6
$\begingroup$

Let $\mathsf{R}$ denote some finitely many relations about finitely many cardinal characteristics (e.g. $\mathfrak{a} \leq \mathfrak{s}$, $\mathfrak{a} < \mathfrak{d} = \mathfrak{r}$, $\mathfrak{b} = \mathfrak{s} \wedge \mathrm{cof}(\mathcal{M}) < \mathrm{non}(\mathcal{N})$ etc). Is there some such statement $\mathsf{R}$ such that

  • $\mathsf{ZFC} + \mathsf{R}$ is consistent, but
  • $\mathsf{ZFC} + \mathsf{R} + \mathfrak{c} < \aleph_\omega$ is inconsistent (or, at the very least, not known to be consistent)?
Improve this question
$\endgroup$
1
  • 4
    $\begingroup$ At one point in time, $\mathfrak d < \mathfrak a$ would have been a good answer to this question. Shelah's original proof of this inequality had $\mathfrak d >\!> \aleph_\omega$, with no obvious way of bringing the size down. But then later Shelah found a non-obvious way of doing it (template forcing) and found a model with $\aleph_2 = \mathfrak d < \mathfrak a = \aleph_3$. It is still an open problem whether $\aleph_1 = \mathfrak d < \mathfrak a$ is possible. $\endgroup$
    – Will Brian
    Commented 14 hours ago

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .