Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 19 (Topics in Set Theory)

In the Lévy hierarchy, a bounded formula in set theory has only quantifiers of the forms $\forall x\in y$ and $\exists x\in y$. A Sigma-one formula in set theory is an existential unbounded quantifier block followed by a bounded formula in set theory, and a Pi-one formula in set theory has a universal unbounded block. A formula is a Delta-one formula modulo ZFC if ZFC proves it equivalent, with the same free variables, both to a ${\Sigma }_1$ formula and to a ${\Pi }_1$ formula. Equivalence modulo the specified theory is part of the definition; it is not necessary that the original string have both syntactic forms. Such formulas satisfy the usual Delta-one absoluteness between transitive models satisfying the relevant axioms. The modulo-ZF analogue is a Delta-one formula in set theory.

A Sierpiński set is an uncountable subset $S$ of $\mathbb{R}$ such that $S\cap N$ is countable for every set $N$ of Lebesgue measure zero. Equivalently it is enough to test Borel null sets, because every null set is contained in a Borel null set. Uncountability and countable intersection with every null set are both required. This is the measure analogue of the category-based Luzin set condition; it does not say that $S$ itself is measurable.

The cumulative-hierarchy reflection principle says that for every finite collection $\Phi$ of formulas and every ordinal $\gamma$, there is an ordinal $\alpha >\gamma$ such that for all $\phi \in \Phi$ and all parameters $a_1,\dots ,a_n\in V_{\alpha }$,Equivalently, all quantifiers on the right are relativized to $V_{\alpha }$. It is a theorem schema of ZF for finite lists of formulas, with reflecting stages arbitrarily high. It does not assert one set-sized stage elementary for every formula at once. By beginning above the ranks of any specified finite parameter list, the parameters can also be required to belong to the reflecting stage.

Let $f$ be given and choose $Y\subseteq \mathbb{R}$ of cardinality ${\aleph }_1$. Each $f(y)$ is countable, so the axiom of choice and infinite cardinal arithmetic giveSince the continuum is larger, choose $x$ outside this union. Then $x\notin f(y)$ for every $y\in Y$. The countable set $f(x)$ cannot contain $Y$, so choose $y\in Y\setminus f(x)$. ThusThe two points are distinct because $x\notin Y$. This proves the countable-valued free-pair criterion and the required direction of the Freiling axiom of symmetry without any measurability assumption on $f$.

Assume the Continuum hypothesis and enumerate $\mathbb{R}=\{r_{\alpha }:\alpha <{\omega }_1\}$. DefineEvery value is countable. For any two indices, say $\alpha \le \beta$, one has $r_{\alpha }\in f(r_{\beta })$, so the two required nonmembership conditions cannot both hold. Including the endpoint in each initial segment also rules out taking the two points equal. Therefore the free-pair assertion implies the negation of Continuum hypothesis. Cantor theorem and choice already give $2^{{\aleph }_0}\ge {\aleph }_1$, so

The Gödel constructible universe theorem gives $\mathrm{Con}(\mathrm{ZFC})\Rightarrow \mathrm{Con}(\mathrm{ZFC}+\mathrm{CH})$. The forcing independence theorem for the Continuum hypothesis gives $\mathrm{Con}(\mathrm{ZFC})\Rightarrow \mathrm{Con}(\mathrm{ZFC}+\neg \mathrm{CH})$. For example, the latter can be obtained by first passing to the constructible universe and then adding sufficiently many Cohen reals. By the equivalence in part (b), these are respectively models of the negation and affirmation of the free-pair assertion. Thus if ZFC is consistent, the assertion is independent of ZFC. The consistency qualification is essential: an inconsistent theory proves every sentence. These are syntactic relative-consistency implications, not a claim that bare consistency supplies a countable transitive model.

The axiom of infinity excludes $V_{\omega }$ and all smaller stages, so the cardinal $\lambda$ is uncountable. Full semantics for second-order logic is decisive: every externally specified functional relation whose ordered pairs lie in $V_{\lambda }$ is an allowed class parameter, even if the relation itself is not an element of $V_{\lambda }$.

If $\mathrm{cf}(\lambda )=\mu <\lambda$, choose an external cofinal function $h:\mu \to \lambda$. Its graph is an allowed class parameter: each input, output and ordered pair has rank of a set below the infinite cardinal $\lambda$, which is a limit ordinal. The domain $\mu$ belongs to $V_{\lambda }$. The second-order form of Axiom schema of replacement would give its range as an element of $V_{\lambda }$. But a cofinal subset of $\lambda$ has rank of a set $\lambda$ and cannot belong to $V_{\lambda }$. Therefore $\mathrm{cf}(\lambda )=\lambda$.

