\in G$, atomless forcing order structure gives incompatible forcing conditions $q,r\le p$. Both cannot belong to the directed filter $G$, so at least one is a strengthening of $p$ in $D$.

All compatibility and extension quantifiers range over the same ground-model set $\mathbb{P}$, so this density argument is also valid internally in the transitive model $M$. Genericity requires $G\cap D\ne \varnothing$, contradicting the definition of $D$. Therefore a generic filter for an atomless order is new:The assumption is essential: a forcing atom determines a ground-model generic filter. For an atom $p$, the conditions compatible with $p$ form a generic filter $G_p\in M$: two such conditions have strengthenings below $p$, which have a common strengthening by the atom property; every dense set has a member below $p$. In a general partial order, $G_p$ need not be the principal filter above $p$.

Put $\kappa ={\aleph }_2^M$ and let $F=\bigcup G$. Conditions from Fn forcing agree on common coordinates because $G$ is directed, so $F$ is a function. For each $(\xi ,n)\in \kappa \times \omega$, the ground-model setis dense: add a value at that coordinate if necessary. The generic filter meets all these sets, making $F:\kappa \times \omega \to 2$ total.

For distinct $\xi ,\eta <\kappa$, the ground-model setis dense. Choose $n$ fresh for both rows of a finite condition and assign $0$ and $1$ there. Genericity therefore makes the rows pairwise different. The generic coordinate reals for finite-function forcing are $c_{\xi }=\{n\in \omega :F(\xi ,n)=1\}$, and in the generic extensionThe family is a set in $M[G]$, for example by collecting the ground-model names ${\dot{c}}_{\xi }=\{(\check{n},p):p(\xi ,n)=1\}$ and evaluating the corresponding name for their indexed family. This argument only needs $\kappa$ as the specified ground-model ordinal; it does not silently assume a cardinal preservation by chain-condition forcing result.

The Rasiowa–Sikorski lemma constructs a generic filter $G_i$ over each $M_i$. Every resulting generic extension remains a countable transitive model of ZFC: there are only countably many ground-model names externally. The order $\mathbb{P}$ has the same elements and ordering at every stage, and atomless forcing order structure is absolute because all its quantifiers are bounded to $\mathbb{P}$. Hence the generic filter for an atomless order is new result gives $G_i\notin M_i$ for every $i$.

The increasing union $N={\bigcup }_{i<\omega }{M}_i$ is transitive and contains $\mathbb{P}$. If it satisfied Axiom of power set for $\mathbb{P}$, there would be a set $A\in N$ withChoose $i$ with $A\in M_i$. The next-stage generic filter $G_i$ belongs to $M_{i+1}\subseteq N$ and is an actual subset of $\mathbb{P}$. This subset assertion is absolute for the transitive set $N$, so $G_i\in A$. Transitivity of $M_i$ and $A\in M_i$ then give $G_i\in M_i$, a contradiction. Thus the power-set failure in an increasing union of generic extensions occurs already at the fixed ground-model order:The stages form an increasing chain, not an elementary chain, so the elementary chain theorem does not assert ZFC for their union.

We give a relative-consistency argument using a finite-function collapse to countable size. The assumed consistency implies that ZFC is consistent. Passing to the constructible universe gives consistency of $\mathsf{ZFC}+V=L$, which also satisfies the Generalized continuum hypothesis. Work in this ground theory, let $\delta ={\omega }_1^L$, and force with finite partial functions $\omega \rightharpoonup \delta$, ordered by reverse inclusion.

The union of a generic filter $H$ is a total surjection $h:\omega \to \delta$: prescribing a new domain coordinate is dense, and putting any specified $\gamma <\delta$ into the range is dense. Hence $\delta$ becomes countable. Set forcing preserves ordinals, and absoluteness of constructible levels implies that the extension has the same constructible universe as the ground model. Its ${\omega }_1^L$ is therefore still the old $\delta$, so it satisfies the stated countability of constructible omega-one condition.

We must also verify GCH after the collapse. The order has ground-model size $\delta$, hence the ${\delta }^{+}$-chain condition. By cardinal preservation by chain-condition forcing, all old cardinals at least ${\delta }^{+}$ survive. Every old ordinal below ${\delta }^{+}$ has size at most $\delta$ and becomes countable, so the old ${\delta }^{+}$ is exactly the new ${\omega }_1$.

