Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 19 (Topics in Set Theory)

Throughout the forcing arguments, a larger condition is stronger, as in the original paper. Ground-model cardinal calculations are internal to the indicated model. All three definition alternatives are supplied in each group that permits choosing two.

An inner model of ZFC is a transitive class containing every ordinal and satisfying all its axioms. Write this class as $A$ with its inherited membership relation. Satisfaction is interpreted by restricting all quantifiers to $A$. In a first-order formulation this is a schema for a definable class, possibly with fixed parameters. The class may equal the whole universe. Transitivity means $x\in y\in A\Rightarrow x\in A$; containing all ordinals rules out treating an arbitrary transitive set model as an inner model.

A strongly inaccessible cardinal is an uncountable cardinal $\lambda$ that is both regular and a strong limit:Regularity excludes expressing $\lambda$ as the supremum of a shorter increasing sequence, while the strong-limit requirement concerns all smaller power sets. Merely being an uncountable regular limit cardinal is the weaker notion of a weakly inaccessible cardinal.

A first-order formula $\phi (x_1,\dots ,x_n)$ is an absolute formula for membership structures $A$ and $B$ if, for every tuple $\vec{a}\in (A\cap B{)}^n$,The same elements interpret the free variables in both structures; bound variables range over their respective domains. Thus absoluteness requires both directions, rather than only preservation of truth from the smaller structure to the larger one.

Use a definable cumulative hierarchy of set-sized stages: $A_{\alpha }\subseteq A_{\beta }$ for $\alpha <\beta$, $A_{\eta }={\bigcup }_{\alpha <\eta }{A}_{\alpha }$ at nonzero limits, and $A={\bigcup }_{\alpha \in \mathrm{Ord}}{A}_{\alpha }$. Class parameters and the hierarchy are fixed definable data. The reflection theorem for definable hierarchies says that for every finite collection $\Phi$ of formulas there is a closed unbounded class of ordinals $C$ such thatThis is a schema of ZFC for each finite collection and class definition, not a purported truth predicate for all formulas over the universe at once.

Close $\Phi$ under subformulas. For each existential subformula $\exists y\psi (y,\vec{x})$ and each tuple in $A_{\alpha }$, if a witness exists in $A$, take the least stage index containing a witness. There are only set-many parameter tuples and finitely many formulas. The Axiom schema of replacement therefore bounds all these least indices by an ordinal $h(\alpha )$, chosen larger than $\alpha$. No definable selection of the witnesses themselves is needed.

Above any prescribed bound choose ${\alpha }_0<{\alpha }_1<\cdots$ with ${\alpha }_{m+1}>h({\alpha }_m)$, and put $\eta ={\sup }_m{\alpha }_m$. Every tuple in $A_{\eta }$ lies in some $A_{{\alpha }_m}$, and each true existential instance for that tuple has a witness in $A_{{\alpha }_{m+1}}\subseteq A_{\eta }$. Induction over the subformulas now proves agreement between $A_{\eta }$ and $A$: atomic formulas use the same membership relation, Boolean operations preserve agreement, and the existential step uses this witness property. Thus reflecting stages are unbounded.

For closedness, suppose reflecting stages have limit $\eta$. Every tuple in $A_{\eta }$ lies in a reflecting stage below $\eta$, and every true existential instance has a witness there. The same subformula induction proves reflection at $\eta$. Hence the reflecting stages for the subformula-closed collection form a closed unbounded class, proving the Lévy reflection theorem in this relative form.

The printed inclusion-and-elementarity assertion is false if transitivity is required of the same submodel. Take the theorem of ZFC that combines the axiom of infinity and the Axiom of power set. Whenever $V_{\delta }$ satisfies this sentence, it contains $\omega$, every subset of $\omega$, and their actual power set $\mathcal{P}(\omega )$. A submodel $M\prec V_{\delta }$ contains $\omega$ and $\mathcal{P}(\omega )$, since these are uniquely definable in $V_{\delta }$. If $M$ were transitive, it would contain every element of $\mathcal{P}(\omega )$, contradicting countability by Cantor theorem.

The corrected conclusion uses an elementary embedding rather than elementary inclusion. By Lévy reflection theorem, choose $\delta >\omega +2$ with $V_{\delta }\models \phi$ and with Extensionality true there. The Downward Lowenheim-Skolem theorem says that an infinite structure in a countable language has a countable elementary substructure. Apply it to obtain a countable $N\prec (V_{\delta },\in )$. The membership relation on $N$ is externally well-founded, and elementarity makes it extensional. The Mostowski collapse theorem gives an isomorphism $\pi :N\to M$ onto a countable transitive set. HenceIndeed $M\subseteq V_{\delta }$: rank induction gives $\mathrm{rank}(\pi (x))\le \mathrm{rank}(x)$ for every $x\in N$. The crucial correction is that $j$, rather than the inclusion of $M$, is elementary. For a formula with free variables, apply this argument to its universal closure.