If $2^{\theta }\ge \lambda$ for a cardinal $\theta <\lambda$, choice supplies an external surjection $q:\mathcal{P}(\theta )\to \lambda$. The domain belongs to $V_{\lambda }$ and the ordered pairs of its graph have ranks below $\lambda$. Applying Axiom schema of replacement to this external functional class would put its range, the ordinal $\lambda$, in $V_{\lambda }$, again impossible. Hence $2^{\theta }<\lambda$ for every $\theta <\lambda$.

ConsequentlyThese are exactly the defining conditions for a strongly inaccessible cardinal. This full second-order replacement rank obstruction would not follow from Henkin semantics, where the available class relations may exclude the external functions just used.

In the usual tail-normalized version, an Ulam matrix on omega-one is a family $A_{\alpha ,n}$ such that, for each $\alpha <{\omega }_1$, the sets indexed by $n<\omega$ partition the tail $(\alpha ,{\omega }_1)$, while for each fixed $n$ the sets indexed by $\alpha <{\omega }_1$ are pairwise disjoint. Empty cells are allowed; harmless bounded-tail variants give the same applications.

For a concrete realization, choose injections $e_{\beta }:\beta \to \omega$ for every countable ordinal $\beta$, and setFor fixed $\alpha$ each $\beta >\alpha$ chooses exactly one $n$; for fixed $n$, injectivity of $e_{\beta }$ prevents one $\beta$ from belonging to two cells. This verifies the two matrix conditions explicitly.

König theorem for cardinal numbers states that if ${\kappa }_i<{\lambda }_i$ for each $i\in I$, then ${\sum }_i{\kappa }_i<{\prod }_i{\lambda }_i$. Its cofinality formulation gives, for every infinite cardinal $\kappa$,Taking a cofinal sequence of smaller cardinals in $\kappa$ and comparing their sum with the product of their successors gives this inequality. Another standard consequence, for infinite $\lambda$ and $\kappa \ge 2$, isIndeed, if $\mu =\mathrm{cf}({\kappa }^{\lambda })\le \lambda$, the preceding inequality at ${\kappa }^{\lambda }$ would contradict $({\kappa }^{\lambda }{)}^{\mu }={\kappa }^{\lambda \mu }={\kappa }^{\lambda }$.

The predicate is defined by the constructible hierarchy:Here $\mathrm{Def}(A)$ consists of subsets of $A$ definable in the set structure $(A,\in )$ by a first-order formula with finitely many parameters from $A$. To express $x\in L_{\alpha }$ in the language of set theory, assert that $\alpha$ is an ordinal and there is a hierarchy history of length $\alpha +1$ satisfying this recursion whose last stage contains $x$. Formulas are coded by natural numbers and truth is the definable satisfaction for a set structure. Transfinite recursion gives a unique history. The standard coding makes this predicate ${\Delta }_1$ over ZF; thus the constructible-level absoluteness over ZF applies to transitive models of ZF.

The bounding number $\mathfrak{b}$ is the least size of an unbounded family in $({\omega }^{\omega },{\le }^{\ast })$. The almost disjointness number $\mathfrak{a}$ is the least size of an infinite maximal almost disjoint family on omega of infinite subsets of $\omega$. Requiring the family to be infinite excludes trivial finite maximal partitions.

A countable family $\{f_i:i<\omega \}$ is bounded by $g(n)=1+{\max }_{i\le n}{f}_i(n)$. Thus ${\aleph }_1\le \mathfrak{b}$.

Suppose an infinite almost disjoint family on omega $\mathcal{A}$ has size less than $\mathfrak{b}$. Choose distinct members $A_n$ and put $C_n=A_n\setminus {\bigcup }_{i<n}{A}_i$. These are infinite and pairwise disjoint. For each $B\in \mathcal{A}$, define $h_B(n)=\max ((B\cap C_n)\cup \{0\})$ whenever the intersection is finite. If $B=A_n$ at the exceptional index $n$, set $h_B(n)=0$. These are the only possible infinite intersections. A single $g$ eventually dominates every $h_B$, because there are fewer than $\mathfrak{b}$ of them. Pick $x_n\in C_n$ with $x_n>g(n)$ and put $X=\{x_n:n<\omega \}$.

