Isomorphic hypergraph duals

Question

Let $H = (V,E)$ be a hypergraph. For $v\in V$ we set $E_v = \{e\in E: v\in E\}$. The dual of $H$ is defined by $H^* =(E, V^*)$ is, where $V^* = \{E_v:v\in V\}$.

We say hypergraphs $H_i=(V_i, E_i)$ for $i=1,2$ are isomorphic, in symbols $H_1\cong H_2$, if there is a bijection $\varphi:V_1\to V_2$ such that $\varphi(e_1)\in E_2$ whenever $e_1\in E_1$, and $\varphi^{-1}(e_2)\in E_1$ whenever $e_2 \in E_2$.

Let $[\omega]^\omega$ denote the set of infinite subsets of $\omega$, and let $E_1, E_2\subseteq [\omega]^\omega$ such that $\bigcup E_i = \omega$ for $i=1,2$. Let $H_i = (\omega, E_i)$ for $i=1,2$.

Question. If $H_1^* \cong H_2^*$, do we necessarily have $H_1\cong H_2$?

Answer 1

Counterexample. Let $E_1=\{e_0,e_1\}$ and $E_2=\{e_0,e_2\}$ where $$e_0=\{0,1\}\cup\{2,4,6,8,\dots\},$$ $$e_1=\{0\}\cup\{3,5,7,9,\dots\},$$ $$e_2=\{0,1\}\cup\{3,5,7,9,\dots\}.$$ Then $(\omega,E_1)\not\cong(\omega,E_2)$, but $(E_1,V_1^*)\cong(E_2,V_2^*)$ since $V_1^*=\{\{e_0\},\{e_1\},\{e_0,e_1\}\}$ and $V_2^*=\{\{e_0\},\{e_2\},\{e_0,e_2\}\}.$

On the other hand, if $H=(V,E)$ is a hypergraph with the property that $E_v\ne E_w$ whenever $v\ne w$, then the dual hypergraph $H^*$ has the same property, and $(H^*)^*\cong H$. Hence, if $H_1$ and $H_2$ are two such hypergraphs, then $H_1^*\cong H_2^*\implies H_1\cong H_2$.