# Non-homeomorphic connected $T_2$-spaces with isomorphic topology poset

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?

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 $$\phi(V) = f^{-1}(V)$$.)

• I'm very curious as to why someone downvoted this. Commented Mar 11 at 13:29

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.)

• Very interesting answer: +1. Commented Nov 10, 2020 at 14:38