For $B$ not among the selected $A_n$, only finitely many $x_n$ can lie in $B$. The same is true for $B=A_j$, with its single exceptional index $j$ ignored. Thus $X$ is infinite and almost disjoint from every member of $\mathcal{A}$, so the family was not maximal. This bounding-to-almost-disjointness inequality proves $\mathfrak{b}\le \mathfrak{a}$.

Finally start with an infinite pairwise disjoint family and use Zorn's lemma to extend it to a maximal almost disjoint family on omega. It is a family of subsets of $\omega$, so has cardinality at most $2^{{\aleph }_0}$. Therefore

Fix $a,\vec{p}\in L$ and a formula $\theta (x,\vec{p})$. The reflection theorem for definable hierarchies supplies a stage $L_{\beta }$ containing these parameters at which the finitely many subformulas of $\theta$ have the same truth values as in $L$, for arguments in $L_{\beta }$.

This reflection can be proved in the ambient ZF without presupposing Axiom schema of separation in $L$: starting at a stage containing the parameters, collect for each relevant existential subformula and every tuple in the current stage the least constructible stage in which a witness occurs, whenever a witness exists in $L$. Ambient Axiom schema of replacement bounds these ordinals. Iterate these witness-stage bounds for $\omega$ steps and take their supremum. Induction on subformulas gives the required finite reflection at the resulting union stage. Only the least stage is chosen, so no ambient axiom of choice is needed.

Since $L_{\beta }$ is transitive and contains $a$, the desired separated subset isIt is definable over $L_{\beta }$ with parameters there, so $b\in \mathrm{Def}(L_{\beta })=L_{\beta +1}\subset L$. Reflection gives $b=\{x\in a:{\theta }^L(x,\vec{p})\}$. Hence ZF proves every Axiom schema of separation instance relativized to $L$. This is a direct Separation proof in the constructible universe, not an appeal to the assertion being proved.

Use the transitive-model convention for “standard” in this assertion: the domain is a transitive set and membership is actual membership. A standard membership model of set theory without transitivity is a weaker notion and would not justify the asserted absoluteness. Ordinalhood is bounded-formula absolute, so the internal ordinals are precisely $A\cap \mathrm{Ord}$. Satisfaction for a set structure is absolute between transitive ZF models containing that structure: the domain, finite parameter tuples, formula codes and the recursive truth clauses are the same. Equivalently one may use the ${\Delta }_1$ absoluteness of the predicate defining $L_{\alpha }$.

Transfinite induction now gives $L_{\alpha }^{\mathbb{A}}=L_{\alpha }$ for each ordinal $\alpha \in A$. At successors both computations use the same definable subsets of the same set structure; at limits they take the same union over all smaller ordinals, which belong to $A$ by transitivity. ThusFor a set-sized model of ordinal height $\theta$, this is $L_{\theta }$. The restriction to actual ordinals makes the printed union precise. Mere well-foundedness of a nontransitive membership substructure would not justify these absoluteness steps; transitivity is the intended standard-model convention.

Let $N$ be any transitive class model of ZF containing every ordinal. By part (b), all its internally computed constructible levels are the actual levels, and each belongs to, and is contained in, $N$. Their union therefore gives $L\subseteq N$. Using the permitted fact that $L\models \mathrm{ZF}$, and that $L$ itself contains every ordinal, we concludeThe word class matters: a set cannot contain every ordinal. Set-sized standard models instead contain their own truncated constructible universe as described in part (b).

Let $E=\{\alpha <\kappa :\mathrm{cf}(\alpha )=\omega \}$. This set is stationary: in any club set, choose a strictly increasing countable sequence and take its supremum, which lies in the club set and has cofinality $\omega$. For each $\alpha \in E$ choose an increasing cofinal sequence $c_{\alpha }:\omega \to \alpha$.

Fix $\beta <\kappa$. For every $\alpha \in E$ above $\beta$, some $c_{\alpha }(n)$ exceeds $\beta$. Partition this stationary tail by the least such $n$. A countable union of nonstationary sets is nonstationary, because fewer than $\kappa$ club sets have club filter completeness, so some cell is stationary. On it the regressive function $\alpha \mapsto c_{\alpha }(n)$ has, by Fodor lemma, a stationary fiber at a value $\gamma >\beta$.

