A monotone countably unbounded function from $\omega^\omega$ to $\omega^{\omega_1}$ For a set $X$ we endow the set $\omega^X$ of all functions from $X$ to $\omega$ with the natural partial order $\le$ defined by $f\le g$ iff $f(x)\le g(x)$ for all $x\in X$. A function $\mu:\omega^\omega\to \omega^X$ is called monotone if for any $f\le g$ in $\omega^\omega$ we have $\mu(f)\le\mu(g)$.

Question. Is there a monotone function $\mu:\omega^\omega\to\omega^{\omega_1}$ which is countably unbounded in the sense that for every countable infinite set $A\subset\omega_1$ and function $f\in\omega^A$ there exist a function $g\in\omega^\omega$ and an infinite set $B\subseteq A$ such that $f{\restriction}_B\le\mu(g){\restriction}_B$?

Remark. By the answer of Johannes Schürz to this question, for every monotone function $\mu:\omega^\omega\to\omega^{\omega_1}$, there exists a countable set $A\subset\omega_1$ and a function $f\in\omega^A$ such that $f\not\le \mu(g){\restriction}A$ for every $g\in\omega^\omega$.
 A: The answer to this problem is negative and follows from

Theorem. For any monotone function $\mu:\omega^\omega\to\omega^{\omega_1}$ there exists a countable infinite set $A\subset\omega_1$ such that for every $f\in\omega^\omega$ the function $\mu(f){\restriction}A$ is bounded.

Proof. For every $\alpha\in\omega_1$ consider the monotone function $\mu_\alpha:\omega^\omega\to\omega$, $\mu_\alpha:f\mapsto\mu(f)(\alpha)$. By Lemma~2.3.5 in this paper, for every $f\in\omega^\omega$ there exists $n\in\omega$ such that $\mu_\alpha[\omega^\omega_{f{\restriction}n}]$ is finite. Here $\omega^\omega_t=\{g\in\omega^\omega:t\subset g\}$.
Let $T_\alpha$ be the set of all (finite) functions $t\in\omega^{<\omega}$ such that $\mu_\alpha[\omega^\omega_t]$ is finite but for any $\tau\in\omega^{<\omega}$ with $\tau\subsetneq t$ the set $\mu_\alpha[\omega^\omega_\tau]$ is infinite. It follows (from the mentioend Lemma 2.3.5) that  for every $f\in\omega^\omega$ there exists a unique $t_f\in T$ such that $t_f\subset f$.
Let $\delta_\alpha(f)=\max\mu_\alpha[\omega^\omega_{t_f}]\ge \mu_\alpha(f)$. It is clear that the function $\delta_\alpha:\omega^\omega\to\omega$ is continuous.
Consider the function $\delta:\omega_1\to C_k(\omega^\omega,\omega)$, $\delta:\alpha\mapsto\delta_\alpha$ and observe that $\delta(\alpha)(f)\ge \mu(f)(\alpha)$ for any $\alpha\in\omega_1$ and $f\in\omega^\omega$.
By Michael's Theorem 11.5 (in Gruenhage's survey), the function space $C_k(\omega^\omega,\omega)$ is an $\aleph_0$-space. In particular, it has a countable network. Using this fact, we can find a sequence of pairwise distinct ordinals $\{\alpha_n\}_{n\in\omega}\subset\omega_1$  such that the sequence  $(\delta_{\alpha_n})_{n\in\omega}$ converges to $\delta_{\alpha_0}$ in the function space $C_k(\omega^\omega,\omega)$.
Consequently, for every $f\in\omega^\omega$ the sequence $(\delta_{\alpha_n}(f))_{n\in\omega}$ converges to $\delta_{\alpha_0}(f)$ and hence is upper bounded by some number $M_f$. Let $A=\{\alpha_n\}_{n\in\omega}$ and observe that for every $n\in\omega$ we have $\mu(f)(\alpha_n)\le\delta_{\alpha_n}(f)\le M_f$, which means that the function $\mu(f){\restriction}A$ is bounded.
