Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 24 / 5

Use the countable-condition collapseordered by reverse inclusion: an extension of a function is a stronger condition. All sizes and conditions here are computed in the ground model $M$.

This forcing is countably closed: the union of a descending countable sequence is a countable partial function. Thus it adds no countable ordinal sequences and preserves ${\omega }_1$. For each $\xi <{\omega }_1$, the conditions whose domains contain $\xi$ are dense; for each $\eta <\kappa$, those whose ranges contain $\eta$ are dense. The union of the generic filter is therefore a surjection from ${\omega }_1$ onto $\kappa$, and $|\kappa {|}^{M[G]}={\aleph }_1$.

Inaccessibility gives ${\kappa }^{{\aleph }_0}=\kappa$: every countable sequence is bounded below the regular $\kappa$, and the strong limit cardinal property bounds the number of sequences at each bound below $\kappa$. Hence $|\mathbb{P}|=\kappa$. The forcing satisfies the ${\kappa }^{+}$-chain condition for forcing, so it preserves every cardinal above $\kappa$. This gives precisely the requested collapse and preservation. The assertion that $\kappa$ is collapsed to ${\aleph }_1$ concerns its new cardinality; the ordinal $\kappa$ itself does not change.

For an arbitrary ground model, this alternative needs an additional arithmetic hypothesis. The requested countable-cover property already prevents any collapse of an infinite ground-model cardinal. If a cardinal $\theta$ were collapsed, there would be a surjection $f:\alpha \to \theta$ with $|\alpha {|}^M<\theta$. The asserted ground-model $F$ would cover $\theta$ by the union of $|\alpha {|}^M$ countable sets, a ground-model set of size at most $\max ({\aleph }_0,|\alpha {|}^M)<\theta$, a contradiction. Thus all infinite cardinals are preserved.

Put $\lambda =({\aleph }_3{)}^M$, which must then remain ${\aleph }_3$. If the extension has continuum $\lambda$, it hasThe ground-model functions from $\omega$ into $\lambda$ remain present and their cardinality cannot be collapsed. NecessarilyFor example, a ground model with continuum larger than ${\aleph }_3$ cannot satisfy the printed request. This is a genuine missing hypothesis, rather than a forcing construction that works for every $M$.

Under the necessary hypothesis, use Cohen forcing to add $\lambda$ reals:The delta-system lemma shows that it has the countable chain condition for forcing: an uncountable family of finite conditions has an uncountable subfamily whose domains form a delta-system and whose values agree on its root, so any two of that subfamily are compatible. It therefore preserves cardinals. Its $\lambda$ coordinate reals are pairwise distinct by dense disagreement requirements. Conversely each nice forcing name for a real uses countably many countable antichains in a forcing order, so the number of such names is at most ${\lambda }^{{\aleph }_0}=\lambda$. HenceFor any ordinal-valued function name, choose for each $\xi <\alpha$ a maximal antichain in a forcing order deciding its value. The countable chain condition for forcing makes the set $F(\xi )$ of possible ordinal values countable in $M$. Pad it with $\omega$ if needed to make it countably infinite. The possible-values lemma for chain-condition forcing gives the stronger pointwise covering statementwhich implies the requested range inclusion. When a condition only forces that the name is such a function, make these choices below that condition; it belongs to the generic filter witnessing the actual function.

Use diamond-sequence forcing. A condition is a sequenceordered by end extension. Its countable descending chains have lower bounds: take their union and, if their lengths approach a new limit, add an arbitrary subset at that last index. Hence it is countably closed and preserves ${\omega }_1$. The union of the generic filter supplies $⟨A_{\xi }:\xi <{\omega }_1⟩$.

