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For each index, collect the ground subsets which some condition forces to be the corresponding diamond guess. Distinct forced values have incompatible witnessing conditions, so the countable chain condition for forcing makes this a countable family. Every ground subset is guessed by these families on a ground stationary set, by testing each ground club set. The countable-family diamond equivalence then yields diamond in the ground model. A generic guess need not itself be a ground subset; only values forced equal to one are collected.

Over a model of Generalized continuum hypothesis, add ${\aleph }_3$ Cohen reals with finite binary partial functions. The Delta-system lemma proves the countable chain condition for forcing. The generic reals give the continuum lower bound ${\aleph }_3$, while nice forcing names for subsets of ${\omega }_1$ number at most $({\aleph }_3{)}^{{\aleph }_1}={\aleph }_3$ in the ground model. Cardinal preservation then gives $2^{{\aleph }_0}=2^{{\aleph }_1}={\aleph }_3$.

The order $\mathrm{Fn}(X,Y,\omega )$ has the countable chain condition for forcing whenever $Y$ is countable. For an uncountable family of conditions, apply the Delta-system lemma to the finite domains. On their common finite root there are only countably many assignments, so thin to an uncountable family agreeing on that root. Any two conditions now have a compatible union. For binary Cohen forcing, there are only finitely many root assignments.

Every uncountable subset of a Knaster forcing has an uncountable pairwise compatible subset. The property implies the countable chain condition for forcing. Finite Cohen forcing has this property by the delta-system lemma and agreement on the finite root. A Knaster forcing times any CCC forcing is CCC

For a cardinal number $\kappa$, ${\mathrm{MA}}_{\kappa }$ says that for every forcing with the countable chain condition for forcing and every family of at most $\kappa$ dense subsets, some filter in an ordered set meets them all. Full Martin axiom asserts this for all $\kappa <2^{{\aleph }_0}$. The strict upper bound is part of the definition.

For a forcing order with the countable chain condition for forcing, a real can be represented by a nice name determined by a countable antichain in a forcing order for each natural-number coordinate. This bounds the number of reals in the extension in terms of the size of the forcing order.

Over a Generalized continuum hypothesis ground model, adding one Cohen real preserves every infinite power $2^{\lambda }={\lambda }^{+}$. For the countable forcing, nice subset forcing names for $\lambda$ number at most $(2^{{\aleph }_0}{)}^{\lambda }=2^{\lambda }$ in the ground model. Old subsets give the matching lower bound and the countable chain condition for forcing preserves cardinals. Starting from the constructible universe, the new real is nonconstructible while the constructible levels remain unchanged.

Forcing constructs new first-order models while controlling which ground-model properties survive. Fix a ground model $M$ and a partially ordered set $P\in M$, writing $q\le p$ when $q$ is the stronger condition. A generic filter is upward closed, directed towards stronger conditions, and meets every ground-model dense subset of a forcing order. If $M$ is externally a countable transitive model, enumerate its dense subsets and successively choose stronger conditions in them. The resulting filter is $M$-generic. External countability is crucial: this is not a claim that the ground model regards its own universe as countable.

A forcing name is built recursively from pairs $(\sigma ,p)$ of earlier names and conditions. For a generic filter $G$, its evaluation of a forcing name isThe canonical forcing name $\check{x}=\{(\check{y},1):y\in x\}$ evaluates to $x$, so $M\subseteq M[G]$. Recursive evaluation also proves transitivity. Name ranks bound the ranks of their evaluations, so an ordinal in the extension is below some ground ordinal and is already a ground ordinal. Thus forcing preserves ordinals.

The central link between syntax and generic interpretation is the forcing theorem. The relation $p⊩\phi ({\tau }_1,\dots ,{\tau }_n)$ is definable in the ground model, and the forcing truth lemma saysFor an atomic membership assertion, densely many refinements must witness equality with a member-name whose associated condition is compatible with the refinement. Equality is obtained by recursive mutual inclusion of names. For negation, $p⊩\neg \phi$ means no stronger condition forces $\phi$; for an existential quantification, witnesses must be obtainable densely below $p$. Induction on name rank establishes the atomic clauses, and induction on formulas establishes definability and the truth lemma. The generic filter meets the relevant dense sets at the existential step. An arbitrary semantic assertion about one chosen generic filter is insufficient: the definable relation must work uniformly for all generic interpretations.