Let $\theta ={\xi }^{\ast }$, and work with the set structure $V_{\theta }\models \mathrm{ZFC}$. Its ordinal height is an infinite limit: a transitive model of ZFC has no largest ordinal, since it can take the successor of each ordinal it contains. Thus $\mathrm{cf}(\theta )\ge \omega$.

Suppose instead that $\mathrm{cf}(\theta )>\omega$. Enumerate all first-order formulas, and close each finite initial collection under subformulas. Applying Lévy reflection theorem inside $V_{\theta }$, choose a strictly increasing sequence ${\beta }_m<\theta$ such that $V_{{\beta }_m}$ agrees with $V_{\theta }$ on the first $m$ collections. Take $\eta ={\sup }_m{\beta }_m<\theta$. The internal rank levels here are the actual rank levels, because $V_{\theta }$ is a transitive rank model.

If $V_{\theta }\models \exists x\psi (x,\vec{a})$ with $\vec{a}\in V_{\eta }$, choose $m$ large enough to include this formula and all the parameters in $V_{{\beta }_m}$. Reflection supplies a witness in $V_{{\beta }_m}\subseteq V_{\eta }$. The Tarski-Vaught test therefore gives $V_{\eta }\prec V_{\theta }$. In particular $V_{\eta }$ satisfies all of ZFC, contrary to the minimality of $\theta$.

ConsequentlyThe countable enumeration is of formulas, not merely axioms: witness closure is what makes the union a model of the entire theory.

The structure $H_{\kappa }$ of hereditarily small sets is transitive. If $\kappa \le \omega$, all its sets are hereditarily finite, so it cannot satisfy Infinity. Therefore the assumed model has $\kappa >\omega$.

Let $\mu <\kappa$ be a cardinal. The ordinal $\mu$ belongs to $H_{\kappa }$, and every actual subset of $\mu$ also belongs to $H_{\kappa }$: its transitive closure has size at most $\max (\mu ,\omega )<\kappa$. The Power set axiom inside $H_{\kappa }$ therefore produces the actual $\mathcal{P}(\mu )$, since all subsets relevant to the internal definition are present. This power set itself belongs to $H_{\kappa }$, soThus $\kappa$ is a strong limit. Its regularity was assumed, and we have proved uncountability. Hence $\kappa$ is strongly inaccessible.

The axiom of constructibility asserts every set belongs to the constructible universe, or $V=L$. Define $L_0=\varnothing$, $L_{\alpha +1}=\mathrm{Def}(L_{\alpha })$, and $L_{\eta }={\bigcup }_{\alpha <\eta }{L}_{\alpha }$ at limit stages. Here $\mathrm{Def}(X)$ consists of subsets of $X$ first-order definable over $(X,\in )$ with finitely many parameters from $X$. Then $L={\bigcup }_{\alpha \in \mathrm{Ord}}{L}_{\alpha }$. The assertion is $\forall x\exists \alpha (x\in L_{\alpha })$, rather than a statement that every set is parameter-free definable.

The Bukovský-Hechler theorem states: if $\kappa$ is singular and there are a cardinal $\rho <\kappa$ and a cardinal $\tau$ such that $2^{\mu }=\tau$ for every cardinal $\mu$ with $\rho \le \mu <\kappa$, thenThus an eventual plateau of the power-set function below a singular cardinal continues at that cardinal. One can see the mechanism by decomposing $\kappa$ into $\mathrm{cf}(\kappa )$ bounded pieces: $2^{\kappa }=(2^{<\kappa }{)}^{\mathrm{cf}(\kappa )}$. On the plateau choose $\mu \ge \mathrm{cf}(\kappa )$, so ${\tau }^{\mathrm{cf}(\kappa )}=(2^{\mu }{)}^{\mathrm{cf}(\kappa )}=2^{\mu }=\tau$.

The singular cardinals hypothesis is the assertion that for every infinite singular cardinal $\kappa$,In particular, for a singular strong-limit cardinal it gives $2^{\kappa }={\kappa }^{+}$. This is a restriction on singular-cardinal exponentiation; it does not assert the full Generalized continuum hypothesis at all cardinals.

