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In this paper, the author discusses colourings of the plane, or in other words, of the underlying graph. I suppose a finite colouring is a colouring using at most $k$ colours for some natural number $k$. However, what do countable and measurable colourings mean? Does the former perhaps mean that any vertex of the graph can get associated with a (potentially unbounded) natural number representing its colour? I have no clue what measurable colourings may be though.

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2 Answers 2

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A countable colouring of the plane $\mathbb{R}^2$ is just a function $c : \mathbb{R}^2 \to \mathbb{N}$, meaning we have countably many distinct colours available to us.

The notion of measurable colouring is a bit ambiguous, it could mean either Borel-measurable or Lebesgue-measurable. A colouring $c : \mathbb{R}^2 \to \mathbf{k}$ or $c : \mathbb{R}^2 \to \mathbb{N}$ is called Borel/Lebesgue-measurable, if each fibre $c^{-1}(n)$ is a Borel respectively Lebesgue-measurable subset of $\mathbb{R}^2$.

An example of a Borel-measurable colouring is taking finitely many lines in $\mathbb{R}^2$, and giving a distinct colour to each resulting component, while giving yet another colour to the lines itself.

Examples for non-Borel-measurable colouring will be weird. For example, take the Cantor middle third set $C$, and give everything in $\mathbb{R}^2 \setminus C \times \{0\}$ colour $0$. Identify each $x \in C \times \{0\}$ first with the corresponding $x' \in 2^\mathbb{N}$, and then view $x'$ as coding a tree $T_x \subseteq \mathbb{N}^{<\omega}$. Now colour $x$ by $1$ iff $T_x$ is well-founded, and by $2$ otherwise.

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  • $\begingroup$ Shouldn't $\mathbf{k}$ be $[k]$? Also, would it be possible to give an example of a colouring which is (and/or one which isn't) Borel-measurable? $\endgroup$
    – J. Schmidt
    Oct 28, 2022 at 15:20
  • $\begingroup$ @J.Schmidt I'm using $\mathbf{k} = \{0,1,\ldots,k-1\}$. I've added examples, although I don't think examples are useful here. $\endgroup$
    – Arno
    Oct 28, 2022 at 16:33
  • $\begingroup$ That non-Borel-measurable colouring example is incomprehensible for me, but thanks for trying. $\endgroup$
    – J. Schmidt
    Oct 29, 2022 at 17:39
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I suppose those are mathematics terms (since the considered graph is infinite).

Countable means you can count those colorings with integers (there is a bijection with $\mathbb{N}$).

Measurable means you can quantify those with a certain notion.

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  • $\begingroup$ To be more precise concerning countable colorings, would it be the colors of the vertices which are in bijection with $\mathbb{N}$? Concerning measurable colorings, I still don't have a clue on which colorings would be considered measurable and which wouldn't... $\endgroup$
    – J. Schmidt
    Oct 28, 2022 at 13:34
  • $\begingroup$ No, I think it is the set of colorings that is countable, not the coloring itself. For the measurability, I don't really know, I haven't really read the paper. $\endgroup$
    – Nathaniel
    Oct 28, 2022 at 13:37
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    $\begingroup$ @Nathaniel It's definitely the set of colours, not the set of colourings, which is countable. $\endgroup$
    – Arno
    Oct 28, 2022 at 14:13

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