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


*

*$e_1\neq e_2\in E$ implies $|e_1\cap e_2| = 1$, and

*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 $|e|<\kappa$?
 A: 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.
A: Encouraged by (Martin?) Goldstern, I resurrect some half-remembered arguments from Emil Artin's Geometric Algebra, although there likely are more recent sources and perhaps cleaner presentations than below.
Dominic's setup and condition 2 implies there is at least one line.  If there is exactly one line, then it is the whole space.  Moving on to the case of there being at least two lines with none of them singleton sets, given a line (warning: notational conflict with question ahead) $E$ there is then a point $p$ which is not on $E$.  The number of lines through $p$ is in 1-1 correspondence with the number $\lambda$ of points on $E$; condition 1 shows there are at most $\lambda$ many lines through $p$, while condition 2 gives at least that many lines.  However, this is true for every point $p$ off of $E$. If there is another line $F$ different from $E$ not containing $p$ (and there is since we have enough points and lines, otherwise $p$ is the only point off of $E$), the same argument shows $F$ has as many points as does $E$.  This further extends to all lines through $p$: each of those has $\lambda$ many points.
Finally we have $\lambda \leq \kappa \leq \lambda \cdot \lambda$. The left inequality follows because $E$ is a proper subset of the space.  The right hand follows because all the lines through $p$ cover the space.  If we are in a set theory with nice cardinal arithmetic, then the left hand side equals the right hand side, and the answer to the question becomes no.  The case involving singleton sets is left to Joel.
Gerhard "More Comfortable With Finitary Logic" Paseman, 2017.12.29.
