TITLE: Infinite projective plane with small edges
QUESTION [7 upvotes]: 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 $|e|<\kappa$?

REPLY [3 votes]: 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.