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Let $G = (V,E)$ be a simple, undirected graph. For $v\in V$ we let $N(v) = \{w \in V: \{v,w\} \in E\}$.

We define the coloring number $\text{Col}(G)$ of the graph $G$ to be the smallest cardinal $\kappa$ such that there is a well-ordering $\leq_{\text{well}}$ on $V$ such that for every vertex $v\in V$ we have $$|N(v) \cap \{w\in V: w \leq_{\text{well}} v\}|< \kappa.$$

Question. Is there an infinite graph $G = (V,E)$ such that $\chi(G)$ is finite but $\text{Col}(G)$ is infinite?

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Take a complete bipartite graph $G=(V_1,V_2,E)$ such that $V_1$ and $V_2$ are infinite. Then $\chi(G)=2$ and $\text{Col}(G)$ is infinite.

Indeed, consider any well-ordering on $V=V_1\cup V_2$. Either there exists a vertex with infinitely many smaller neighbors, or there exists an infinite path $(v_1,v_2,\dotsc)$ with increasing vertices. In the first case, $\text{Col}(G)$ is clearly infinite. In the second case, for any positive integer $n$, the vertex $v_{2n}$ has at least $n$ neighbors smaller than $v_{2n}$, namely $v_1,v_3,\dotsc,v_{2n-1}$. Therefore $\text{Col}(G)$ is infinite.

P.S. Thanks to @lambda for fixing this argument.

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    $\begingroup$ There is clearly no such path if for instance the well-ordering is such that $V_1$ is an initial segment. I think it's probably true that there is either such a path or a vertex with infinitely many neighbours ordered before it. $\endgroup$ – lambda Jun 20 at 17:27
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    $\begingroup$ @lambda: You are right, thanks for the correction. When I wrote up my argument I was secretly assuming that $\text{Col}(G)$ is finite, which then guarantees the infinite path readily. I updated my post accordingly. $\endgroup$ – GH from MO Jun 20 at 21:24
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    $\begingroup$ Not only the coloring number of $G$ but even the list chromatic number (choice number) $\chi_\ell(G)$ is infinite. $\endgroup$ – bof Jun 20 at 23:33
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    $\begingroup$ Any $n$-regular graph has coloring number $n+1$ since the last vertex in the ordering is preceded by $n$ of its neighbors. Since your graph $G$ contains $K_{n,n}$ as a subgraph, its coloring number is greater than $n$ for every $n$, so it's infinite. If a locally finite example were desired, you could just take a disjoint union of $K_{n,n}$. $\endgroup$ – bof Jun 21 at 6:22
  • $\begingroup$ @bof: Good point. I was sure my proof was not the simplest one! $\endgroup$ – GH from MO Jun 21 at 6:39

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