Cardinal-preserving forcing Created 2026-10-05 Updated 2026-10-06
A forcing order preserves cardinals over a ground model if every ground-model cardinal number remains a cardinal in every generic extension. This assertion and any sufficient chain condition for forcing are evaluated inside the ground model. External countability of a countable transitive model does not make all of its forcing orders internally ccc.
Generic extension Created 2026-09-24 Updated 2026-10-03
If is a generic filter over a countable transitive model , the generic extension consists of the interpretations by of all forcing names in .
A worldly cardinal permits a countable transitive model of ZFC and a transitive model of ZFC with , , and . First collapse a countable elementary substructure of . Then collapse an -sized Skolem hull containing every countable ordinal and every member of , as well as itself. The transitive collapse fixes transitive subsets lemma preserves and those ordinals.
Past exam of the mathematics course of the University of Cambridge 2004 iii Paper 24 3 Solution Created 2026-10-03 Updated 2026-10-14
Forcing constructs new first-order models while controlling which ground-model properties survive. Fix a ground model and a partially ordered set , writing when is the stronger condition. A generic filter is upward closed, directed towards stronger conditions, and meets every ground-model dense subset of a forcing order. If is externally a countable transitive model, enumerate its dense subsets and successively choose stronger conditions in them. The resulting filter is -generic. External countability is crucial: this is not a claim that the ground model regards its own universe as countable.
A forcing name is built recursively from pairs of earlier names and conditions. For a generic filter , its evaluation of a forcing name isThe canonical forcing name evaluates to , so . Recursive evaluation also proves transitivity. Name ranks bound the ranks of their evaluations, so an ordinal in the extension is below some ground ordinal and is already a ground ordinal. Thus forcing preserves ordinals.
The central link between syntax and generic interpretation is the forcing theorem. The relation is definable in the ground model, and the forcing truth lemma saysFor an atomic membership assertion, densely many refinements must witness equality with a member-name whose associated condition is compatible with the refinement. Equality is obtained by recursive mutual inclusion of names. For negation, means no stronger condition forces ; for an existential quantification, witnesses must be obtainable densely below . Induction on name rank establishes the atomic clauses, and induction on formulas establishes definability and the truth lemma. The generic filter meets the relevant dense sets at the existential step. An arbitrary semantic assertion about one chosen generic filter is insufficient: the definable relation must work uniformly for all generic interpretations.
This explains preservation of ZFC. Extensionality and foundation follow from transitivity; canonical names supply infinity. Pair and union names give pairing and union. For separation, restrict a name's members to conditions forcing the defining formula. For replacement, the definable forcing relation lets ground axiom schema of collection bound a set of names for all possible witnesses to the unique outputs; evaluating their collected name gives the range. For power set, let name the given set and let be the set of its member-names. Every subset of has an equivalent name drawn from , by the truth lemma, so a ground set of candidate names collects all its subsets. Finally choose a ground well-order of the relevant names; its evaluated image surjects onto the desired set, and choosing the first name for each value well-orders that set. This is why ordinary forcing over a choice model preserves choice.
For a concrete example, Cohen forcing consists of finite partial functions , ordered by extension. The generic union is total because deciding each coordinate is dense. For every ground real , the set of conditions disagreeing with somewhere is dense, so . This construction changes the universe by adding a genuinely new real without changing its ordinals.
The size and compatibility of conditions determine what else changes. With the countable chain condition for forcing, an ordinal-valued name has only countably many possible values at each coordinate: take a maximal antichain deciding that value. Therefore the possible range of a name for a map with infinite ground domain of cardinality lies in a ground set of size at most . This prevents cardinal collapse and changes of uncountable regular cofinalities. Forcing by finite partial functions is ccc: the delta-system lemma thins any uncountable family of finite domains to a delta-system, and a further thinning makes their values agree on its finite root. Two remaining conditions are compatible. Distinct coordinates add distinct reals.
For example, start with GCH and take . This forcing has size and adds distinct reals. A nice name for a real uses countably many antichains, and each antichain is countable. The ground arithmetic bounds the number of such names by . Hencegiving a relative-consistency example of the failure of CH. The ground choice of GCH is available through the constructible universe; no claim that every starting model has this arithmetic is needed.
By contrast, countably closed forcing adds no new countable sequences of ground ordinals. Given a name, recursively strengthen a condition to decide each coordinate, then take a lower bound of the descending sequence. That condition decides the entire sequence. Closure and chain conditions thus control different features; neither should be silently substituted for the other.
