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A forcing order preserves cardinals over a ground model if every ground-model cardinal number remains a cardinal in every generic extension. This assertion and any sufficient chain condition for forcing are evaluated inside the ground model. External countability of a countable transitive model does not make all of its forcing orders internally ccc.

Let $\kappa =({\omega }_2{)}^M$ be the fixed ground ordinal. We prove that $S$ remains stationary in this ordinal. The chain condition for forcing ensures that $\kappa$ remains regular: possible values of each coordinate of an ordinal-valued name form a ground set of size less than $\kappa$, by a maximal deciding forcing antichain. A hypothetical cofinal map with domain below $\kappa$ would have its range covered by fewer than $\kappa$ such small sets, hence bounded by regularity in $M$.

Let $\dot{C}$ name a club set in $\kappa$, and take $p\in G$ forcing this. For each $\alpha <\kappa$, choose in $M$ a maximal forcing antichain above $p$ deciding the least point of $\dot{C}$ strictly above $\alpha$. Its set of possible values has size less than $\kappa$, so choose a ground bound $h(\alpha )<\kappa$ larger than all of them. The ground setis a club set by the usual countable closure iteration and regularity. For every $\delta \in D$, $p$ forces $\dot{C}\cap \delta$ unbounded in $\delta$, and closedness forces $\delta \in \dot{C}$. Hence $p⊩\check{D}\subseteq \dot{C}$. In $M$, choose $\delta \in S\cap D$; that same ordinal belongs to $S\cap C$ in the extension. Stationarity at the fixed ground $\kappa$ is preserved.

There is an important qualification to the printed ${\omega }_2$ notation. If it is recomputed internally as $({\omega }_2{)}^{M[G]}$, the assertion is false without preservation of smaller cardinals. Finite partial maps from $\omega$ to $({\omega }_1{)}^M$ form a forcing of size $({\aleph }_1{)}^M$, hence have the $({\aleph }_2{)}^M$-chain condition, but collapse $({\omega }_1{)}^M$ to countable. Then $({\omega }_2{)}^M=({\omega }_1{)}^{M[G]}$, while $({\omega }_2{)}^{M[G]}$ is larger. The old $S$ is bounded in this new ${\omega }_2$, so cannot be stationary there. The proved statement uses the fixed ground ordinal, or alternatively requires the lower-cardinal preservation needed to retain its aleph index.

Use the countable-condition collapseordered by reverse inclusion: an extension of a function is a stronger condition. All sizes and conditions here are computed in the ground model $M$.

This forcing is countably closed: the union of a descending countable sequence is a countable partial function. Thus it adds no countable ordinal sequences and preserves ${\omega }_1$. For each $\xi <{\omega }_1$, the conditions whose domains contain $\xi$ are dense; for each $\eta <\kappa$, those whose ranges contain $\eta$ are dense. The union of the generic filter is therefore a surjection from ${\omega }_1$ onto $\kappa$, and $|\kappa {|}^{M[G]}={\aleph }_1$.

Inaccessibility gives ${\kappa }^{{\aleph }_0}=\kappa$: every countable sequence is bounded below the regular $\kappa$, and the strong limit cardinal property bounds the number of sequences at each bound below $\kappa$. Hence $|\mathbb{P}|=\kappa$. The forcing satisfies the ${\kappa }^{+}$-chain condition for forcing, so it preserves every cardinal above $\kappa$. This gives precisely the requested collapse and preservation. The assertion that $\kappa$ is collapsed to ${\aleph }_1$ concerns its new cardinality; the ordinal $\kappa$ itself does not change.

Fix $p$ and a name $\dot{f}$ such that $p$ forces a two-coloring of $[\lambda {]}^{<\omega }$. For each ground-model finite set $a\subseteq \lambda$, record its full forcing decision patternThere are at most $3^{|\mathbb{P}|}<\lambda$ such patterns, since $\lambda$ is a strong limit cardinal and $|\mathbb{P}|<\lambda$. For finite $\mathbb{P}$ the number of patterns is finite, also below $\lambda$. Encode the patterns by an ordinal $\beta <\lambda$ and apply the assumed finite-subset partition property to obtain $H\subseteq \lambda$ of size $\lambda$ such that $F$ is constant on $[H{]}^n$ for every $n<\omega$.

For each $n$, choose one $a_n\in [H{]}^n$. Conditions below $p$ deciding $\dot{f}(a_n)$ are dense. Whenever one such condition decides the value, equality of patterns means it forces that same value at every $a\in [H{]}^n$. A generic filter containing $p$ meets this dense set for each $n$, so $f$ is constant on each $[H{]}^n$ in the extension. The constants may differ with $n$, exactly as required.

The forcing has size below the regular $\lambda$, hence satisfies the $\lambda$-chain condition for forcing and preserves its cardinality. The ground-model set $H$ consequently still has size $\lambda$. Since the argument works for every name and condition,This is small forcing preservation of finite-subset partition properties; no closure assumption on the forcing is required.

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 $M$.

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 $\kappa$.

If $M$ regards $\mathbb{P}$ as satisfying the $\kappa$-chain condition, then forcing with $\mathbb{P}$ preserves every cardinal and cofinality at least $\kappa$. If $M$ regards $|\mathbb{P}|$ as $\mu$, then every antichain has size at most $\mu$, so $\mathbb{P}$ has the ${\mu }^{+}$-chain condition. It follows that every cardinal and cofinality at least ${\mu }^{+}$ is preserved in a $\mathbb{P}$-generic extension.