It turns out I made a mistake in my last post on filters.
We had ended up in the situation where I claimed that the topological filter spaces, that is, the filters spaces that actually arise from a topological space, are those that satisfy the extra axiom:
- if (Fi)i in I is any family of filters on X such that Fi → x for every i in I, then F → x where F is the intersection of the Fis.
Certainly, any topological space satisfies this extra axiom. The problem is in the converse implication.
The filter spaces that satisfy the above extra axiom are the pretopological filter spaces. Every topological filter space is pretopological, but the converse is wrong. Here is a counterexample, given to me by Frédéric Mynard yesterday. Let X=R2. Given a point x=(s,t) in X, draw a small cross centered at (s,t), with arms of length 2. Formally, for a>0, let Cx be the set of points of the form (s+δ,t) or (s,t+δ) where -1 < δ < 1. Say that a filter F converges to x=(s,t) if and only if Cx is an element of F. One checks easily that this is a pretopological notion of convergence.
If it were topological, then it would be the notion of convergence for Top(X). Let us elucidate the topology of the latter. The opens of Top(X) are those subsets U such that for every element x of U, every filter that converges to x contains U, i.e., every filter that contains Cx also contains U. Using the filter of all supersets of Cx, one sees that the opens are those subsets that are “neighborhoods” of all of their elements, where U is a “neighborhood” of x if and only if U contains Cx. So, if U is a non-empty open subset, let x be one if its elements. The whole of Cx is included in U. In particular, every point y obtained by translating x horizontally along a distance < 1 is in U. Applying the same argument with y, and moving vertically, now, we can reach any point z in the open square centered at x with side length 2, while staying in U. Repeating the argument, we can actually reach any point in X while staying in U. So U=X, meaning that the topology of Top(X) is indiscrete. In particular, every filter converges to any point in Top(X). This is really far from our original notion of convergence!
And therefore, certainly, X is pretopological, but not topological.
What happens in this counterexample can be generalized to the following. Call a neighborhood system on a set X the data of a family Nx of subsets of X containing x, for each point x in X. (The notion is due to Felix Hausdorff .) Given any neighborhood system, one can define a notion on convergence in the usual way: a filter F converges to x if and only if Nx is included in F. It is an easy exercise to show that this is a notion of convergence, and that it is always pretopological.
Conversely, given a pretopological notion of convergence →, one obtains a neighborhood system in the exact same way we (erroneously) tried to prove that pretopological notions of convergence were topological at the end of part II. Consider the family of all filters that converge to x, and take their intersection. Call this intersection filter Nx. The extra axiom defining pretopologies implies that Nx → x, and then that a filter converges to x iff it contains Nx. So neighborhood systems and pretopologies are essentially the same thing. (To make this precise, you will have to strengthen the definition of a neighborhood system so that Nx is a filter, not just a family of subsets. That is no essential difference, since every family of sets gives rise to a unique smallest filter containing it.)
What went wrong last time? Let me cite: “From Nx, we retrieve a topology by declaring open any set U that is a neighborhood of each of its points, i.e., such that U is in Nx for every x in U: in particular, Nx is really the filter of neighborhoods of x: our filter space is indeed topological.” Of course, this gives you a topology, but Nx will in general be a superset of the set of all neighborhoods of x, and not necessarily equal to it. In Frédéric’s example, Nx would be the filter generated by Cx, that is, the set of all supersets of Cx. In Top(X), there is only one neighborhood of x: the whole space X itself. This is a much smaller set of neighborhoods!
So how can we characterize those filter spaces that are topological?
You need yet one more axiom, which essentially says that limits of limits are limits. This is probably a bit too complex to be put here, and is best defined using convergence of ultrafilters instead of filters. In short, there is a space UX of all ultrafilters on a given filter space X. UX can be given a natural notion of convergence by saying that an ultrafilter of ultrafilters A (in UUX!) converges to an ultrafilter a (in UX) if and only if for every element u of A, for every element u of A, there is an element a of u and an element x of u such that a converges to x. One can also flatten out an ultrafilter of ultrafilters by the so-called Kowalsky sum operation µX: for A in UUX, µX(A) is the collection of subsets u of X such that u# is in A… where u# is the set of all ultrafilters a (in UX ) such that u is in a. (This is the so-called multiplication operation of the U monad.)
Oops… don’t worry if you don’t understand, that is precisely why I don’t want to explain. I’m not sure I understand too much of it either.
Anyway, to finish the story, a filter space is topological if and only if for every ultrafilter of ultrafilters A that converges (in UX) to some ultrafilter a that itself converges (in X) to x, then µX(A) converges to x. This was proved by Dirk Hofmann and Walter Tholen .
As Walter Tholen mentioned last Saturday at the categorical topology session I was at, the novelty here is that this new axiom, together with the axiom that the ultrafilter at x converges to x (the first axiom of filter spaces), are enough to define exactly the topological filter spaces. All the other axioms, including the fact that if a filter contains a filter that converges to x, then it already converges to x (the second axiom of filter spaces), and mainly that intersections of filters that converge to a point again converge to this point (pretopologies) are all redundant.
— Jean Goubault-Larrecq (January 21st, 2014)