Write $\mathfrak m_k$ for the Martin's axiom number for $k$-Knaster, i.e., for the smallest size of a family of dense subsets of some $k$-Knaster poset for which there is no generic filter. (A poset is $k$-Knaster if every uncountable set of conditions has an uncountable $k$-linked subset, i.e., any $k$ conditions in this subset have a common lower bound.)

Write $\mathfrak m^c$ for the Martin's axiom number for precaliber (i.e., as above, replacing "$k$-linked" by "centered": every finite subset has a lower bound).

Clearly $\mathfrak m_k\le \mathfrak m_{k+1}\le\mathfrak m^c$.

In a recent paper arXiv:1904.02617 we show (as a side result) that consistently $\mathfrak m_k=\aleph_1$ for all $k$ while $\mathfrak m^c=\lambda<\mu=2^{\aleph_0}$ for any desired regular uncountable $\lambda<\mu$.

Question: We suspect that this result is known, and that a known proof might even use the same (nameless?) forcing notion that we use (we call it $ P_{\text{cal},\lambda}$). Does anybody have a reference?

Details for our forcing:

  • It is well known (e.g., Barnett, Fund. Math. 141, 1992) that Cohen reals add a witness for $\mathfrak m_k=\aleph_1$ which is preserved under further $k+1$-Knaster extensions. So whenever we have a FS ccc iteration which is $\ell$-Knaster for all $\ell$ and adds Cohens (at the beginning, say), the resulting model will satisfy $\mathfrak m_k=\aleph_1$ for all $k$.
  • If we additionally make sure that by bookkeeping we force with all small ($<\lambda$) precaliber forcings, we get $\mathfrak m^c\ge \lambda$.
  • To ensure $\mathfrak m^c\le \lambda$, we initially force with $P_{\text{cal},\lambda}$ (defined in 5.1 of the linked paper) which adds $Q_{\text{cal}}$ (defined in 5.3) witnessing $\mathfrak m^c\le \lambda$, and furthermore this witnessing-property of $Q_{\text{cal}}$ is preserved in FS iterations of small precaliber extensions. (For our paper we additionally need that the iterations may also contain additional iterands which are $(\sigma,k)$-linked for all $k$.)

Your Answer

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

Browse other questions tagged or ask your own question.