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This question is "take 2" of this older one, following a suggestion of Francois Dorais. 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)

Clearly we must have $\mathfrak{b}=\mathfrak{d}$ in order for these to yield bornomorphic structures on $\mathcal{N}$, and Francois Dorais showed at the above-linked question that the converse holds as well. However, the "tame maps" version of the question is still open:

Suppose $\mathfrak{b}=\mathfrak{d}$. Must there be a Borel $i:\mathcal{N}\rightarrow\mathcal{N}$ 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 $\mathbb{D}$?

It's not even clear to me (per James Hanson's comment below) what the answer is assuming $\mathsf{V=L}$.

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  • $\begingroup$ Do you know if the answer is yes assuming $V=L$? $\endgroup$ Jun 19, 2023 at 4:01
  • $\begingroup$ @JamesHanson Nope. I briefly thought I saw an absoluteness argument giving a negative answer in all cases, but that broke down. $\endgroup$ Jun 19, 2023 at 4:25

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A bornomorphic map $i\colon\mathcal{N}\to\mathcal{N}$ for these bornologies cannot be Borel.

First, I shall use $f<^\infty g$ to mean that $(\forall m\in\mathbb{N})(\exists n>m)f(n)<g(n)$, and $f<^*g$ to mean that $(\exists m\in\mathbb{N})(\forall n>m)f(n)<g(n)$.

For $f\in\mathcal{N}$, let $\mathbb{D}_f=\{g\in\mathcal{N}\mid g<^* f\}$, and let $\mathbb{E}_f=\{g\in\mathcal{N}\mid g<^\infty f\}$. Since the $\mathbb{D}_f$ and $\mathbb{E}_f$ generate their respective ideals, $i$ is a bornomorphism if and only if for all $f\in\mathcal{N}$ there is $a(f)\in\mathcal{N}$ such that $i``\mathbb{D}_f\subseteq\mathbb{E}_{a(f)}$ and $u(f)\in\mathcal{N}$ such that $i^{-1}(\mathbb{E}_f)\subseteq\mathbb{D}_{u(f)}$.

We shall focus on the $u$ case. Another way of saying this is for all $g,f\in\mathcal{N}$, if $i(g)<^\infty f$ then $g<^*u(f)$.

Adjoin to the universe a Cohen generic function $c\colon\mathbb{N}\to\mathbb{N}$, and let $d=\tilde{i}(c)$, where $\tilde{i}$ is the Borel function with the same code as $i$. Since ground model elements of $\mathcal{N}$ are a $<^\infty$-dominating family in the extension, there must be $f\in\mathcal{N}$ in the ground such that $d<^\infty f$. Let $h=u(f)$. Then, by our assumption, in the ground model we get

$$(\forall x\in\mathcal{N})(i(x)<^\infty f\implies x<^*h)$$

This is a $\Pi^1_1$-statement involving only $f$, $h$, and (a code for) $i$, so it remains true in our extension. In particular, generic cohen real $c$ is satisfies $c<^*g$. However, Cohen generic functions $\mathbb{N}\to\mathbb{N}$ are never dominated by ground model reals.


This proof is a small adjustment of Andreas Blass's Theorem 4.15 in Combinatorial Cardinal Characteristics of the Continuum.

Blass, Andreas, Combinatorial cardinal characteristics of the continuum, Foreman, Matthew (ed.) et al., Handbook of set theory. In 3 volumes. Dordrecht: Springer (ISBN 978-1-4020-4843-2/hbk; 978-1-4020-5764-9/ebook). 395-489 (2010). ZBL1198.03058.

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    $\begingroup$ @WillBrian What good timing, I had just finished making such an edit as you commented! You are correct, statement a is wrong and not needed; we only need that the ground model reals are an $<^\infty$-dominating family. Thank you for confirming statement b. $\endgroup$ Jun 22, 2023 at 13:38

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