Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 128 / 3

The forcing theorem has two parts. The definability lemma says that for every formula $\phi$, the relationis definable in $M$. The truth lemma says that if $G$ is generic over $M$, then

Assume the forcing relation and the forcing theorem have been constructed for $\phi (x,\vec{y})$. Defineto mean thatis dense below $p$. This definition is first-order over $M$, so the definability lemma is preserved.

Suppose $p\in G$ forces the existential statement. Genericity below $p$ gives $q\in G\cap D_p$ and a name $\sigma$ with $q⊩\phi (\sigma ,\vec{\tau })$. The truth lemma for $\phi$ yieldsso the existential statement is true. Conversely, if $M[G]\models \exists x\phi (x,{\vec{\tau }}^G)$, choose a name $\sigma$ for a witness. The truth lemma for $\phi$ gives $q\in G$ with $q⊩\phi (\sigma ,\vec{\tau })$, and then $q⊩\exists x\phi (x,\vec{\tau })$. This proves both directions of the forcing theorem for the existential formula.

Conditions in $G$ are compatible finite functions, so their union $F=\bigcup G$ is a function $({\omega }_1{)}^M\rightharpoonup ({\omega }_1{)}^M$. For each $\alpha <{\omega }_1$, the setis dense: choose a normal function extending $p$ and add its value at $\alpha$. Genericity makes $F$ total.

If $\alpha <\beta$, choose a condition in the filter extending conditions that decide both values. It is contained in a normal function on an ordinal, so $F(\alpha )<F(\beta )$. Thus $F$ is strictly increasing.

It remains to prove continuity. For every limit $\delta <{\omega }_1$ and $\gamma <{\omega }_1$, let $D_{\delta ,\gamma }$ contain the conditions $p$ such that $\delta \in \mathrm{dom}p$ and eitheror there is some $\alpha <\delta$ in $\mathrm{dom}p$ with $p(\alpha )>\gamma$. This set is dense. Given $p$, extend it to a normal function $f$ and add $(\delta ,f(\delta ))$; if $f(\delta )>\gamma$, continuity of $f$ supplies an $\alpha <\delta$ with $f(\alpha )>\gamma$, which may also be added.

Now fix $\gamma <F(\delta )$. Since $G$ meets $D_{\delta ,\gamma }$, compatibility with the condition deciding $F(\delta )$ rules out the first alternative and gives $\alpha <\delta$ with $F(\alpha )>\gamma$. Therefore values below $\delta$ are cofinal in $F(\delta )$. Strict increase supplies the reverse bound, soHence $F$ is normal on $({\omega }_1{)}^M$ in $M[G]$.