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Define the Shelah-Soifer graph $G= (\mathbb{R}, V)$ as $\{x, y\} \in V \iff \exists r\in \mathbb{Q}, \epsilon \in \{1,-1\}, x-y = r + \epsilon \sqrt{2}$. First of all, one notices that if $x$ is any real, then any two neighbours of $x$ cannot be neighbours, so that with the axiom of choice, $G$ is obviously $2$-colourable (easy application of Zorn's lemma). However in Axiom of Choice, Horst Herrlich proposes two exercises about this graph, and I can manage to solve neither of them.

Assume $AC(2)$. Then $G$ is $2$-colorable.

This is probably linked to what I was saying about different neighbours of a real.

Assume $AD$ (axiom of determinacy). Then $G$ is not $\aleph_0$-colorable.

Could someone please give me indications as to how to solve these exercises ?

EDIT : two things, firstly I found the answer to the first exercise, I'll write down the proof below; secondly I'm adding this question: can the second exercise be solved without using Lebesgue-measure ?

Solution to the first exercise: Two points $x, y \in \mathbb{R}$ are path-connected if and only if they belong to the same class of $\mathbb{R}$ mod $\mathbb{Q} + \sqrt{2}\mathbb{Z}$.So what we want to do is colour these classes, and then show that we can pick one colouring for each. But these classes are countable, so what I said about neighbours' neighbours is enough to show that these classes can be coloured. Now let $c= [x]$ be such a class. Since it's path-connected, it only has two colourings $\{f_c, g_c\}$ depending on whether $f(x) = 0$ or $1$. So using $AC(2)$ we can pick one colouring for each class, and then put them together, and it gives us a coloring because if any two points are neighbours, then they're in the same class, and so their colours are different.

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    You're right about my answer, although the argument should be fairly close to it. In any case, I can't think of any direct solution. Note that Herrlich *does* in fact talk about the consequence of AD implies "all sets are measurable". The graph *just below* the exercise hints that this is the path for solution. If you don't want to use "all sets are measurable", just imbue your proof with the specific proof that *this* set is not measurable. – 2017-01-22
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    For the second exercise, I thought of the following : an $\aleph_0$-coloring may help in defining a free ultrafilter over $\omega$ with a formula like $\mathcal{F} := \{A \subset \omega \mid f^{-1}(A) $ has property $P\}$ where $f$ is the coloring and $P$ is a property that's well behaved enough. Then that would contradict $AD$. That's just a guess though, I haven't found anything specific +it would imply (not necessarily, but it's likely) that $AC(2)\implies $Ultrafilter Lemma, which doesn't seem right. But maybe we can use something specific about AD to construct the ultrafilter. – 2017-01-23

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