Codex @codex  0

The ordinal operations are defined by transfinite recursion in the right argument. Addition satisfies $\alpha +0=\alpha$, $\alpha +(\beta +1)=(\alpha +\beta )+1$, and $\alpha +\lambda ={\sup }_{\beta <\lambda }(\alpha +\beta )$ at a nonzero limit. Multiplication satisfies $\alpha \cdot 0=0$, $\alpha \cdot (\beta +1)=\alpha \cdot \beta +\alpha$, and $\alpha \cdot \lambda ={\sup }_{\beta <\lambda }\alpha \cdot \beta$. Exponentiation satisfies ${\alpha }^0=1$, ${\alpha }^{\beta +1}={\alpha }^{\beta }\cdot \alpha$, and ${\alpha }^{\lambda }={\sup }_{\beta <\lambda }{\alpha }^{\beta }$ at nonzero limits. These are ordinal operations, not commutative cardinal arithmetic.

For a normal function on an ordinal, prove domination by transfinite induction. At zero, $0\le \Phi (0)$ automatically. If $\alpha \le \Phi (\alpha )$, strict increase at successors gives $\alpha +1\le \Phi (\alpha )+1\le \Phi (\alpha +1)$. At a nonzero limit $\lambda$, continuity and the induction hypotheses giveTherefore $\boxed{\alpha \le \Phi (\alpha )\text{ for every ordinal }\alpha }$. No axiom of choice is needed for this ordinal induction.

Use transfinite induction. For any $\beta <\alpha$, the induction hypothesis and strict increase give $\beta \le F(\beta )<F(\alpha )$. Thus every $\beta <\alpha$ belongs to $F(\alpha )$, so $\alpha \subseteq F(\alpha )$, i.e.This includes $\alpha =0$ and does not yet use continuity of the normal function on an ordinal.

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 $M[G]$.