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 $\mu(f)\le\mu(g)$.



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> **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 $f{\restriction}\_B\le\mu(g){\restriction}\_B$?
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**Remark.** By the answer of Johannes Schürz to [this question](https://mathoverflow.net/q/361970/61536), 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 monotone countably unbounded function from $\omega^\omega$ to $\omega^{\omega_1}$
The answer to this problem is negative and follows from



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> **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.
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*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](https://arxiv.org/pdf/1607.07978.pdf), 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](https://www.sciencedirect.com/science/article/pii/B9780444865809500136)), 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.


