9
$\begingroup$

A countably complete ideal $I$ on a set $Z$ ideal is c.c.c. when there is no uncountable family of pairwise disjoint $I$-positive subsets of $Z$. If such an ideal exists, then there exists a weakly Mahlo cardinal $\kappa$ and a $\kappa$-complete, c.c.c. ideal $J$ on $\kappa$. An example of such an ideal is the collection of measure zero subsets of a real-valued measurable cardinal.

The existence of such an ideal, which is nowhere maximal ("ultra"), requires a large continuum. The standard proof of this is as follows. Build a tree of subsets of $\kappa$, with root $\kappa$. Suppose by induction that each level of the tree up to level $\alpha$ is a partition of $\kappa$. For every $J$-positive member of the partition, partition this into two disjoint $J$-positive sets. For members in $J$, don't split them. This gives you level $\alpha+1$ and a finer partition. At limit stages, take intersections. By the c.c.c., we must have all members of the partition in $J$ at level $\omega_1$, and no cofinal branches. There are $2^{<\omega_1} = 2^\omega$ many measure zero sets at that stage, so we have $\kappa$ equals a union of continuum many sets from $J$, so $2^\omega \geq \kappa$.

My questions is: Does this process of building the tree terminate before $\omega_1$? If it does not, then by the c.c.c., some subtree of what we've constructed is a normal Suslin tree. So in some cases, we know the process does halt at a countable stage:

(a) If we start with a measurable $\kappa$ and force MA$_\kappa$ with the Solovay-Tennenbaum forcing, we get no Suslin trees, and a c.c.c. ideal on $\kappa$.

(b) By a result of Laver, we can start with a model of MA$_{\omega_2}$ and a measurable above, and then force a real-valued measurable cardinal by random real forcing, keeping SH.

So the obvious question is, is there any way to get a c.c.c., $\kappa$-complete ideal $J$ on $\kappa$, where we can find a Suslin tree embedded in the algebra of $J$-positive sets? Can this be done by adding measurably many Cohen reals?

$\endgroup$
9
$\begingroup$

Such trees must have countable height because every $x \in \kappa$ has to leave the tree at some level below $\omega_1$ so that by ccc-ness of the ideal the tree dies at a countable level.

$\endgroup$
  • 2
    $\begingroup$ Very nice argument. $\endgroup$ – Joel David Hamkins Jan 31 '14 at 4:01
  • 1
    $\begingroup$ Fremlin also has this interesting remark (due to Silver) on trees and rvms: If $\kappa$ is rvm the there is no $\kappa$-Aronszajn tree. Let me also advertise one of his great open questions: Suppose there is a total $\kappa$-additive probability measure over $\kappa$. Must there be a non (Lebesgue) null set of size $\kappa$ all of whose uncountable subsets are non null. $\endgroup$ – Ashutosh Jan 31 '14 at 4:25
  • $\begingroup$ This is problem EG(b) (worth 14 pounds) mentioned here: essex.ac.uk/maths/people/fremlin/problems.pdf For smaller cardinals, this is due to Gitik and Shelah $\endgroup$ – Ashutosh Jan 31 '14 at 4:39

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.