If ${\cal C}$ is a collection of subsets of a set $X$, we associate to ${\cal C}$ a graph $G\_{\cal C} = (V,E)$ where $V = {\cal C}$ and $$E = \big\{\{A,B\}: A\neq B\in {\cal C} \land A\cap B \neq \emptyset\big\}.$$


If $G$ is a simple, undirected graph, we define its *intersection number* $\iota(G)$ to be the smallest cardinal $\kappa$ such that there is a collection ${\cal C}$ of subsets of $\kappa$ such that $G\_{\cal C} \cong G$.


For any cardinal $\kappa>0$ we define $\log(\kappa) = \min\{\mu\in\kappa\cup\{\kappa\}: 2^\mu \geq \kappa\}.$


If $G=(V,E)$ is infinite, is it true that $\iota(G) \in \{\;\log(|V|), |V| \;\}$? 



Intersection number of infinite graphs
I think the answer is no, at least not without some additional assumptions. Suppose that $|\omega| < |\omega\_1| < \mathfrak c$ and let $G$ be the union of the complete graph on $\mathfrak c$ and an independent set of size $|\omega\_1|$. As you mention in the comments, the clique can be represented as an intersection graph of subsets of $\omega$, so we can represent $G$ using a ground set of size $|\omega\_1| < \mathfrak c$. But we certainly need a ground set of size at least $|\omega\_1| > |\omega| = \log \mathfrak c$.


