For any integer $k>1$ we say a hypergraph $H=(\omega,E)$ where $E\subseteq {\cal P}(\omega)$ is $k$-*regular* if $|e|=k$ for all $e\in E$. Moreover, we say $H$ is *linear* if $|e\_1\cap e\_2|\leq 1$ for all $e\_1\neq e\_2 \in E$.


Zorn's lemma shows that whenever $(\omega,E)$ is linear and $k$-regular, there is $E'\supseteq E$ such that $(\omega,E')$ has the same properties and is maximal with respect to $k$-regularity and linearity (that is, whenever $e\in [\omega]^k\setminus E'$ we have that $(\omega, E'\cup \{e\})$ is no longer linear, where $[\omega]^k$ denotes the collection of subsets of $\omega$ with cardinality $k$).


If $\kappa>0$ is a cardinal and $H=(V,E)$ we say $c:V\to \kappa$ is a *(hypergraph) colouring* if the restriction $c\restriction\_e$ is non-constant whenever $e\in E$ and $|e|>1$. The smallest cardinal $\kappa$ such that there is a colouring $c:V\to \kappa$ is called the *chromatic number* of $H$ and denoted by $\chi(H)$.


**Question.** Whenever $(\omega, E\_1)$ and $(\omega,E\_2)$ are $k$-regular for $k>2$ and maximal linear, do we have $\chi(\omega,E\_1)= \chi(\omega,E\_2)$?



Chromatic number of maximal linear $k$-regular hypergraphs on $\omega$
The answer is negative. Suppose $3\le k\lt\omega$. As I showed in my answer to [this question](https://mathoverflow.net/questions/363267/chromatic-number-of-regular-linear-hypergraphs-on-omega/363283), there is a linear $k$-hypergraph $(\omega,E)$ with chromatic number $\aleph\_0$; of course it can be extended to a maximal linear $k$-hypergraph on $\omega$, which will still have chromatic number $\aleph\_0$. (By a $k$-hypergraph I mean a hypergraph whose edges are $k$-element sets.) I will now show how to construct a maximal linear $k$-hypergraph with chromatic number $2$.


Let $\omega=X\cup Y$ where $X$ and $Y$ are disjoint infinite sets. Choose sets $E\subseteq[X]^{k-1}$ and $F\subseteq[Y]^{k-1}$ so that $(X,E)$ and $(Y,F)$ are maximal linear $(k-1)$-hypergraphs. Enumerate the sets $E$ and $F$ without repetition as $E=\{e\_n:n\in\omega\}$ and $F=\{f\_n:n\in\omega\}$. For $n\in\omega$ choose recursively $x\_n\in X\setminus(e\_1\cup\cdots\cup e\_n\cup\{x\_1,\dots,x\_{n-1}\})$ and $y\_n\in Y\setminus(f\_1\cup\cdots\cup f\_n\cup\{y\_1,\dots,y\_{n-1}\})$. Let $\hat E=\{e\_n\cup\{y\_n\}:n\in\omega\}$ and $\hat F=\{f\_n\cup\{x\_n\}:n\in\omega\}$.


It can easily be verified that $\hat H=(\omega,\hat E\cup\hat F)$ is a linear $k$-hypergraph, and $\chi(H)=2$. Extend $\hat H$ to a maximal linear $k$-hypergraph $H$. (If $k=3$ then $E=[X]^2$, $F=[Y]^2$, and $\hat H$ is already maximal, so $H=\hat H$ in this case.) Because of the maximality of $E$ and $F$, each edge of $H$ meets both $X$ and $Y$, so $H$ is still $2$-colorable.


For a simple concrete example of a $2$-colorable maximal linear $3$-hypergraph, take $H=(\omega,E\cup F)$ where
$$E=\{\{x,y,z\}\in[\omega]^3:x\ne y,\ x\equiv y\equiv0\pmod2,\ z=x+y+1\}$$
and
$$F=\{\{x,y,z\}\in[\omega]^3:x\ne y,\ x\equiv y\equiv1\pmod2,\ z=x+y\}.$$


