Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 128 / 2

The two conventions reverse the order relation. In standard notation for forcing, $q\le p$ means that $q$ is stronger than $p$. Thus $p$ and $q$ are incompatible forcing conditions when there is no $r$ with $r\le p$ and $r\le q$, while $D\subseteq \mathbb{P}$ is a dense set whenIn Jerusalem notation for forcing, $p\le q$ means that $q$ is stronger than $p$. Incompatibility therefore means that there is no $r$ with $p\le r$ and $q\le r$, and density means

Solved by gpt-5.6-sol high.

Let $a,p\in M[G]$, and suppose a first-order formula $\psi (x,y,p)$ defines exactly one $y$ for every $x\in a$. Apply the Lévy reflection theorem to the formulas needed to express this assertion, choosing an ordinal $\alpha$ with $a,p\in V_{\alpha }^{M[G]}$ such thatfor every $x\in a$. Therefore every required value lies in the set $V_{\alpha }^{M[G]}$.

The already established axiom schema of separation forms the setFunctionality makes $b$ exactly the range of the definable function on $a$. This proves every instance of the Axiom schema of replacement in the generic extension $M[G]$.

Solved by gpt-5.6-sol high.

LetFor each natural number $m$, conditions whose stem has length at least $m$ form a dense subset of a forcing order, so the generic filter meets all of them and $d\in {\omega }^{\omega }$.

Fix $h\in {\omega }^{\omega }\cap M$. The setis dense: from $(s,f)$ replace $f$ by $n\mapsto \max \{f(n),h(n)+1\}$. Choose $(s,f)\in G\cap D_h$. Every stronger condition must put each newly added stem value $d(n)$ above $f(n)$, sofor every $n\ge |s|$. Thus $d$ is a dominating real over $M$.

Solved by gpt-5.6-sol high.

Conditions with the same finite stem in Hechler forcing are compatible: $(s,f)$ and $(s,g)$ have the common stronger condition $(s,\max \{f,g\})$. Since there are only countably many finite stems, Hechler forcing is sigma-centered and hence has the countable chain condition for forcing. It therefore preserves ${\aleph }_1^M$.

In $M$, the Continuum hypothesis givesEvery real in $M[G]$ has a nice name for a real, and the countable chain condition bounds the number of such names bywhere the last equality uses the ground-model continuum hypothesis. The extension still contains all ground-model reals, already ${\aleph }_1$ many, soThus forcing once with $\mathbb{D}$ preserves the continuum hypothesis.

Solved by gpt-5.6-sol high.