Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 121 / 3

Use standard notation for forcing, so $q\le p$ means that $q$ is stronger. Let $\check{p}$ be the canonical forcing name for the ground-model condition $p$. DefineHere $q\perp p$ means that they are incompatible forcing conditions. The checks and their collection are formed by recursion and Axiom schema of replacement in $M$, and the displayed set is selected by axiom schema of separation, so this is a forcing name in $M$.

By evaluation of a forcing name,If $p\in G$, directedness of the generic filter gives a common stronger condition for $p$ and each $q\in G$, so $p$ is absent from this value.

Conversely, for fixed $p\in \mathbb{P}$ the setis a dense subset of a forcing order belonging to $M$. A condition incompatible with $p$ is already in it, and one compatible with $p$ has a common extension below $p$. Genericity supplies $q\in G\cap D_p$. If $p\notin G$, upward closure of $G$ rules out $q\le p$, hence $q\perp p$. ThereforeThis forcing name for the complement of a generic filter works without a separativity assumption on the order.

We prove the implication from existence of a witness in the extension to a condition forcing the existential statement. If $M[G]\models \exists x\phi (x)$, choose a witness $a\in M[G]$ and a forcing name $\sigma \in M$ with $\mathrm{val}(\sigma ,G)=a$. The assumed instance of the truth lemma supplies $p\in G$ withThe syntactic forcing relation is monotone under stronger conditions, so every $q\le p$ also forces this same named instance. In particular, the conditions below $p$ having some forced witness are dense below a forcing condition $p$. The existential clause of syntactic forcing therefore givesOnly the truth lemma assumed for individual named instances and the recursive existential clause were used.

For the reverse implication, suppose $p\in G$ and $M\models p{⊩}^{\ast }\exists x\phi (x)$. By the existential clause of syntactic forcing, the setis dense below $p$; the existential quantifier ranges over forcing names of $M$. Definability of the syntactic forcing relation and axiom schema of separation put $D$ in $M$.

To apply genericity, augment it to the globally dense setA condition compatible with $p$ has an extension below $p$, then one in $D$, and an incompatible condition already lies in $D^{\prime }$. Thus $D^{\prime }$ is a dense subset of a forcing order in $M$. The generic filter meets it. Since $p\in G$, directedness prevents $G$ from containing a condition incompatible with $p$, so some $q\in G\cap D$ exists.

There is consequently a name $\sigma \in M$ with $M\models q{⊩}^{\ast }\phi (\sigma )$. The assumed truth lemma gives $M[G]\models \phi (\mathrm{val}(\sigma ,G))$, and henceTogether with part (a), this completes the existential step of the forcing theorem without assuming the desired existential truth lemma in advance.

A Delta-system is a family $\mathcal{A}$ of sets with a fixed root $r$ such thatThe Delta-system lemma says that every uncountable family of finite sets has an uncountable subfamily forming a Delta-system. The root may be empty. Finiteness is a hypothesis on the family to which this lemma is applied, rather than a requirement in the general definition of a Delta-system.

A cardinal-preserving forcing over $M$ leaves every ground-model cardinal number a cardinal in every generic extension. In the present notation, for every $M$-generic filter $G$ and every ordinal $\kappa \in M$,Set forcing preserves ordinals, and an old bijection witnessing that an ordinal is not a cardinal remains available. Thus this condition also means that the two models have exactly the same cardinals. Cardinality here is computed inside the respective models, not from their external sizes.

The countable chain condition for forcing says that every antichain in a forcing order is a countable set. Equivalently,In this question the assertion is evaluated inside $M$. Although every subset of the countable set $M$ is externally countable, that observation does not establish the internal chain condition: $M$ need not contain an enumeration of an antichain.

Assume $M$ satisfies the countable chain condition for forcing. We prove cardinal preservation by chain-condition forcing using the possible-values lemma for chain-condition forcing, and give the argument explicitly for ccc.

Let $\kappa$ be an uncountable cardinal of $M$. If it ceased to be a cardinal, some ordinal $\mu <\kappa$ would admit a surjection onto $\kappa$ in an extension. By the forcing theorem, there would be a condition $p$ and a forcing name $\dot{f}$ such thatInside $M$, for each $\xi <\mu$ choose a maximal antichain in a forcing order below $p$ deciding the ordinal value of $\dot{f}(\xi )$. Conditions deciding that value are dense below $p$. The chain condition makes the chosen antichain countable in $M$, so the corresponding set $B_{\xi }\subseteq \kappa$ of possible decided values is countable in $M$.

A generic filter containing $p$ meets the downward closure of every such maximal antichain, so its value $f(\xi )$ belongs to $B_{\xi }$. Therefore its whole range lies in the ground-model setThe last inequality is infinite cardinal arithmetic and uses only that $\kappa$ is uncountable and $\mu <\kappa$; no regularity of $\kappa$ is needed. Some ordinal in $\kappa \setminus B$ exists already in $M$ and cannot occur in the range, contradicting the forced surjectivity.

Finite cardinals cannot collapse, and $\omega$ cannot become finite: a finite domain has finite image, and transitive models agree on the natural numbers. Old non-cardinals cannot become cardinals because their old bijections persist. Thus

Let $s_p$ be the finite stem of a condition $p=(n_p,s_p,A_p)$, and defineThe stems of two conditions in the generic filter agree on their common domain, because they have a common stronger extension. Hence this union is a function. For every $\ell \in \omega$, the set of conditions with $n_p\ge \ell$ is dense: append values from the nonempty infinite reservoir until the desired length is reached. Genericity therefore makes $f$ total on $\omega$. The generic stems, and thus their union, are available in $M[G]$.

Fix $g\in M\cap {\omega }^{\omega }$ and $K\in \omega$. In $M$ form the setWe show that it is a dense subset of a forcing order. Given $p=(n,s,A)$, put $k=\max (n,K)$. Fill the new positions $n,\dots ,k-1$ with any fixed element of $A$, and choose $a\in A$ with $a>g(k)$ for the new position $k$. This is possible because an infinite subset of $\omega$ is unbounded. Let $t$ be the resulting stem of length $k+1$, and keep the reservoir $A$ unchanged. ThenEvery new stem value came from the old reservoir, exactly as required by the extension relation. The construction is performed in $M$, so $D_{g,K}\in M$ and is internally dense.

Genericity gives a condition in $G\cap D_{g,K}$. Its witnessing coordinate remains fixed in every later stem and hence in $f$, so some $k\ge K$ satisfies $g(k)<f(k)$. Since this holds for every $K$, there are infinitely many such $k$. As $g$ was arbitrary,Thus this infinite-reservoir stem forcing produces an unbounded real over a model, which is precisely the stipulated meaning of bounding $M$.