• Corpus ID: 119647751

Madness and regularity properties

  title={Madness and regularity properties},
  author={Haim Horowitz and Saharon Shelah},
  journal={arXiv: Logic},
Starting from an inaccessible cardinal, we construct a model of $ZF+DC$ where there exists a mad family and all sets of reals are $\mathbb Q$-measurable for $\omega^{\omega}$-bounding sufficiently absolute forcing notions $\mathbb Q$. 

The Ramsey property implies no mad families

If all collections of infinite subsets of N have the Ramsey property, then there are no infinite maximal almost disjoint (mad) families, and the implication is proved in Zermelo–Fraenkel set theory with only weak choice principles.

On the non-existence of $$\kappa $$-mad families

Starting from a model with a Laver-indestructible supercompact cardinal $\kappa$, we construct a model of $ZF+DC_{\kappa}$ where there are no $\kappa$-mad families.

Transcendence bases, well-orderings of the reals and the axiom of choice

We prove that $ZF+DC+"$there exists a transcendence basis for the reals$"+"$there is no well-ordering of the reals$"$ is consistent relative to $ZFC$. This answers a question of Larson and Zapletal.


The technology reduces many questions about ZF implications between consequences of the Axiom of Choice to natural ZFC forcing problems.

On measure and category

We show that under ZF + DC, even if every set of reals is measurable, not necessarily every set of reals has the Baire property. This was somewhat surprising, as for the Σ21 set the implication holds.

A model of set-theory in which every set of reals is Lebesgue measurable*

We show that the existence of a non-Lebesgue measurable set cannot be proved in Zermelo-Frankel set theory (ZF) if use of the axiom of choice is disallowed. In fact, even adjoining an axiom DC to ZF,

A barren extension

It is shown that provided ω→(ω)ω, a well-known Boolean extension adds no new sets of ordinals. Under an additional assumption, the same extension preserves all strong partition cardinals. This fact

Can you take Toernquist's inaccessible away?

We prove that ZF + DC + ”There are no mad families” is equiconsistent with ZFC.

E-mail address: haim.horowitz@mail.huji.ac.il (Saharon Shelah

    Einstein Institute of Mathematics Edmond J. Safra campus,
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