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

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 $\kappa$,