# A model where the uniformity of the meager ideal is strictly below the almost disjointness number

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!).

Thanks!

• 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. – Will Brian Mar 19 at 13:22
• shelah.logic.at/files/95437/700.pdf – Will Brian Mar 19 at 13:24
• @WillBrian Great thanks for the reference I will look into it! – Corey Bacal Switzer Mar 19 at 16:29
• This paper deals with your first question question.Your second question seems trickier, because using Hechler forcing in the template construction will likely yield $\text{add}(\mathcal{M})=\text{cof}(\mathcal{M})$. – Johannes Schürz Mar 19 at 20:27
• @JohannesSchürz thanks! I'll check it out – Corey Bacal Switzer Mar 22 at 14:57