Seminar on Filters and Topology

This seminar was held in #mathematics on EFnet on 13th of November 2004 at 13:00 EDT by Kit.

It covered limits of filters, Tychonoff's theorem, Nets, Topological groups.

Seminar log

[13:02] (Moduli) Okay, it's my great pleasure to introduce our speaker, Kit, who's an analyst from Cambridge. He'll be speaking about Filters and Topology
[13:02] (Moduli) all yours :)
[13:03] (Kit) ok then. So, Filters.
[13:03] (Kit) First warning, as Moduli says (and no doubt you all know), I'm an analyst. :) As such, this talk will have a bit of an analytic flavour to it.
[13:03] (Kit) The first half won't actually require any analysis, but I'll use some ideas for motivation.
[13:04] (Kit) The second you'll need to know a little bit of analysis - some basic notions of analysis will be needed. Things like cauchy sequences, uniform continuity, total boundedness.
[13:04] (Kit) Oh. Completeness of course.
[13:05] (Kit) So. Filters are a notion of limits.
[13:05] (Kit) In a metric space we've got sequences, and can use them to prove a lot of neat results that on the surface of it haven't really got anything to do with sequences.
[13:05] (Kit) You can of course do a lot of these results without ever touching a sequence (indeed, I often do, 'cause I'm weird), but they're often a good way of looking at things.
[13:06] (Kit) The definition of a sequence x_n converging to a point x is a purely topological one.
[13:06] (Kit) You say x_n -> x if for every neighbourhood U of x, x_n is eventually in U.
[13:07] (Kit) Problem: What you can use sequences for screws up horribly in a general topological space.
[13:08] (Kit) e.g. compactness in a metric space is equivalent to saying 'every sequence has a convergent subsequence' - this is sequential compactness. In a general topological space sequential compactness and compactness are completely independent.
[13:08] (Kit) There's a natural generalisation of a sequence to topological spaces. These are called nets. Basically a sequence is a function from N to X. You can replace N with what is known as a directed set, and this works.
[13:09] (Kit) You can get a lot of theory to work using nets.
[13:10] (Kit) Oh. Before I go any further, I should establish some notation.
[13:10] (Kit) A neighbourhood U of a point x is *not* an open set containing x. U is a neighbourhood of x if there is some open set V such that x \in V <= U.
[13:10] (Kit) This has various advantages. Among other things the theory I'm about to develop wouldn't work if we didn't define it this way. :)
[13:11] (Kit) We denote the set of neighbourhoods of a point by N_x
[13:12] (Kit) So. Lets look at our sequence again. x_n -> x means that for every U in N_x, x_n is eventually in N_x. That is to say there is some N such that for all n >= N, x_n \in U.
[13:12] (Kit) We can rephrase this as follows:
[13:12] (Kit) Define F_x_n = { U <= X : x_n is eventually in U }
[13:13] (Kit) Then x_n -> x iff N_x <= F_x_n
[13:13] (Kit) (This should be clear - it's literally a restatement of the definitions).
[13:13] (Kit) So we can restate convergence of a sequence purely in terms of collections of sets.
[13:14] (Kit) Lets investigate some common properties of these two sets:
[13:14] (Kit) Neither contains the empty set.
[13:14] (Kit) If either one contains a set U, and U <= V then it also contains V.
[13:14] (Kit) And they're closed under finite intersections.
[13:15] (Kit) Oh. They're also non-empty (equivalently they contain X).
[13:15] (Kit) We call such a set a filter.
[13:15] (Kit) Filters will prove to be in some sense the 'right' objects we need to talk about limits in a topological space.
[13:15] (Kit) Motivated by what we had from that sequence, we define convergence of a filter as follows:
[13:16] (Kit) A filter F tends to a point x (written, unsurprisingly, F -> x) if N_x <= F
[13:17] (Kit) It should be relatively clear that if F -> x and F <= G then G -> x, just by the definitions.
[13:17] (Kit) Also, if you have a subsequence x_n_k then F_x_n <= F_x_n_k
[13:18] (Kit) Because if x_n is eventually in U then clearly x_n_k must also be.
[13:18] (Kit) So the analogue of taking subsequences is extending to a larger filter.
[13:18] (Kit) Now that we've got voices, is anyone actually listening to me? :)
[13:19] (wli) yes
[13:19] (Kit) Good good. :)
[13:19] (Mgo) yes
[13:19] (lsmsrbls) no
[13:19] (fante7) yes, I am
[13:19] (Kit) Anyway, a lot of natural concepts turn out to have expressions in terms of filters.
