# Is there any class of ideals for which $\mathfrak{b}(\mathcal I)\neq\mathfrak{b}$?

Consider the Baire space $$\mathbb{N}^\mathbb{N}$$. A natural pre-order $$\leq^*$$ on $$\mathbb{N}^\mathbb{N}$$ is defined by $$f\leq^*g$$ if and only if $$f(n)\leq g(n)$$ for all but finitely many $$n$$. A subset $$A$$ of $$\mathbb{N}^\mathbb{N}$$ is said to be bounded if there is a $$g\in\mathbb{N}^\mathbb{N}$$ such that $$f\leq^*g$$ for all $$f\in A$$. Let $$\mathfrak{b}$$ be the minimum cardinality of an unbounded subset of $$\mathbb{N}^\mathbb{N}$$.

Let $$\mathcal{I}$$ be an ideal on $$\mathbb{N}$$. For any $$f,g\in\mathbb{N}^\mathbb{N}$$, we write $$f\leq_\mathcal{I} g$$ if and only if $$\{n\in\mathbb{N} : f(n)>g(n)\}\in\mathcal{I}$$. Note that $$\leq_{\bf{Fin}}=\leq^*$$, where $$\bf{Fin}$$ is the ideal on $$\mathbb N$$ containing all finite subsets of $$\mathbb N$$. A subset $$A$$ of $$\mathbb{N}^\mathbb{N}$$ is said to be $$\mathcal{I}$$-bounded if there is a $$g\in\mathbb{N}^\mathbb{N}$$ such that $$f\leq_\mathcal{I} g$$ for all $$f\in A$$. Let $$\mathfrak{b}(\mathcal{I})$$ be the minimum cardinality of a $$\leq_\mathcal{I}$$-unbounded subset of $$\mathbb{N}^\mathbb{N}$$.

Is there any class of ideals for which $$\mathfrak{b}(\mathcal I)\neq\mathfrak{b}$$?

In the trivial case that $$\mathcal{I}$$ is a principal ideal on a cofinite set, then the whole partial order $$\mathbb{N}^{\mathbb{N}}/\mathcal{I}$$ $$\frak{b}(\mathcal{I})$$ is countable, and in this case $$\frak{b}(\mathcal{I})=\omega\neq\frak{b}$$. So if one allows this kind of ideal, the answer is affirmative.

One may want therefore to disallow those cases, and consider only ideals for which $$\frak{b}(\mathcal{I})$$ will be uncountable. The answer for this version of the question is that it is independent of ZFC.

Consistently negative. If the continuum hypothesis holds, of course, then both $$\frak{b}$$ and $$\frak{b}(\mathcal{I})$$ will be $$\omega_1$$, and hence equal to each other.

Consistently positive. Meanwhile, if $$\mathcal{I}$$ is a maximal ideal, then the order relative to that ideal becomes linear, the same as the corresponding ultrapower, and the bounding number $$\frak{b}(\mathcal{I})$$ is the cofinality of that ultrapower, the same as $$\frak{d}(\mathcal{I})$$, since in a linear order unbounded is the same as dominating.

But evidently it is a theorem of Canjar and Roitman that it is consistent with ZFC that $$\frak{b}\ll\frak{d}$$ and yet every regular cardinal between them arises as the cofinality of an ultrafilter on $$\mathbb{N}$$. The model is simply to add a large number of Cohen reals over any model of ZFC.

See further discussion and results in the paper of Andreas Blass and Heike Mildenberger, in which that result is stated as theorem 1.

Blass, Andreas; Mildenberger, Heike, On the cofinality of ultrapowers, J. Symb. Log. 64, No. 2, 727-736 (1999). ZBL0930.03060. arxiv:9611210

• I expect that Andreas will likely be able to shed some additional useful light on the situation, and I would encourage him to post a further answer. Aug 9 at 14:10
• Is there any ideal $\mathcal I$ on $\mathbb N$ for which $\mathfrak{b}<\mathfrak{b}(\mathcal I)$? Aug 10 at 13:53
• Yes, this can occur, and this is part of the content of the theorem of Canjar and Roitman that I mention. There are models of ZFC in which $\frak{b}$ is much less than $\frak{d}$, and every regular cardinal $\kappa$ between them arises as $\kappa=\frak{b}(\mathcal{I})$ for some maximal ideal $\mathcal{I}$. Meanwhile, it is consistent that $\frak{b}=\frak{c}$, in which case there are no larger cardinals than $\frak{b}$ below the continuum, so in those models, there is no such $\mathcal{I}$. Aug 10 at 15:03
• I am learning set theory. The proof (of Theorem 1) is not included in the mentioned paper. I need to understand how for every regular cardinal $\kappa$ between $\mathfrak{b}$ and $\mathfrak{d}$ we can find a maximal ideal $\mathcal I$ for which $\kappa=\mathfrak{b}(\mathcal I)$. It will be helpful if you elaborate this. Aug 11 at 12:03