Given a [hypergraph](https://en.wikipedia.org/wiki/Hypergraph) $H=(V,E)$ we let its intersection graph $I(H)$ be defined by $V(I(H)) = E$ and $E(I(H)) = \{\{e,e'\}: (e\neq e'\in E) \land (e\cap e'\neq \emptyset)\}$.


A *linear hypergraph* is a hypergraph $H=(V,E)$ such that every edge has at least $2$ elements and for all $e\neq e'\in E$ we have $|e\cap e'|\leq 1$. 


It turns out that for every simple undirected graph $G$, finite or infinite, there is a linear hypergraph $H$ such that $I(H)\cong G$. The *linear intersection number* $\ell(G)$ of a graph $G$ is the smallest cardinal $\kappa$ such that there is a linear hypergraph $H=(\kappa,E)$ such that $I(H)\cong G$.


**Question.** Is there an infinite graph $G$ such that $\ell(G) < \chi(G)$?


**Note.** For finite graphs, the answer to this question is [not known](https://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Faber%E2%80%93Lov%C3%A1sz_conjecture#Equivalent_formulations).



Linear intersection number and chromatic number for infinite graphs
The answer is no. This is equivalent to stating that for any linear hypergraph $H=(\kappa,E)$ with $\kappa$ infinite (note that for $\kappa$ finite, $G=I(H)$ would be finite), we have $\chi(I(H))\leq\kappa$. This follows immediately once we show $|V(I(H))|=|E|\leq\kappa$.


For $\alpha<\kappa$ let $E\_\alpha=\{e\in E:\alpha\in e\}$. If we remove $\alpha$ from every element of $E\_\alpha$ we get a family of disjoint subsets of $\kappa$. By considering their least elements, it's clear there are at most $\kappa$-many of them, i.e. $|E\_\alpha|\leq\kappa$. Therefore
$$|E|\leq\sum\_{\alpha<\kappa}|E\_\alpha|\leq\kappa^2=\kappa.$$


