What are examples of non-homeomorphic connected $T_2$-spaces $(X_i,\tau_i)$ for $i=1,2$ such that the posets $(\tau_1, \subseteq)$ and $(\tau_2,\subseteq)$ are order-isomorphic?
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0 asked Nov 10, 2020 at 13:16
Dominic van der Zypen
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2 Answers
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1 There aren't any. Hausdorff spaces are sober spaces. If $X, Y$ are sober, then every frame map $\mathcal{O}(Y) \to \mathcal{O}(X)$, i.e., every poset map between their topologies that preserves finite meets and arbitrary joins, arises from a uniquely determined continuous map $X \to Y$. It follows that a poset isomorphism $\mathcal{O}(X) \cong \mathcal{O}(Y)$, being a frame isomorphism, arises from a homeomorphism between the spaces.
(Just to give slightly more detail: for a sober space $X$, the points of $X$ are in natural bijection with frame maps $\mathcal{O}(X) \to \mathcal{O}(1)$ where the codomain is the topology on a one-point space. Thus a frame map $\phi: \mathcal{O}(Y) \to \mathcal{O}(X)$ induces, via composition with frame maps $\mathcal{O}(X) \to \mathcal{O}(1)$, a function $f: X \to Y$, and is itself of the form $\phi(V) = f^{-1}(V)$.)
- $\begingroup$ I'm very curious as to why someone downvoted this. $\endgroup$Todd Trimble– Todd Trimble2024-03-11 13:29:33 +00:00Commented Mar 11, 2024 at 13:29
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1 Here's an answer without familiarity with locales, which I started after reading Todd Trimble's answer (so his answer is the right one to accept).
Let $X$ be a $\mathsf{T}_1$ topological space, $\tau_X$ the lattice of open subsets, and $\Phi_X$ the opposite lattice, which identifies to the lattice of closed subsets. Let's reconstruct $X$ from $\Phi_X$.
Denote by $0$ "zero" the unique minimal element in $\Phi_X$. Let $\Phi_X^\min$ be the set of minimal elements in $\Phi_X\smallsetminus\{0\}$. Let $i$ be the map $x\mapsto\{x\}$. Since $X$ is $\mathsf{T}_1$, $i$ is a well-defined injective map $X\to\Phi_X$, and its image is exactly $\Phi_X^\min$. (This already retrieves the cardinal of $X$.)
Now we wish to retrieve the topology. Namely, I claim that for $K\subset\Phi_X^\min$, $i^{-1}(K)$ is closed if and only if there exists $F\in\Phi_X$ such that $K=\{Z\in\Phi_X^\min\,:\,Z\le F\}$.
Indeed, suppose that $i^{-1}(K)$ is closed: define $F_K=i^{-1}(K)$ (so $K=i(F_K)$): then $\{Z\in\Phi_X:Z\le F_K\}=\{\{z\}:z\in F_K\}=i(F_K)=K$. Conversely, suppose $K=\{Z\in\Phi_X:Z\le F\}$ for some $F\in\Phi_X$. So $K=\{\{z\}:z\in F\}=i(F)$, so $i^{-1}(K)=F$ is closed.
Hence, for any $\mathsf{T}_1$ topological spaces $X,Y$, every isomorphism $\tau_X\to\tau_Y$ is induced by a unique homeomorphism $X\to Y$. This also shows that the automorphism group of $\Phi_X$ is canonically isomorphic to the self-homeomorphism group of $X$.
This works without assuming $X$ to be sober. For example, it applies for the cofinite topology, $\Phi_X$ consisting of $X$ and its finite subsets, which is non-sober as soon as $X$ is infinite. (Todd's answer also encompasses non-bijective maps, which I didn't address; soberness is then probably important. Also, there are non-$\mathsf{T}_1$ sober spaces.)
- 1$\begingroup$ Very interesting answer: +1. $\endgroup$Todd Trimble– Todd Trimble2020-11-10 14:38:56 +00:00Commented Nov 10, 2020 at 14:38