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

  • 1
    $\begingroup$ 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. $\endgroup$ – 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 $i\in I$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.