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 $f:\omega^\omega\to\omega^X$ is called *monotone* if for any $\alpha\le\beta$ in $\omega^\omega$ we have $f(\alpha)\le f(\beta)$ in $\omega^X$.



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> **Question.** Is there a monotone function $f:\omega^\omega\to\omega^{\omega\_1}$ such that for every countable set $A\subset\omega\_1$ and every $\alpha\in\omega^A$ there exists $\beta\in\omega^\omega$ such that $\alpha\le f(\beta){\restriction}\_A$? 
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**Remark.** By Proposition 2.4.1(2) in [this paper](https://arxiv.org/pdf/1607.07978.pdf), for every monotone function $f:\omega^\omega\to\omega^{\omega\_1}$ there exists $\alpha\in\omega^{\omega\_1}$ such that for every $\beta\in\omega^\omega$ we have $\alpha\not\le f(\beta)$. 


**PS.** I learned this question from Jerzy Kąkol who arrived to it studying $\mathfrak G$-bases in locally convex spaces.



A monotone countably cofinal function from $\omega^\omega$ to $\omega^{\omega_1}$
The answer is no, if you require that the monotone function $F \, \colon \, \omega^\omega \rightarrow \omega^{\omega\_1}$ is total. I will use fairly standard notation, i.e. $f,g \in \omega^\omega$, $\alpha, \beta \in \omega\_1$ and $k,m,n \in \omega$.


The proof works as follows: Towards a contradiction assume that such a function $F$ exists.


First we will construct a sequence $(a\_n)\_{n \in \omega} \in \omega^\omega$ and find an ordinal $\beta < \omega\_1$ such that $$\forall n \in \omega \,\, \exists f \in \prod\_{m \leq n} a\_m \times \prod\_{m > n} \omega \, \colon \, F(f)(\beta) \geq n \, ,$$ where $\prod\_{m \leq n} a\_m \times \prod\_{m > n} \omega =\{ f \in \omega^\omega \, \colon \, \forall m \leq n \,\, f(m) < a\_m \}$. The idea behind this is that we can require a bound for finitely many values of an input function $f$ and still make $F(f)(\beta)$ arbitrarily large.


In the second step we construct a sequence $(b\_n)\_{n \in \omega} \in \omega^\omega$ such that $(b\_n)\_{n \in \omega} \geq (a\_n)\_{n \in \omega}$, and a sequence $(f\_n)\_{n \in \omega}$ such that $$\forall n \in \omega \, \colon \, f\_n \in \omega^\omega \, \land \,(b\_m)\_{m \in \omega} \geq f\_n \, \land \, F(f\_n)(\beta) \geq n \,.$$ But this leads to a contradiction, since the monotonicity of $F$ implies that $\forall n \in \omega \, \colon \, F((b\_m)\_{m \in \omega})(\beta) \geq n$.


**First step:**


We will construct $(a\_n)\_{n\in \omega}$ by induction. For the base case $n=0$ we claim that $$\exists a\_0 \in \omega \,\exists \alpha\_0 \in \omega\_1 \, \forall A \in [\omega\_1 \setminus\alpha\_0]^{\leq \aleph\_0} \, \colon \, F[\prod\_{m = 0} a\_0 \times \prod\_{m>0} \omega] \,\text{is cofinal in} \, \omega^A\, .$$ This means that already $F\restriction \prod\_{m = 0} a\_0 \times \prod\_{m>0} \omega$ dominates $\omega^A$ for every countable $A \subseteq \omega\_1$ above $\alpha\_0$.


