English

A solution to Roitman's problem

Logic 2015-02-23 v3

Abstract

We answer Question~3.2 from Shelah \cite{Sh:666}: Given a maximal almost disjoint (mad) family A\mathcal A of size 1\aleph_1, we construct a forcing Q(A){\mathbb Q}(\mathcal A) that has Axiom A, is ωω{}^\omega \omega-bounding, preserves selective ultrafilters, has the 2\aleph_2-properness isomorphism condition (p.i.c.), and destroys the mad family A\mathcal A. We develop a new construction technique for partial orders, combining ladder systems for ω1\omega_1 with trees of normed creatures. Countable support iteration of the new kind of iterands solves Roitman's problem in the case of d=1d=\aleph_1 and also simultaneously the open question about the relative consistency of u=1<au = \aleph_1 < a: It is consistent relative to ZFC that there is a dominating set of size 1\aleph_1 and a selective ultrafilter with character 1\aleph_1 and the minimal size of a mad family is 2\aleph_2, like the continuum.

Keywords

Cite

@article{arxiv.1404.7343,
  title  = {A solution to Roitman's problem},
  author = {Heike Mildenberger},
  journal= {arXiv preprint arXiv:1404.7343},
  year   = {2015}
}

Comments

20 pages Lemma 3.2 is flawed The author thanks Jindrich Zapletal

R2 v1 2026-06-22T04:01:44.604Z