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 2013 iii Paper 59 1 b Solution Created 2026-10-03 Updated 2026-10-07
A proposed assignment for 2UN-SAT can be checked in polynomial time, including checking the syntactic restriction, so the language belongs to NP.
For an explicit reduction, parse any Boolean formula into a Boolean circuit over binary AND (logical conjunction), binary OR (logical disjunction) and unary NOT (negation), and apply the gate equivalences in part (a), asserting its output. Every gate clause has at most two positive literals: the AND clauses have respectively one, one and one; the OR clauses have one, one and two; and the NOT clauses have two and zero. The unit output/constant clauses also obey the restriction. Hence the resulting formula is an instance of 2UN-SAT.
The Tseitin transformation is linear in the gate description, and its auxiliary variables enforce the gate values rather than relaxing their relation to the inputs. Therefore the original formula is satisfiable if and only if the transformed restricted formula is satisfiable. This is a polynomial-time many-one reduction from SAT, whose hardness follows from the Cook-Levin theorem. ConsequentlyThe condition limits positive literals, without limiting total clause length.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 59 3 b Solution Created 2026-10-03 Updated 2026-10-07
For each fixed input length, take the complete truth table of the language. For every accepted string , form the mintermAn OR of these minterms equals the indicator function, since precisely at . This is the truth-table upper bound for circuit size.
Generate the negated input wires once and share them. With accepted strings, use at most binary AND gates and binary OR gates, besides the negations. Empty truth tables use a constant-zero Boolean circuit; length zero is handled by a constant Boolean circuit. ThusThis is an existence bound for a nonuniform circuit family, even when the language is undecidable. It supplies no algorithm for computing the truth tables of an arbitrary language.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 59 3 c Solution Created 2026-10-03 Updated 2026-10-07
The AND-NOT circuit value problem belongs to P: validate the Boolean circuit and evaluate its gates in a topological ordering. Evaluation takes polynomial time even if its description is not already in a topological ordering.
For P-completeness we use logspace many-one reductions. Start from the supplied circuit value problem over the usual AND (logical conjunction), OR (logical disjunction) and NOT (negation) basis. Retain AND and NOT gates, and replace every OR gate by the De Morgan's laws gadgetThis adds only a constant number of gates per old gate, preserves its truth value for all inputs and keeps the Boolean circuit acyclic. If the format includes constant source nodes, replace them by additional input nodes assigned fixed bits zero and one; these are inputs, not disallowed gates. Larger fan-in gates can first be replaced by binary trees of their inputs.
To output the new description, keep the old gate index and a constant-size gadget position, rescan old references when necessary, and assign consistent new indices to each gadget's terminal output. These counters and references occupy bits; the input assignment is copied with any constant-source bits appended. Thus the construction is a logspace many-one reduction preserving acceptance. Since Boolean circuit Value is P-complete under such reductions,
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 121 1 iv Solution Created 2026-10-03 Updated 2026-10-05
First prove set-theoretic absoluteness for syntactically bounded formulas in set theory by induction on their construction. Atomic equality and membership use the actual equality and membership relations in both transitive classes. Boolean operations preserve agreement.
For the bounded existential step, take and suppose . Its witness is an actual member of . Since is a transitive class, ; the induction hypothesis then gives . The converse uses the same witness in . The bounded universal step follows similarly because both structures quantify over exactly the actual members of , or follows by negation from the existential step. HenceNo ZF axioms are needed for this syntactic statement beyond the ambient meaning of a transitive class. For a ZF-equivalent bounded formula , choose a bounded equivalent . The assumption that both structures model ZF gives and . Combining these with the displayed equivalence proves the asserted set-theoretic absoluteness of .
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 121 2 iii Solution Created 2026-10-03 Updated 2026-10-05
Use relative constructible level recognition. The finite coding conditions matter here: merely writing “every set is relatively constructible” inside an arbitrary transitive set does not ensure that its computations are correct.
Let be a single finite conjunction expressing finite relation closure for set-theoretic coding. Concretely require the empty set, pairing, set union, set difference, Cartesian products, and the following uniform relation operations: for each finite ordinal and each set , the set of finite tuples exists; relations on these tuple domains can be complemented, intersected, projected along a coordinate and pulled back along finite coordinate maps; the equality and membership relations restricted to exist. These are finitely many first-order closure assertions with quantified, not an axiom schema. Each operation is specified by its usual elementwise membership equivalence. Transitivity makes the operations correct externally. Finite tuple domains are also correct: every individual finite tuple is already present by pairing and union.
These conditions make satisfaction for a set structure absolute. For a fixed coded first-order formula, compute its truth relations on by induction: equality and membership give the atomic cases, relative complement gives negation, intersection gives logical conjunction, and projection gives existential quantification. All required truth tables exist by and agree with the actual ones. Thus the first-order assertion is absolute whenever belong to a transitive structure satisfying : require that each member of has such a finite formula-and-parameter definition, and that each coded definition contributes a member of . Codes are finite objects; no truth predicate for the ambient universe is being used.
Write for the coded relative constructible stage assertion: is an ordinal and there is a function with domain , with , , at nonzero limit ordinals, and . Under , any such internal code is correct by transfinite induction. Restrictions of a code to shorter domains exist by the relation operations. Define the one-free-variable first-order formulaAll displayed abbreviations expand into first-order formulas of the membership language.
Suppose is transitive, , and . Let be the externally defined set of indices of stage codes in . It contains , is downward closed by restricting codes, and has no largest member by the last conjunct. Thus is a nonzero limit ordinal . Correctness of stage codes gives for , hence by transitivity. Conversely the exhaustion conjunct puts every in some such level. Therefore
For the converse, every with nonzero limit satisfies : each listed operation on parameters from one level is definable at finitely many later levels, still below . Moreover stage histories appear below every limit constructible level. Here is the essential limit step of that lemma. If the histories for are available in , the correct predicate defines their graph of endpoints as a subset of . No endpoint with can belong to , because would then give , contradicting Axiom of foundation. The defined graph therefore has domain exactly . Adjoining its final pair takes finitely many more stages. Together with finite successor extensions, this proves by transfinite induction that every belongs to for some finite .
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 121 3 i Solution Created 2026-10-03 Updated 2026-10-05
Use standard notation for forcing: means that is stronger. A generic filter is nonempty, upward closed and downward directed, and meets every dense subset of a forcing order in the ground model. Names below are forcing names in , with pairs ordered as .
The semantic forcing relation isHere is the evaluation of a forcing name. Countability of supplies such generic filters through every condition by the Rasiowa–Sikorski lemma.
Define the syntactic forcing relation by mutual well-founded recursion on the forcing names for the atomic cases, then induction on the first-order formula. Its membership clause isFor equality, first abbreviateand set exactly when both inclusions hold. Each recursive call lowers the rank of at least one forcing name without raising the other. This makes the mutual recursion well-founded.
For a basis of connectives consisting of logical conjunction, negation and existential quantification, the remaining clauses areThe last line is the existential clause of syntactic forcing; witnesses need only occur densely, rather than be forced by itself with one preselected name. Other connectives are defined by logical abbreviations. The recursion gives a definable relation inside for each fixed first-order formula, with quantification over the class of its names. If has no greatest element, use canonical forcing names , which still evaluate to . It also proves monotonicity: strengthening a condition preserves what it forces. The forcing theorem identifies the two relations and supplies the truth lemma.