Density of Ramsey subsets of $\omega$ 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:

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.

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 $d(S)>0$?
 A: 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.
A: If we define $\pi: [\mathbb N]^2 \rightarrow \{0,1\}$ randomly (say each $\pi(a,b)$ is determined by a coin flip), then almost surely there is no set $S$ with $d(S) > 0$ that is Ramsey for $\pi$. In fact, it is almost surely true that every $S$ with $d(S) > 0$ contains an induced isomorphic copy of the randomly colored infinite graph.
Even more: for a random coloring $\pi$ of $[\mathbb N]^2$, there is almost surely no set $S$ with $\sum_{n  \in S \setminus \{0\}} \frac{1}{n} = \infty$ that is Ramsey for $\pi$. In fact, it is almost surely true that every such $S$ contains an induced copy of every coloring of every finite graph. (But in this case, "finite" cannot be improved to "infinite" as above.)
These results can be found in Section 2 of my paper "Which subsets of the infinite random graph look random?" (Mathematical Logic Quarterly 64 (2018), pp. 478-486), available here.
