Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 24 / 6

The forcing is cardinal-preserving over $M$ when every ground-model cardinal remains a cardinal in every generic extension:Forcing preserves ordinals, and an ordinal that was not a cardinal already has a ground-model bijection with a smaller ordinal, which persists. Thus this definition gives equality of the cardinal ordinals in the two models.

The forcing definability lemma states that for each formula $\phi (v_1,\dots ,v_n)$ there is a first-order formula ${\mathrm{Force}}_{\phi }$, depending on $\phi$ but not on the chosen names, such that for $p\in P$ and forcing names ${\tau }_i\in M$,The order relation of $\mathbb{P}$ is included in its set code. Thus each instance of the forcing relation is definable inside the ground model, allowing the uses of ground-model axiom schema of separation and replacement in the forcing arguments.

The forcing is $\kappa$-closed in $M$ when every descending sequence of conditions of length less than $\kappa$, belonging to $M$, has a lower bound in $M$:We use the convention that $q\le p$ means that $q$ is stronger. In particular, ${\omega }_1$-closed forcing is countably closed. Internal quantification over the sequences is essential when $M$ is externally countable.

This is the Rasiowa–Sikorski lemma. Put $p_0=p$ and recursively chooseusing density. SetIt contains $p$, is upward closed toward weaker conditions, and is directed: if $p_i\le q$ and $p_j\le r$, then $p_{\max (i,j)}$ is a common stronger condition in $G$. It meets each $D_n$ through $p_{n+1}$. Thus $G$ is the required generic filter for the prescribed countable family of dense sets. For a countable transitive model, externally enumerate its dense sets to obtain a generic filter over the model by this same argument.

The PDF's middle equality is $({\omega }^2{)}^M=({\omega }^2{)}^{M[G]}$, concerning ordinal exponentiation. The TeX incorrectly turns the exponent into a subscript; preserving ${\omega }_2$ would be incompatible with forcing CH over many ground models.

Let $A=(\mathcal{P}(\omega ){)}^M$ and use the countable-condition collapse of $A$ onto ${\omega }_1$:ordered by extension. Countable unions give lower bounds, so closed forcing adds no short ordinal sequences; in particular it adds no reals and preserves ${\omega }_1$. Dense requirements ensure that the union of the generic filter is a total surjection $g:{\omega }_1\to A$. ThusFinally forcing preserves ordinals, and the recursion defining ordinal multiplication and exponentiation is absolute. Both models compute ${\omega }^2$ as the same order type of $\omega$ successive copies of $\omega$. This proves the printed middle equality too. Higher cardinals may collapse, which is allowed by the actual PDF.

In the generic extension from the preceding construction, a surjection from ${\omega }_1$ onto all reals gives $2^{{\aleph }_0}\le {\aleph }_1$. Cantor theorem gives the reverse inequality, so the Continuum hypothesis holds.

For the exact consistency-strength statement, existence of a countable transitive model is stronger than bare consistency of ZFC. One can formalize the forcing proof syntactically, or use the constructible universe theorem: ZFC proves that its constructible universe satisfies ZFC and the Generalized continuum hypothesis. If there is any model of ZFC, its internal constructible universe is therefore a model of ZFC with CH. By the Godel completeness theorem,This deduction does not silently replace consistency by existence of a transitive model.

For each $\xi <\kappa$, choose a maximal antichain in a forcing order below $p$ deciding a witness $\eta >\xi$ in $\dot{C}$. Write the decided values as ${\eta }_{\xi ,q}$, one for each condition $q$ in that antichain. The $\kappa$-chain condition for forcing gives fewer than $\kappa$ possible values. Since $\kappa$ is regular in $M$, chooseThe possible-values lemma for chain-condition forcing ensuresIn $M$ take the club set of limit closure pointsIt is unbounded by repeatedly closing any starting ordinal under $g$ for countably many steps, and closed by taking increasing limits. For $\delta \in D$, the forced displayed property gives points of $\dot{C}$ arbitrarily high below $\delta$. Its forced closedness yields $\delta \in \dot{C}$. ThereforeThis proves the ground-model club containment lemma. The name on the right is $\dot{C}$, as printed in the PDF.

Fix $p$ and a name $\dot{f}$ such that $p$ forces a two-coloring of $[\lambda {]}^{<\omega }$. For each ground-model finite set $a\subseteq \lambda$, record its full forcing decision patternThere are at most $3^{|\mathbb{P}|}<\lambda$ such patterns, since $\lambda$ is a strong limit cardinal and $|\mathbb{P}|<\lambda$. For finite $\mathbb{P}$ the number of patterns is finite, also below $\lambda$. Encode the patterns by an ordinal $\beta <\lambda$ and apply the assumed finite-subset partition property to obtain $H\subseteq \lambda$ of size $\lambda$ such that $F$ is constant on $[H{]}^n$ for every $n<\omega$.

For each $n$, choose one $a_n\in [H{]}^n$. Conditions below $p$ deciding $\dot{f}(a_n)$ are dense. Whenever one such condition decides the value, equality of patterns means it forces that same value at every $a\in [H{]}^n$. A generic filter containing $p$ meets this dense set for each $n$, so $f$ is constant on each $[H{]}^n$ in the extension. The constants may differ with $n$, exactly as required.

The forcing has size below the regular $\lambda$, hence satisfies the $\lambda$-chain condition for forcing and preserves its cardinality. The ground-model set $H$ consequently still has size $\lambda$. Since the argument works for every name and condition,This is small forcing preservation of finite-subset partition properties; no closure assumption on the forcing is required.