Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 121 / 3

For a condition $p\in \mathbb{P}$, a set $D\subseteq \mathbb{P}$ is dense below a forcing condition $p$ whenThus every extension of $p$ can be strengthened into $D$.

A set $G\subseteq \mathbb{P}$ is a filter when it is upward closed toward weaker conditions and downward directed: if $p\in G$ and $p\le q$, then $q\in G$, while any $p,q\in G$ have some $r\in G$ with $r\le p,q$. It is a generic filter over $M$ when

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 first that $p{⊩}^{\ast }\phi$. If $G$ is any generic filter over $M$ containing $p$, the forcing theorem gives $M[G]\models \phi$. Hence $p⊩\phi$ in the semantic forcing relation.

Conversely, suppose $p/{⊩}^{\ast }\phi$. By the stated density equivalence for the syntactic forcing relation, some $r\le p$ satisfies $r{⊩}^{\ast }\neg \phi$. The Rasiowa–Sikorski lemma supplies a generic filter $G$ over the countable model $M$ containing $r$, and upward closure then gives $p\in G$. The forcing theorem yields $M[G]\models \neg \phi$, so $p/⊩\phi$ semantically. Therefore

Fix names $\dot{x},{\dot{a}}_1,\dots ,{\dot{a}}_n\in M$ and a first-order formula $\phi$. Using the syntactic forcing relation, form in $M$ the nameThis is a set by the axiom schema of separation in $M$. If $(\tau ,r)\in \dot{y}$ and $r\in G$, then the forcing theorem gives both ${\tau }^G\in {\dot{x}}^G$ and $\phi ({\tau }^G,{\dot{a}}_1^G,\dots ,{\dot{a}}_n^G)$. Conversely, if $z\in {\dot{x}}^G$ satisfies $\phi$, choose $(\tau ,q)\in \dot{x}$ with $q\in G$ and ${\tau }^G=z$. The truth direction of the forcing theorem supplies $s\in G$ forcing $\phi (\tau ,{\dot{a}}_1,\dots ,{\dot{a}}_n)$; directedness of $G$ gives $r\in G$ below both $q$ and $s$, so $(\tau ,r)\in \dot{y}$.

ThusEvery instance has such a witness, so separation in a generic extension proves $M[G]\models$ Separation.

Let $F=\bigcup G$, the generic union for the Finite-condition Lévy collapse, and fix an infinite cardinal number $\lambda <\kappa$ in $M$. For every $n\in \omega$, the set of conditions defining a value at $(\lambda ,n)$ is dense: extend a condition at that fresh coordinate with any value below $\lambda$. Hence $F_{\lambda }(n)=F(\lambda ,n)$ is a total function $\omega \to \lambda$.

For every $\beta <\lambda$, the set of conditions assigning the value $\beta$ at some fresh coordinate $(\lambda ,n)$ is also dense. Genericity therefore makes $F_{\lambda }$ surjective. Thusso $\lambda$ is a countable set in $M[G]$. Finite cardinals are already countable.

Assume that $\kappa$ is regular and uncountable in $M$, as required for this claim. If an antichain $A\in M$ had size $\kappa$, apply the delta-system lemma at a regular uncountable cardinal inside $M$ to the finite domains of its conditions. After thinning, their domains form a delta-system with finite root $R$.

There are fewer than $\kappa$ possible restrictions to $R$, because each coordinate $(\alpha ,n)\in R$ has fewer than $\kappa$ possible values. Since $\kappa$ is a regular cardinal, we can thin again to $\kappa$ conditions agreeing on $R$. Any two now have disjoint domains outside $R$ and agree on $R$, so their union is a common extension. This contradicts that $A$ is an antichain. ThereforeIf the word “regular” is allowed to include $\kappa =\omega$, the printed claim needs the additional hypothesis that $\kappa$ is uncountable; the finite-condition order can have an infinite antichain when $\kappa =\omega$.

Suppose first that $\kappa$ is regular and uncountable in $M$. Part (b) and cardinal preservation by chain-condition forcing show that $\kappa$ remains a cardinal number in $M[G]$, while part (a) makes every infinite ground-model cardinal below $\kappa$ countable. Every ordinal below $\kappa$ has ground-model cardinality below $\kappa$ and is therefore countable in the extension. Thus $\kappa$ is the least uncountable ordinal there:

Conversely, suppose $M[G]\models \kappa ={\aleph }_1$ and $\kappa$ were singular in $M$. Let $\mu ={\mathrm{cf}}^M(\kappa )<\kappa$ and take in $M$ a cofinal function $f:\mu \to \kappa$. Part (a) makes $\mu$ countable in $M[G]$, while the same $f$ remains cofinal there. This would give $\kappa$ countable cofinality, contradicting first uncountable ordinal is regular. Hence $\kappa$ was regular in $M$, and for the intended uncountable $\kappa$,