This is a follow-up question to an [older question](https://mathoverflow.net/questions/401774/putting-omega-in-two-boxes).


Let $\alpha \in \big(\omega\cup\{\omega\}\big) \setminus \{0,1\}$ be an ordinal. We say that a function $f: \omega \to \alpha$ is *fair* if $$|f^{-1}(\{j\})| = \aleph\_0$$ for all $j\in\alpha$.


We call a set ${\frak F}$ of fair functions $f:\omega\to\alpha$ *equalising for $\alpha$* if for all $a,b\in\omega$ there is $f\in{\frak F}$ with $f(a) = f(b)$. Let $B\_\alpha$ be the minimum cardinality that an set equalising for $\alpha$ can have. [This post](https://mathoverflow.net/a/401781/8628) establishes $B\_2 \leq 3$.


**Question.** Can an explicit value for $B\_\alpha$ be given for $\alpha \in \big(\omega\cup\{\omega\}\big) \setminus \{0,1\}$? I am particularly interested in $B\_\omega$.



Putting $\omega$ into $\alpha$ boxes where $\alpha \in \big(\omega\cup\{\omega\}\big)\setminus\{0,1\}$
$B\_\alpha = 3$ for every $\alpha \in (\omega \cup \{\omega\}) \setminus \{0,1\}$. Here is a construction that shows $B\_\alpha \le 3$.


1. Let $g : \mathbb{N} \to \alpha$ be any function such that $g^{-1}(\{j\})$ is infinite for all $j \in \alpha$.
2. Let $h$ be any bijection from $\omega$ to $\{0,1,2\} \times \mathbb{N}$.
3. For $i \in \{0,1,2\}$ define $t\_i : \{0,1,2\} \times \mathbb{N} \to \alpha$ as follows: $$t\_i(r,s) = \begin{cases} g(s) & \text{if } i=r \\ 0 & \text{otherwise} \end{cases}$$
and define $f\_i : \omega \to \alpha$ by $f\_i = t\_i \circ h$.


The functions $f\_i$ are fair. In fact, $f\_i^{-1}(\{j\}) = h^{-1} (t\_i^{-1} (\{j\}))$, which is an infinite set because $h$ is a bijection and $t\_i^{-1}(\{j\})$ contains $\{i\} \times g^{-1}(\{j\})$. The set $\{f\_0,f\_1,f\_2\}$ is equalising for $\alpha$ because for all $a,b \in \omega$, if $h(a) = (r\_a,s\_a)$ and $h(b) = (r\_b,s\_b)$, then for any $i \not\in \{r\_a,r\_b\}$ we have $f\_i(a)=f\_i(b)=0$.


The reasoning used to show $B\_2 > 2$ in the answer to [this question](https://mathoverflow.net/a/401781/8049) extends to show $B\_\alpha > 2$ for all $\alpha > 1$. Hence $B\_\alpha = 3$ for all $\alpha \in (\omega \cup \{\omega\}) \setminus \{0,1\}$.