This explains preservation of ZFC. Extensionality and foundation follow from transitivity; canonical names supply infinity. Pair and union names give pairing and union. For separation, restrict a name's members to conditions forcing the defining formula. For replacement, the definable forcing relation lets ground axiom schema of collection bound a set of names for all possible witnesses to the unique outputs; evaluating their collected name gives the range. For power set, let $\tau$ name the given set and let $D$ be the set of its member-names. Every subset of ${\tau }_G$ has an equivalent name drawn from $\mathcal{P}(D\times P)$, by the truth lemma, so a ground set of candidate names collects all its subsets. Finally choose a ground well-order of the relevant names; its evaluated image surjects onto the desired set, and choosing the first name for each value well-orders that set. This is why ordinary forcing over a choice model preserves choice.

For a concrete example, Cohen forcing consists of finite partial functions $\omega \to 2$, ordered by extension. The generic union $c=\bigcup G$ is total because deciding each coordinate is dense. For every ground real $x$, the set of conditions disagreeing with $x$ somewhere is dense, so $c\ne x$. This construction changes the universe by adding a genuinely new real without changing its ordinals.

The size and compatibility of conditions determine what else changes. With the countable chain condition for forcing, an ordinal-valued name has only countably many possible values at each coordinate: take a maximal antichain deciding that value. Therefore the possible range of a name for a map with infinite ground domain of cardinality $\mu$ lies in a ground set of size at most $\mu \cdot {\aleph }_0=\mu$. This prevents cardinal collapse and changes of uncountable regular cofinalities. Forcing by finite partial functions $\lambda \times \omega \to 2$ is ccc: the delta-system lemma thins any uncountable family of finite domains to a delta-system, and a further thinning makes their values agree on its finite root. Two remaining conditions are compatible. Distinct coordinates add distinct reals.

For example, start with GCH and take $\lambda ={\aleph }_2$. This forcing has size $\lambda$ and adds $\lambda$ distinct reals. A nice name for a real uses countably many antichains, and each antichain is countable. The ground arithmetic ${\lambda }^{{\aleph }_0}=\lambda$ bounds the number of such names by $\lambda$. Hencegiving a relative-consistency example of the failure of CH. The ground choice of GCH is available through the constructible universe; no claim that every starting model has this arithmetic is needed.

By contrast, countably closed forcing adds no new countable sequences of ground ordinals. Given a name, recursively strengthen a condition to decide each coordinate, then take a lower bound of the descending sequence. That condition decides the entire sequence. Closure and chain conditions thus control different features; neither should be silently substituted for the other.

Further applications include changing cofinalities, adding branches through trees, and constructing prescribed continuum patterns. If the aim is to violate choice, an ordinary generic extension of a choice model will not do: one restricts to a suitable symmetric collection of names, checking the axioms again. These examples also mark a limitation: forcing proves relative consistency, not absolute consistency of the background theory. The countable-transitive-model picture gives a concrete semantic construction; the definable forcing relation and Boolean-valued formulation supply the corresponding formal consistency arguments without asserting that a transitive ground model can be proved to exist.

Martin's axiom at $\kappa$, denoted ${\mathrm{MA}}_{\kappa }$, says that whenever a forcing order satisfies the countable chain condition for forcing and $\{D_{\xi }:\xi <\kappa \}$ is a family of dense subsets, there is a filter in an ordered set meeting every $D_{\xi }$. Equivalently, allow any family of at most $\kappa$ dense subsets. The filter in an ordered set is directed towards stronger common extensions and closed towards weaker conditions. Full Martin's axiom requires ${\mathrm{MA}}_{\kappa }$ for every $\kappa <2^{{\aleph }_0}$. The bound below the continuum is part of the usual full axiom, explaining its role in part (iv)(b).

Prune an ${\aleph }_1$-Suslin tree as in part (a), then use its nodes as forcing conditions, with extensions stronger. Two conditions are compatible exactly when comparable, so the absence of uncountable tree antichains is the forcing countable chain condition for forcing. For each $\alpha <{\omega }_1$, the set $D_{\alpha }$ of nodes of height at least $\alpha$ is dense, by well-pruned set-theoretic tree.

If $2^{{\aleph }_0}>{\aleph }_1$, full Martin's axiom includes ${\mathrm{MA}}_{{\aleph }_1}$. It would provide a filter in an ordered set meeting all these dense subsets of a forcing order. Directedness makes that filter in an ordered set a chain in a partial order, and meeting every $D_{\alpha }$ makes its heights unbounded, producing an uncountable branch. This contradicts the Suslin-tree property. A Suslin set-theoretic tree together with failure of Continuum hypothesis therefore implies failure of Martin's axiom. This is the Suslin-tree obstruction to Martin's axiom.

Use the standard normal set-theoretic tree convention: a unique root, extensions at every higher level, splitting into at least two successors, and tree with unique limits. The small-level and height assumptions already give an ${\omega }_1$-tree, while the given tree antichain condition gives the countable chain condition for forcing. We only need to exclude an uncountable branch.

