Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 121 / 3 / iv

For the product forcing order, let $G_0$ and $G_1$ be the coordinate projections of $H$:These are nonempty upward-closed, directed filters in an ordered set. Clearly $H\subseteq G_0\times G_1$. If $p\in G_0$ and $q\in G_1$, choose $(p,q^{\prime })\in H$ and $(p^{\prime },q)\in H$. Directedness supplies $(r,s)\in H$ below both, so $(r,s)\le (p,q)$. Upward closure of $H$ gives $(p,q)\in H$. Therefore the projection of a product-generic filter satisfiesNo greatest conditions in the factors are required for this argument.

Let $D\in M$ be a dense subset of a forcing order $\mathbb{P}$. Then $D\times \mathbb{Q}\in M$ is dense in the product forcing order: below $(p,q)$, choose $p^{\prime }\le p$ in $D$ and retain $q$. The generic filter $H$ meets it, so its first projection $G_0$ meets $D$. Together with the filter properties proved above, this givesThe same argument proves that $G_1$ is $\mathbb{Q}$-generic over $M$, but genericity over the larger generic extension $M[G_0]$ needs the next argument.

We prove mutual genericity for product forcing. Let $D\in M[G_0]$ be a dense subset of $\mathbb{Q}$, and take a forcing name $\dot{D}\in M$ evaluating to $D$. By the forcing theorem, some $p_0\in G_0$ forces that $\dot{D}$ is a dense subset of the canonical forcing name for $\mathbb{Q}$.

In $M$ defineThis set is dense in the product forcing order. Given $(p,q)$, the incompatible case is immediate. Otherwise first strengthen $p$ below $p_0$. The forced density assertion and the existential clause of syntactic forcing supply a further $r\le p,p_0$ and a ground-model $q^{\prime }\le q$ with $r⊩{\check{q}}^{\prime }\in \dot{D}$. To justify choosing a ground-model $q^{\prime }$, a name forced to lie in $\check{\mathbb{Q}}$ can densely be made equal to some ${\check{q}}^{\prime }$ by the atomic membership clause; strengthen to that equality and use the forced order comparison. Thus $(r,q^{\prime })\in E$ lies below $(p,q)$.

The generic filter $H$ meets $E$. It cannot meet the first part, because its first projection contains $p_0$ and is directed. Hence there is $(p,q)\in H$ with $p\in G_0$ and $p⊩\check{q}\in \dot{D}$. Soundness of the forcing theorem gives $q\in D$, and the projection gives $q\in G_1$. Since every such $D$ is met,This establishes the stronger property, rather than merely meeting dense ground-model subsets.