Put $\theta =\mathrm{cf}(\delta )$. There is a strictly increasing cofinal map $a:\theta \to \delta$. If $\mathrm{cf}(\theta )=\nu <\theta$, choose a cofinal map $b:\nu \to \theta$. Then $a\circ b$ is cofinal in $\delta$, contradicting the definition of $\theta$ as its least cofinal order type. Since always $\mathrm{cf}(\theta )\le \theta$, we obtainFor a nonzero limit ordinal this is an infinite regular cardinal. If the convention includes zero among limit ordinals, its cofinality is zero and the identity is immediate there.

First the increasing sequence is cofinal in $\kappa$. Write $\mu ={\sup }_{\xi <\delta }{\kappa }_{\xi }$. Strict increase gives $|\delta |\le \mu$ when the sequence is infinite, so its cardinal sum is $\max (|\delta |,\mu )=\mu$. The given sum therefore implies $\mu =\kappa$ and $|\delta |\le \kappa$.

For any function $f:\lambda \to \kappa$, choose an index ${\xi }_t$ with $f(t)<{\kappa }_{{\xi }_t}$ for every $t<\lambda$. There are at most $\lambda <\mathrm{cf}(\delta )$ such indices, so they are bounded below $\delta$. Enlarging the bound if needed gives an index $\xi$ with $\mathrm{ran}(f)\subseteq {\kappa }_{\xi }$. ThusConversely each summand is at most ${\kappa }^{\lambda }$, and $|\delta |\le \kappa \le {\kappa }^{\lambda }$ because $\lambda \ne 0$. Infinite cardinal multiplication consequently givesCombining the inequalities proves ${\kappa }^{\lambda }={\sum }_{\xi <\delta }{\kappa }_{\xi }^{\lambda }$.

For the intended infinite regular cardinal $\kappa$, set $S_0(\delta )=0$. Suppose $0<\zeta <{\kappa }^{+}$ and the functions for its predecessors have been constructed. Since $|\zeta |\le \kappa$, choose a surjection $e_{\zeta }:\kappa \to \zeta$ and defineThe supremum is below $\kappa$: it uses fewer than $\kappa$ ordinals below $\kappa$, and $\kappa$ is regular. This defines a function $\kappa \to \kappa$ at each stage of the recursion.

For any $\alpha <\zeta$, choose $\eta$ with $e_{\zeta }(\eta )=\alpha$. Whenever $\delta >\eta$, the displayed supremum includes $S_{\alpha }(\delta )+1$, so $S_{\alpha }(\delta )<S_{\zeta }(\delta )$. Therefore the recursion yields a long chain under eventual domination of length ${\kappa }^{+}$.

Take the graphs $G_{\zeta }=\{(\delta ,S_{\zeta }(\delta )):\delta <\kappa \}$. Each has size $\kappa$, and any two meet in fewer than $\kappa$ points, because their functions are eventually strictly ordered. Fix a bijection $b:\kappa \times \kappa \to \kappa$ and let $A_{\zeta }=b[G_{\zeta }]\in [\kappa {]}^{\kappa }$.

Given $I\in [{\kappa }^{+}{]}^{\kappa }$, enumerate it as $\{{\zeta }_{\eta }:\eta <\kappa \}$ and trim the sets in this order:The deleted part is a union of fewer than $\kappa$ pairwise intersections, each of size less than $\kappa$. Regularity ensures its size is less than $\kappa$. If $\nu <\eta$, then $A_{{\zeta }_{\eta }}^{\ast }$ is disjoint even from the full earlier $A_{{\zeta }_{\nu }}$, hence from its trimmed version. Thus all the retained sets are pairwise disjoint and each lost fewer than $\kappa$ elements. This provesThe argument is the essential disjointness of small subfamilies of a regular-cardinal almost disjoint family.

For a nonzero limit ordinal $\delta$, a subset $C\subseteq \delta$ is a club set if it is unbounded in $\delta$ and contains every one of its limit points below $\delta$. A subset $S\subseteq \delta$ is stationary ifThis defines stationarity in a limit ordinal. The regular uncountable case is the usual stationary set setting; results such as Fodor lemma require that additional hypothesis, rather than an arbitrary limit ordinal.

A set-theoretic tree is a partial order for which the strict predecessors of every node are well-ordered by the tree order. The height of a node is that predecessor order type, and $T_{\alpha }$ is the set of nodes of height $\alpha$. Under the ordinary height-and-width definition, a kappa-tree satisfies

For an arbitrary cardinal in this definition one must distinguish it from additional conventions such as being well-pruned, normal or splitting. Splitting means that every node has two incompatible extensions; it is not implied by the height-and-width clauses. This distinction is material in Question 5(ii)(a).

