So what does it mean when we say ${k_\beta}$ “almost works”? We’ll start with a lemma:

Lemma 1 For a fixed ${\nu\in^{<\omega}\mu}$, the set ${\{\alpha<\beta:\nu=\eta_\alpha\upharpoonright k_\beta(\alpha)+1\}}$ contains at most one element from each set ${A_{n+1}^\beta\setminus A^\beta_n}$, and hence is at most countable.

Proof: Suppose by way of contradiction what ${\alpha}$ and ${\gamma}$ are distinct members of ${A^\beta_{n+1}\setminus A^\beta_n}$ for which

$\displaystyle \nu=\eta_\alpha\upharpoonright (k_\beta(\alpha)+1)=\eta_\gamma\upharpoonright (k_\beta(\gamma)+1). \ \ \ \ \ (1)$

Then

$\displaystyle f^\beta_{n+1}(\alpha)=\eta_\alpha(k_\beta(\alpha))=\nu(k_\beta(\alpha))=\nu(k_\beta(\gamma))=\eta_\gamma(k_\beta(\gamma))=f^\beta_{n+1}(\gamma), \ \ \ \ \ (2)$

and this contradicts the fact that ${f^\beta_{n+1}}$ is a transversal for ${\{x_\epsilon:\epsilon\in A^\beta_{n+1}\}}$. $\Box$

Now given ${\alpha<\beta}$, we are going to define ${E(\alpha)}$ to be those ${\gamma<\beta}$ for which ${k_\beta}$ has failed to work, that is,

$\displaystyle E(\alpha)=\{\gamma<\beta:\max\{k_\beta(\alpha),k_\beta(\gamma)\}<\Delta(\alpha,\gamma)\}. \ \ \ \ \ (3)$

Lemma 2 The set ${E(\alpha)}$ is at most countable.

Proof: If not, find ${k^*<\omega}$ for which the set ${B:=\{\gamma\in E(\alpha):k_\beta(\gamma)=k^*\}}$ is uncountable, and set

$\displaystyle \nu=\eta_\alpha\upharpoonright k^*+1. \ \ \ \ \ (4)$

Then for each ${\gamma\in B}$, we have

$\displaystyle \eta_\gamma\upharpoonright k_\beta(\gamma)+1=\eta_\alpha\upharpoonright k^*+1=\nu, \ \ \ \ \ (5)$

which contradicts the preceding lemma. $\Box$

So for a given ${\alpha<\beta}$, the function ${k_\beta}$ will “disjointify” ${A_\alpha}$ from all but countably many ${A_\gamma}$ with ${\gamma<\beta}$: the function ${k_\beta}$ “almost works”.

Notice that ${\gamma\in E(\alpha)}$ if and only if ${\alpha\in E(\gamma)}$, so that we can define a graph ${\Gamma}$ on ${\beta}$ by connecting ${\alpha}$ and ${\gamma}$ if and only if ${\gamma\in E(\alpha)}$. By the preceding lemma, every vertex of the graph has at most countably many edges coming out of it. An easy argument tells us that the connected components of our graph ${\Gamma}$ are at most countable as well.

Now if ${\alpha}$ and ${\gamma}$ are members of different connected components of ${\Gamma}$ then ${k_\beta}$ will disjointify ${A_\alpha}$ and ${A_\gamma}$. But each connected component of ${\Gamma}$ can be disjointified easily (by induction) as it is at most countable.

Thus, it is straightforward now to “correct” ${k_\beta}$ to a function ${h_\beta}$ which will work everywhere.