Chromatic number of rainbow hypergraphs Let $H=(V,E)$ be a 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 $\chi(H)=k$?
 A: 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.
