A *linear hypergraph* is a pair $\pi=(X, L)$ where $X\neq \emptyset$ is a set and $L\subseteq {\cal P}(X)$ has the following properties:


1. for $e\in L$ we have $|e|\geq 2$;
2. if $e\_1\neq e\_2 \in L$ then $|e\_1\cap e\_2|\leq 1$.


We set $X(\pi)=X$ and $L(\pi)=L$. The *graph* $G\_\pi$ *associated to* a linear hypergraph $\pi$ is given by $G=(V,E)$ where $V = L$ and $E = \{\{e\_1, e\_2\} \subseteq L: e\_1\neq e\_2\text{ and } e\_1\cap e\_2\neq \emptyset\}$. It turns out that for any graph $G$ there is a linear hypergraph $\pi$ such that $G\cong G\_\pi$. For any graph $G$ the we set $$\ell(G) := \text{min}\{|X(\pi)|:\pi \text{ is a linear hypergraph such that } G\_{\pi} \cong G\}$$ and call this the *linear intersection number* of $G$. (For infinite graphs, this concept is boring: $\ell(G) = |V(G)|$ for infinite graphs.)


For any graph $G$ let $\tau(G)$ be the minimum cardinality of a [vertex cover](http://en.wikipedia.org/wiki/Vertex_cover) of $G$. 


Does $\tau(G) \leq \ell(G)$ hold for all graphs $G$?



Linear intersection number and vertex covering number
Let $H$ be a graph on $n$ vertices, thought of as the hypergraph in the question. Then a vertex cover in $G\_H$ is precisely a subgraph of $H$ that contains all but at most one edge at each vertex of $H$. Such a subgraph is obtained from $H$ by deleting a subgraph of maximum degree at most $1$, which contains at most $n/2$ edges. So any graph $H$ with more than $3n/2$ edges will provide a counterexample.


