Let $H=(V,E)$ be a [hypergraph](https://en.wikipedia.org/wiki/Hypergraph). For $v\in V$ we let $v^\* = \{e\in E:v\in e\}$. We define the *dual* of $H$ by $H^\*= (E, V^\*)$ where $V^\* = \{v^\*: v\in V\}$. We say that a $H$ is *self-dual* if $H \cong H^\*$.


Is there a self-dual hypergraph $H = (\omega, E)$ with the following property?



> 
> Whenever $e\_0\neq e\_1 \in E$ we have both $e\_0 \not\subseteq e\_1$ and $|e\_0| \neq |e\_1|$.
> 
> 
> 



Self-dual hypergraph on $\omega$
Expanding my comments into an answer: by rephrasing this as a question of constructing a symmetric $\omega\times\omega$ matrix with the same properties, we can give an explicit construction. For convenience's sake I'm going to start numbering at 0. Let $e\_0 = \{v\_0, v\_1\}$ and for $n\geq 1$ let $e\_n = \{v\_i\} \cup \{v\_j: f(n)\lt j\leq f(n+1)\}$, where $f(n)$ is the sum-of-integers function $f(n) = \frac12n(n+1)$ and $i$ is the unique number such that $f(i)\lt n\leq f(i+1)$ — in other words, $i$ is chosen to make the generated matrix symmetric.


This hypergraph has the nice property of being connected: for any two vertices there's some chain of edges that connects them to each other. In fact, it's a tree; for any two vertices there's exactly one chain of edges connecting them, or from a different perspective for any edge $e\_i$ there's exactly one edge $e\_j$ with $j\lt i$ and $e\_i\cap e\_j\neq\emptyset$. By allowing for more overlap in one half of the matrix, we can get 'greater' connectivity: for instance, take $e\_0=\{v\_0, v\_1, v\_2\}$, $e\_1 = \{v\_0, v\_2, v\_3, v\_4\}$, $e\_2 = \{v\_0, v\_1, v\_4, v\_5, v\_6, v\_7\}$, etc.