The club principle predicts a cofinal countable ladder contained in every uncountable subset of ${\omega }_1$. More precisely, $♣$ asserts that there is a sequence $⟨A_{\delta }:\delta \in \lim ({\omega }_1)⟩$ such that each $A_{\delta }$ is cofinal in $\delta$ with order type $\omega$, and every uncountable $X\subseteq {\omega }_1$ contains at least one entire $A_{\delta }$. It predicts a countable ladder contained in $X$, rather than predicting $X\cap \delta$ exactly. The latter is the stronger kind of guessing in the diamond principle.

For every $\beta <{\omega }_1$, choose an injection $e_{\beta }:\beta \to \omega$. Define the Ulam matrix on omega-one byFor a fixed $\alpha$, every $\beta >\alpha$ belongs to exactly one of these sets, so their union is the entire tail $(\alpha ,{\omega }_1)$. Its complement is the countable ordinal $\alpha +1$. For distinct $\alpha ,{\alpha }^{\prime }$ and fixed $n$, membership in both sets would give $e_{\beta }(\alpha )=e_{\beta }({\alpha }^{\prime })=n$ for some $\beta$ above both, contradicting injectivity. This verifies both requested properties.

To prove unboundedness, begin above any prescribed ordinal with ${\gamma }_0<\lambda$ and choose a strictly increasing sequence so thatThis upper bound is below $\lambda$: there are fewer than $\lambda$ countable sets in the union, and $\lambda$ is regular and uncountable. Put $\delta ={\sup }_m{\gamma }_m<\lambda$. For any $\alpha <\delta$, choose $m$ with $\alpha <{\gamma }_m$; then $B(\alpha )\subseteq {\gamma }_{m+1}\subseteq \delta$. Thus $\delta \in E$.

For closedness, suppose $\delta <\lambda$ is a limit point of $E$. Given $\alpha <\delta$, choose $\eta \in E$ with $\alpha <\eta <\delta$. Then $B(\alpha )\subseteq \eta \subseteq \delta$. Therefore $\delta \in E$ as well. Hence $E$ is a club set in $\lambda$. This is the club of closure points for countable set-valued functions.

Write ${◊}_S^{\prime }$ for the countable-family assertion in the PDF; its prime is not a superscript $c$. A single stationary diamond at a regular cardinal sequence gives such families by taking singletons, so ${◊}_S\Rightarrow {◊}_S^{\prime }$.

Conversely enumerate each countable family as $E_{\alpha }(n)$, $n<\omega$, padding finite families and allowing the empty set as a default. Fix a bijection $\pi :\lambda \times \omega \to \lambda$. The previous part applied to $B(\xi )=\{\pi (\xi ,n):n<\omega \}$ gives a club set $D$ on which $\pi [\alpha \times \omega ]\subseteq \alpha$. Define, for each $n$, a single candidate sequenceSuppose no candidate sequence witnesses ${◊}_S$. For each $n$ choose $X_n\subseteq \lambda$ and a club set $C_n$ such that $D_{\alpha }^n\ne X_n\cap \alpha$ for every $\alpha \in S\cap C_n$. Code all the counterexamples intoThe countable-family hypothesis guesses $X$ on a stationary set. Choose a guessing $\alpha$ in the club set $D\cap {\bigcap }_n{C}_n$, and choose $n$ with $E_{\alpha }(n)=X\cap \alpha$. For every $\xi <\alpha$, closure under $\pi$ givesThus $D_{\alpha }^n=X_n\cap \alpha$, contradicting $\alpha \in C_n$. At least one candidate is a diamond sequence, provingThis is the countable-family diamond equivalence, for every regular uncountable $\lambda$ and stationary $S\subseteq \lambda$.

There is a cofinal branch. We give a proof that does not require distinct limit-level nodes to have different predecessor chains.

The set $S=\{\alpha <\lambda :\mathrm{cf}(\alpha )=\kappa \}$ is stationary. Indeed a strictly increasing continuous $\kappa$-sequence in any club set has supremum in that club set, below $\lambda$, of cofinality $\kappa$. At each $\alpha \in S$, there are fewer than $\kappa$ pairs of nodes on $T_{\alpha }$. For every pair whose predecessor chains below $\alpha$ differ, choose a height where they differ. Since $\mathrm{cf}(\alpha )=\kappa$, all these heights are bounded by some ${\beta }_{\alpha }<\alpha$. Consequently nodes in $T_{\alpha }$ having the same predecessor at ${\beta }_{\alpha }$ have identical predecessor chains below $\alpha$.

