Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 128 / 3 / d / Solution

The forcing $\mathbb{P}=\mathrm{Fn}({\aleph }_1^N,2,{\aleph }_1^N)$ is ${\aleph }_1^N$-closed, so it adds no reals:

By contrast, ${\aleph }_1^M$ is countable in $N$. The union of the $\mathbb{Q}$-generic filter is a total binary functionFix in $N$ a bijection $e:\mathbb{N}\to {\aleph }_1^M$. Then $r(n)=h(e(n))$ is a real in $N[H]$. It is not in $N$: for any ground-model real $r_0$ and any $p\in \mathbb{Q}$, some coordinate of ${\aleph }_1^M$ is outside $\mathrm{dom}p$, and extending there forces $r$ to differ from $r_0$. Hence $\mathbb{Q}$ adds a new real, and

Solved by gpt-5.6-sol high.