If such a branch existed, its heights would be unbounded, since each initial segment contains only countably many nodes. Fill in predecessors to obtain its node $b_{\alpha }$ at every level. At each successor step choose a successor $s_{\alpha }$ of $b_{\alpha }$ different from $b_{\alpha +1}$. For $\alpha <\beta$, the node $s_{\beta }$ extends the branch successor $b_{\alpha +1}$, and so is incompatible with $s_{\alpha }$. Thus $\{s_{\alpha }:\alpha <{\omega }_1\}$ is an uncountable tree antichain, a contradiction.

Therefore the set-theoretic tree is ${\aleph }_1$-Suslin. The splitting part of normality matters: a single chain in a partial order of height ${\omega }_1$ would satisfy the tree antichain condition but not the conclusion if one used a weakened definition of normality allowing no splitting.

Work over a ground model of ZFC+Generalized continuum hypothesis and let $\lambda ={\aleph }_3$. Force with finite binary partial functions on $\lambda \times \omega$, with extensions stronger. The Delta-system lemma thins any uncountable family of finite domains to an uncountable family with one common root. Only finitely many binary assignments on that root occur, so two conditions agree there and their union is a common extension. Hence the forcing has the countable chain condition for forcing and preserves cardinals and cofinalities.

The generic union yields $\lambda$ distinct reals. Totality at each coordinate is dense, and for two different coordinates it is dense to assign different values at some unused natural-number position. Thus the extension satisfies $2^{{\aleph }_0}\ge \lambda$.

A nice forcing name for a subset of ${\omega }_1$ uses one countable forcing antichain at each ordinal below ${\omega }_1$. Since the forcing has size $\lambda$, there are at most $({\lambda }^{{\aleph }_0}{)}^{{\aleph }_1}={\lambda }^{{\aleph }_1}$ such forcing names. Ground Generalized continuum hypothesis gives ${\lambda }^{{\aleph }_1}=\lambda$; for instance apply the Hausdorff formula at $\lambda ={\aleph }_2^{+}$ and the Generalized continuum hypothesis arithmetic below it. Therefore $2^{{\aleph }_1}\le \lambda$ in the extension. Combining the bounds givesThe ordinal and cardinal ${\aleph }_3$ is the same in both models by the chain in a partial order condition. The forcing theorem formalizes this construction as the requested relative-consistency implication. A countable transitive ground model is a convenient presentation, not an additional consequence silently derived from mere consistency. This is the Cohen forcing two-level continuum plateau.

Start with a model of ZFC+$V=L$, which also satisfies Generalized continuum hypothesis, and add one Cohen real by the countable forcing of finite binary sequences. Its countable chain condition for forcing preserves cardinals. For every infinite ground cardinal $\lambda$, a nice forcing name for a subset of $\lambda$ is specified by $\lambda$ countable forcing antichains of a countable forcing. The number of such forcing names is at mostin the ground model. The ground subsets already supply the preserved lower bound ${\lambda }^{+}$. Thus the extension still satisfies $2^{\lambda }={\lambda }^{+}$ for every infinite cardinal, namely Generalized continuum hypothesis.

The Cohen real is not in the ground model: for each ground real it is dense to disagree at a new coordinate. Forcing leaves the ordinals unchanged, and constructible levels are absolute, so the extension has the same constructible universe as the ground model. The new real is therefore not constructible. We obtainThe forcing theorem makes this a relative-consistency construction. Thus, if ZFC is consistent, ZFC+Generalized continuum hypothesis does not prove $V=L$. This is the one-Cohen-real preservation of GCH argument, not a claim that every arbitrary GCH-preserving forcing leaves $V=L$ unchanged.

For an arbitrary ground model, this alternative needs an additional arithmetic hypothesis. The requested countable-cover property already prevents any collapse of an infinite ground-model cardinal. If a cardinal $\theta$ were collapsed, there would be a surjection $f:\alpha \to \theta$ with $|\alpha {|}^M<\theta$. The asserted ground-model $F$ would cover $\theta$ by the union of $|\alpha {|}^M$ countable sets, a ground-model set of size at most $\max ({\aleph }_0,|\alpha {|}^M)<\theta$, a contradiction. Thus all infinite cardinals are preserved.

Put $\lambda =({\aleph }_3{)}^M$, which must then remain ${\aleph }_3$. If the extension has continuum $\lambda$, it hasThe ground-model functions from $\omega$ into $\lambda$ remain present and their cardinality cannot be collapsed. NecessarilyFor example, a ground model with continuum larger than ${\aleph }_3$ cannot satisfy the printed request. This is a genuine missing hypothesis, rather than a forcing construction that works for every $M$.