Towards a contradiction assume that the claim is wrong. Therefore, we can construct a sequence $(A\_n)\_{n \in \omega}$ such that $A\_n \in [\omega\_1]^{\leq \aleph\_0}$ and $\sup A\_n < \min A\_{n+1}$, and a sequence of functions $(f\_n)\_{n \in \omega}$ such that $f\_n \in \omega^{A\_n}$ and $F\restriction \prod\_{m = 0} n \times \prod\_{m>0} \omega$ does not dominate $f\_n$. But if we define $B:= \bigcup\_{n \in \omega} A\_n \in [\omega\_1]^{\leq \aleph\_0}$ and $f:= \bigcup\_{n \in \omega} f\_n \in \omega^B$, we reach a contradiction, since noting that $\omega^\omega = \bigcup\_{n \in \omega} \prod\_{m = 0} n \times \prod\_{m>0} \omega$, there cannot exist a $g \in \omega^\omega$ such that $F(g) \restriction B \geq f$.


Assume inductively that $a\_0,..,a\_n$ and increasing $\alpha\_0,...,\alpha\_n$ have already been defined, such that $\forall A \in [\omega\_1 \setminus\alpha\_n]^{\leq \aleph\_0} \, \colon \, F[\prod\_{m \leq n} a\_m \times \prod\_{m>0} \omega] \,\text{is cofinal in} \, \omega^A$. With a similar argument as before, and again noting that $$\prod\_{m \leq n} a\_m \times \prod\_{m>n} \omega = \bigcup\_{k \in \omega} \prod\_{m \leq n} a\_m \times \prod\_{m=n+1} k \times \prod\_{m>n+1} \omega$$ we can show that $$\exists a\_{n+1} \in \omega \, \exists \alpha\_{n+1} \in \omega\_1 \setminus \alpha\_n \, \forall A \in [\omega\_1 \setminus\alpha\_{n+1}]^{\leq \aleph\_0} \, \colon$$ $$F[\prod\_{m \leq n} a\_m \times \prod\_{m=n+1} a\_{n+1} \times \prod\_{m>n+1} \omega] \,\text{is cofinal in} \, \omega^A \, .$$ So we can find $a\_{n+1}$ and $\alpha\_{n+1}$ as required.


Now let $\beta > \sup\_{n \in \omega} \alpha\_n$. It follows from our construction of the $(a\_n)\_{n \in \omega}$ that $$\forall n \in \omega \,\, \exists f \in \prod\_{m \leq n} a\_m \times \prod\_{m > n} \omega \, \colon \, F(f)(\beta) \geq n \, .$$


**Second step:**


Again, we will construct $(b\_n)\_{n \in \omega}$ and $(f\_n)\_{n \in \omega}$ by induction. The base case $n=0$ is quite easy: Set $b\_0=a\_0$ and pick any $f\_0 \in \prod\_{m = 0} b\_0 \times \prod\_{m>0} \omega$. Then $F(f\_0)(\beta)\geq 0$ trivially holds.


Assume inductively that $b\_0,...,b\_n$ and $f\_0,...,f\_n$ have already been constructed such that $$\forall m \leq n \, \colon \, a\_m \leq b\_m \, \land \, f\_m \in \prod\_{m \leq n} b\_m \times \prod\_{m>n} \omega \land \, F(f\_m)(\beta) \geq m \, .$$


By using what we have shown in the first step, we can now find $f\_{n+1} \in \prod\_{m \leq n} b\_m \times \prod\_{m>n} \omega$ such that $F(f\_{n+1})(\beta) \geq n+1$. We set $b\_{n+1}:= \max( \max\_{m \leq n+1} f\_m(n+1), a\_{n+1})+1$. It follows that $$\forall m \leq n+1 \, \colon \, a\_m \leq b\_m \, \land \, f\_m \in \prod\_{m \leq n+1} b\_m \times \prod\_{m>n+1} \omega \land \, F(f\_m)(\beta) \geq m \, .$$


But now we have reached a contradiction, since $\forall n \in \omega \, \colon \, (b\_m)\_{m \in \omega} \geq f\_n$, and therfore the monotonicity (and totality) of $F$ implies that $\forall n \in \omega \, \colon \, F((b\_m)\_{m \in \omega})(\beta) \geq n$.


