Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 128 / 2

The Fn forcing order isIt is ordered by reverse inclusion:so a stronger condition specifies more values. Its maximal, or weakest, element is the empty function $\varnothing$.

For a regular cardinal $\kappa$,Again $q\le p$ means $q\supseteq p$, and the maximal element is the empty function. The regularity of $\kappa$ ensures that the union of a descending sequence of fewer than $\kappa$ conditions still has domain of cardinality below $\kappa$ whenever the conditions form a compatible increasing chain of partial functions.

Let $A\subseteq \mathrm{Fn}(I,J)$ be uncountable. Apply the Delta-system lemma to the finite sets $\mathrm{dom}p$ for $p\in A$. After passing to an uncountable subset $A^{\prime }$, there is a fixed finite root $R$ such thatfor distinct $p,q\in A^{\prime }$. Because $J$ is countable and $R$ is finite, there are only countably many functions $R\to J$. A further uncountable subset $A^{\prime \prime }\subseteq A^{\prime }$ therefore has the same restriction to $R$.

Any two conditions in $A^{\prime \prime }$ agree on the intersection of their domains, so their union is a common stronger condition. Thus every uncountable family contains two compatible conditions, and no uncountable antichain exists. Therefore

Let $c=\bigcup G$ and $d(n)=c(2n)$. DefineThis is a filter: restrictions of finite pieces of $d$ remain in $H$, and the union of two members is a common stronger condition.

To prove genericity, take a dense set $D\in M$. Let $E_D$ consist of conditions $q\in \mathrm{Fn}(\omega ,2)$ for which some $p\in D$ satisfiesThe set $E_D$ is dense. Indeed, given $q$, first read its finitely many assigned even coordinates as a condition $p_0$ on $\omega$. Choose $p\le p_0$ in $D$, and extend $q$ by setting $q(2n)=p(n)$ at the remaining coordinates of $p$. Since $E_D\in M$, the generic filter $G$ meets it. For $q\in G\cap E_D$, the corresponding $p\in D$ is a finite subfunction of $d$, so $p\in H\cap D$.

Thus $H$ meets every dense subset belonging to $M$. Moreover, each singleton $\{(n,d(n))\}$ belongs to $H$, and hence