[13:20] (Kit) Before I tell you about them, I'll need to sort out some minor technicalities.
[13:20] (Kit) How do we produce filters? We've only got two ways so far - neighbourhoods, and sequences.
[13:20] (Kit) One of the perks of filters over other methods is that they're really quite easy to produce. We'll need the following lemma:
[13:21] (Kit) Let C be a collection of sets such that given U_1, ... , U_n \in C then intersection_{i = 1 to n} U_i != \emptyset. There is a unique smallest filter F such that C <= F
[13:22] (Kit) Proof: Define F = { V : \exists U_1, ... , U_n \in C such that \int_{i = 1}^n U_i <= V }
[13:22] (Kit) F is clearly a filter and any filter containing C must contain F.
[13:23] (Kit) This is trivial, but turns out to be surprisingly useful.
[13:24] (Kit) Oh, by the way, we say such a collection C has the finite intersection property.
[13:24] (Kit) Analagously to the definition of a base for a topological space, we define a filter base: A collection C is a filter base if it doesn't contain the empty set and for any U, V in C we have W <= U n V for some W \in C.
[13:24] #mathematics: mode change '+vvvv KimJ XdeltaX DaMancha mlabrum' by Galois!scythe@DOMINIA.MIT.EDU
[13:25] (Kit) This is precisely the condition that you need for the filter generated by C to be of the form { U : \exists V \in C s.t. V <= U }
[13:25] (monochrom) ! So this essentially generates a filter from generators by throwing in finite intersections, right?
[13:25] (Kit) Right.
[13:25] (Kit) Well, throwing in finite intersections and taking supersets.
[13:25] (Kit) (Technically I think you're meant to just say ! and then wait for me to call on you. I'm not terribly bothered though. :)
[13:26] (monochrom) ! Also, will the "finite intersection property" be analogous to "being algebraic"?
[13:26] (Kit) err. I have no idea. :)
[13:26] (Kit) What's 'being algebraic' meant to mean?
[13:26] (monochrom) Nevermind. I have forgotten the technical details. :)
[13:26] (Kit) ok. No problem. :)
[13:27] (Kit) Right. Now all that's set up, lets actually prove something vaguely interesting. :)
[13:27] (Kit) Limits of filters can be reasonably badly behaved. In particular take the indiscrete topology on a set. Then every filter converges to every point.
[13:27] (Kit) Which is annoying. :)
[13:28] (Kit) When do filters have unique limits?
[13:28] (Kit) The answer turns out to be that filters have unique limits in X exactly when X is hausdorff.
[13:28] (Kit) One way is easy. Suppose X is hausdorff, F -> x and F -> y with x != y.
[13:29] (Kit) Then there are disjoint neighbourhoods U of x and V of y.
[13:29] (Kit) But these must both be in F.
[13:29] (Kit) So U n V is in F.
[13:29] (Kit) But this is the empty set. So F isn't a filter. Contradiction.
[13:30] (Kit) Now. Suppose X isn't hausdorff. Pick x and y distinct such that x and y don't have disjoint neighbourhoods.
[13:30] (Kit) Consider N_x u N_y
[13:30] (Kit) I claim this has the finite intersection property.
[13:31] (Kit) Proof: Any finite intersection of things in this will be of the form U n V with U in N_x and V in N_y. This is because both N_x and N_y are filters.
[13:31] (Kit) But by hypothesis x and y don't have disjoint neighbourhoods, so U n V != \emptyset
[13:31] (Kit) Thus we can generate a filter F from N_x u N_y, and so F -> x and F -> y.
[13:32] (Kit) So 'hausdorff' is exactly the requirement that limits are unique.
[13:33] (Kit) This, incidentally, is where my semi-serious comment that 'all reasonable spaces are hausdorff' comes from. As an analyst I'm generally speaking concerned with limits, so I'd really quite like them to be unique. :)
[13:33] (Kit) Filters do work in non-hausdorff spaces, they're just a little less well behaved.
[13:34] (Kit) ok. Filters and functions.
[13:34] (Kit) Oh. Before I move on to the next part, any questions on what we've had so far?
[13:35] (Kit) ok. Guess not. :)
[13:35] (Kit) Filters and functions.
[13:35] (Kit) Suppose we have a filter F on a space X, and a function f : X -> Y
[13:35] (Kit) We'd like to be able to form a filter f(F)
[13:36] (Kit) After all, most of what concerns us in topology is continuous functions.
[13:36] (Kit) Now, the obvious definition f(F) = { f(U) : U \in F } doesn't work - it's not a filter.