The Fodor lemma states that a regressive function on a stationary subset of a regular uncountable cardinal is constant on a stationary subset. Applying it here, ${\beta }_{\alpha }$ has a constant value $\beta$ on a stationary subset $S^{\prime }$. Choose $t_{\alpha }\in T_{\alpha }$ for each $\alpha \in S^{\prime }$. The level $T_{\beta }$ has fewer than $\kappa <\lambda$ nodes. Partitioning $S^{\prime }$ according to the predecessor of $t_{\alpha }$ at height $\beta$, one fiber $S^{\prime \prime }$ is stationary, since the union of fewer than $\lambda$ nonstationary sets is nonstationary. Let its common predecessor be $t$.

For $\alpha <{\alpha }^{\prime }$ in $S^{\prime \prime }$, the predecessor of $t_{{\alpha }^{\prime }}$ at level $\alpha$ and $t_{\alpha }$ have the same predecessor $t$ at level $\beta$. Their chains below $\alpha$ therefore agree. For each $\nu <\lambda$, choose $\alpha \in S^{\prime \prime }$ above $\nu$ and let $b_{\nu }$ be the predecessor of $t_{\alpha }$ at level $\nu$. The preceding comparison makes $b_{\nu }$ independent of that choice. The nodes $b_{\nu }$ form a chain through every level:This proves the uniformly narrow regular-height tree branch theorem. The uniform bound below the smaller regular $\kappa$ is stronger than merely bounding each level below $\lambda$.

A forcing name for $\mathbb{P}$ is a set of ordered pairs $(\sigma ,p)$ where $p\in P$ and $\sigma$ is itself a forcing name. This recursive definition is made well-founded by assigning the forcing name rank $\sup \{\mathrm{nrk}(\sigma )+1:(\sigma ,p)\in \tau \}$. Names over the ground model are those names belonging to $M$; conditions remain ground-model objects. A name describes which recursively interpreted elements are activated by the generic filter.

With this paper's order convention, larger conditions are stronger. A generic filter $G\subseteq P$ over $M$ is nonempty, closed toward weaker conditions, and directed toward stronger conditions: if $p,q\in G$, some $r\in G$ satisfies $r\ge p,q$. It also meets every $D\in M$ which $M$ regards as dense in $\mathbb{P}$.

For a countable transitive $M$, such a filter containing any prescribed condition can be built by enumerating its dense sets and successively choosing stronger conditions in them. The definition does not require $G\notin M$: for an atomic forcing a generic filter may already belong to $M$.

A subset $E\subseteq P$ is dense above a forcing condition $p$ ifAll displayed strengthenings use the PDF convention. This only requires density in the cone of extensions of $p$; it need not be dense throughout $P$, nor downward or upward closed.

The generic extension is $M[G]=\{{\tau }^G:\tau \in M\text{ is a forcing name}\}$. It is transitive: if $x\in {\tau }^G$, an active pair in $\tau$ supplies a subname $\sigma \in M$ with $x={\sigma }^G$. Ground sets belong to it by their canonical names. Without assuming a weakest condition, use $\check{a}=\{(\check{b},p):b\in a,p\in P\}$; nonemptiness of $G$ gives ${\check{a}}^G=a$.

Induction on forcing name rank gives $\mathrm{rank}({\tau }^G)\le \mathrm{nrk}(\tau )$. The right side is an ordinal of $M$. If $\alpha \in M[G]$ is an ordinal, its rank equals $\alpha$, so it is at most a ground-model ordinal. Transitivity of $M$ then implies $\alpha \in M$. Conversely every ground ordinal remains the same ordinal in the transitive extension, since membership is unchanged. HenceThis proves forcing preserves ordinals directly from ranks, without assuming cardinal preservation.

We use only the definability and truth clauses of the forcing theorem. For every formula, its forcing relation on names is definable inside $M$; forcing is preserved by strengthening; and $M[G]\models \phi ({\vec{\tau }}^G)$ exactly when some condition in $G$ forces $\phi (\vec{\tau })$. These clauses do not presuppose the Separation axiom we are proving.

Let $a={\tau }^G$ and let ${\vec{\sigma }}^G$ be the parameters in the desired instance. In $M$, form the nameThis is a set in $M$ by its Separation axiom and forcing definability, using the subnames of $\tau$ and $P$ as a set-sized bound.

If $x\in {\rho }^G$, some $q\in G$ activates $(\nu ,q)$ and $q\ge p$ activates $(\nu ,p)\in \tau$. Thus $x={\nu }^G\in a$, and the forcing theorem gives $\phi (x,{\vec{\sigma }}^G)$. Conversely, if $x\in a$ satisfies that formula, choose an active pair $(\nu ,p)\in \tau$ with $x={\nu }^G$ and a condition $r\in G$ forcing the formula. Directedness gives $q\in G$ stronger than both $p$ and $r$. Monotonicity puts $(\nu ,q)\in \rho$, so $x\in {\rho }^G$.

