I first asked the question below at math.stackexchange.com ( <https://math.stackexchange.com/questions/920442/number-of-points-in-an-intersecting-linear-hypergraph> ) but somebody suggested I ask it in mathoverflow.net instead. Here's the question:


An *intersecting linear hypergraph* is a pair $H=(P,\mathcal{L})$ where $P\neq\emptyset$ is a finite set and $\mathcal{L}$ is a collection of subsets of $P$ such that


* every member of $\mathcal{L}$ has at least 2 elements, and
* if $l\_1, l\_2 \in \mathcal{L}$ then there is $p\in P$ such that $l\_1 \cap l\_2 = \{p\}$.


I am convinced that in any intersecting linear hypergraph $H=(P,\mathcal{L})$ we have at least as many points as lines (i.e. $|P| \geq |\mathcal{L}|$). How can this be proved (if it is correct at all)? And if it is correct, does $|P| \geq |\mathcal{L}|$ also hold if $P$ is infinite?



Number of points in an intersecting linear hypergraph
The usual proof of Fisher's inequality is very finite. I will call things in $\mathcal{P}$ points and those in $\mathcal{L}$ lines although the reverse might be more natural. Let $L=|\mathcal{L}|,P=|\mathcal{P}|$ and let line $\ell\_i$ have $c\_i+1$ points. Consider $A$, the $L \times P$ $0,1$ matrix giving incidence between the two sets. It has rank at most $\min(L,P).$ Now $B=AA^T$ is an $L \times L$ matrix with all off diagonal entries $1$ and diagonal entries $1+c\_i$. It is easy to find the determinant of $B$ and that it is positive. In fact it is $\Pi c\_i(1+\Sigma\frac{1}{c\_i})$. So $A$ has rank at least that of $B$ which is $L.$ Hence $\min(P,L) \ge L.$ The case $P=L$ is quite interesting but is another story.


In the infinite case I think it always happens that $P=L$ if we eliminate unneeded points, however this is not that special. For $\mathcal{P}$ infinite we would consider that the set $\mathcal{P}\_2$ of pairs from $\mathcal{P}$ has the same size as $\mathcal{P}$ itself. But $|\mathcal{L}|$ is certainly no larger $|\mathcal{P}\_2$ since any pair of points from a line $\ell$ determines it uniquely.


