This is a related question to an [older one](https://mathoverflow.net/questions/362115/are-complete-regular-linear-hypergraphs-on-omega-isomorphic).


If $H\_i = (V\_i, E\_i)$ are [hypergraphs](https://en.wikipedia.org/wiki/Hypergraph) for $i=1,2$ then we say they are *isomorphic* if there is a bijection $f: V\_1 \to V\_2$ such that for $A \subseteq V\_1$ we have $$A\in E\_1 \text{ if and only if } f(A) \in E\_2.$$
We say that $H=(\omega, E)$ is a *strongly complete regular linear* hypergraph on $\omega$ if


1. $|e| = \aleph\_0$ for all $e\in E$,
2. $e\_1\neq e\_2\in E \implies |e\_1\cap e\_2| = 1$, and
3. for all $m,n\in\omega$ there is $e\in E$ such that $\{m,n\}\subseteq e$.


**Question.** If $H\_i = (\omega, E\_i)$ are strongly complete regular linear hypergraphs for $i = 1,2$, are $H\_1$ and $H\_2$ necessarily isomorphic?



Are strongly complete regular linear hypergraphs on $\omega$ isomorphic?
If $H\_1= P^2(\mathbb{Q})$ (the projective plane defined on $\mathbb{Q}$ with the points as vertices and lines as hyper-edges), and $H\_2$ the [Moulton plane](https://en.wikipedia.org/wiki/Moulton_plane) defined on $\mathbb{Q}$, then $H\_1$ and $H\_2$ are both strongly complete regular linear hypergraphs but they are not isomorphic.


The reason is that, if $H\_1$ and $H\_2$ are isomorphic, Desargues' theorem would hold on both of them or neither of them, but Desargues' theorem holds on $H\_1$ and not $H\_2$.