[13:36] (Kit) Specifically the problem is that it's not closed under taking supersets.
[13:36] (Kit) It is however a filter base.
[13:36] (Kit) So we define f(F) to be the filter generated by the filter base { f(U) : U \in F }
[13:37] (Kit) Lets look at the definition of local continuity at x.
[13:37] (monochrom) !
[13:38] (Kit) monochrom: Yes?
[13:38] (monochrom) what is the significance of "filter base"? Does it guarantee the finite-intersection property?
[13:39] (Kit) Yes. Like I said, a set C is a filter base if the following two hold: It doesn't contain the empty set, and given any U, V in C there is some W in C with W <= U n V
[13:40] (Kit) So if U n V = emptyset then neccesarily W = emptyset
[13:40] (Kit) Which we've disallowed.
[13:40] (monochrom) Gotcha.
[13:40] (Kit) Hmm. I guess that's not quite enough.
[13:40] (Kit) But you can generalise to the intersection of n sets by induction
[13:41] (Kit) And then the filter generated takes a nicer form in terms of C (the set { U : \exists V in C with V <= U } )
[13:42] (Kit) Ok. Continuity at x.
[13:42] (Kit) f is continuous at x if for every neighbourhood of f(x), U, there is some neighbourhood V of x such that f(V) <= U
[13:43] (Kit) If you stare at this for a moment you will see that this is exactly the requirement that f(N_x) <= N_f(x)
[13:43] (Kit) Or with a bit of fiddling, whenever F -> x, f(F) -> f(x)
[13:43] (fante7) ! Do you also require that a filter base is nonempty?
[13:43] (Kit) fante7: Err. Yes. Sorry.
[13:43] (Kit) Whenever I generate a filter by a set I'm probably assuming that set is non-empty.
[13:44] (Kit) Although it's still *true* that there's a unique minimal filter containing the empty-set. It's just the set {X}
[13:44] (Kit) So I guess in principle we could have an empty filter-base. But things break down a bit if you do, so lets require that not to be the case.
[13:45] (Kit) ok. So it looks like we're on the right track with generalising sequences. Compare the metric space case: f is continuous at x iff whenever x_n -> x we have f(x_n) -> f(x)
[13:46] (Kit) Lets look at compactness now.
[13:46] (Kit) Metric space case: A space X is compact iff every sequence has a convergent subsequence.
[13:46] (Kit) So what we'd hope was true is the following: A space X is compact iff every filter can be extended to a convergent filter.
[13:47] (Kit) And yay, it is true. :)
[13:47] (Kit) Here's a proof:
[13:47] (Kit) We'll prove this to be equivalent to the following formulation of compactness: A space X is compact iff every collection of closed sets with the finite intersection property has non-empty intersection.
[13:48] (Kit) Suppose C were such a collection.
[13:48] (lsmsrbls) !
[13:48] (Kit) lsmsrbls: Yes?
[13:48] (lsmsrbls) What do you mean by "extended"?
[13:49] (Kit) Oh. Sorry. If F <= G then G extends F.
[13:50] (Kit) So 'every filter can be extended to a convergent filter' means 'every filter is contained in some convergent filter'
[13:50] (Kit) So. We suppose every filter can be extended to a convergent filter, and want to show compactness.
[13:50] (Kit) Let C be a collection of closed sets with the finite intersection property.
[13:50] (Kit) These generate a filter F.
[13:51] (Kit) And we know we can extend F to a convergent filter. Say F <= G and G -> x
[13:51] (Kit) I claim that x must lie in the intersection of C.
[13:51] (Kit) Pick some K \in C. Then both N_x and K are in G, and G is a filter. Thus for any U \in N_x, U n K != \emptyset
[13:52] (Kit) So every neighbourhood of x intersects K. But K is closed. Hence we must in fact have x \in K.
[13:52] (Kit) Thus x is in every element of C, so x is in the intersection of C. Hence the intersection of C is non-empty.
[13:52] (Kit) So X is compact.
[13:53] (Kit) Conversely, suppose X is compact.
[13:53] (Kit) Let F be some filter.
[13:53] (Kit) We form the set { cl(U) : U \in F }
[13:53] (Kit) (cl(U) is the closure of U)
[13:54] (Kit) This is a set of closed sets with the finite intersection property. So it has non-empty intersection. Say x is in the intersection.
[13:54] (Kit) F u N_x has the finite intersection property.
