9
$\begingroup$

Background/Motivation. We know that some of the usefulness of Martin's Axiom lies in giving certain "smallness" properties to sets of size less than continuum, e.g. we have that for all infinite cardinals $ \lambda < 2^{\aleph_0}, 2^{\lambda} = 2^{\aleph_0}$; also all sets of reals of cardinality less than $2^{\aleph_0}$ are Lebesgue measure zero sets.

But I note at the end of the article "Between Martin's Axiom and Souslin's Hypothesis" (by K. Kunen and F. Tall) that the topological characterization of MA restricted to compact hereditarily separable spaces is consistent with having $\aleph_1 < 2^{\aleph_0} < 2^{\aleph_1}$. Which leads me to wonder...

Question: Is there any known restriction/modification of MA ( + not-CH, of course) which is consistent with existence of a non-Lebesgue measurable set of reals of cardinality less than $2^{\aleph_0}$?

(The proof of small sets being Lebesgue null (assuming MA) in Jech's set theory book seems so simple and compelling, it would make an affirmative answer rather intriguing.)

$\endgroup$
5
$\begingroup$

Adding on to Andreas's comment above:

In Section 3 of Chapter XVIII of "Proper and Improper forcing", Shelah proves a very general preservation theorem. "Application 3.8" on page 912 deals with preserving positive outer measure, and in Claim 3.8C(2) he pins down the (very technical) condition he needs to prove that a countable support iteration preserves the property of "being of positive outer measure".

This is an extension of the Judah-Shelah result mentioned by Blass, as in the paper

The Kunen-Miller chart (Lebesgue measure, the Baire property, Laver reals and preservation theorems for forcing) J. Symbolic Logic 55 (1990), no. 3, 909–927,

Judah and Shelah prove an iteration theorem strong enough to prove that the ground model reals have positive outer measure after an $\omega_2$-length iteration of Laver forcing. That particular situation is covered by the more general theorem from Proper and Improper Forcing.

$\endgroup$
8
$\begingroup$

It seems that most of the (widely used) restrictions of Martin's Axiom makes non-measurable sets big, but if you restrict Martin's Axiom to measure algebras, then you can have small non-null sets. The model obtained (from a CH model) by forcing with the measure algebra of type $\omega_2$ is an example of universe where MA(measure algebras) + non CH holds and there is a non-null set of size $\omega_1$.

$\endgroup$
  • $\begingroup$ This is interesting. My impression is that measure algebras are often involved in constructing models where the continuum is real-valued measurable (i.e. enormous) and where there is a non-null $\aleph_1$ size set; whereas strengthening (the full) MA to get Martin's Maximum for instance gives us $2^{\aleph_0} = {\aleph_2}$. (I am still wondering about the significance of these things.) $\endgroup$ – RedRover Nov 8 '15 at 16:10
  • $\begingroup$ The fact that models where continuum is real valued measurable (rvm) have non null set of size $\aleph_1$ is not an accident: Gitik and Shelah proved that if continuum is rvm, then there is a Sierpinski set of size $\theta$ for every $\theta$ less than continuum. It is open if there must also be a continuum sized Sierpinski set. $\endgroup$ – Ashutosh Nov 8 '15 at 20:52
7
$\begingroup$

Shelah's oracle cc forcing can be used to construct models of $2^{\aleph_0}=\aleph_2$ which have a non-meager set of reals of size $\aleph_1$.

A variation of this for measure exists, also due to Shelah, in [Sh669], "Non Cohen oracle cc". I haven't checked the details, but this should allow it to construct models of $\neg$CH that have a non-measurable set of size $\aleph_1$ with a bit more flexibility than just by forcing with the measure algebra. The corresponding forcing axiom, if consistent, would be exactly what you are looking for.

$\endgroup$
  • 2
    $\begingroup$ Another way to get a bit more flexibility is Laver forcing. Judah and Shelah have shown that the ground model has outer measure 1 in the extension, but, in contrast to forcing with a measure algebra, you get dominating reals. I would expect that you can combine Laver and random forcing and still have a small set of outer measure 1. $\endgroup$ – Andreas Blass Nov 8 '15 at 11:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.