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.
Cofinality-preserving forcing 2026-10-06
A forcing is cofinality-preserving over a transitive model if every generic extension has the same cofinality for every ground ordinal. The values are compared as ordinals, which forcing preserves.
Evaluation of a forcing name Created 2026-10-05 Updated 2026-10-07
Interpret a forcing name recursively by . The collection of interpreted ground-model names is the generic extension.
Forcing name 2026-10-05
Forcing theorem Created 2026-09-24 Updated 2026-10-05
The forcing theorem identifies truth in a generic extension with the forcing relation: a formula is true in exactly when some condition in forces it.
Forcing truth lemma 2026-10-06
A formula holds in a generic extension exactly when some condition in the generic filter forces it for names of its parameters.
An uncountable set of ground-model elements in a generic extension by countable forcing contains an uncountable ground-model subset. Partition membership witnesses according to the countably many forcing conditions.
Past exam of the mathematics course of the University of Cambridge 2006 iii Paper 25 6 Solution Created 2026-10-03 Updated 2026-10-14
Let be an externally countable model of ZFC, and let . In form the forcing orderThis is the finite-function collapse to countable size. For every and , the internal setsare dense: extend a finite function by assigning a missing input, and use a fresh input to put any required value in its range.
Externally enumerate all the dense subsets of which recognizes as dense. There are only countably many, because is countable. Recursively choose in the th such dense set and let be the upward closure of this descending chain. This produces a generic filter over . By the forcing theorem, its generic extension satisfies ZFC. The canonical forcing name is forced to be a function with domain and range : the dense sets and ensure exactly these assertions. ThereforeIts new first uncountable ordinal is accordingly different from .
For a transitive , ordinary evaluation of names gives the familiar literal extension. The question only assumes a countable model, so one must also cover externally ill-founded models. In that case use the quotient of the internally defined forcing names by forced equality modulo , with membership defined by forced membership. The internal forcing identities and the forcing theorem hold in and give a well-defined first-order structure satisfying every standard ZFC axiom. Check names embed the original membership structure, so identify their images with . Its domain is a quotient of a subset of the countable domain of , hence is externally countable. Thus the required extension exists even without assuming transitivity. It is a membership extension, not an elementary extension: the assertion that the parameter is uncountable changes truth value.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 19 4 ii a Solution Created 2026-10-03 Updated 2026-10-07
The generic extension is . It is transitive: if , an active pair in supplies a subname with . Ground sets belong to it by their canonical names. Without assuming a weakest condition, use ; nonemptiness of gives .
Induction on forcing name rank gives . The right side is an ordinal of . If is an ordinal, its rank equals , so it is at most a ground-model ordinal. Transitivity of then implies . 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.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 19 5 i a Solution Created 2026-10-03 Updated 2026-10-07
The valuation of a forcing name is defined recursively byThe recursion is on forcing name rank. Only pairs with an active condition in 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.
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 19 5 i c Solution Created 2026-10-03 Updated 2026-10-06
A forcing preserves cofinalities over 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.
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 19 6 i a Solution Created 2026-10-03 Updated 2026-10-06
A forcing collapses a cardinal over if it forces that the ground ordinal is no longer a cardinal: in the extension there is a bijection from some smaller ordinal onto . Equivalently there is a surjection onto 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.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 24 5 b Solution Created 2026-10-03 Updated 2026-10-06
Fix forcing names for the set and the parameters. Define a nameOnly appearing in and are needed, so this is a subset of a ground-model set. The forcing definability lemma makes its defining predicate a formula of . Ground-model axiom schema of separation therefore gives .
Conversely, if satisfies this formula, choose with and . The forcing truth lemma supplies forcing the formula for these names. Directedness gives with . Then , and . ThusThis proves the instance of the axiom schema of separation in the generic extension, without assuming that instance there in order to construct the name.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 24 6 a i Solution Created 2026-10-03 Updated 2026-10-06
The forcing is cardinal-preserving over when every ground-model cardinal remains a cardinal in every generic extension:Forcing preserves ordinals, and an ordinal that was not a cardinal already has a ground-model bijection with a smaller ordinal, which persists. Thus this definition gives equality of the cardinal ordinals in the two models.
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 2017 iii Paper 121 3 iii Solution Created 2026-10-03 Updated 2026-10-05
We prove power set in a generic extension by bounding possible subnames in the ground model, rather than presupposing the desired power set in . Let , and let . In formEvery is a forcing name. Its value is a subset of : an active pair has for some , so implies and .
Now take any with , and choose a forcing name with . By axiom schema of separation and definability of the syntactic forcing relation,The atomic membership truth lemma for forcing shows . One inclusion follows immediately from its soundness direction. For the other, if , choose with and . The truth direction supplies forcing ; strengthen within below and to obtain an active pair in representing .
Finally the ground-model forcing nameexists by Axiom schema of replacement and the ground-model power set axiom. Since is nonempty, all contribute their values, andThis set belongs to by the definition of a generic extension. The construction uses all conditions in the outer pairs and therefore does not require a greatest condition in .
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 121 3 iv b Solution Created 2026-10-03 Updated 2026-10-05
Let be a dense subset of a forcing order . Then is dense in the product forcing order: below , choose in and retain . The generic filter meets it, so its first projection meets . Together with the filter properties proved above, this givesThe same argument proves that is -generic over , but genericity over the larger generic extension needs the next argument.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 121 4 iv Solution Created 2026-10-03 Updated 2026-10-05
Put and let . Conditions from Fn forcing agree on common coordinates because is directed, so is a function. For each , the ground-model setis dense: add a value at that coordinate if necessary. The generic filter meets all these sets, making total.
For distinct , the ground-model setis dense. Choose fresh for both rows of a finite condition and assign and there. Genericity therefore makes the rows pairwise different. The generic coordinate reals for finite-function forcing are , and in the generic extensionThe family is a set in , for example by collecting the ground-model names and evaluating the corresponding name for their indexed family. This argument only needs as the specified ground-model ordinal; it does not silently assume a cardinal preservation by chain-condition forcing result.
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 3 iv a Solution Created 2026-10-03 Updated 2026-10-05
A cardinal-preserving forcing over leaves every ground-model cardinal number a cardinal in every generic extension. In the present notation, for every -generic filter and every ordinal ,Set forcing preserves ordinals, and an old bijection witnessing that an ordinal is not a cardinal remains available. Thus this condition also means that the two models have exactly the same cardinals. Cardinality here is computed inside the respective models, not from their external sizes.