A *linear hypergraph* is a pair $\pi=(\{1,\ldots,n\}, L)$ where $n\in\mathbb{N}$, $n\geq 2$ and $L\subseteq {\cal P}(X)$ has the following properties:

- for $e\in L$ we have $|e|\geq 2$;
- if $e_1\neq e_2 \in L$ then $|e_1\cap e_2|\leq 1$.

We call $\pi = (\{1,\ldots,n\}, L)$ **projective** if for all $x, y\in \{1,\ldots,n\}$ there is $e\in L$ such that $\{x,y\}\subseteq e$. The *graph* $G_\pi$ *associated to* a linear hypergraph $\pi$ is given by $G=(V,E)$ where $V = L$ and $E = \{\{e_1, e_2\} \subseteq L: e_1\neq e_2\text{ and } e_1\cap e_2\neq \emptyset\}$.

Question.Is there a non-projective hypergraph $\pi=(\{1,\ldots,n\}, L)$ such that $G_\pi$ cannot be colored with $n-1$ colors?

projectiveby using $E$, which is only defined later on when you introduce the graph $G_{\pi}$. That's why I thought you might mean $e \in L$ instead $e \in E$. $\endgroup$ – Yuichiro Fujiwara Feb 8 '17 at 10:38projectiveshould be $L$, as in the definition of $\pi = (\{1,\ldots, n\}, L)$. $\endgroup$ – Dominic van der Zypen Feb 8 '17 at 15:00