Let $I_n=\{\gamma <\kappa :\{\alpha \in E:c_{\alpha }(n)=\gamma \}\text{ is stationary}\}$. The preceding argument says ${\bigcup }_n{I}_n$ is unbounded in $\kappa$. Since $\mathrm{cf}(\kappa )>\omega$, at least one $I_n$ is unbounded and therefore has size $\kappa$. Its fibers are pairwise disjoint stationary sets. Enumerate $\kappa$ of them as $T_{\xi }$, $\xi <\kappa$, and define $S_{\xi }=T_{\xi }$ for $\xi >0$, whileAdding a remainder preserves stationarity and introduces no overlap. ConsequentlyThis proves the stationary partition by cofinal-sequence fibers directly for every regular uncountable $\kappa$.

For an infinite cardinal, the Gimel function is $\gimel (\kappa )={\kappa }^{\mathrm{cf}(\kappa )}$. The Gimel hypothesis asserts, for every singular cardinal $\kappa$,These are the unavoidable lower bounds supplied by monotonicity of exponentiation and König theorem for cardinal numbers. The hypothesis imposes the least allowed value at singular cardinals; it does not constrain the continuum function on regular cardinals to their successors. Thus it is weaker than Generalized continuum hypothesis. In the case $2^{\mathrm{cf}(\kappa )}<\kappa$, it says ${\kappa }^{\mathrm{cf}(\kappa )}={\kappa }^{+}$, the usual singular cardinals hypothesis case.

Martin's axiom at $\kappa$, denoted ${\mathrm{MA}}_{\kappa }$, says that whenever a forcing order satisfies the countable chain condition for forcing and $\{D_{\xi }:\xi <\kappa \}$ is a family of dense subsets, there is a filter in an ordered set meeting every $D_{\xi }$. Equivalently, allow any family of at most $\kappa$ dense subsets. The filter in an ordered set is directed towards stronger common extensions and closed towards weaker conditions. Full Martin's axiom requires ${\mathrm{MA}}_{\kappa }$ for every $\kappa <2^{{\aleph }_0}$. The bound below the continuum is part of the usual full axiom, explaining its role in part (iv)(b).

A well-pruned set-theoretic tree of height $\kappa$ has the property that every node $t$ extends to every higher level below $\kappa$: if $\mathrm{ht}(t)<\beta <\kappa$, there is $u$ of height $\beta$ with $t{<}_{T}u$. Equivalently, the heights of extensions of every node are unbounded in $\kappa$, since taking predecessors then gives an extension at any prescribed intermediate level. This is stronger than merely having no terminal nodes. For a kappa-tree we use the usual regular uncountable height cardinal and levels of size less than that cardinal.

The Beth numbers form a strictly increasing continuous sequence: strictness comes from Cantor theorem, and at a limit $\delta$ one has ${\beth }_{\delta }={\sup }_{\xi <\delta }{\beth }_{\xi }$. A cofinal sequence in $\delta$ of length $\mathrm{cf}(\delta )$ therefore induces a cofinal sequence in ${\beth }_{\delta }$, giving $\mathrm{cf}({\beth }_{\delta })\le \mathrm{cf}(\delta )$.

Conversely, let $\{a_i:i<\mathrm{cf}({\beth }_{\delta })\}$ be cofinal in ${\beth }_{\delta }$. For each $i$ choose ${\xi }_i<\delta$ with $a_i<{\beth }_{{\xi }_i}$. The indices must be cofinal in $\delta$; otherwise all $a_i$ would be bounded by a single smaller Beth number. Thus $\mathrm{cf}(\delta )\le \mathrm{cf}({\beth }_{\delta })$, andThis is the cofinality of a continuous cardinal hierarchy at a nonzero limit index; the limit hypothesis is essential to the supremum argument.

We give an explicit Gimel recursion for cardinal exponentiation. First recover the continuum function by induction on infinite cardinals. At a regular $\kappa$,because $2^{\kappa }\le {\kappa }^{\kappa }\le (2^{\kappa }{)}^{\kappa }=2^{\kappa }$. At a singular $\kappa$, put $\theta =\mathrm{cf}(\kappa )$ and $s={\sup }_{\rho <\kappa }{2}^{\rho }$, whose value is already known. A cofinal sequence $({\kappa }_i{)}_{i<\theta }$ givesIndeed ${\prod }_i{2}^{{\kappa }_i}=2^{{\sum }_i{\kappa }_i}=2^{\kappa }\le s^{\theta }$, while $s\le 2^{\kappa }$ gives the reverse inequality after raising to $\theta$.

