Let $H=(V,E)$ be a [hypergraph](http://en.wikipedia.org/wiki/Hypergraph), and $\kappa$ be a cardinal. We say that a map $c:V \to \kappa$ is a *coloring* if the restriction $c\restriction\_e$ is non-constant whenever $e\in E$ and $|e|\geq 2$. The smallest cardinal $\kappa$ such that there is a coloring $c:V\to\kappa$ is said to be the *chromatic number* of $H$, and we denote it by $\chi(H)$.


Let $\omega$ denote the set of non-negative integers. We say that a hypergraph $H=(\omega,E)$ is a *rainbow hypergraph* if every member of $E$ is finite, and if for every $n\in\omega\setminus\{0,1\}$ there is exactly one $e\in E$ with $|e|=n$ (that is, for all $n\in\omega\setminus\{0,1\}$ we have $|\{e\in E: |e| = n\}|=1$).


There are many rainbow hypergraphs with chromatic number $2$ (for instance, any hypergraph in which the edges are pairwise disjoint.)


**Question.** Given $k\in (\omega\cup\{\omega\})\setminus \{0,1\}$, is there a rainbow hypergraph $H=(\omega,E)$ with $\chi(H)=k$?



Chromatic number of rainbow hypergraphs
In fact, every rainbow hypergraph has chromatic number $2$.


Let $H=(V,E)$ be a rainbow hypergraph, $E=\{e\_2,e\_3,\dots\}$, $|e\_n|=n$. Consider a random coloring $c:V\to\{0,1\}$, let $A$ be the event that $c$ is not a proper coloring of $H$, and let $A\_n$ be the event that $c$ is constant on $e\_n$. Then
$$P(A)=P\left(\bigcup\_{n=2}^\infty A\_n\right)\lt\sum\_{n=2}^\infty P(A\_n)=\sum\_{n=2}^\infty\frac1{2^{n-1}}=1,$$
so proper $2$-colorings exist and $\chi(H)=2$.


**P.S.** Here is an alternative argument which even proves a slightly stronger result: a rainbow graph remains $2$-colorable if one more edge is added arbitrarily.; i.e., there are now two edges of size $2$ and one edge of size $n$ for each integer $n\gt2$.


Let $H=(V,E)$ be a hypergraph, $E=\{e\_1,e\_2,e\_3,\dots\}$ where $|e\_n|=\max(n,2)$. We color the vertices sequentially, coloring $2$ vertices at each step, one red and the other blue; this is done in such a way that after the $n^\text{th}$ step is completed (if not sooner) the edge $e\_n$ contains at least one vertex of each color. This can always be done, unless the elements of $e\_n$ have all been given the same color before the $n^\text{th}$ step; but that can't happen because $|e\_n|\ge n$ and each color has only been used $n-1$ times before the $n^\text{th}$ step.


