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

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 $\kappa$,