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

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 $M[G]$.