Further applications include changing cofinalities, adding branches through trees, and constructing prescribed continuum patterns. If the aim is to violate choice, an ordinary generic extension of a choice model will not do: one restricts to a suitable symmetric collection of names, checking the axioms again. These examples also mark a limitation: forcing proves relative consistency, not absolute consistency of the background theory. The countable-transitive-model picture gives a concrete semantic construction; the definable forcing relation and Boolean-valued formulation supply the corresponding formal consistency arguments without asserting that a transitive ground model can be proved to exist.
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 19 1 ii c Solution Created 2026-10-03 Updated 2026-10-06
The Gödel constructible universe theorem gives . The forcing independence theorem for the Continuum hypothesis gives . 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.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 24 6 b iii Solution Created 2026-10-03 Updated 2026-10-06
In the generic extension from the preceding construction, a surjection from onto all reals gives . 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.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 24 6 b i Solution Created 2026-10-03 Updated 2026-10-06
This is the Rasiowa–Sikorski lemma. Put and recursively chooseusing density. SetIt contains , is upward closed toward weaker conditions, and is directed: if and , then is a common stronger condition in . It meets each through . Thus 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.
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 121 1 e Solution Created 2026-10-03 Updated 2026-10-06
It suffices to obtain a model of ZFC without weakly inaccessible cardinals. Starting with any model of ZFC, pass to its constructible universe, which satisfies ZFC and the Generalized continuum hypothesis. If it has no inaccessible cardinal, use that model. Otherwise pass to its rank segment at its least inaccessible cardinal . This segment satisfies ZFC, retains the Generalized continuum hypothesis, and has no inaccessible cardinals. Under the Generalized continuum hypothesis, every weakly inaccessible cardinal is strongly inaccessible: if and is a limit cardinal, then . Thus in either case the resulting model has no weakly inaccessible cardinals.
Work inside . If is a nonzero limit ordinal, let . It is an uncountable limit cardinal. If it were regular, it would be a weakly inaccessible cardinal, which is impossible in . It is therefore singular, and the singular cardinal enumeration is continuous at this index:The cofinality of an increasing ordinal supremum now givesfor every nonzero limit ordinal in . Hence satisfies the negation of the proposed existential assertion. By the soundness theorem for first-order logic, consistency of ZFC prevents ZFC from proving that assertion. The model construction is a relative-consistency argument; it does not assume that consistency alone supplies a countable transitive model. Here, as usual, a limit ordinal excludes zero.
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 121 5 b Solution Created 2026-10-03 Updated 2026-10-06
False as printed: separativity alone does not guarantee a new generic filter. A separative forcing order means that if , some is incompatible with . The one-point order is separative vacuously. Its only generic filter is , and this belongs to . Thus no exists in this example.
The natural missing hypothesis is that the nonempty order is an atomless forcing order. Under that hypothesis the intended argument works. Since is a countable transitive model, enumerate externally its dense subsets as . Choose with . Thenis a directed upward-closed filter in an ordered set and meets every . This is the Rasiowa–Sikorski lemma, giving a generic filter over .
If , the set difference would belong to . It is dense: a condition outside is already in ; a condition in has two incompatible stronger conditions, at least one of which cannot be in the directed filter . Genericity would require , a contradiction. ThusThis uses generic filter for an atomless order is new. A forcing atom is precisely the obstruction illustrated by the counterexample; separativity does not remove it.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 121 4 v Solution Created 2026-10-03 Updated 2026-10-05
The Rasiowa–Sikorski lemma constructs a generic filter over each . Every resulting generic extension remains a countable transitive model of ZFC: there are only countably many ground-model names externally. The order has the same elements and ordering at every stage, and atomless forcing order structure is absolute because all its quantifiers are bounded to . Hence the generic filter for an atomless order is new result gives for every .
The increasing union is transitive and contains . If it satisfied Axiom of power set for , there would be a set withChoose with . The next-stage generic filter belongs to and is an actual subset of . This subset assertion is absolute for the transitive set , so . Transitivity of and then give , a contradiction. Thus the power-set failure in an increasing union of generic extensions occurs already at the fixed ground-model order:The stages form an increasing chain, not an elementary chain, so the elementary chain theorem does not assert ZFC for their union.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 121 1 v a Solution Created 2026-10-03 Updated 2026-10-05
We construct the two transitive models together, taking care that the later collapse really retains the earlier model. First, a worldly cardinal is greater than : ZFC in contains the full power set , and the axiom of choice there gives a bijection between it and an ordinal . That power set is externally uncountable by Cantor theorem, so . Infinity rules out the finite cardinals and before this argument is applied.
