In a Hausdorff space $X$, it's a known theorem that the lattice structure on the open sets $\mathcal{O}(X)$ determine the topology of $X$ up to homeomorphism (actually, the Hausdorff condition can even be related a little bit to sober spaces, but I don't want too many unfamiliar concepts). Two questions:
- Is there an elementary proof that doesn't get into 'lattice theory' proper, filters, sober spaces etc.? The result is very intuitive but I'm not sure what the cleanest way to define the homeomorphism is. Obviously we need to define when two lattices are isomorphic, but can we avoid other theory?
- When is the same true of the lattice of closed sets? Of the poset on the closed + connected subsets?
Thanks!
Re: $1$, I don't see how you'd hope for a proof avoiding some lattice theory. After all, a homeomorphism crucially involves points, and the only way points are coded into the lattice of opens of a sober space is via filters. To do anything "point-flavored" with the lattice of opens, you need in some sense to talk about those filters.
Re: $2$, as noted in the comments the lattice of closed sets carries exactly the same information as the lattice of open sets (they're duals). Restricting attention to connected sets is a bit more interesting. Of course for most spaces we lose too much information, but among "nice" spaces there are some distinctions we can draw. For example, if $m\not=n$ then the lattice of connected open subsets of $\mathbb{R}^m$ is not homeomorphic to the lattice of connected open subsets of $\mathbb{R}^n$ (think about covering dimension).