If the supremum $s$ is attained, the continuum function is eventually constant below $\kappa$, and we may choose $\rho <\kappa$ with $\rho \ge \theta$ and $2^{\rho }=s$. Then $s^{\theta }=(2^{\rho }{)}^{\theta }=s$. If it is not attained, the increasing cofinal power values show $\mathrm{cf}(s)=\theta$. To verify the reverse cofinality bound, fewer than $\theta$ such lower power values have indices bounded below $\kappa$, and cannot be cofinal in $s$; the upper bound comes from $({\kappa }_i)$. ConsequentlyEvery singular-stage value is therefore determined by the previously computed powers and the given Gimel function.

Now fix an infinite exponent $\lambda$ and recurse on the infinite base $\kappa$. If $2\le \kappa \le \lambda$, then ${\kappa }^{\lambda }=2^{\lambda }$, already known. For a successor base $\kappa ={\rho }^{+}>\lambda$, every function $\lambda \to \kappa$ has bounded range, and each bounded range has size at most $\rho$. HenceThis is the Hausdorff formula for cardinal exponentiation. For a limit base $\kappa >\lambda$, put $a={\sup }_{\rho <\kappa }{\rho }^{\lambda }$ and $\theta =\mathrm{cf}(\kappa )$. If $\theta >\lambda$, all ranges are bounded and counting over those bounds gives ${\kappa }^{\lambda }=a$ (here $a\ge \kappa$).

If $\theta \le \lambda$, then ${\kappa }^{\lambda }=a^{\theta }$. To see the nontrivial upper bound, use a cofinal sequence of bounds ${\kappa }_i$. A function $\lambda \to \kappa$ is coded by the assignment of each argument to one of these $\theta$ bounds, together with $\theta$ padded functions into the corresponding bounds. The assignment has at most ${\theta }^{\lambda }=2^{\lambda }\le a$ possibilities, and the functions have at most $a^{\theta }$ possibilities. Conversely $a\le {\kappa }^{\lambda }$ and $({\kappa }^{\lambda }{)}^{\theta }={\kappa }^{\lambda }$, giving the lower bound. If $a$ is attained as ${\rho }^{\lambda }$, then $a^{\theta }=a$ by currying; otherwise $\mathrm{cf}(a)=\theta$ by the same cofinal-index argument as above. ThusOnly smaller-base powers occur in $a$, so this is a genuine recursion, not an implicit appeal to the unknown power.

Finally, finite positive exponents give ${\kappa }^n=\kappa$ for infinite $\kappa$; finite bases at infinite exponents satisfy $m^{\lambda }=2^{\lambda }$ for $m\ge 2$. The cases with base $0$ or $1$, exponent $0$, or both arguments finite are elementary, with $0^0=1$ under the empty-function convention. The Gimel function therefore determines both requested class functions completely.

For a regular uncountable kappa-tree, keep exactly the nodes whose extensions have unbounded heights:This set is predecessor-closed. At any level $\alpha$, if no node survived, the extension heights above each of its fewer than $\kappa$ nodes would be bounded. Regularity would give a single bound for their union, contradicting the height of the original set-theoretic tree. Thus every level of $T^{\ast }$ is nonempty and still has size less than $\kappa$.

If $t\in T^{\ast }$ and $\beta >\mathrm{ht}(t)$, consider its extensions at level $\beta$. If none survived, fewer than $\kappa$ bounded extension sets would again bound every extension of $t$, a contradiction. Therefore a surviving level-$\beta$ extension exists. HenceThis unbounded-extension kernel of a regular tree uses regularity essentially. If one permits singular-height set-theoretic trees in the term “$\kappa$-tree”, the unrestricted assertion is false: take fewer than $\kappa$ disjoint branches with lengths cofinal in a singular $\kappa$. The levels are small and the height is $\kappa$, but no node has unbounded extensions. A common root can be added without creating a well-pruned subtree. The usual regular-height convention is therefore the one used here.

Prune an ${\aleph }_1$-Suslin tree as in part (a), then use its nodes as forcing conditions, with extensions stronger. Two conditions are compatible exactly when comparable, so the absence of uncountable tree antichains is the forcing countable chain condition for forcing. For each $\alpha <{\omega }_1$, the set $D_{\alpha }$ of nodes of height at least $\alpha$ is dense, by well-pruned set-theoretic tree.

