# 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 $$(X_1, \tau_1)\cong (X_2, \tau_2)$$?

• If $X_1$ and $X_2$ are non-homeomorphic countably infinite spaces each having exactly one non-isolated point, then they have the same dense subsets. – Anonymous Jul 27 at 11:42

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 $$i\in I$$.