Do escaping sets "uniformly" cover dominating sets under determinacy?

Question

For $\mathbb{A},\mathbb{B}\subseteq\mathcal{P}(\omega^\omega)$, say $\mathbb{A}$ spreads onto $\mathbb{B}$ iff there is some $F:\omega^\omega\rightarrow\omega^\omega$ such that for all $X\in\mathbb{A}$ the set $F[X]=\{F(r):r\in X\}$ is in $\mathbb{B}$.

Let $\mathbb{D},\mathbb{E}$ be the sets of dominating, escaping families of reals respectively. Trivially $\mathbb{D}$ spreads onto $\mathbb{E}$ via the identity function, since every dominating family is escaping. The converse however is more complicated, even in $\mathsf{ZFC}$: $\mathbb{E}$ obviously can't spread onto $\mathbb{D}$ if $\mathfrak{b}<\mathfrak{d}$, but $\mathsf{CH}$ implies that $\mathbb{E}$ does spread onto $\mathbb{D}$ per vzoltan's answer to an old question of mine.

I'm curious what happens in $\mathsf{ZF+DC+AD}$:

Under $\mathsf{ZF+DC+AD}$, does $\mathbb{E}$ spread onto $\mathbb{D}$?

Answer 1

Here's something that seems to work. I can add more details if needed.

Suppose that every set of reals has the property of Baire. Then every function from $\omega^{\omega} \to \omega^{\omega}$ is continuous on a comeager set. Note that every comeager set is unbounded. Fix $F \colon \omega^{\omega} \to \omega^{\omega}$ and let $C$ be a comeager set on which $F$ is continuous. We will find a dense open sets $D_{i}$ ($i \in \omega$) and $y \in \omega^{\omega}$ such that $F(x)$ fails to dominate $y$ whenever $x \in C \cap \bigcap_{i \in \omega}D_{i}$. Let $\langle (i_{n}, \sigma_{n}) : n \in \omega \rangle$ be an enumeration of $\omega \times \omega^{<\omega}$. In the $n$th stage of the construction choose $y \upharpoonright k_{n}$ for some $k_{n} \in \omega$ and also some $\tau_{n} \in \omega^{<\omega}$ extending $\sigma_{n}$ such that $F(x) \upharpoonright k_{n}$ is the same sequence $\rho_{n}$ for all $x \in C$ extending $\tau_{n}$, and $\{ j < k_{n} : \rho_{n}(j) < y(j)\}$ has size at least $i_n$. Then we can let $D_{i}$ be the set of $x \in \omega^{\omega}$ extending any of the sets $\tau_{n}$ where $i_{n} = i$.

The $\mathbb{P}_{\mathrm{max}}$ claim has to be modified somewhat, since $\mathbb{P}_{\mathrm{max}}$ needs AD$^{+}$ to work, which is not known to follow from AD + DC. Forcing over a model of AD$^{+}$, $\mathbb{P}_{\mathrm{max}}$ produces a model of ZFC which is $\Pi_{2}$ maximal for the powerset of $\omega_{1}$ by (very roughly speaking) forming a generic direct limit of all countable models, subject to agreement about stationarity for subsets of $\omega_{1}$. There is a natural $\mathbb{P}_{\mathrm{max}}$ variation for each $\Sigma_{2}$ sentence, where each model fixes a witness to the sentence, and a model in the direct limit passes its witness on to stronger conditions. This doesn't always succeed, but for the statement $\mathfrak{b} = \aleph_{1}$ it does succeed, producing (without adding reals) a model of $\mathfrak{b} = \aleph_{1}$ in which every $\Pi_{2}$ sentence for $\mathcal{P}(\omega_{1})$ holds which can be provably forced to hold along with $\mathfrak{b} = \aleph_{1}$. For instance, the statement $\mathfrak{d} > \aleph_{1}$. This is discussed at a general level in the last section of my article from the Handbook of Set Theory, and in more detail in the Shelah-Zapletal paper "Canonical models for $\aleph_{1}$-combinatorics".

Having produced such a model, one can then take an unbounded family $B$ of size $\aleph_{1}$, find a real $y$ which is not dominated by $F[B]$, and then find the unbounded set of $x$ such that $F(x)$ does not dominate $y$ back in the ground model.

There is an intermediate argument one could run using the fact that, under AD$^{+}$, true $\Sigma^{2}_{1}$ sentences have Suslin, co-Suslin witnesses. This is at the heart of the $\mathbb{P}_{\mathrm{max}}$ argument. For the current problem, continuity on a comeager set works as a substitute for being Suslin and co-Suslin.

Morally, anything that is consistent with large cardinals should be forceable over a determinacy model without adding reals.