Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 121 3 v Solution Created 2026-10-03 Updated 2026-10-05
Let be the finite stem of a condition , and defineThe stems of two conditions in the generic filter agree on their common domain, because they have a common stronger extension. Hence this union is a function. For every , the set of conditions with is dense: append values from the nonempty infinite reservoir until the desired length is reached. Genericity therefore makes total on . The generic stems, and thus their union, are available in .
Fix and . In form the setWe show that it is a dense subset of a forcing order. Given , put . Fill the new positions with any fixed element of , and choose with for the new position . This is possible because an infinite subset of is unbounded. Let be the resulting stem of length , and keep the reservoir unchanged. ThenEvery new stem value came from the old reservoir, exactly as required by the extension relation. The construction is performed in , so and is internally dense.
Genericity gives a condition in . Its witnessing coordinate remains fixed in every later stem and hence in , so some satisfies . Since this holds for every , there are infinitely many such . As was arbitrary,Thus this infinite-reservoir stem forcing produces an unbounded real over a model, which is precisely the stipulated meaning of bounding .