Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 19 / 4

A forcing name for $\mathbb{P}$ is a set of ordered pairs $(\sigma ,p)$ where $p\in P$ and $\sigma$ is itself a forcing name. This recursive definition is made well-founded by assigning the forcing name rank $\sup \{\mathrm{nrk}(\sigma )+1:(\sigma ,p)\in \tau \}$. Names over the ground model are those names belonging to $M$; conditions remain ground-model objects. A name describes which recursively interpreted elements are activated by the generic filter.

With this paper's order convention, larger conditions are stronger. A generic filter $G\subseteq P$ over $M$ is nonempty, closed toward weaker conditions, and directed toward stronger conditions: if $p,q\in G$, some $r\in G$ satisfies $r\ge p,q$. It also meets every $D\in M$ which $M$ regards as dense in $\mathbb{P}$.

For a countable transitive $M$, such a filter containing any prescribed condition can be built by enumerating its dense sets and successively choosing stronger conditions in them. The definition does not require $G\notin M$: for an atomic forcing a generic filter may already belong to $M$.

A subset $E\subseteq P$ is dense above a forcing condition $p$ ifAll displayed strengthenings use the PDF convention. This only requires density in the cone of extensions of $p$; it need not be dense throughout $P$, nor downward or upward closed.

The generic extension is $M[G]=\{{\tau }^G:\tau \in M\text{ is a forcing name}\}$. It is transitive: if $x\in {\tau }^G$, an active pair in $\tau$ supplies a subname $\sigma \in M$ with $x={\sigma }^G$. Ground sets belong to it by their canonical names. Without assuming a weakest condition, use $\check{a}=\{(\check{b},p):b\in a,p\in P\}$; nonemptiness of $G$ gives ${\check{a}}^G=a$.

Induction on forcing name rank gives $\mathrm{rank}({\tau }^G)\le \mathrm{nrk}(\tau )$. The right side is an ordinal of $M$. If $\alpha \in M[G]$ is an ordinal, its rank equals $\alpha$, so it is at most a ground-model ordinal. Transitivity of $M$ then implies $\alpha \in M$. Conversely every ground ordinal remains the same ordinal in the transitive extension, since membership is unchanged. HenceThis proves forcing preserves ordinals directly from ranks, without assuming cardinal preservation.

We use only the definability and truth clauses of the forcing theorem. For every formula, its forcing relation on names is definable inside $M$; forcing is preserved by strengthening; and $M[G]\models \phi ({\vec{\tau }}^G)$ exactly when some condition in $G$ forces $\phi (\vec{\tau })$. These clauses do not presuppose the Separation axiom we are proving.

Let $a={\tau }^G$ and let ${\vec{\sigma }}^G$ be the parameters in the desired instance. In $M$, form the nameThis is a set in $M$ by its Separation axiom and forcing definability, using the subnames of $\tau$ and $P$ as a set-sized bound.

If $x\in {\rho }^G$, some $q\in G$ activates $(\nu ,q)$ and $q\ge p$ activates $(\nu ,p)\in \tau$. Thus $x={\nu }^G\in a$, and the forcing theorem gives $\phi (x,{\vec{\sigma }}^G)$. Conversely, if $x\in a$ satisfies that formula, choose an active pair $(\nu ,p)\in \tau$ with $x={\nu }^G$ and a condition $r\in G$ forcing the formula. Directedness gives $q\in G$ stronger than both $p$ and $r$. Monotonicity puts $(\nu ,q)\in \rho$, so $x\in {\rho }^G$.

ThereforeEvery requested instance of separation in a generic extension follows.

For a maximal forcing antichain $A\in M$, the set $D_A=\{q:\exists a\in A(q\ge a)\}$ is dense. Maximality ensures each condition is compatible with some $a\in A$, and a common strengthening lies in $D_A$. A generic filter meets $D_A$ and hence, by closure toward weaker conditions, meets $A$. Directedness makes any two of its conditions compatible, so it contains at most one member of an forcing antichain. Thus $|G\cap A|=1$.

Conversely, suppose this equality holds for every ground-model maximal forcing antichain. For any dense $D\in M$, use Choice in $M$ to select an forcing antichain $A\subseteq D$ maximal among forcing antichains contained in $D$. It is maximal in $P$ as well: a condition incompatible with every member of $A$ would have an extension in $D$ still incompatible with every member, enlarging $A$. The hypothesis gives a member of $G\cap A\subseteq G\cap D$. Thus $G$ meets every ground dense set and is generic.

This proves the maximal-antichain criterion for genericity in both directions.

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.