Suppose $X\neq \emptyset$ is a set. Let $\tau\_1, \tau\_2$ be Hausdorff topologies on $X$ with the property that the partially ordered sets $(\tau\_1,\subseteq)$ and $(\tau\_2,\subseteq)$ are order-isomorphic.


Does this imply that $(X,\tau\_1)\cong (X,\tau\_2)$ as topological spaces?




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If Hausdorffness is dropped, then there are easy counterexamples, for instance $X = \omega$ and $\tau\_1 = \{\emptyset,\{0\},\omega\}, \tau\_2 = \{\emptyset, \omega\setminus\{0\},\omega\}$.



$T_2$-spaces with order-isomorphic topologies
If $h:(\tau\_1, \subseteq) \to (\tau\_2, \subseteq)$ is an order isomorphism.


Define $M(\tau\_1) = \{O \in \tau\_1: O \neq X \land \forall O' \in \tau\_1: O \subseteq O' \implies O' = X\}$, the "submaximal" elements.


Because the definition is purely order theoretical, $h[M(\tau\_1)] = M(\tau\_2)$.


Then all $O \in M(\tau\_1)$ are of the form $X\setminus \{p\}$ for some $p \in X$ when $X$ is a $T\_1$ space: if $|X \setminus O| \ge 2$ pick $x$ in this complement, and $O \subseteq X\setminus \{x\} \subseteq X$, where the middle set is open by $T\_1$-ness, this shows $O \notin M(\tau\_1)$. So $\forall O \in M(\tau\_1): |X \setminus O| =1$. Also, all sets of the form $X\setminus \{p\}$ are in $M(\tau\_1)$.


So the above implies that if both topologies are $T\_1$, we can define $f: X \to X$ by 
$\{f(x)\} = X\setminus h(M\setminus \{x\})$.


One can now check that $f$ is the required homeomorphism, using that $h$ is an order isomorphism.