ThereforeEvery requested instance of separation in a generic extension follows.

For a maximal forcing antichain $A\in M$, the set $D_A=\{q:\exists a\in A(q\ge a)\}$ is dense. Maximality ensures each condition is compatible with some $a\in A$, and a common strengthening lies in $D_A$. A generic filter meets $D_A$ and hence, by closure toward weaker conditions, meets $A$. Directedness makes any two of its conditions compatible, so it contains at most one member of an forcing antichain. Thus $|G\cap A|=1$.

Conversely, suppose this equality holds for every ground-model maximal forcing antichain. For any dense $D\in M$, use Choice in $M$ to select an forcing antichain $A\subseteq D$ maximal among forcing antichains contained in $D$. It is maximal in $P$ as well: a condition incompatible with every member of $A$ would have an extension in $D$ still incompatible with every member, enlarging $A$. The hypothesis gives a member of $G\cap A\subseteq G\cap D$. Thus $G$ meets every ground dense set and is generic.

This proves the maximal-antichain criterion for genericity in both directions.

Let $\kappa =({\omega }_2{)}^M$ be the fixed ground ordinal. We prove that $S$ remains stationary in this ordinal. The chain condition for forcing ensures that $\kappa$ remains regular: possible values of each coordinate of an ordinal-valued name form a ground set of size less than $\kappa$, by a maximal deciding forcing antichain. A hypothetical cofinal map with domain below $\kappa$ would have its range covered by fewer than $\kappa$ such small sets, hence bounded by regularity in $M$.

Let $\dot{C}$ name a club set in $\kappa$, and take $p\in G$ forcing this. For each $\alpha <\kappa$, choose in $M$ a maximal forcing antichain above $p$ deciding the least point of $\dot{C}$ strictly above $\alpha$. Its set of possible values has size less than $\kappa$, so choose a ground bound $h(\alpha )<\kappa$ larger than all of them. The ground setis a club set by the usual countable closure iteration and regularity. For every $\delta \in D$, $p$ forces $\dot{C}\cap \delta$ unbounded in $\delta$, and closedness forces $\delta \in \dot{C}$. Hence $p⊩\check{D}\subseteq \dot{C}$. In $M$, choose $\delta \in S\cap D$; that same ordinal belongs to $S\cap C$ in the extension. Stationarity at the fixed ground $\kappa$ is preserved.

There is an important qualification to the printed ${\omega }_2$ notation. If it is recomputed internally as $({\omega }_2{)}^{M[G]}$, the assertion is false without preservation of smaller cardinals. Finite partial maps from $\omega$ to $({\omega }_1{)}^M$ form a forcing of size $({\aleph }_1{)}^M$, hence have the $({\aleph }_2{)}^M$-chain condition, but collapse $({\omega }_1{)}^M$ to countable. Then $({\omega }_2{)}^M=({\omega }_1{)}^{M[G]}$, while $({\omega }_2{)}^{M[G]}$ is larger. The old $S$ is bounded in this new ${\omega }_2$, so cannot be stationary there. The proved statement uses the fixed ground ordinal, or alternatively requires the lower-cardinal preservation needed to retain its aleph index.

The valuation of a forcing name is defined recursively byThe recursion is on forcing name rank. Only pairs with an active condition in $H$ contribute elements, and their first coordinates are evaluated in the same filter. Thus a name is a ground-model set, while its value is a set in the generic extension.

A nice forcing name for a subset of a ground-model set $a$, often an ordinal, has the formwhere each $A_x$ is an antichain in $\mathbb{Q}$ and the entire construction belongs to $M$. Its value consists of the $x$ for which $H$ meets $A_x$. The antichains need not be maximal. Under the countable chain condition they are countable in $M$, which makes nice forcing names useful for counting possible subsets in extensions.

For a countable transitive ground model, the semantic forcing relation isThe forcing theorem identifies this with the recursively defined relation inside $M$ and makes it definable there. In particular, if $r\ge q$ is stronger, then $q⊩\phi$ implies $r⊩\phi$. The names are interpreted in each relevant generic filter, not replaced by their value in one fixed extension in the definition.

Under the ordinary height-and-width definition of an ${\aleph }_1$-tree, the printed claim is false. The chain $({\omega }_1,\le )$ has one node at each level, every antichain has size at most one, and it is itself an uncountable chain.

