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A forcing order has the countable chain condition when every antichain in a forcing order is countable. Such forcing preserves cardinals and cofinalities at least ${\aleph }_1$.

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.

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.

For each $\xi <\kappa$, choose a maximal antichain in a forcing order below $p$ deciding a witness $\eta >\xi$ in $\dot{C}$. Write the decided values as ${\eta }_{\xi ,q}$, one for each condition $q$ in that antichain. The $\kappa$-chain condition for forcing gives fewer than $\kappa$ possible values. Since $\kappa$ is regular in $M$, chooseThe possible-values lemma for chain-condition forcing ensuresIn $M$ take the club set of limit closure pointsIt is unbounded by repeatedly closing any starting ordinal under $g$ for countably many steps, and closed by taking increasing limits. For $\delta \in D$, the forced displayed property gives points of $\dot{C}$ arbitrarily high below $\delta$. Its forced closedness yields $\delta \in \dot{C}$. ThereforeThis proves the ground-model club containment lemma. The name on the right is $\dot{C}$, as printed in the PDF.

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

The forcing order $\mathbb{P}$ has the chain condition for forcing at $\kappa$ when every antichain in a forcing order has cardinality below $\kappa$:When this assertion is evaluated inside $M$, both the quantified subsets and their cardinalities are those of $M$.

Let $\lambda \ge \kappa$ be a cardinal number of $M$. If forcing collapsed $\lambda$, some condition would force a surjective function $\dot{f}:\mu \to \lambda$ for an ordinal $\mu <\lambda$. Replacing $\mu$ by its ground-model cardinality lets us assume that $\mu$ is a cardinal.

For each $\xi <\mu$, choose in $M$ a maximal antichain in a forcing order deciding $\dot{f}(\xi )$. The chain condition for forcing makes its size less than $\kappa$, so the set $B_{\xi }$ of possible values of $\dot{f}(\xi )$ has size below $\kappa$. This is the possible-values lemma for chain-condition forcing. Every interpreted range is contained in $B={\bigcup }_{\xi <\mu }{B}_{\xi }$. If $\mu <\kappa$, the fact that $\kappa$ is a regular cardinal gives $|B|<\kappa \le \lambda$; if $\kappa \le \mu <\lambda$, infinite cardinal arithmetic gives $|B|\le \mu <\lambda$. Either way $B$ cannot contain the range of a surjection onto $\lambda$, a contradiction. By cardinal preservation by chain-condition forcing, $\mathbb{P}$ preserves every cardinal at least $\kappa$.

Suppose $\mathbb{P}$ has the $\kappa$-chain condition and $\dot{f}$ is forced to be a function from an ordinal $\mu$ to an ordinal $\lambda$. For each $\xi <\mu$, a maximal antichain in a forcing order deciding $\dot{f}(\xi )$ has size below $\kappa$. Hence the ground model has a set $B_{\xi }\subseteq \lambda$ of size below $\kappa$ containing every possible value of $\dot{f}(\xi )$.