For any cardinal $\alpha \in \omega\cup \{\omega\}$, let $[\omega]^\alpha$ denote the collection of subsets of $\omega$ having cardinality $\alpha$.


A *linear hypergraph* $H=(V,E)$ is a [hypergraph](https://en.wikipedia.org/wiki/Hypergraph) such that whenever $e\neq e\_1\in E$ we have $|e\cap e\_1|\leq 1$.


A *coloring* of a hypergraph $H=(V,E)$ is a map $c:V \to \alpha$, where $\alpha \neq \varnothing$ is a cardinal, such that for all $e\in E$ with $|e|>1$ we have that the restriction $c{\restriction}\_e$ is non-constant. We denote by $\chi(H)$ the smallest cardinal such that there is a coloring from $V$ to that cardinal.


If $\alpha \in (\omega\cup\{\omega\})\setminus \{0,1,2\}$, is there a linear hypergraph $H = (\omega, E)$ with $E\subseteq [\omega]^\alpha$ and $\chi(H)=\aleph\_0$?



Chromatic number of regular linear hypergraphs on $\omega$
For $\alpha=\omega$ the answer is no. If $H=(\omega,E)$ is a linear hypergraph, then $E$ is countable; if $H=(\omega,E)$ is any hypergraph with $E\subseteq[\omega]^\omega$ and $E$ countable, then $\chi(H)\le2$.


For $3\le\alpha\lt\omega$ the answer is yes. It will be convenient to define the hypergraph on a countable vertex set $V\ne\omega$. Let $H=(V,E)$ where $V=[\omega]^{\alpha-1}$ and $E=\{[A]^{\alpha-1}:A\in[\omega]^\alpha\}\subseteq[V]^\alpha$. Then $H$ is a linear hypergraph, and $\chi(G)=\aleph\_0$ because $\omega\to(\alpha)^{\alpha-1}\_n\ (n\lt\omega)$ by the finite Ramsey theorem.


**P.S.** $\omega\to (m)^r\_n$ is Rado's "[arrow notation](https://en.wikipedia.org/wiki/Infinitary_combinatorics#Ramsey_theory_for_infinite_sets)" for "partition relations"; it means that, for any $n$-coloring of the $r$-element subsets of $\omega$, there is an $m$-element subset of $\omega$ whose $r$-element subsets all have the same color.