Here is the intended argument under the additional splitting convention. Any uncountable chain in a tree with countable levels is unbounded in height, because a bounded set of levels below ${\omega }_1$ has only countably many nodes. Its predecessor closure therefore gives a cofinal branch $B$. At each node on $B$, splitting supplies an extension off $B$, incompatible with a later node on $B$. Recursively for $\xi <{\omega }_1$, choose such an off-branch node $s_{\xi }$, then move sufficiently far along $B$ that all subsequent choices are above a branch node incompatible with $s_{\xi }$. At a limit stage the previous countably many heights are bounded below ${\omega }_1$, so the recursion continues. The $s_{\xi }$ are pairwise incompatible, an uncountable antichain.

Thus a splitting ${\aleph }_1$-tree with only countable antichains has no uncountable chains. A splitting convention must be stated; the raw tree hypothesis alone does not suffice.

Use the equivalent tree formulation of the Suslin hypothesis: there is no normal, well-pruned Suslin tree of height ${\omega }_1$, with countable levels and no uncountable chains or antichains. This is equivalent to the linear-order formulation that every complete dense order without endpoints satisfying the countable chain condition for a linear order is separable.

If such a tree $T$ existed, use its nodes as forcing conditions, with a higher extension stronger. It is CCC because its antichains are countable. For every $\alpha <{\omega }_1$, the set $D_{\alpha }=\{t:\mathrm{ht}(t)\ge \alpha \}$ is dense because the tree is well-pruned. The assertion ${\mathrm{MA}}_{{\aleph }_1}$ is precisely that a CCC forcing and a family of at most ${\aleph }_1$ dense sets admit a filter meeting all of them.

Apply it to these dense sets. A directed filter in a tree is a chain: two compatible nodes are comparable, since both lie among the well-ordered predecessors of a common extension. Meeting every $D_{\alpha }$ makes this chain cofinal, contrary to the defining absence of uncountable chains in a Suslin tree. Consequently

Start with a ground model containing a normal splitting Suslin tree $T$, and force with $\mathrm{Add}(\omega ,{\omega }_2)$, the finite partial functions from ${\omega }_2\times \omega$ to $2$. Such a starting model is relatively consistent with ZFC: the constructible universe has $◊$, and $◊$ implies the existence of a Suslin tree.

This Cohen forcing is Knaster. In an uncountable family of finite conditions, the delta-system lemma yields an uncountable family whose domains have one common root; thinning to common values on that finite root makes its members pairwise compatible. If $P$ is Knaster and $T$ is CCC, the product $P\times T$ is CCC: thin any uncountable family to pairwise compatible $P$ coordinates, then use CCC to find two compatible $T$ coordinates. Thus $P$ forces that the old $T$ is still CCC A name for an uncountable antichain would otherwise produce an uncountable antichain in this product by deciding its nodes.

Countable levels and the tree's splitting property persist, and the splitting argument from the previous part rules out a new cofinal branch in a CCC tree. Therefore the old $T$ is still Suslin. Meanwhile the forcing adds at least ${\aleph }_2$ distinct reals, with all cardinals preserved, so $2^{{\aleph }_0}\ge {\aleph }_2$.

We obtain a model with a Suslin tree and $\neg \mathrm{CH}$. Hence existence of a Suslin tree does not imply the Continuum hypothesis. This uses Knaster forcing preserves Suslin trees, not a claim that arbitrary CCC forcing preserves them.

For uncountable regular $\kappa$, let $\theta =\mathrm{otp}(C_{\zeta })$. Since $C_{\zeta }$ is a club set in $\zeta$, $\mathrm{cf}(\theta )=\mathrm{cf}(\zeta )=\kappa$, so $\theta \ge \kappa$.

If $\theta >\kappa$, its cofinality ensures $\theta >\kappa +\omega$: ordinals strictly between $\kappa$ and $\kappa +\omega$ are successors, and $\kappa +\omega$ has cofinality $\omega$. In the continuous increasing enumeration $c$ of $C_{\zeta }$, take $\gamma =c(\kappa +\omega )$. It is a limit point of $C_{\zeta }$ and has cofinality $\omega <\kappa$. Coherence gives $C_{\gamma }=C_{\zeta }\cap \gamma$, with order type $\kappa +\omega$ and size $\kappa$. This contradicts the size clause for ordinals of cofinality below $\kappa$. ThusThe printed hypotheses need this qualification. At $\kappa =\omega$, choose $C_{\zeta }=\zeta$ for every nonzero countable limit ordinal. These club sets are coherent, and the small-cofinality clause is vacuous. But at $\zeta ={\omega }^2$ their order type is ${\omega }^2$, not $\omega$. The standard square principle includes an order-type bound $\mathrm{otp}(C_{\zeta })\le \kappa$, which also repairs this countable case.

