Let $H\_i = (V\_i, E\_i)$ be [hypergraphs](https://en.wikipedia.org/wiki/Hypergraph) for $i=1,2$. Then we say that $H\_1\cong H\_2$ if there is a bijection $\varphi:V\_1\to V\_2$ such that $A\in E\_1$ implies $\varphi(A) \in E\_2$ and $B\in E\_2$ implies $\varphi^{-1}(B)\in E\_1$.


Is there a collection $\cal C$of pairwise non-isomorphic hypergraphs on $\omega$ with $|{\cal C}| = 2^{2^{\aleph\_0}}$?



Non-isomorphic hypergraphs on $\omega$
The answer is yes. 


Consider the collection $\mathcal C$ of hypergraphs of the following form. They have underlying set $\omega$ as the vertices, the natural numbers. The finite edges in the hypergraph are all and only the sets of the form $\{0,1,\ldots,n\}$. And then the hypergraph can have any desired collection of infinite edge sets. 


Since there are $2^{\aleph\_0}$ many infinite subsets of $\omega$, it follows that there are $2^{2^{\aleph\_0}}$ many members of $\mathcal C$. And I claim that they are all pairwise non-isomorphic. To see this, note first that $\{0\}$ is the only edge with one member, and so the isomorphism must fix $0$. Similarly, $\{0,1\}$ is the only edge with two members, and so the isomorphism must fix $1$. And so on. The finite edges force the isomorphism to fix every individual vertex, and so non-identical members of $\mathcal C$ will be non-isomorphic.



> 
> **EDIT**   Let $\ F:=\ \{\{1\ \ldots\ n\}: n\in\mathbb N\},\ $ and
>  $\ D:=\{X\in 2^\mathbb N : |X|=|\mathbb N|\}. $ Then define:
>  $$ C\ :=\ \{ F\cup G:\ G\subseteq D\} $$
>  The Dominic's isomorphisms are reduced to just one identity ismorphism per selected hypergraph, i.e. there are no non-trivial Domininc's isomorphisms between members of $\ C$.
> 
> 
> 


