Certainly, any topological space satisfies this extra axiom.\u00a0 The problem is in the converse implication.<\/p>\n
Pretopologies.<\/strong><\/p>\n The filter spaces that satisfy the above extra axiom are the pretopological<\/em> filter spaces.\u00a0 Every topological filter space is pretopological, but the converse is wrong.\u00a0 Here is a counterexample, given to me by Fr\u00e9d\u00e9ric Mynard<\/a> yesterday.\u00a0 Let X<\/em>=R2<\/sup>.\u00a0 Given a point x<\/em>=(s<\/em>,t<\/em>) in X<\/em>, draw a small cross centered at (s<\/em>,t<\/em>), with arms of length 2.\u00a0 Formally, for a<\/em>>0, let Cx<\/sub><\/em> be the set of points of the form (s<\/em>+\u03b4,t<\/em>) or (s<\/em>,t+\u03b4<\/em>) where -1 <\u00a0\u03b4<\/em> < 1. Say that a filter F<\/em> converges to x<\/em>=<\/em>(s<\/em>,t<\/em><\/em>) if and only if\u00a0Cx<\/sub><\/em> is an element of F<\/em>.\u00a0 One checks easily that this is a pretopological notion of convergence.<\/p>\n If it were topological, then it would be the notion of convergence for Top<\/em><\/em>(X<\/em><\/em>).\u00a0 Let us elucidate the topology of the latter.\u00a0 The opens of Top<\/em><\/em>(X<\/em><\/em>) are those subsets\u00a0U<\/em> such that for every element x<\/em> of U<\/em>, every filter that converges to x<\/em> contains U<\/em>, i.e., every filter that contains Cx<\/sub><\/em> also contains U<\/em>.\u00a0 Using the filter of all supersets of Cx<\/sub><\/em>, one sees that the opens are those subsets that are “neighborhoods” of all of their elements, where U is a “neighborhood” of x<\/em> if and only if U<\/em> contains Cx<\/sub><\/em>.\u00a0 So, if U<\/em> is a non-empty open subset, let x<\/em> be one if its elements.\u00a0 The whole of\u00a0Cx<\/sub><\/em> is included in U<\/em>.\u00a0 In particular, every point y <\/em>obtained by translating x<\/em> horizontally along a distance < 1 is in U<\/em>.\u00a0 Applying the same argument with y<\/em>, and moving vertically, now, we can reach any point z<\/em> in the open square centered at x<\/em> with side length 2, while staying in U.\u00a0 Repeating the argument, we can actually reach any point in X<\/em> while staying in U<\/em>.\u00a0 So U=<\/em>X, meaning that the topology of Top<\/em><\/em>(X<\/em><\/em>) is indiscrete<\/em>.\u00a0 In particular, every filter converges to any point in Top<\/em><\/em>(X<\/em><\/em>).\u00a0 This is really far from our original notion of convergence!<\/p>\n And therefore, certainly, X<\/em> is pretopological, but not topological.<\/p>\n Neighborhood systems.<\/strong><\/p>\n What happens in this counterexample can be generalized to the following.\u00a0 Call a neighborhood system<\/em> on a set X<\/em> the data of a family\u00a0Nx<\/sub><\/em> of subsets of X<\/em> containing x<\/em>, for each point x<\/em> in X<\/em>.\u00a0 (The notion is due to Felix Hausdorff [1].)\u00a0 Given any neighborhood system, one can define a notion on convergence in the usual way: a filter F<\/em> converges to x<\/em> if and only if\u00a0Nx<\/sub><\/em> is included in F<\/em>.\u00a0 It is an easy exercise to show that this is a notion of convergence, and that it is always pretopological.<\/p>\n Conversely, given a pretopological notion of convergence \u2192<\/em>, 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<\/a>.\u00a0 Consider the family of all filters that converge to x<\/em>, and take their intersection.\u00a0 Call this intersection filter Nx<\/sub><\/em>.\u00a0 The extra axiom defining pretopologies implies that Nx<\/sub><\/em> \u2192<\/em> x<\/em>, and then that a filter converges to x<\/em> iff it contains Nx<\/sub><\/em>.\u00a0 So neighborhood systems and pretopologies are essentially the same thing.