Coloring the line You probably know the Hadwiger-Nelson problem, which states:

What is the minimal number $k$ such that there is a $k$-coloring of the plane such that no two points at distance $1$ have the same color?

This problem is open. There are known upper ($k=7$) and lower ($k=4$) bounds.
My question is: is the same problem for the line solved? I can see the obvious bound below given by $2$ and $3$ as a bound above, given by the coloring by intervals of length $\frac{2}{3}$. But is there a $2$-coloring of the line solving the problem?
 A: A colouring of the line using just two colours would be to colour the intervals $[n,n+1)$, where $n\in \mathbb Z$ alternatively using the two colours. Clearly, no two points of distance $1$ have the same colour. 
A: Sure! For $x, y \in \mathbb{R}$, consider a relation $a \sim b \iff a - b \in \mathbb{Z}$ -- that is, two numbers are in the relation if they differ by an integer. This is an equivalence relation, and it's easy to see that its equivalence classes are of the form $[x] = \{ x + k: k \in \mathbb{Z}\}$. Consider the set of equivalence classes $A = \mathbb{R}/\sim$. Pick a choice function $f: A \to \mathbb{R}$ that from each equivalence class picks a representative, i.e. satisfies $x = [f(x)]$: it exists by the axiom of choice. Now we'll construct a coloring, that is, a function $g: \mathbb{R} \to \{0, 1\}$: for $a \in \mathbb{R}$, consider $x = f([a])$. Since $[x] = [a]$, $a = x + k$ for some integer $k$. Put $g(a) = k \mod 2$. I'll leave it for you to show that $g$ is actually the coloring we want.
