Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

A set $X\subseteq 2^\omega$ of reals is of strong measure zero (smz) if $X+M\not=2^\omega$ for every meager set $M$. (This is a theorem of Galvin-Mycielski-Solovay, but for the question I am going to ask we may as well take it as a definition.)

A set $Y$ is strongly meager (sm) if $Y+N\not=2^\omega$ for every Lebesgue null set $N$.

The Borel conjecture (BC) says that every smz set is countable; the dual Borel conjecture (dBC) says that every sm set is countable.

In Laver's model (obtained by a countable support iteration of Laver reals of length $\aleph_2$) the BC holds. Same for the Mathias model.

In a paper that I (with Kellner+Shelah+Wohofsky) just sent to arxiv.org, we claim that it is not clear if Laver's model satisfies the dBC.

QUESTION: Is that correct? Or is it perhaps known that Laver's model has uncountable sm sets?

Additional remark 1: Bartoszynski and Shelah (MR 2020043) proved in 2003 that in Laver's model there are no sm sets of size continuum ($\aleph_2$). (The MR review states that the paper proves that the sm sets are exactly $[\mathbb R] ^{\le \aleph_0}$. This is obviously a typo in the review.)

Additional remark 2: If many random reals are added to Laver's model (either during the iteration, or afterwards), then BC still holds, but there will be sm sets of size continuum, so dBC fails in a strong sense.

share|improve this question
The paper mentioned in the question is arxiv.org/abs/1105.0823 –  Andres Caicedo May 8 '11 at 13:54
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.