If $2^{{\aleph }_0}>{\aleph }_1$, full Martin's axiom includes ${\mathrm{MA}}_{{\aleph }_1}$. It would provide a filter in an ordered set meeting all these dense subsets of a forcing order. Directedness makes that filter in an ordered set a chain in a partial order, and meeting every $D_{\alpha }$ makes its heights unbounded, producing an uncountable branch. This contradicts the Suslin-tree property. A Suslin set-theoretic tree together with failure of Continuum hypothesis therefore implies failure of Martin's axiom. This is the Suslin-tree obstruction to Martin's axiom.

The partition relation $\kappa \to (\lambda {)}_{\beta }^{\alpha }$ means that for every coloring $c:[\kappa {]}^{\alpha }\to \beta$, there is $H\subseteq \kappa$ with $|H|=\lambda$ such that $c$ is constant on $[H{]}^{\alpha }$. Here $[X{]}^{\alpha }$ denotes the subsets of $X$ of cardinality $\alpha$, and the subscript is the number of colors. The domain consists of subsets of size $\alpha$, not ordered tuples, and the definition also makes sense for infinite arity.

The stationary-indexed club principle ${♣}_S$ asserts a sequence $⟨A_{\delta }:\delta \in S\cap \mathrm{Lim}({\omega }_1)⟩$ with each $A_{\delta }\subset \delta$ cofinal of order type $\omega$, such that for every uncountable $X\subseteq {\omega }_1$ some $\delta \in S$ satisfies $A_{\delta }\subseteq X$. It predicts a cofinal ladder contained in the target, rather than predicting the entire initial segment of the target as diamond does. For a stationary $S$ with successor members, only its stationary limit-ordinal part is relevant; countable limit ordinals have cofinality $\omega$. This is the restricted version of the club principle.

A tree with unique limits has no distinct nodes at a limit level with the same history. Precisely, for a limit ordinal $\delta$ and $s,t\in T_{\delta }$,Here $s\upharpoonright \beta$ denotes the unique predecessor of $s$ of height $\beta$. This is uniqueness of a limit node when it exists, not a requirement that every cofinal chain in a partial order below a limit level acquire a limit node.

Use the standard normal set-theoretic tree convention: a unique root, extensions at every higher level, splitting into at least two successors, and tree with unique limits. The small-level and height assumptions already give an ${\omega }_1$-tree, while the given tree antichain condition gives the countable chain condition for forcing. We only need to exclude an uncountable branch.

If such a branch existed, its heights would be unbounded, since each initial segment contains only countably many nodes. Fill in predecessors to obtain its node $b_{\alpha }$ at every level. At each successor step choose a successor $s_{\alpha }$ of $b_{\alpha }$ different from $b_{\alpha +1}$. For $\alpha <\beta$, the node $s_{\beta }$ extends the branch successor $b_{\alpha +1}$, and so is incompatible with $s_{\alpha }$. Thus $\{s_{\alpha }:\alpha <{\omega }_1\}$ is an uncountable tree antichain, a contradiction.

Therefore the set-theoretic tree is ${\aleph }_1$-Suslin. The splitting part of normality matters: a single chain in a partial order of height ${\omega }_1$ would satisfy the tree antichain condition but not the conclusion if one used a weakened definition of normality allowing no splitting.

Fix a diamond principle sequence $D_{\alpha }\subseteq \alpha$. Construct a normal splitting set-theoretic tree of height ${\omega }_1$ with countable levels. At successors give every node two successors. At a countable limit stage $\alpha$, the set-theoretic tree below $\alpha$ is countable. Choose countably many cofinal branches through it covering all its nodes, and put one node at level $\alpha$ above each distinct chosen branch. This preserves extension to all higher levels and tree with unique limits.

Arrange a coding of each level into the ordinal block $[\omega \alpha ,\omega (\alpha +1))$. On the club set of limit fixed points of $\alpha \mapsto \omega \alpha$, the nodes coded below $\alpha$ are exactly the nodes of height below $\alpha$. At a limit stage, if $D_{\alpha }$ codes a maximal tree antichain of the current set-theoretic tree below $\alpha$, require every chosen branch to meet it. This is possible: for any starting node $t$, maximality provides a comparable tree antichain member; if above $t$, first extend to it, and if below $t$, it has already been met. Then extend along a sequence of heights cofinal in $\alpha$. If the prediction is not a maximal tree antichain, use the ordinary covering branches. Thus every level is countable and the construction remains normal.

