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

The forcingis ${\aleph }_1^M$-closed, so by closed forcing it adds no new real numbers. Its generic union can be viewed as a sequenceof old reals. For every $r\in \wp (\mathbb{N}{)}^M$, conditions asserting that some unused row equals $r$ are dense: assigning all countably many values of that row is a legitimate condition. Thus the generic sequence surjects ${\aleph }_1^M$ onto the old set of reals.

In $M$, that set had cardinality ${\aleph }_2^M$. The forcing therefore collapses ${\aleph }_2^M$ to ${\aleph }_1^M$, while adding no reals and preserving ${\aleph }_1^M$. Consequently

Solved by gpt-5.6-sol high.