A name for a subset of a fixed ground-model ordinal $\lambda$ can be chosen as a set of pairs $(\check{\xi },p)$ with $\xi <\lambda$ and $p$ a condition: use an antichain deciding membership at each coordinate. Thus the number of such names in the ground model is at most $2^{|\lambda \times \mathbb{P}|}$. For subsets of $\omega$ this is $2^{\delta }={\delta }^{+}$ by ground-model GCH. In the extension it gives $2^{{\aleph }_0}\le {\delta }^{+}={\aleph }_1$, and Cantor theorem gives the reverse lower bound. For every old cardinal $\lambda \ge {\delta }^{+}$ the bound isin the ground model. Both $\lambda$ and ${\lambda }^{+}$ remain cardinals, so again the upper bound and Cantor theorem give the extension's equality $2^{\lambda }={\lambda }^{+}$. These are all its uncountable cardinals. This proves GCH preservation by a finite-function collapse in the required case.

The forcing relative-consistency theorem now yieldsThe construction in fact only needs consistency of ZFC. This argument uses the formal inner-model and forcing consistency theorems; it does not infer the existence of a countable transitive model merely from consistency.

For the reverse implication, suppose $p\in G$ and $M\models p{⊩}^{\ast }\exists x\phi (x)$. By the existential clause of syntactic forcing, the setis dense below $p$; the existential quantifier ranges over forcing names of $M$. Definability of the syntactic forcing relation and axiom schema of separation put $D$ in $M$.

To apply genericity, augment it to the globally dense setA condition compatible with $p$ has an extension below $p$, then one in $D$, and an incompatible condition already lies in $D^{\prime }$. Thus $D^{\prime }$ is a dense subset of a forcing order in $M$. The generic filter meets it. Since $p\in G$, directedness prevents $G$ from containing a condition incompatible with $p$, so some $q\in G\cap D$ exists.

There is consequently a name $\sigma \in M$ with $M\models q{⊩}^{\ast }\phi (\sigma )$. The assumed truth lemma gives $M[G]\models \phi (\mathrm{val}(\sigma ,G))$, and henceTogether with part (a), this completes the existential step of the forcing theorem without assuming the desired existential truth lemma in advance.

Use standard notation for forcing, so $q\le p$ means that $q$ is stronger. Let $\check{p}$ be the canonical forcing name for the ground-model condition $p$. DefineHere $q\perp p$ means that they are incompatible forcing conditions. The checks and their collection are formed by recursion and Axiom schema of replacement in $M$, and the displayed set is selected by axiom schema of separation, so this is a forcing name in $M$.

By evaluation of a forcing name,If $p\in G$, directedness of the generic filter gives a common stronger condition for $p$ and each $q\in G$, so $p$ is absent from this value.

Conversely, for fixed $p\in \mathbb{P}$ the setis a dense subset of a forcing order belonging to $M$. A condition incompatible with $p$ is already in it, and one compatible with $p$ has a common extension below $p$. Genericity supplies $q\in G\cap D_p$. If $p\notin G$, upward closure of $G$ rules out $q\le p$, hence $q\perp p$. ThereforeThis forcing name for the complement of a generic filter works without a separativity assumption on the order.

Assume $M$ satisfies the countable chain condition for forcing. We prove cardinal preservation by chain-condition forcing using the possible-values lemma for chain-condition forcing, and give the argument explicitly for ccc.

Let $\kappa$ be an uncountable cardinal of $M$. If it ceased to be a cardinal, some ordinal $\mu <\kappa$ would admit a surjection onto $\kappa$ in an extension. By the forcing theorem, there would be a condition $p$ and a forcing name $\dot{f}$ such thatInside $M$, for each $\xi <\mu$ choose a maximal antichain in a forcing order below $p$ deciding the ordinal value of $\dot{f}(\xi )$. Conditions deciding that value are dense below $p$. The chain condition makes the chosen antichain countable in $M$, so the corresponding set $B_{\xi }\subseteq \kappa$ of possible decided values is countable in $M$.

A generic filter containing $p$ meets the downward closure of every such maximal antichain, so its value $f(\xi )$ belongs to $B_{\xi }$. Therefore its whole range lies in the ground-model setThe last inequality is infinite cardinal arithmetic and uses only that $\kappa$ is uncountable and $\mu <\kappa$; no regularity of $\kappa$ is needed. Some ordinal in $\kappa \setminus B$ exists already in $M$ and cannot occur in the range, contradicting the forced surjectivity.

Finite cardinals cannot collapse, and $\omega$ cannot become finite: a finite domain has finite image, and transitive models agree on the natural numbers. Old non-cardinals cannot become cardinals because their old bijections persist. Thus

Let $s_p$ be the finite stem of a condition $p=(n_p,s_p,A_p)$, and defineThe stems of two conditions in the generic filter agree on their common domain, because they have a common stronger extension. Hence this union is a function. For every $\ell \in \omega$, the set of conditions with $n_p\ge \ell$ is dense: append values from the nonempty infinite reservoir until the desired length is reached. Genericity therefore makes $f$ total on $\omega$. The generic stems, and thus their union, are available in $M[G]$.

