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Consider the following bornologies $\mathbb{D},\mathbb{E}$ on the set $\mathcal{N}$ of all functions from $\mathbb{N}$ to $\mathbb{N}$:

  • $\mathbb{D}=\{A: \exists f\in\mathcal{N}\forall g\in A\exists m\in\mathbb{N}\forall n>m(f(n)>g(n))\}$. (Dominatable sets)

  • $\mathbb{E}=\{A: \exists f\in\mathcal{N}\forall g\in A\forall m\in\mathbb{N}\exists n>m(f(n)>g(n))\}$. (Escapable sets)

My question is when, set-theoretically speaking, these yield equivalent (= "bornomorphic") bounded structures on $\mathcal{N}$:

Is it consistent that $\mathfrak{b}=\mathfrak{d}$ but there is not a bijection $i:\mathcal{N}\rightarrow\mathcal{N}$ such that the $i$-image of each set in $\mathbb{D}$ is in $\mathbb{E}$ and the $i$-preimage of each set in $\mathbb{E}$ is in $\mathbb{D}$?

Here $\mathfrak{b}$ and $\mathfrak{d}$ are the cardinal characteristics corresponding to escaping(/bounding) and domination: the minimal cardinalities of sets of functions not dominated/escaped by any individual function, respectively. Clearly in order for $\mathbb{D}$ and $\mathbb{E}$ to be equivalent as bornologies we need $\mathfrak{b}=\mathfrak{d}$; I'm curious whether any additional information is captured by considering the bornological structure present.

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  • $\begingroup$ An interesting variation: do such "bornological equivalences" exist assuming CH? $\endgroup$ Commented Apr 10, 2022 at 8:01
  • $\begingroup$ @FrançoisG.Dorais I think $\mathsf{CH}$ lets you perform a sort of "greedy algorithm" to construct a bornomorphism (countable sets being unproblematic in any way). $\endgroup$ Commented Apr 10, 2022 at 17:32
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    $\begingroup$ Maybe more information is captured by considering Borel isomorphisms? $\endgroup$ Commented Apr 12, 2022 at 7:52
  • $\begingroup$ @FrançoisG.Dorais I'd be interested in that follow-up question; would you like to ask it? (I feel a bit hesitant to "double-dip" reputation from the same basic idea.) $\endgroup$ Commented Apr 12, 2022 at 16:14
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    $\begingroup$ @FrançoisG.Dorais I gave in to temptation. $\endgroup$ Commented Jun 14, 2023 at 6:03

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Note that $\mathfrak{b}=\mathfrak{d}$ is equivalent to the existence of a $<^\ast$-increasing sequence $(f_\alpha)_{\alpha<\mathfrak{d}}$ which is cofinal in $(\mathcal{N},{<^\ast})$, where $f <^\ast g \iff \exists n\,\forall m \geq n\,{f(m) < g(m)}$.

For $\alpha<\mathfrak{d}$, let $$\begin{aligned} D_\alpha &= \{ g \in \mathcal{N} \mid g <^\ast f_\alpha \}, & E_\alpha &= \{ g \in \mathcal{N} \mid f_\alpha \not<^\ast g \}. \end{aligned}$$ Note that $D$ is dominatable iff $D \subseteq D_\alpha$ for some $\alpha < \mathfrak{d}$, and that $E$ is escapable iff $E \subseteq E_\alpha$ for some $\alpha < \mathfrak{d}$.

With some thinning of the cofinal sequence $(f_\alpha)_{\alpha<\mathfrak{d}}$ if necessary, we can make sure that for every $\alpha < \mathfrak{d}$ the sets $$\begin{aligned} &D_\alpha \setminus {\textstyle\bigcup_{\beta<\alpha} D_\beta}, & &E_\alpha \setminus {\textstyle\bigcup_{\beta<\alpha} E_\beta}, \end{aligned}$$ each have size $\mathfrak{c}$. Working level-by-level, we can construct a bijection $\mathcal{N} \leftrightarrow \mathcal{N}$ which restricts to a bijection $D_\alpha \leftrightarrow E_\alpha$ for every $\alpha<\mathfrak{d}$. Such a bijection shows that the two bornologies are equivalent.

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