To prove the diamond principle, let a condition force that $\dot{X}\subseteq {\omega }_1$ and that $\dot{C}$ is a club set. Below any such condition build $p_n$ and strictly increasing countable ordinals ${\delta }_n$ so that ${\delta }_n>{\delta }_{p_n}$, $p_{n+1}$ forces ${\delta }_n\in \dot{C}$, decides $\dot{X}\cap {\delta }_n$, and has length past ${\delta }_n$. Unboundedness supplies ${\delta }_n$; countable closure allows deciding all the bits below it. The construction can be carried out in $M$.

Let $\delta ={\sup }_n{\delta }_n={\sup }_n{\delta }_{p_n}$. The union of the conditions has entries exactly below $\delta$ and decides a ground-model set $x=\dot{X}\cap \delta$. Extend that union by setting $A_{\delta }=x$. This condition forces $\delta \in \dot{C}$ by closure, and $A_{\delta }=\dot{X}\cap \delta$. The conditions giving a correct guess inside any named club set are therefore dense. ThusNo ground-model Continuum hypothesis is needed; this forcing is allowed to collapse higher cardinals.

Fix forcing names $\tau ,{\tau }_1,\dots ,{\tau }_n\in M$ for the set $w$ and the parameters. Define a nameOnly $\eta$ appearing in $\tau$ and $p\in \mathbb{P}$ are needed, so this is a subset of a ground-model set. The forcing definability lemma makes its defining predicate a formula of $M$. Ground-model axiom schema of separation therefore gives $\sigma \in M$.

If $(\eta ,p)\in \sigma$ with $p\in G$, the accompanying $q\ge p$ belongs to $G$, so ${\eta }_G\in {\tau }_G$. The forcing theorem gives $\phi ({\eta }_G,{\tau }_G,({\tau }_1{)}_G,\dots ,({\tau }_n{)}_G)$.

Conversely, if $u\in {\tau }_G$ satisfies this formula, choose $(\eta ,q)\in \tau$ with $q\in G$ and ${\eta }_G=u$. The forcing truth lemma supplies $r\in G$ forcing the formula for these names. Directedness gives $p\in G$ with $p\le q,r$. Then $(\eta ,p)\in \sigma$, and $u\in {\sigma }_G$. ThusThis proves the instance of the axiom schema of separation in the generic extension, without assuming that instance there in order to construct the name.

Countability of the forcing and uncountability of $Y$ are understood internally in $M$ and $M[G]$, respectively. This matters because the model itself is externally countable. Choose a name $\dot{Y}\in M$ and a ground-model set $A\in M$ containing every ground-model element that can occur in $Y$. Such an $A$ exists: the rank of $\dot{Y}$ bounds the ranks of its values, so a sufficiently high $V_{\theta }^M$ contains $Y\subseteq M$.

For $p\in P$, ground-model axiom schema of separation and the forcing definability lemma giveBy the forcing truth lemma,If every $X_p$ for $p\in G$ were countable in $M$, it would remain countable in $M[G]$, where $G\subseteq P$ is countable. Then $Y$ would be countable, contrary to the hypothesis. Thus some $p\in G$ has $X_p$ uncountable in $M$; otherwise its ground-model enumeration would still enumerate it in the extension. Let $X=X_p$. Every one of its members is forced by $p$ into $Y$, soThe countable chain condition for forcing also preserves its uncountability. Separativity is not needed for this particular argument. This is the ground-model uncountable subset lemma for countable forcing.

The finite binary-function Cohen forcing $\mathrm{Fn}(\omega ,2,<\omega )$ is countable in $M$. It preserves ${\omega }_1$, and the countable levels, height, and normal extensions of the ground-model tree remain unchanged.

If the extension contained an uncountable tree antichain, apply the ground-model uncountable subset lemma for countable forcing to obtain an uncountable $X\in M$ contained in it. Incomparability in the fixed ground-model tree is absolute, so $M$ already regards $X$ as an uncountable tree antichain, contradicting that $T$ is Suslin in $M$. Similarly, a new cofinal branch has an uncountable ground-model subset. Comparability is absolute, so this would be an uncountable chain in the ground-model tree, again impossible.

Therefore countable forcing preserves Suslin trees, and in particular