If $H\_i = (V\_i, E\_i)$ are [hypergraphs](https://en.wikipedia.org/wiki/Hypergraph) 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 $(X\_1, \tau\_1)\cong (X\_2, \tau\_2)$?



Isomorphic hypergraphs of dense sets of Hausdorff spaces
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 $i\in I$.