Fix $g\in M\cap {\omega }^{\omega }$ and $K\in \omega$. In $M$ form the setWe show that it is a dense subset of a forcing order. Given $p=(n,s,A)$, put $k=\max (n,K)$. Fill the new positions $n,\dots ,k-1$ with any fixed element of $A$, and choose $a\in A$ with $a>g(k)$ for the new position $k$. This is possible because an infinite subset of $\omega$ is unbounded. Let $t$ be the resulting stem of length $k+1$, and keep the reservoir $A$ unchanged. ThenEvery new stem value came from the old reservoir, exactly as required by the extension relation. The construction is performed in $M$, so $D_{g,K}\in M$ and is internally dense.

Genericity gives a condition in $G\cap D_{g,K}$. Its witnessing coordinate remains fixed in every later stem and hence in $f$, so some $k\ge K$ satisfies $g(k)<f(k)$. Since this holds for every $K$, there are infinitely many such $k$. As $g$ was arbitrary,Thus this infinite-reservoir stem forcing produces an unbounded real over a model, which is precisely the stipulated meaning of bounding $M$.

A set $G\subseteq \mathbb{P}$ is a filter when it is upward closed toward weaker conditions and downward directed: if $p\in G$ and $p\le q$, then $q\in G$, while any $p,q\in G$ have some $r\in G$ with $r\le p,q$. It is a generic filter over $M$ when

Suppose first that $p{⊩}^{\ast }\phi$. If $G$ is any generic filter over $M$ containing $p$, the forcing theorem gives $M[G]\models \phi$. Hence $p⊩\phi$ in the semantic forcing relation.

Conversely, suppose $p/{⊩}^{\ast }\phi$. By the stated density equivalence for the syntactic forcing relation, some $r\le p$ satisfies $r{⊩}^{\ast }\neg \phi$. The Rasiowa–Sikorski lemma supplies a generic filter $G$ over the countable model $M$ containing $r$, and upward closure then gives $p\in G$. The forcing theorem yields $M[G]\models \neg \phi$, so $p/⊩\phi$ semantically. Therefore

Let $c=\bigcup G$ and $d(n)=c(2n)$. DefineThis is a filter: restrictions of finite pieces of $d$ remain in $H$, and the union of two members is a common stronger condition.

To prove genericity, take a dense set $D\in M$. Let $E_D$ consist of conditions $q\in \mathrm{Fn}(\omega ,2)$ for which some $p\in D$ satisfiesThe set $E_D$ is dense. Indeed, given $q$, first read its finitely many assigned even coordinates as a condition $p_0$ on $\omega$. Choose $p\le p_0$ in $D$, and extend $q$ by setting $q(2n)=p(n)$ at the remaining coordinates of $p$. Since $E_D\in M$, the generic filter $G$ meets it. For $q\in G\cap E_D$, the corresponding $p\in D$ is a finite subfunction of $d$, so $p\in H\cap D$.

Thus $H$ meets every dense subset belonging to $M$. Moreover, each singleton $\{(n,d(n))\}$ belongs to $H$, and hence

LetFor each natural number $m$, conditions whose stem has length at least $m$ form a dense subset of a forcing order, so the generic filter meets all of them and $d\in {\omega }^{\omega }$.

Fix $h\in {\omega }^{\omega }\cap M$. The setis dense: from $(s,f)$ replace $f$ by $n\mapsto \max \{f(n),h(n)+1\}$. Choose $(s,f)\in G\cap D_h$. Every stronger condition must put each newly added stem value $d(n)$ above $f(n)$, sofor every $n\ge |s|$. Thus $d$ is a dominating real over $M$.

With the convention that $p\le q$ means that $p$ is stronger, a set $G\subseteq \mathbb{P}$ is a generic filter over $M$ when it is a filter and meets every dense set $D\subseteq \mathbb{P}$ with $D\in M$. Explicitly, $G$ is upward closed toward weaker conditions, every two members have a common stronger member in $G$, and $G\cap D\ne \varnothing$ for every such $D$.

Let $M_{n+1}=M_n[G_n]$ for a fixed atomless forcing order $\mathbb{P}$, and $N={\bigcup }_{n<\omega }{M}_n$. If a set $A\in N$ were its internal power set of $\mathbb{P}$, choose $n$ with $A\in M_n$. The next generic filter $G_n$ belongs to $N$, so $G_n\in A$; transitivity of $M_n$ implies $G_n\in M_n$, contradicting generic filter for an atomless order is new. The chain is increasing but need not be an elementary chain.

The principal filter in an ordered set generated by $p$ contains exactly the elements above $p$. It is upward closed, and $p$ is a common lower bound in the filter for any two of its members. With standard notation for forcing, these are the conditions weaker than $p$. The principal filter need not be a generic filter: a dense set can require a strict strengthening. The compatible-condition filter in a forcing atom determines a ground-model generic filter includes strengthenings as well as weakenings and can therefore be strictly larger.