How does one prove existence decomposition of $R^2$ for countable many subsets $A(n)$
 s
.t.$\forall$ $x,y$ $\epsilon$ $A(n)$ $|x-y|$ is nonrational?


I tried thinking of $R^2$ as infinite tree with $2^\omega$ levels of cardinality $2^\omega$ but I don't see how to make this nonrationality condition hold.



Countable coloring of a plane
This is a result of Erdos: Problems and results in chromatic graph theory, Proof Techniques in Graph Theory (Proc. Second Ann Arbor Graph Theory Conf., Ann Arbor, Mich., 1968) , pp. 27--35, Academic Press, New York, 1969, see p. 32. Available at the Renyi Institute's collection of Erdos papers: 
<http://www.renyi.hu/~p_erdos/1969-13.pdf> 


Here is the proof. Let $X$ be the graph on the plane where two points are 
joined iff their distance is rational. What we want to prove is that the 
graph has countable chromatic number. Instead, we prove that it has a 
well ordering in which each point is joined into finitely many smaller 
(by the well ordering) points. This done, it is easy to good color the 
vertices via a transfinite recursion along the well ordering. 


We prove the above statement for every $X\subseteq R^2$, by tranfinite 
induction on $|X|$. The case when $X$ is countable is obvious: order $X$ 
as the natural numbers. Assume that $\kappa=X$ is uncountable, and the statement 
is proved for every set $Y$ of cardinality less than $\kappa$. 


If $a\neq b$ are point of the plane, then let $F(a,b)$ denote the set 
of points in rational distance from both $a$ and $b$. It is easy to see 
that $F(a,b)$ is always countable. 


We can exhibit $X$ as the increasing, continuous union of subsets 
$\{X\_\alpha:\alpha<\kappa\}$ such that $X\_0=\emptyset$, 
$|X\_\alpha|<\kappa$ and each $X\_\alpha$ is closed under $F$. 
One way of seeing this is to enumerate 
$X$ as $\{x\_\alpha:\alpha<\kappa\}$ and let $X\_\alpha$ be the closure 
of $\{x\_\beta:\beta<\alpha\}$ under the operation $F$.  

Now $X$ can partitioned into the difference sets $X\_{\alpha+1}-X\_\alpha$.
Each set $X\_{\alpha+1}-X\_\alpha$ possesses a well ordering $\prec\_\alpha$ as 
required and we define the well ordering of $X$ as follows: 
$x\prec y$ iff either $x\in X\_{\alpha+1}-X\_\alpha$ and 
$y\in X\_{\beta+1}-X\_\beta$ with $\alpha<\beta$ 
or else $x \prec\_\alpha y$ for some $\alpha$. 


We have to show that $\prec$ is as wanted, so let $a$ be a point. 
$a\in X\_{\alpha+1}-X\_\alpha$ for some $\alpha<\kappa$.  

We count those points $x$ which are joined to $a$ and $x\prec a$. 
Of these, only finitely many are in the $X\_{\alpha+1}-X\_\alpha$, as 
those precede $a$ by $\prec\_\alpha$. The remaining points are in $X\_\alpha$. 
But there can be only at most one point in $X\_\alpha$ in rational 
distance from $a$, as otherwise, if say, $p,q\in X\_\alpha$ are in rational 
distance from $a$, then $a\in F(p,q)\subseteq X\_\alpha$, a contradiction. 
Altogether, we get finitely many plus one=finitely many points 
preceding $a$ and in rational distance from it. 


