I'm looking for a model satisfying the inequality described in the title. Recall that the uniformity of the meager ideal, denoted $\operatorname{non}(\mathcal M)$ (or $\operatorname{non}(\mathcal B)$) is the least size of a non meager set of reals and the almost disjointness number, denoted $\mathfrak{a}$ is the least size of an infinite, almost disjoint family. The reverse inequality, $\mathfrak{a} < \operatorname{non}(\mathcal M)$ holds in many well studied models, for example the random model. I seem to remember reading that $\operatorname{non}(\mathcal M) < \mathfrak{a}$ is also consistent though I can't seem to find the reference now. Therefore my question is:

Is it consistent that $\operatorname{non}(\mathcal M) < \mathfrak{a}$ if so, what model does this hold in?

As a remark, such a model is necessarily a model of $\mathfrak{b} < \mathfrak{a}$ which I do know to be consistent in several models (though none of the constructions are particularly easy compared to those for $\mathcal{a} < \operatorname{non}(\mathcal M)$). Therefore a related question is in which (if any) of the models for $\mathfrak{b} < \mathfrak{a}$ has $\operatorname{non}(\mathcal M)$ been computed?

I would be also extremely interested in a model of $\mathfrak{b} < \operatorname{non}(\mathcal M) < \mathfrak{a}$ though I wouldn't be surprised if this consistency was not known (please correct me if I'm mistaken!).


  • $\begingroup$ Shelah proved the consistency of both $\mathfrak{d} < \mathfrak{a}$ and the consistency $\mathfrak{u} < \mathfrak{a}$. (Same paper, but two different models.) I would not be surprised if one of these models had $\mathrm{non}(\mathcal M) < \mathfrak{a}$ as well. If no one is able to answer your question outright, you might want to try computing $\mathrm{non}(\mathcal M)$ in these models, or try to modify the techniques from that paper. $\endgroup$ – Will Brian Mar 19 at 13:22
  • $\begingroup$ shelah.logic.at/files/95437/700.pdf $\endgroup$ – Will Brian Mar 19 at 13:24
  • $\begingroup$ @WillBrian Great thanks for the reference I will look into it! $\endgroup$ – Corey Bacal Switzer Mar 19 at 16:29
  • 2
    $\begingroup$ This paper deals with your first question question.Your second question seems trickier, because using Hechler forcing in the template construction will likely yield $\text{add}(\mathcal{M})=\text{cof}(\mathcal{M})$. $\endgroup$ – Johannes Schürz Mar 19 at 20:27
  • $\begingroup$ @JohannesSchürz thanks! I'll check it out $\endgroup$ – Corey Bacal Switzer Mar 22 at 14:57

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