[13:55] (Kit) This is because for any U \in F, we know x \in cl(U)
[13:55] (Kit) So every neighbourhood V of x has non-empty intersection with U.
[13:55] (Kit) So F u N_x generates some filter G with F <= G and G -> x.
[13:55] (Kit) Yay.
[13:56] (Kit) Now. There's a cool thing which you can do with filters that you can't do with sequences.
[13:57] (Kit) You can take a subsequence of a subsequence, and kepe on going. However you can only do this countably many times, because sequences are countable objects.
[13:57] (Kit) keep
[13:57] (Kit) With filters you can extend them a lot more than countably many times, but eventually you get to a point where you can't extend them any further.
[13:58] (Kit) We say a filter F is an ultrafilter if whenever G is a filter with F <= G then F = G.
[13:58] (Kit) Theorem: Ultrafilters exist, and given any filter F there is some ultrafilter G with F <= G
[13:58] (Kit) Proof: Standard zorn's lemma argument.
[13:59] (Kit) Sorry to those who haven't seen zorn's lemma. I forgot to mention that would be needed. If you haven't seen it, just take it on faith. :)
[13:59] (Kit) It won't be mentioned again except in the proof of this theorem.
[13:59] (Galois) !
[13:59] (Kit) Galois: Yes?
[13:59] (Galois) I thought there was some intuitionistic sheaf theoretic construction of ultrafilters that didn't use axiom of choice
[14:00] (Kit) I doubt it. The existence of free ultrafilters is independent of ZFC.
[14:00] (Kit) err
[14:00] (Kit) Sorry
[14:00] (Kit) Independent of ZF
[14:00] (Galois) okay thanks for the clarification
[14:00] (Kit) (A free ultrafilter is an ultrafilter that isn't of the form { U : a \in U } for some a )
[14:00] (Kit) Warning attached to that: Don't try and construct a free ultrafilter explicitly. You'll get nowhere and you'll hurt your brain, 'cause it's not possible. :)
[14:01] (Kit) No problem.
[14:01] (Kit) There's another characterisation of ultrafilters: F is an ultrafilter iff for every U <= X, either U \in F or U^c \in X.
[14:01] (Kit) I'll leave this as an exercise for the interested reader. :)
[14:02] (lsmsrbls) !
[14:02] (Kit) (By the way, this part is taking longer than I thought it would. Sorry about that)
[14:02] (Kit) lsmsrbls: Yes?
[14:02] (lsmsrbls) What is c?
[14:02] (Kit) complement
[14:02] (lsmsrbls) thank you.
[14:02] (Kit) ok. In light of this, we can reformulate compactness as follows: X is compact iff every ultrafilter converges.
[14:03] (Kit) Proof: If X is compact then every ultrafilter can be extended to a convergent filter, but ultrafilters don't have proper extensions so it must have converged.
[14:03] (Kit) If every ultrafilter converges, you can extend any F to an ultrafilter, so every filter has a convergent extension. Thus X is compact.
[14:04] (Kit) We're now going to use this concept to actually prove something non-trivial: Tychonoff's theorem.
[14:04] (Kit) First, let me introduce the notion of a product filter.
[14:04] (Kit) Let X_a be an indexed collection of sets with F_a a filter on X_a.
[14:05] (Kit) We define a filter \prod F_a to be the filter on \prod X_a generated by sets of the following form:
[14:05] (Kit) \prod U_a with U_a \in F_a and all but finitely many of the U_a being equal to X_a
[14:05] (Kit) Assuming our X_a are non-empty, this clearly has the finite intersection property because all the F_a are filters.
[14:06] (Galois) ! this looks like the restricted direct product on adeles
[14:06] (Kit) Galois: It might be. I'm not terribly familiar with adeles.
[14:07] (Kit) It's analagous to the definition of the product topology.
[14:08] (Kit) Sorry, I'm feeling a bit useless with all the "Err, I don't know" answers I'm having to give. :) My algebra isn't the best, so some of the connections are a tad opaque to me.
[14:08] (Galois) sorry, go on
[14:08] (Kit) Another way of looking at the product filter is as follows:
[14:09] (Kit) If pi_i : \prod X_a -> X_i is the i'th projection map, the product filter is the smallest filter F such that for all i, pi_i(F) = F_i
[14:09] (Kit) Proof: Exercise.
[14:09] (Kit) So now we have a nice result.
[14:10] (Kit) Let F be a filter on \prod X_a
[14:10] (Kit) F -> x iff for every i, pi_i (F) -> pi_i(x)
[14:10] (Kit) i.e. the product topology is the topology of pointwise convergence.
