For any set $X$ let $[X]^2=\{\{x,y\}:x\neq y \in X\}$. The starting point of this question is the following statement that follows from a more general theorem by Ramsey:



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> If $\pi:[\omega]^2\to\{0,1\}$ is any map, then here is an infinite set $S\subseteq \omega$ such that the restriction $\pi|\_{[S]^2}$ is constant.
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We call an infinite set $S$ with the above property a *Ramsey set* for the map $\pi:[\omega]^2\to\{0,1\}$. For $A\subseteq \omega$ we define its upper density $d(A)\in [0,1]$ by $$d(A) = \lim\sup\_{n\to\infty}\frac{|A\cap\{1,\ldots,n\}|}{n}.$$


**Question.** Given a map $\pi:[\omega]^2\to\{0,1\}$, is there necessarily a Ramsey set $S\subseteq \omega$ with $d(S)>0$?



Density of Ramsey subsets of $\omega$
Decompose $\omega$ into the disjoint union of the sets $I\_k$ where $I\_k=[k!,(k+1)!-1]$. Let $f(x,y)$ be 1 if $x,y$ are in distinct intervals, otherwise 0. It is easy to see that each homogeneous set for 1 is finite, for 0 has zero density.


