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If $\kappa$ is a regular cardinal and $\mathbb{P}$ has the $\kappa$-chain condition, then forcing with $\mathbb{P}$ preserves every cardinal number and cofinality at least $\kappa$. Indeed, a proposed surjection $f:\mu \to \lambda$ with $\mu <\lambda$ has its range contained in ${\bigcup }_{\xi <\mu }{B}_{\xi }$ from the possible-values lemma for chain-condition forcing. The regularity of $\kappa$ and infinite cardinal arithmetic make this union have cardinality below the ground-model cardinal $\lambda$, a contradiction.

With $\kappa$ an uncountable regular cardinal, finite partial functions on $\kappa \times \omega$ assigning a value below the first coordinate collapse every ground ordinal below $\kappa$ to countable size. The Delta-system lemma gives the $\kappa$ chain in a partial order condition; the possible-values lemma for chain-condition forcing then preserves the regularity of $\kappa$. Therefore $\kappa$ becomes the extension ${\omega }_1$. Strong inaccessibility is enough but is not needed for this identification.

Choose a name $\dot{f}\in M$ for $f$ and a condition $p\in G$ forcing that it is a function $\alpha \to \beta$. The ordinals $\alpha ,\beta$ belong to $M$, by forcing preserves ordinals. For each $\zeta <\alpha$, choose in $M$ a maximal forcing antichain $A_{\zeta }$ in the cone above $p$, every member deciding $\dot{f}(\zeta )$ as an ordinal below $\beta$. Conditions deciding an ordinal-valued name are dense, by the forcing theorem.

Let $y(\zeta )$ be the set of values decided by members of $A_{\zeta }$. The chain condition gives $|y(\zeta ){|}^M<\kappa$, and Choice and Replacement in $M$ assemble all these sets into a function $y\in M$ with domain $\alpha$. Since $p\in G$, genericity ensures that $G$ meets each deciding forcing antichain above $p$; equivalently enlarge it to a global maximal forcing antichain by conditions incompatible with $p$. Its chosen value is the actual $f(\zeta )$. ThereforeThis is the possible-values lemma for chain-condition forcing. All size bounds in its construction are internal to the ground model.

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.

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 $M$.

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 $\lambda \ge \kappa$ 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 $\kappa$.