Here is the full chain-condition verification. Let $A$ be a maximal tree antichain in the final set-theoretic tree. For every node $t$, choose a witness $a_t\in A$ comparable with $t$. There is a club set of countable limit stages $\alpha$ closed under these witness choices: starting from any bound, repeatedly bound the heights of witnesses for all the countably many nodes below the current stage, and take the supremum after countably many steps. At such an $\alpha$, $A\cap T_{<\alpha }$ is already maximal in $T_{<\alpha }$.

View $A$ as a subset of ${\omega }_1$ through the coding. Diamond gives stationarily many stages with $D_{\alpha }=A\cap \alpha$. Choose one also in the witness-closure club set and the coding club set. The construction at that stage seals this very tree antichain: every node of level $\alpha$ extends one of its members below $\alpha$, and so does every node at a later level. No such node can itself belong to $A$, since it is comparable with an earlier member of $A$. HenceEvery tree antichain extends to a maximal one, so the set-theoretic tree has no uncountable tree antichain. Its normal splitting also excludes uncountable branches by part (ii). It is therefore a Suslin tree. By the standard Suslin-tree characterization of Suslin hypothesis, diamond implies failure of Suslin hypothesis. The decisive step is antichain sealing by diamond, with maximality below a correctly guessed club set stage verified explicitly.

A kappa-filtration is an increasing continuous sequence with union $A$ and $|A_{\alpha }|<\kappa$ at every stage. Intersect $S$ with the club set of nonzero limit ordinals. For each remaining $\alpha$, continuity gives $A_{\alpha }={\bigcup }_{\beta <\alpha }{A}_{\beta }$, so choose $r(\alpha )<\alpha$ with $f(\alpha )\in A_{r(\alpha )}$. This is a regressive function. Fodor lemma gives a stationary subset $S_1$ and a fixed $\beta$ such that all these values lie in $A_{\beta }$.

Since $|A_{\beta }|<\kappa$, partition $S_1$ into fewer than $\kappa$ fibers of $f$. If every fiber were nonstationary, choose a club set avoiding each one. Their intersection is club set by regularity, contradicting stationarity of $S_1$. Thus one fiber is stationary, andThis proves the filtration form of Fodor lemma; continuity at limit stages, the small size of each stage, and regularity of $\kappa$ are all used.

A forcing name is built by well-founded recursion: every member of a $\mathbb{P}$-name $\tau$ is a pair $(\sigma ,p)$ with $p\in \mathbb{P}$ and $\sigma$ itself a $\mathbb{P}$-name of lower forcing name rank. In the ground model all such forcing names form the recursively defined class $V^{\mathbb{P}}$. Its evaluation by a generic filter isThe recursion is on forcing name rank, not on the forcing order. Forcing names need not have a unique evaluation across different generics, and many different forcing names may have the same evaluation.

The forcing truth lemma says that for a formula $\phi$ and ground-model forcing names ${\tau }_1,\dots ,{\tau }_n$,The forcing relation is the ground-model relation furnished by the forcing theorem. The witnessing condition must belong to the particular generic filter, not merely to the forcing order.

A forcing preserves cofinalities over $M$ if every generic extension hasThe two values are compared as ordinals, using preservation of ordinals by forcing. Equivalently the forcing asserts that no ground ordinal acquires a smaller cofinality. The assertion concerns all ordinal cofinalities, not just that one specified cardinal remains uncountable.

Let $x={\tau }_G$ be any set in the extension. The set $D=\{\sigma :\exists p((\sigma ,p)\in \tau )\}$ belongs to $M$. Ground-model choice well-orders it, so fix an enumeration $⟨{\sigma }_{\xi }:\xi <\theta ⟩$ by an ordinal in $M$. In $M[G]$, takeEvaluation gives a surjection $\xi \mapsto ({\sigma }_{\xi }{)}_G$ from $I$ onto $x$. This is a set function by the ZF part of the forcing theorem; no extension choice is needed. For each $y\in x$, take its least preimage ordinal. Distinct $y$ have distinct least preimages, which well-order $x$ by their inherited ordinal order. Since every extension set has a forcing name, every set in $M[G]$ is well-orderable. ThereforeThis choice preservation by well-ordered names tolerates repeated evaluations, because the least-index step removes them canonically.

Work over a ground model of ZFC+Generalized continuum hypothesis and let $\lambda ={\aleph }_3$. Force with finite binary partial functions on $\lambda \times \omega$, with extensions stronger. The Delta-system lemma thins any uncountable family of finite domains to an uncountable family with one common root. Only finitely many binary assignments on that root occur, so two conditions agree there and their union is a common extension. Hence the forcing has the countable chain condition for forcing and preserves cardinals and cofinalities.

