= 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:
$$
M[G]\models\kappa=\aleph_1.
$$
Conversely, suppose $M[G]\models\kappa=\aleph_1$ and $\kappa$ were singular in $M$. Let $\mu=\operatorname{cf}^M(\kappa)<\kappa$ 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 $\kappa$,
$$
\boxed{\kappa\text{ is regular in }M\quad\Longleftrightarrow\quad M[G]\models\kappa=\aleph_1.}
$$
Back to article page