\u00a0 (To make this precise, you will have to strengthen the definition of a neighborhood system so that Nx<\/sub><\/em> is a filter, not just a family of subsets.\u00a0 That is no essential difference, since every family of sets gives rise to a unique smallest filter containing it.)<\/p>\n What went wrong last time?\u00a0 Let me cite: “From Nx<\/sub><\/em>, we retrieve a topology by declaring open any set U<\/em> that is a neighborhood of each of its points, i.e., such that U<\/em> is in Nx<\/sub><\/em> for every x<\/em> in U<\/em>: in particular, Nx<\/sub><\/em> is really the filter of neighborhoods of x<\/em>: our filter space is indeed topological.”\u00a0 Of course, this gives you a topology, but Nx<\/sub><\/em> will in general be a superset<\/em> of the set of all neighborhoods of x<\/em>, and not necessarily equal to it.\u00a0 In Fr\u00e9d\u00e9ric’s example, Nx<\/sub><\/em> would be the filter generated by Cx<\/sub><\/em>, that is, the set of all supersets of Cx<\/sub><\/em>.\u00a0 In Top<\/em><\/em>(X<\/em><\/em>), there is only one neighborhood of x<\/em>: the whole space X<\/em> itself.\u00a0 This is a much smaller set of neighborhoods!<\/p>\n And topologies?<\/strong><\/p>\n So how can we characterize those filter spaces that are topological?<\/p>\n You need yet one more axiom, which essentially says that limits of limits are limits.\u00a0 This is probably a bit too complex to be put here, and is best defined using convergence of ultrafilters instead of filters.\u00a0 In short, there is a space U<\/strong>X <\/em>of all ultrafilters on a given filter space X<\/em>.\u00a0 U<\/strong>X <\/em>can be given a natural notion of convergence by saying that an ultrafilter of ultrafilters A<\/span><\/em> (in U<\/strong>U<\/strong>X<\/em>!) converges to an ultrafilter a (in U<\/strong>X<\/em>) if and only if for every element u<\/span><\/em> of A<\/span>, <\/em>for every element u<\/em> of A<\/em>, there is an element a<\/em><\/span> of u<\/em><\/span> and an element\u00a0x<\/em> of u<\/em> such that a<\/em><\/span> converges to x<\/em>.\u00a0 One can also flatten out an ultrafilter of ultrafilters by the so-called Kowalsky sum<\/em> operation \u00b5X<\/sub><\/em>: for A<\/em><\/span> in U<\/strong>U<\/strong>X<\/em>, \u00b5X<\/sub><\/em>(A<\/em><\/span>) is the collection of subsets u<\/em> of X<\/em> such that u<\/em>#<\/sup> is in A<\/span><\/em>… where\u00a0u<\/em>#<\/sup> is the set of all ultrafilters a<\/em><\/span> (in U<\/strong>X <\/em>) such that u<\/em> is in a<\/em><\/span>.\u00a0 (This is the so-called multiplication operation of the U<\/strong> monad.)<\/p>\n Oops… don’t worry if you don’t understand, that is precisely why I don’t want to explain.\u00a0 I’m not sure I understand too much of it either.<\/p>\n Anyway, to finish the story, a filter space is topological if and only if for every ultrafilter of ultrafilters A<\/em><\/span> that converges (in U<\/strong>X<\/em>) to some ultrafilter a<\/em><\/span> that itself converges (in X<\/em>) to x<\/em>, then \u00b5X<\/sub><\/em>(A<\/em><\/span>) converges to x<\/em>.\u00a0 This was proved by Dirk Hofmann and Walter Tholen [2].<\/p>\n 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<\/em> converges to x<\/em> (the first axiom of filter spaces), are enough to define exactly the topological filter spaces.\u00a0 All the other axioms, including the fact that if a filter contains a filter that converges to x<\/em>, then it already converges to x<\/em> (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.<\/p>\n \u2014 Jean Goubault-Larrecq<\/a>\u00a0(January 21st, 2014) 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, … Continue reading ![\"jgl-2011\"](\"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/wp-content\/uploads\/2016\/08\/jgl-2011.png\")
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