For any set $X$, let $[X]^2 = \big\{\{a,b\}: a, b\in X \land a\neq b\big\}$. Let $\kappa$ be an infinite cardinal. Is there a set ${\cal E} \subseteq {\cal P}([\kappa]^2)$ such that


1. for all $E \in {\cal E}$ the simple undirected graph $(\kappa,E)$ is connected,
2. if $E\_1\neq E\_2\in {\cal E}$ then the graphs $(\kappa,E\_1)$ and $(\kappa, E\_2)$ are non-isomorphic, and
3. $|{\cal E}| = 2^\kappa$


?



Mutually non-isomorphic connected graphs on $\kappa$ points
Under the axiom of choice, the answer is yes. Interpret $\kappa$ as its initial ordinal. We take the vertex to consist of distinct vertices $A\_\alpha,B\_\alpha,C\_\alpha,D\_\alpha$ for each $\alpha<\kappa$. In each of the graphs, $B\_\alpha$ is connected to $A\_\alpha$ and $C\_\alpha$ and possibly to $D\_\alpha$, so that $B\_\alpha$ always hes degree $2$ or $3$. $D\_\alpha$ will only be connected to $B\_\alpha$ if anything, so will have degree $0$ or $1$. We connect $A\_\alpha$ to $C\_\beta$ for $\alpha<\beta$. For any $S\subseteq\kappa$ let $G\_S$ be the graph in which an edge between $B\_\alpha$ and $D\_\alpha$ is present iff $\alpha\in S$.


If I'm not mistaken, then we can show by transfinite induction of $\alpha$ that in any isomorphism between $G\_S,G\_{S'}$, $A\_\alpha$ goes to $A\_\alpha$, $B\_\alpha$ to $B\_\alpha$ and $C\_\alpha$. But since $B\_\alpha$ has degree $3$ in $G\_S$ iff $\alpha\in S$, it follows that this can truly be an isomorphism only if $S=S'$.