Under the necessary hypothesis, use Cohen forcing to add $\lambda$ reals:The delta-system lemma shows that it has the countable chain condition for forcing: an uncountable family of finite conditions has an uncountable subfamily whose domains form a delta-system and whose values agree on its root, so any two of that subfamily are compatible. It therefore preserves cardinals. Its $\lambda$ coordinate reals are pairwise distinct by dense disagreement requirements. Conversely each nice forcing name for a real uses countably many countable antichains in a forcing order, so the number of such names is at most ${\lambda }^{{\aleph }_0}=\lambda$. HenceFor any ordinal-valued function name, choose for each $\xi <\alpha$ a maximal antichain in a forcing order deciding its value. The countable chain condition for forcing makes the set $F(\xi )$ of possible ordinal values countable in $M$. Pad it with $\omega$ if needed to make it countably infinite. The possible-values lemma for chain-condition forcing gives the stronger pointwise covering statementwhich implies the requested range inclusion. When a condition only forces that the name is such a function, make these choices below that condition; it belongs to the generic filter witnessing the actual function.

Countability of the forcing and uncountability of $Y$ are understood internally in $M$ and $M[G]$, respectively. This matters because the model itself is externally countable. Choose a name $\dot{Y}\in M$ and a ground-model set $A\in M$ containing every ground-model element that can occur in $Y$. Such an $A$ exists: the rank of $\dot{Y}$ bounds the ranks of its values, so a sufficiently high $V_{\theta }^M$ contains $Y\subseteq M$.

For $p\in P$, ground-model axiom schema of separation and the forcing definability lemma giveBy the forcing truth lemma,If every $X_p$ for $p\in G$ were countable in $M$, it would remain countable in $M[G]$, where $G\subseteq P$ is countable. Then $Y$ would be countable, contrary to the hypothesis. Thus some $p\in G$ has $X_p$ uncountable in $M$; otherwise its ground-model enumeration would still enumerate it in the extension. Let $X=X_p$. Every one of its members is forced by $p$ into $Y$, soThe countable chain condition for forcing also preserves its uncountability. Separativity is not needed for this particular argument. This is the ground-model uncountable subset lemma for countable forcing.

Use standard notation for forcing: $q\le p$ means that $q$ is stronger than $p$. A closed forcing order is $\lambda$-closed when every decreasing sequence $⟨p_{\xi }:\xi <\delta ⟩$ with $\delta <\lambda$ has a common lower bound:The chain condition for forcing isAn antichain in a forcing order consists of pairwise incompatible forcing conditions; incompatibility means having no common stronger extension. In particular, the ${\aleph }_1$-chain condition is the countable chain condition for forcing. These properties and their sequences or antichains are computed inside the ground model $M$.

Inside $M$, the finite-function order has the countable chain condition for finite-function forcing. In an uncountable family of binary conditions there are uncountably many distinct finite domains, since each domain supports only finitely many conditions. These domains would have an uncountable Delta-system subfamily by the Delta-system lemma. There are only finitely many assignments to the common finite root, so thin further to an uncountable family agreeing there. Any two such conditions have a union that is again a function and is a common stronger extension. Thus no uncountable antichain in a forcing order exists.

Here is the relevant preservation argument, rather than an appeal to the condition alone. Fix $p\in G$ forcing that a forcing name $\dot{h}$ is a function $\omega \to {\alpha }_1$. For each $n$, choose inside $M$ a maximal antichain in a forcing order below $p$ deciding $\dot{h}(n)$. By the countable chain condition for forcing, this antichain is countable in $M$, so the possible values form a countable $B_n\subseteq {\alpha }_1$. The union $B={\bigcup }_n{B}_n$ is countable in $M$, hence bounded below its regular ${\omega }_1={\alpha }_1$. The interpreted $h$ has range contained in $B$, by maximality and the dense-below generic meeting lemma. It therefore cannot be a surjection onto ${\alpha }_1$.

If ${\alpha }_1$ were not a cardinal number in $M[G]$, it would be equinumerous with a smaller ordinal, which was countable already in $M$; composing with that ground-model enumeration would give a surjection $\omega \to {\alpha }_1$. This is impossible. ThereforeThis is the possible-values lemma for chain-condition forcing specialized to ${\omega }_1$. Countability and regularity in this proof are computed inside $M$, not inferred from the external countability of $M$.

The countable chain condition for forcing says that every antichain in a forcing order is a countable set. Equivalently,In this question the assertion is evaluated inside $M$. Although every subset of the countable set $M$ is externally countable, that observation does not establish the internal chain condition: $M$ need not contain an enumeration of an antichain.

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