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

Use standard notation for forcing: $q\le p$ means that $q$ is stronger. A generic filter is nonempty, upward closed and downward directed, and meets every dense subset of a forcing order in the ground model. Names below are forcing names in $M$, with pairs ordered as $(\text{name},\text{condition})$.

The semantic forcing relation isHere ${\tau }^K=\mathrm{val}(\tau ,K)$ is the evaluation of a forcing name. Countability of $M$ supplies such generic filters through every condition by the Rasiowa–Sikorski lemma.

Define the syntactic forcing relation by mutual well-founded recursion on the forcing names for the atomic cases, then induction on the first-order formula. Its membership clause isFor equality, first abbreviateand set $p{⊩}^{\ast }\sigma =\tau$ exactly when both inclusions hold. Each recursive call lowers the rank of at least one forcing name without raising the other. This makes the mutual recursion well-founded.

For a basis of connectives consisting of logical conjunction, negation and existential quantification, the remaining clauses areThe last line is the existential clause of syntactic forcing; witnesses need only occur densely, rather than be forced by $p$ itself with one preselected name. Other connectives are defined by logical abbreviations. The recursion gives a definable relation inside $M$ for each fixed first-order formula, with quantification over the class of its names. If $\mathbb{P}$ has no greatest element, use canonical forcing names $\check{a}=\{(\check{b},p):b\in a,p\in \mathbb{P}\}$, which still evaluate to $a$. It also proves monotonicity: strengthening a condition preserves what it forces. The forcing theorem identifies the two relations and supplies the truth lemma.

This is the forward implication of the atomic membership truth lemma for forcing. Suppose ${\sigma }^G\in {\tau }^G$. By the evaluation of a forcing name, choose $(\rho ,r)\in \tau$ with $r\in G$ and ${\sigma }^G={\rho }^G$. The assumed equality truth lemma supplies $s\in G$ with $s{⊩}^{\ast }\sigma =\rho$.

Directedness of the generic filter gives $p\in G$ with $p\le r,s$. For every $q\le p$, monotonicity of the syntactic forcing relation gives $q{⊩}^{\ast }\sigma =\rho$, and $q\le r$. Thus $q$ itself witnesses the required membership density below $p$. HenceOnly the equality truth lemma stipulated in the source and the recursive membership clause have been used.

Conversely, suppose $p\in G$ and $p{⊩}^{\ast }\sigma \in \tau$. The syntactic forcing relation says thatis dense below a forcing condition $p$. This is a set in $M$ by definability of forcing and axiom schema of separation. The dense-below generic meeting lemma gives $s\in G\cap D$: augment $D$ by all incompatible forcing conditions with $p$ to obtain a globally dense ground-model set, and use $p\in G$ to exclude the incompatible part.

Choose the witnessing $(\rho ,r)\in \tau$. Since $s\le r$ and $s\in G$, upward closure gives $r\in G$. The assumed equality truth lemma gives ${\sigma }^G={\rho }^G$, while $r\in G$ gives ${\rho }^G\in {\tau }^G$. Combining both directions,This completes the atomic membership truth lemma for forcing; no separate assumption of the membership truth lemma was made.

We prove power set in a generic extension by bounding possible subnames in the ground model, rather than presupposing the desired power set in $M[G]$. Let $x={\tau }^G$, and let $S=\{\rho :\exists r((\rho ,r)\in \tau )\}$. In $M$ formEvery $\nu \in B$ is a forcing name. Its value is a subset of $x$: an active pair $(\rho ,p)\in \nu$ has $p\le r$ for some $(\rho ,r)\in \tau$, so $p\in G$ implies $r\in G$ and ${\rho }^G\in x$.

Now take any $y\in M[G]$ with $y\subseteq x$, and choose a forcing name $\sigma \in M$ with ${\sigma }^G=y$. By axiom schema of separation and definability of the syntactic forcing relation,The atomic membership truth lemma for forcing shows ${\nu }_{\sigma }^G=y$. One inclusion follows immediately from its soundness direction. For the other, if $a\in y\subseteq x$, choose $(\rho ,r)\in \tau$ with $r\in G$ and ${\rho }^G=a$. The truth direction supplies $p\in G$ forcing $\rho \in \sigma$; strengthen within $G$ below $p$ and $r$ to obtain an active pair in ${\nu }_{\sigma }$ representing $a$.

Finally the ground-model forcing nameexists by Axiom schema of replacement and the ground-model power set axiom. Since $G$ is nonempty, all $\nu \in B$ contribute their values, andThis set belongs to $M[G]$ by the definition of a generic extension. The construction uses all conditions in the outer pairs and therefore does not require a greatest condition in $\mathbb{P}$.

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.