# Two topologies over the same set which are carried to each other by a lattice isomorphism induce homeomorphic spaces?

This is a question that I've been turning in my head and haven't been able to come up with a way to proceed to either prove it or construct a counter-example:

Let $X$ be a set and $\mathfrak{T}_X$ the set of all topologies over $X$. $\mathfrak{T}_X$ forms a complete lattice when ordered by set-inclusion $\subseteq$. Let $f$ be an automorphism of $\mathfrak{T}_X$ (as in a complete lattice isomorphism). Does it follow that for any topology $\mathcal{T}$ of $X$ that $(X,\mathcal{T})$ and $(X,f(\mathcal{T}))$ are homeomorphic topological spaces?

EDIT

I've added the Category Theory tag, as I feel like someone with more precision might be able to turn this into a question about a directed system in a small category and an endo-functor which preserves all limits.

• To save someone else the search for a minimal counter-example, I tested the topologies over a three element set, and the statement is true for it. – Robert Wolfe Mar 20 '14 at 7:53
• Maybe ams.org/journals/tran/1966-122-02/S0002-9947-1966-0190893-2/… helps? It seems like infraspaces and ultraspaces are preserved under $f$, and all topologies are infima of ultraspaces or suprema of infraspaces. – Henno Brandsma Mar 20 '14 at 18:53
• @Bryan: I'm pretty sure no finite $X$ will be a counterexample. – tomasz Mar 20 '14 at 19:49
• Your definition of an automorphism is a little bit ambiguous. Do you mean lattice automorphisms (order reflecting mappings) or just order endomorphisms? It's different whether you consider the bijectivity of the order relation (considered as set) or just the bijectivity of the base set. The latter allows strict set inclusions of the order relations. I'd guess you are considering lattice-automorphisms (these are the stronger ones). – Keinstein Mar 20 '14 at 21:12
• I've edited the question to be more clear. – Robert Wolfe Apr 7 '14 at 3:41

Automorphisms of the lattice $L={\mathfrak T}(X)$ of topologies on arbitrary set $X$ are classified as follows (see here):
1. If $X$ is infinite or has cardinality $\le 2$, then every automorphism $\phi$ of $L$ is induced by a bijection $f_\phi: X\to X$. In particular, we obtain a natural homeomorphism of the topological spaces $$f: (X,\tau)\to (X, \phi(\tau)), \tau\in {\mathfrak T}(X).$$
2. If $X$ has finite cardinality $\ge 3$, then $Aut(L)$ is the direct product of its subgroup induced by bijections $X\to X$ as above and the order 2 group $Z_2$ whose generator is induced by the automorphism $\theta\in Aut(L)$ swapping subsets of $X$ and their complements. The latter automorphism (and its composition with an automorphism induced by a bijection), of course, will not induce a homeomorphism $(X,\tau)\to (X, \theta(\tau))$ for general $\tau$.
Edit. As a specific example, take any finite set $X$ of cardinality $\ge 3$ and its topology $\tau=\{\emptyset, \{x\}, X\}$ where $x$ is a certain element of $X$. Then $\theta(\tau)= \{\emptyset, X\setminus \{x\}, X\}$. It is clear that $(X,\tau)$ and $(X, \theta(\tau))$ are not homeomorphic.
See also a nice (although dated) survey of the properties of the lattice $L$.
• Thank you for these references. I was more worried about $\theta$ being a lattice isomorphism as I thought I had considered all possible ones in my computation for a 3-element set, but I was wrong. It's interesting that $\theta$ isn't an automorphism for infinite base sets. – Robert Wolfe Apr 11 '14 at 20:47