By the Downward Lowenheim-Skolem theorem, choose a countable elementary substructure . Its inherited membership is a well-founded relation, and elementarity transfers extensionality from . Apply the Mostowski collapse theorem to obtain a countable transitive model of ZFC. Because is a countable transitive set, it is a hereditarily countable set and its rank of a set is below . In particular, .
Now take the Skolem hull in of the seedCall it . The seed has external cardinality and lies inside . The hull is an elementary substructure of with : closure under countably many finitary Skolem functions gives the upper bound, and gives the lower bound. Let be its Mostowski collapse theorem map. Then is a transitive model of ZFC.
The transitive collapse fixes transitive subsets lemma gives for every , since is transitive. Since also , the collapse sends that element to itself. Thus , and transitivity of yields the required inclusion:This is the construction of nested transitive models from a worldly cardinal; the next two parts verify its size and ordinal requirements.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 121 1 v b Solution Created 2026-10-03 Updated 2026-10-05
For the model constructed in part (a), the collapsing map is a bijection. The Downward Lowenheim-Skolem theorem supplied countable , so is externally countable. It is infinite because it satisfies the axiom of infinity. ConsequentlyThis is external cardinality. A countable transitive model still regards many of its own sets and ordinals as uncountable, because external enumerations need not belong to it.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 121 2 iv Solution Created 2026-10-03 Updated 2026-10-05
We give a relative-consistency argument using a finite-function collapse to countable size. The assumed consistency implies that ZFC is consistent. Passing to the constructible universe gives consistency of , which also satisfies the Generalized continuum hypothesis. Work in this ground theory, let , and force with finite partial functions , ordered by reverse inclusion.
The union of a generic filter is a total surjection : prescribing a new domain coordinate is dense, and putting any specified into the range is dense. Hence becomes countable. Set forcing preserves ordinals, and absoluteness of constructible levels implies that the extension has the same constructible universe as the ground model. Its is therefore still the old , so it satisfies the stated countability of constructible omega-one condition.
We must also verify GCH after the collapse. The order has ground-model size , hence the -chain condition. By cardinal preservation by chain-condition forcing, all old cardinals at least survive. Every old ordinal below has size at most and becomes countable, so the old is exactly the new .
A name for a subset of a fixed ground-model ordinal can be chosen as a set of pairs with and a condition: use an antichain deciding membership at each coordinate. Thus the number of such names in the ground model is at most . For subsets of this is by ground-model GCH. In the extension it gives , and Cantor theorem gives the reverse lower bound. For every old cardinal the bound isin the ground model. Both and remain cardinals, so again the upper bound and Cantor theorem give the extension's equality . These are all its uncountable cardinals. This proves GCH preservation by a finite-function collapse in the required case.
The forcing relative-consistency theorem now yieldsThe construction in fact only needs consistency of ZFC. This argument uses the formal inner-model and forcing consistency theorems; it does not infer the existence of a countable transitive model merely from consistency.
Past exam of the mathematics course of the University of Cambridge 2025 iii Paper 128 1 b Solution Created 2026-09-24 Updated 2026-10-03
Let be a finite subset of , and let be the finite collection of axioms of occurring in . Choose the finite fragment supplied by the hypothesis. The Lévy reflection theorem gives a level satisfying ; the Downward Lowenheim-Skolem theorem gives a countable elementary substructure of that level, and the Mostowski collapse theorem turns it into a countable transitive model of . By the assumed extension property, is contained in a countable transitive model of , so satisfies .
This argument is formalizable over for each finite . Hence, if is consistent, every finite subset of is consistent. The compactness theorem now gives
Rasiowa–Sikorski lemma Created 2026-10-03 Updated 2026-10-07
If is a countable transitive model, , and , enumerate the dense subsets of belonging to and recursively choose a decreasing sequence starting below that meets each one. Its upward closure is a generic filter over containing .
Semantic forcing relation Created 2026-10-03 Updated 2026-10-06
For a countable transitive model , semantic forcing declares when every generic filter over containing gives .