Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 19 / 6

A forcing collapses a cardinal $\kappa$ over $M$ if it forces that the ground ordinal $\kappa$ is no longer a cardinal: in the extension there is a bijection from some smaller ordinal onto $\kappa$. Equivalently there is a surjection onto $\kappa$ from an ordinal whose ground cardinality is smaller. Forcing preserves the ordinal itself; collapse changes cardinality, not the identity of the ordinal. If the assertion is about one particular generic extension, the witnessing collapse is interpreted there; a forcing-wide assertion means it is forced by the weakest condition.

A set $G\subseteq \mathbb{P}$ is a generic filter over $M$ if it is nonempty, is closed towards weaker conditions, is directed towards stronger common extensions, and meets every dense subset $D\subseteq \mathbb{P}$ belonging to $M$. Density means that every condition has a stronger extension in $D$. Genericity is relative to $M$, not to every dense subset of a forcing order in the ambient universe. Under the printed weaker-first convention in Question 5, closure is towards smaller conditions and directedness is towards larger ones; in the standard stronger-first notation these inequalities reverse.

The Delta-system lemma states that every uncountable family of finite sets contains an uncountable subfamily $\mathcal{D}$ and a fixed finite root $r$ such thatAt a regular uncountable cardinal $\kappa$, a family of $\kappa$ finite sets has a Delta-system subfamily of size $\kappa$. The root may be empty. Neither version asserts that all sets in the original family have the same intersection.

Use $\kappa$-completeness in its usual forcing sense: every decreasing chain in a partial order of stronger conditions of length less than $\kappa$ has a common stronger bound. Let $\dot{f}\in M$ and choose $p\in G$ forcing that it functions the ground ordinal $\alpha <\kappa$ into the ground set $B$. Below any stronger condition $r$, recursively decide each value of $\dot{f}$ in order. At a successor step the deciding conditions are dense, and at a limit stage use $\kappa$-completeness. After all $\alpha$ steps take another common bound. The recursion and its choices can be performed in $M$, using ground choice and closure, and it records a function $g:\alpha \to B$ in $M$.

Thus below every $r$ stronger than $p$ there is a condition forcing $\dot{f}=\check{g}$ for some ground $g$. The set $D$ of such whole-function deciding conditions belongs to $M$ and is dense below $p$. Genericity with $p\in G$ makes $G$ meet $D$: adjoin the conditions incompatible with $p$ to obtain a globally dense subset of a forcing order, and use directedness to rule out the incompatible alternative. A condition in $G\cap D$ then gives $f=g\in M$.

The reverse inclusion follows because ground functions remain functions with the same domain and values. ThereforeThis closed forcing adds no short ground-valued sequences argument needs density of complete decisions. A single arbitrarily constructed lower bound need not belong to $G$, and would not by itself prove the claim.

Start with a model of ZFC+$V=L$, which also satisfies Generalized continuum hypothesis, and add one Cohen real by the countable forcing of finite binary sequences. Its countable chain condition for forcing preserves cardinals. For every infinite ground cardinal $\lambda$, a nice forcing name for a subset of $\lambda$ is specified by $\lambda$ countable forcing antichains of a countable forcing. The number of such forcing names is at mostin the ground model. The ground subsets already supply the preserved lower bound ${\lambda }^{+}$. Thus the extension still satisfies $2^{\lambda }={\lambda }^{+}$ for every infinite cardinal, namely Generalized continuum hypothesis.

The Cohen real is not in the ground model: for each ground real it is dense to disagree at a new coordinate. Forcing leaves the ordinals unchanged, and constructible levels are absolute, so the extension has the same constructible universe as the ground model. The new real is therefore not constructible. We obtainThe forcing theorem makes this a relative-consistency construction. Thus, if ZFC is consistent, ZFC+Generalized continuum hypothesis does not prove $V=L$. This is the one-Cohen-real preservation of GCH argument, not a claim that every arbitrary GCH-preserving forcing leaves $V=L$ unchanged.

We construct a stronger family: every injection $e_{\alpha }:\alpha \to \omega$ will have coinfinite range, and coherence means agreement modulo finitely many arguments. Begin with the empty function. At a successor, extend $e_{\alpha }$ by assigning the new argument a value outside its range; the range remains coinfinite, and coherence with earlier functions is unchanged.

At a countable limit $\alpha$, take $0={\alpha }_0<{\alpha }_1<\cdots$ cofinal in $\alpha$. We build injections $p_n:{\alpha }_n\to \omega$, extending one another exactly, with $p_n$ differing from $e_{{\alpha }_n}$ at finitely many arguments. Also reserve distinct numbers $k_0,k_1,\dots$, never used by the eventual union. After constructing $p_n$, choose $k_n$ outside its range and different from the earlier reserved numbers; its range is coinfinite since it differs finitely from that of $e_{{\alpha }_n}$.

To extend to ${\alpha }_{n+1}$, first use $e_{{\alpha }_{n+1}}$ on the new arguments while keeping $p_n$ on the old ones. Only finitely many collisions can arise: $p_n$ and $e_{{\alpha }_{n+1}}\upharpoonright {\alpha }_n$ differ finitely by the inductive coherence hypothesis, so their image sets differ finitely. The candidate is otherwise injective. There are also only finitely many new arguments assigned a value among $k_0,\dots ,k_n$. Reassign these finitely many bad arguments to distinct fresh values outside the candidate image and the reserved finite set. Infinitely many fresh values are available because the candidate image differs only finitely from the coinfinite image of $e_{{\alpha }_{n+1}}$. The resulting $p_{n+1}$ is injective, extends $p_n$, omits the reserved numbers, and differs finitely from $e_{{\alpha }_{n+1}}$.

Put $e_{\alpha }={\bigcup }_n{p}_n$. It is injective and omits all $k_n$, so its range is coinfinite. For each $\beta <\alpha$, choose $n$ with $\beta \le {\alpha }_n$. The restriction $e_{\alpha }\upharpoonright \beta$ is $p_n\upharpoonright \beta$, which differs finitely from $e_{{\alpha }_n}\upharpoonright \beta$, and hence from $e_{\beta }$. Transfinite recursion now givesThese coherent coinfinite injections into omega solve the requested problem. Reserving infinitely many omitted values is important: a union of injections with individually coinfinite ranges need not itself have coinfinite range.

Use the displayed family to form the coherent-injection Aronszajn tree. A node at level $\beta$ is a restriction $e_{\alpha }\upharpoonright \beta$ for some $\alpha \ge \beta$. Every such node differs only finitely from $e_{\beta }$. There are countably many finite subsets of the countable domain $\beta$ and countably many assignments of natural-number values to each, so there are only countably many possible finite modifications. Hence each level is countable. It is nonempty because it contains $e_{\beta }$.

Every shorter restriction of a node is again a node, and its predecessors have order type its domain ordinal. Thus this is a set-theoretic tree of height ${\omega }_1$. If it had an uncountable chain in a partial order, its domain heights would be unbounded in ${\omega }_1$, since the levels below any countable height contain only countably many nodes. The union of that chain in a partial order would be an injection ${\omega }_1\to \omega$, impossible. ThereforeThe countable-level proof uses coherence, whereas the no-branch proof uses injectivity; the two features play different roles.