Fix $\delta <{\kappa }^{+}$ with $\mathrm{cf}(\delta )>\omega$. The limit points of $C_{\delta }$ form a club set in $\delta$. For each such point $\gamma$, coherence givesAs $\gamma$ increases through those limit points, these order types strictly increase. Hence the club set contains at most one member of a fixed fiber $E_{\xi }$. Removing an initial segment past that member leaves a club set disjoint from $E_{\xi }$. Therefore $E_{\xi }\cap \delta$ is not stationary. This proves each $E_{\xi }$ is non-reflecting, by the nonreflection of fixed order-type fibers.

For uncountable regular $\kappa$, part (a) gives type $\kappa$ at cofinality $\kappa$, and the size clause gives type less than $\kappa$ at smaller cofinalities. All nonzero limit ordinals below ${\kappa }^{+}$ fall into one of these cases. ThusThe index set $\kappa +1$ has cardinality $\kappa$, so this is a disjoint union of $\kappa$ non-reflecting subsets. Empty fibers can be omitted; if nonempty pieces are required, a fiber of size ${\kappa }^{+}$ can be split further, since nonreflection is inherited by subsets. Such a fiber exists because $\kappa$ sets of size at most $\kappa$ cannot cover ${\kappa }^{+}$ points.

At $\kappa =\omega$, the nonreflection definition is vacuous for every subset of ${\omega }_1$, since no smaller ordinal has uncountable cofinality. The partition conclusion is therefore still true by an arbitrary countable partition, although the particular fibers from the weak printed square assumptions need not cover all limit ordinals. Under the standard order-type-bounded square definition, the displayed fiber partition works in this case too.

The forcing theorem has two central clauses. Definability: for each first-order formula $\phi$, the relation $p⊩\phi (\vec{\tau })$ on conditions and ground-model names is first-order definable in $M$. Truth lemma: for every $M$-generic $G$ and every tuple of names in $M$,Forcing is monotone under strengthening and agrees with the generic-extension semantics. Together with the generic model construction, $M[G]$ is a transitive model of ZFC containing $M$ with the same ordinals. The definability and truth clauses are the auxiliary parts used when proving individual extension axioms.

The canonical forcing name $\dot{a}=\check{a}$ is defined recursively byIf a weakest condition $1_P$ is provided, one may instead pair each $\check{b}$ only with $1_P$. The all-conditions version works for an arbitrary nonempty forcing without that extra convention. Induction on rank gives ${\check{a}}^G=a$ for every generic filter, since each element $b\in a$ is activated by some condition of the nonempty $G$. Thus all ground-model objects have canonical names.

A precise form of the generalized delta-system lemma is this. Let $\mu <\theta$ be infinite cardinals, with $\theta$ regular and uncountable, and suppose $|[\alpha {]}^{<\mu }|<\theta$ for every $\alpha <\theta$. Every family $\mathcal{A}$ of $\theta$ distinct sets, each of cardinality less than $\mu$, contains a subfamily $\mathcal{B}$ of size $\theta$ and a set $r$ such thatThe members form a delta-system with root $r$. A sufficient usual arithmetic hypothesis is ${\alpha }^{\mu }<\theta$ for all infinite cardinals $\alpha <\theta$. The finite-set case $\mu =\omega$ needs only regular uncountable $\theta$, since finite subsets of each $\alpha <\theta$ form a family of size less than $\theta$. This is the form used for the finite-support collapse below.

Choose a name $\dot{f}\in M$ for $f$ and a condition $p\in G$ forcing that it is a function $\alpha \to \beta$. The ordinals $\alpha ,\beta$ belong to $M$, by forcing preserves ordinals. For each $\zeta <\alpha$, choose in $M$ a maximal forcing antichain $A_{\zeta }$ in the cone above $p$, every member deciding $\dot{f}(\zeta )$ as an ordinal below $\beta$. Conditions deciding an ordinal-valued name are dense, by the forcing theorem.

Let $y(\zeta )$ be the set of values decided by members of $A_{\zeta }$. The chain condition gives $|y(\zeta ){|}^M<\kappa$, and Choice and Replacement in $M$ assemble all these sets into a function $y\in M$ with domain $\alpha$. Since $p\in G$, genericity ensures that $G$ meets each deciding forcing antichain above $p$; equivalently enlarge it to a global maximal forcing antichain by conditions incompatible with $p$. Its chosen value is the actual $f(\zeta )$. ThereforeThis is the possible-values lemma for chain-condition forcing. All size bounds in its construction are internal to the ground model.