The generic union yields $\lambda$ distinct reals. Totality at each coordinate is dense, and for two different coordinates it is dense to assign different values at some unused natural-number position. Thus the extension satisfies $2^{{\aleph }_0}\ge \lambda$.

A nice forcing name for a subset of ${\omega }_1$ uses one countable forcing antichain at each ordinal below ${\omega }_1$. Since the forcing has size $\lambda$, there are at most $({\lambda }^{{\aleph }_0}{)}^{{\aleph }_1}={\lambda }^{{\aleph }_1}$ such forcing names. Ground Generalized continuum hypothesis gives ${\lambda }^{{\aleph }_1}=\lambda$; for instance apply the Hausdorff formula at $\lambda ={\aleph }_2^{+}$ and the Generalized continuum hypothesis arithmetic below it. Therefore $2^{{\aleph }_1}\le \lambda$ in the extension. Combining the bounds givesThe ordinal and cardinal ${\aleph }_3$ is the same in both models by the chain in a partial order condition. The forcing theorem formalizes this construction as the requested relative-consistency implication. A countable transitive ground model is a convenient presentation, not an additional consequence silently derived from mere consistency. This is the Cohen forcing two-level continuum plateau.

All cardinalities in this part are first computed in $M$. The forcing has size $\kappa$. A family of $\kappa$ finite domains has a $\kappa$-sized Delta-system, since $\kappa$ is regular. There are fewer than $\kappa$ possible value assignments on its finite root: each coordinate $(\alpha ,n)$ allows fewer than $\kappa$ values. Regularity lets us thin to two conditions, indeed $\kappa$ many, with identical root assignments. Their union is a condition, proving the $\kappa$-chain condition.

Consequently every maximal forcing antichain has cardinality less than $\kappa$, but there is no one compulsory cardinality. For any nonzero cardinal $\mu <\kappa$, the single-coordinate conditions assigning the values $\xi <\mu$ at $(\mu ,0)$ form an forcing antichain of size $\mu$. It is maximal: a condition already assigning that coordinate is compatible with its matching value, and a condition not assigning it is compatible with every allowed value. Thus every such size occurs, including singleton maximal forcing antichains. These are the maximal-antichain sizes in the finite Lévy collapse.

The paper writes $p\le q$ for $p\subseteq q$. We use this printed weaker-first convention: $q$ extends $p$ and is stronger. Compatibility and generic meeting arguments below always refer to common extensions, so do not accidentally reverse the convention.

For every infinite $\alpha <\kappa$, the generic union at coordinate $\alpha$ gives a function $g_{\alpha }:\omega \to \alpha$. The requirement to assign $(\alpha ,n)$ is dense for each $n$, and the requirement to use any specified value $\beta <\alpha$ is dense by assigning it at a fresh natural-number position. Therefore $g_{\alpha }$ is a surjection, and every ordinal below $\kappa$ becomes countable.

The $\kappa$-chain condition preserves the regularity of $\kappa$. For a direct verification, a forcing name for a function from some $\mu <\kappa$ into $\kappa$ has fewer than $\kappa$ possible values at each coordinate, using a maximal deciding forcing antichain. The union of these possible-value sets has size less than $\kappa$ by regularity and is bounded in $\kappa$. No such function can be cofinal. In particular $\kappa$ remains uncountable, while every smaller ordinal is countable. HenceThis is the finite Lévy collapse to omega-one.

In $M$, take $T={\bigcup }_{\alpha <\kappa }{}^{\alpha }2$ ordered by extension. Strong inaccessibility gives $|T_{\alpha }|=2^{|\alpha |}<\kappa$. In $M[H]$, each of these ground levels is therefore countable, while the height $\kappa$ is ${\omega }_1$. Thus the unchanged ground set-theoretic tree is an ${\aleph }_1$-tree in the extension.

Every ground binary function of length $\kappa$ still yields a distinct cofinal branch through this set-theoretic tree. There are $(2^{\kappa }{)}^M\ge ({\kappa }^{+}{)}^M$ such branches. The $\kappa$-chain condition preserves cardinals at and above $\kappa$, so $({\kappa }^{+}{)}^M={\aleph }_2^{M[H]}$. The ground branch family still has at least that cardinality. ConsequentlyThis Kurepa tree from an inaccessible binary tree uses ground-model levels and branches; it does not claim the full binary set-theoretic tree newly computed in the extension has countable levels.

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.