Let $\kappa$ be an infinite cardinal. We say $E\subseteq {\cal P}(\kappa)$ is an *infinite projective plane* on $\kappa$ if


1. $e\_1\neq e\_2\in E$ implies $|e\_1\cap e\_2| = 1$, and
2. whenever $n\neq m\in \kappa$, there is $e\in E$ with $\{n,m\}\subseteq e$.


Is it possible to find an infinite projective plane $E$ on an infinite cardinal $\kappa$ such that for all members of $E$ we have $|e|<\kappa$?



Infinite projective plane with small edges
**Update.** Here is a new simpler answer that works for all regular
$\kappa$, including $\kappa=\omega$. And I have omitted the use
of Fodor's lemma, using instead merely the pigeon-hole principle.


Suppose that $\kappa$ is infinite and we have a projective plane on
$\kappa$ many points, with all lines of size less than $\kappa$.
Since there are $\kappa$ many pairs of points, there must be
$\kappa$ many distinct lines. Enumerate them as $\langle
e\_\alpha\mid\alpha<\kappa\rangle$. When $\kappa=\omega$, this is a
simple $\omega$-sequence $e\_0$, $e\_1$, and so on.


Consider the line $e\_0$, which has fewer than $\kappa$ many points.
For each line $e\_\gamma$ for $\gamma\neq 0$, let $\alpha\_\gamma$ be
the unique element of $e\_\gamma\cap e\_0$. Since $\kappa$ is
regular, it follows by the pigeon-hole principle that there is a
subset $S\subset\kappa$ of size $\kappa$ with constant value
$\alpha\_\gamma=\alpha$ for $\alpha\in S$. Since $\alpha$ cannot be
on all the lines, there is some line $e\_i$ with $\alpha\notin e\_i$.
For $\gamma>i$, let $\beta\_\gamma$ be the unique element of
$e\_\gamma\cap e\_i$. Since $e\_i$ has size less than $\kappa$, it
follows again by the pigeon-hole principle that there is some
$S'\subset S$ of size $\kappa$ with constant value
$\beta\_\gamma=\beta$ for all $\gamma\in S'$.


Note that $\alpha\neq\beta$, since $\alpha\notin e\_i$, while
$\beta\in e\_i$. But note also that $\alpha,\beta\in e\_\gamma$ for
all $\gamma\in S'$, which means that all these lines have at least
two points in common, contrary to the requirement (1) that any two
lines intersect in exactly one point. $\Box$


The case of singular $\kappa$ seems to remain open.


