We call $E\subseteq {\cal P}(\omega)$ a *Fano-like plane* if


1. for all $x,y\in \omega$ there is $e\in E$ with $\{x,y\}\subseteq e$,
2. whenever $e\_1\neq e\_2\in E$ we have $|e\_1\cap e\_2|=1$, and
3. $|e|>1$ for all $e\in E$.


There are Fano-like planes in which not all edges (members of $E$) have the same cardinality: consider the "pseudo-pencil" for $E =\{\omega\setminus\{0\}\}\cup\big\{\{0,i\}: i \in \omega\setminus\{0\}\big\}$.


Is there a Fano-like plane $E\subseteq {\cal P}(\omega)$ such that $|e|>2$ for all $e\in E$, and not every member of $E$ has the same cardinality?



Fano-like planes on $\omega$
No. Suppose $E\subseteq\mathcal P(\omega)$ is a Fano-like plane. If $e\in E$ and $x\in\omega\setminus e$, then there is a bijection between the edges containing $x$ and the points in $e$. Therefore, given two edges $e\_1,e\_2$ and a point $x\notin e\_1\cup e\_2$, there is a bijection between the points of $e\_1$ and the points of $e\_2$. Therefore, if there are two edges $e\_1$ and $e\_2$ with different cardinalities, then we must have $e\_1\cup e\_2=\omega$. Now choose a point $x\in e\_1\setminus e\_2$ and a point $y\in e\_2\setminus e\_1$, and let $e\_3$ be the edge containing $e\_1$ and $e\_2$. then $e\_3=\{x,y\}$ is an edge containing only two points. In fact, it's easy to see that, if the edges don't all have the same cardinality, then there is only one edge $e$ with $|e|\gt2$.


