Let $[\omega]^\omega$ denote the set of infinite subsets of $\omega$. 


Let $S\subseteq [\omega]^\omega$. We say that a map $c:\omega \to \{0,\ldots,n-1\}$ is a *coloring for $S$ with $n$ colors* if for all $s\in S$ the restriction $c|\_s$ of $c$ to $s$ is non-constant.


What is an example of a set $S\subseteq [\omega]^\omega$ such that $S$ has a coloring with $3$ colors, but not with $2$ colors?



Subset of $[\omega]^\omega$ that can be "colored" with $3$, but not $2$ colors
Partition $\omega$ into three infinite subsets $A\_0,A\_1,A\_2$. Let $S$ consists of subsets which intersects precisely two of the $A\_i$ at infinitely many elements. It can obviously be $3$-colored. Suppose there was a $2$-coloring, with color classes $c\_0,c\_1$. Then either $c\_0$ or $c\_1$ contains infinitely many elements of some two of $A\_0,A\_1,A\_2$, say $c\_0\cap A\_0,c\_0\cap A\_1$ are infinite. Then $c\_0\cap(A\_0\cup A\_1)\in S$ is monochromatic.


This can be generalized to a hypergraph $S\subseteq[\omega]^\omega$ with chromatic number $n$: partition $\omega$ into $A\_0,\dots,A\_{n-1}$ and let $S$ consists of subsets which intersect two of those at infinitely many elements.


