Isomorphic hypergraphs of dense sets of Hausdorff spaces If $H_i = (V_i, E_i)$ are hypergraphs for $i = 1,2$ , we say that they are isomorphic if there is a bijection $f:V_1 \to V_2$ such that for all $e\subseteq V_1$ we have $e\in E_1$ if and only if $f(e)\in E_2$.
If $(X,\tau)$ is a topological space, we let the dense set hypergraph ${\cal D}(X,\tau)$ be the collection of all dense subsets of $X$ with respect to $\tau$.
Note that for $X = \{0,1,2\}$ let $\tau_1 = \{\varnothing, \{1\}, X\}$ and $\tau_2 = \tau_1 \cup\big\{\{1,2\}\big\}$. We have $(X,\tau_1)\not\cong(X,\tau_2)$, but ${\cal D}(X,\tau_1)$ and ${\cal D}(X,\tau_2)$ are isomorphic (they are even equal).
Question. Let $(X_i, \tau_i)$ be Hausdorff spaces for $i=1,2$. If ${\cal D}(X_1,\tau_1)$ and ${\cal D}(X_1,\tau_2)$ are isomorphic, does this imply that $(X_1, \tau_1)\cong  (X_2, \tau_2)$?
 A: Counterexamples abound. Here are a few of them.
Theorem. If $\emptyset\ne X\subseteq\mathbb R$ and $X\subseteq\operatorname{cl}(\operatorname{int}(X))$, then $\mathcal D(X)\cong\mathcal D(\mathbb R)$.
Proof. Construct an infinite sequence of pairwise disjoint open intervals $I_n$ so that $\bigcup_{n=1}^\infty I_n$ is a dense subset of $X$, and $|X\setminus\bigcup_{n=1}^\infty I_n|=2^{\aleph_0}$. Then a set $D\subseteq X$ is dense in $X$ if and only if $D\cap I_n$ is dense in $I_n$ for each $n$.
Thus, if $X'$ is another set satisfying the same hyotheses, with an analogous sequence of intervals $I'_n$, then a bijection $f:X\to X'$ which maps each $I_n$ homeomorphically onto the corresponding $I'_n$ is an isomorphism from $\mathcal D(X)$ to $\mathcal D(X')$.
P.S. With a somewhat more complicated argument one can prove:
Theorem. If $X$ and $Y$ are nonempty Polish spaces with no isolated points, then $\mathcal D(X)\cong\mathcal D(Y)$.
The idea is to construct a family $(U_i:i\in I)$ of open subsets of $X$ and a family $(V_i:i\in I)$ of open subsets of $Y$ so that:
(1) a set $D\subseteq X$ is dense in $X$ if and only if $D\cap U_i\ne\emptyset$ for each $i\in I$;
(2) a set $D\subseteq Y$ is dense in $Y$ if and only if $D\cap V_i\ne\emptyset$ for each $i\in I$;
(3) there is a bijection $f:X\to Y$ such that $f[U_i]=V_i$ for each $i\in I$.
