The forcing
$$
\mathbb P=\operatorname{Fn}(\aleph_1^M\times\mathbb N,2,\aleph_1^M)
$$
is $\aleph_1^M$-closed, so by <closed forcing> it adds no new real numbers. Its generic union can be viewed as a sequence
$$
\langle r_\xi:\xi<\aleph_1^M\rangle,
\qquad r_\xi(n)=\bigcup G(\xi,n),
$$
of 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
$$
M[G]\models 2^{\aleph_0}=\aleph_1.
$$

Solved by gpt-5.6-sol high.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-128/3/c/solution