Let $G=(L,R,E)$ be a complete bipartite graph, such that $|L|=\aleph\_0$ and $|R|=\kappa$. I'd like to show that if $\kappa<2^{\aleph\_0}$ then the list-chromatic number of $G$ can't be more than $\aleph\_0.$


Intuitively I find this correct, but I couldn't find a way to prove this. Is it really true that $\chi\_l(G) = \aleph\_0$? 



How to prove a certain connection between the list-chromatic number of a bipartite graph and a cardinal?
From among the $2^{\aleph\_0}$ different ways to color the vertices of $L$ (from the given lists), you can choose a family of $2^{\aleph\_0}$ *almost disjoint* colorings. (Two colorings are almost disjoint if they have only finitely many colors in common.)



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> For each vertex $v\in V(G)=L\cup R$, let $C(v)$ be the list of colors associated with $v$; we assume that $|C(v)|=\aleph\_0$. Let $L=\{v\_n:n\lt\omega\}$ and let $C\_n=C(v\_n)$. Choose a color $c\in C\_0$; then choose two distinct colors $c\_0,c\_1\in C\_1\setminus \{c\_0\}$; then choose four distinct colors $c\_{00},c\_{01},c\_{10},c\_{11}\in C\_2\setminus\{c,c\_0,c\_1\}$; and so on. In this way we build a full binary tree of height $\omega$: the nodes of the tree are distinct colors, the nodes on level $n$ are colors in $C\_n$, each branch is a list-coloring of $L$, there are $2^{\aleph\_0}$ branches, and two different branches have only finitely many colors in common.
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At least one of those colorings (in fact all but $\kappa$ of them) can be extended to a proper vertex coloring  

of $G$.



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> For each vertex $w\in R$, its list $C(w)$, being infinite, is contained in at most one of those $2^{\aleph\_0}$ almost disjoint colorings of $L$. Since $|R|\lt2^{\aleph\_0}$, we can choose a coloring of $L$ which does not contain any of the lists $C(w),w\in R$. Clearly such a coloring can be extended.
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That's how you show that $\chi\_\ell(G)\le\aleph\_0$. To see that $\chi\_\ell(G)\ge\aleph\_0$ observe that $\chi\_\ell(G)\ge\chi\_{\ell}(K\_{\aleph\_0,\aleph\_0})\ge\chi\_\ell(K\_{n,n^n})=n+1$ for each $n\in\mathbb N$.



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> It's clear that $\chi\_\ell(K\_{n,n^n})\le n+1$. To show that $\chi\_\ell(K\_{n,n^n})\gt n$, assign pairwise disjoint lists of size $n$, call them $C\_1,C\_2,\dots,C\_n$, to the vertices of $L$; then assign the $n^n$ different transversals of the family $\{C\_1,C\_2,\dots,C\_n\}$ to the $n^n$ vertices of $R$.
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