[14:11] (Kit) Which reminds me. I'm not going to prove this (although it's relatively straightforward), but the filters that converge on a topology uniquely characterise it.
[14:11] (Kit) Basically because a topology is uniquely characterised by the collection of neighbourhood filters.
[14:12] (Kit) So. Proof. One way is easy, because the projection maps are continuous and I've already proved that if F -> x and f is continuous then f(F) -> f(x)
[14:12] (Kit) So lets prove the other way.
[14:12] (Kit) Suppose pi_i (F) -> pi_i (x) for each i.
[14:13] (Kit) It's enough to show that for a collection of filters F_i -> x_i, \prod F_i -> x
[14:13] (Kit) This is because we know that \prod pi_i (F) <= F
[14:13] (Kit) By the alternate characterisation of the product I gave.
[14:13] (Kit) But looking at the definition of the product topology, it's clear that N_x = \prod N_x_i
[14:14] (Kit) So if N_x_i <= F_i then N_x = \prod N_x_i <= \prod F_i, because N_x_i <= F_i
[14:14] (Kit) Which proves the desired result.
[14:14] (Kit) Phew. :)
[14:15] (Kit) Corollary: Tychonoff's theorem.
[14:15] (Kit) Proof:
[14:15] (Kit) Let F be an ultrafilter on \prod X_a
[14:15] (Kit) Then it's clear from the second characterisation that pi_i (F) is an ultrafilter.
[14:16] (Kit) So because the X_a were compact, pi_i (F) converges.
[14:16] (Kit) Say to x_a
[14:16] (Kit) err. X_i
[14:16] (Kit) Argh. x_i
[14:16] (Kit) Note: If the X were hausdorff, we've got a unique choice of x_i because limits are unique.
[14:17] (Kit) But if they're not, then we need to make a choice.
[14:17] (Kit) This isn't interesting unless you care about set theory, like me. :)
[14:17] (Kit) So we have F -> x
[14:17] (Kit) So every ultrafilter on \prod X_a converges. Thus \prod X_a is compact.
[14:17] (Kit) Yay. :)
[14:18] (Kit) So after all this work we've got a nice short proof of tychonoff's theorem. So at least *something* useful comes out of it.
[14:18] (Kit) Phew.
[14:18] (Kit) ok. Now I'm going to give a little bit of an overview.
[14:18] (Kit) Like I said, there are a number of ways you can do generalised limits.
[14:19] (Kit) In some sense filters are the 'minimal' such way.
[14:19] (Kit) i.e. a filter contains exactly the neccesary amount of information you need to talk about convergence - no more, no less.
[14:19] (Kit) This makes them kindof useful, even if you're never actually interested in filters themself. Given most sensible ways of talking about convergence, it's clear how we'd attach a filter to them.
[14:20] (Kit) So in the same way that even if you never care about anything that's not a metric space you should probably still learn topology, even if you're only interested in nets filters are a useful thing to know about.
[14:21] (Kit) Anyway, thus ends the pure topology part of this seminar. About half an hour later than I'd intended. :)
[14:21] (Kit) I'm going to carry on to talk about some other useful applications, leading up to some stuff about topological groups. The next part will need some more analysis than what we've got so far.
[14:21] (Kit) Any questions?
[14:22] (monochrom) ! Just a remark. I think using filters to characterise topologies is beautiful.
[14:22] (Kit) Yeah, I like it too. I'd intended to mention more on it, but I'm already running over on this section and it's non-crucial. :)
[14:23] (Kit) If there are no more questions I'm going to take a 5 minute break before I move on. My hands are tired. :)
[14:26] Nick change: metateck -> meta_10s
[14:27] (Kit) Alright. Part 2. :)
[14:28] (Kit) What follows *could* be done in the general setting of uniformities. In the same way that a topological space is the most general thing you can talk about something being continuous, a uniformity is the most general thing in which you can talk about something being uniformly continuous.
[14:28] (Kit) I'm not going to do this.
[14:28] (Kit) It would take too much time to introduce the general theory of uniformities, and it's kindof not pretty.
[14:29] (Kit) What I'm going to do is I'm going to talk about metric spaces, and then show how we can extend what I've been saying to work on topological groups.
[14:29] (Kit) So I'll be able to introduce notions of a 'complete topological group' and prove an analogue of the uniform extension theorem for topological groups.
[14:30] (Kit) But first I'm going to answer the phone. :)
[14:32] (Kit) Sorry. Back now.
[14:33] (Kit) Ok. We've got a notion of a 'cauchy sequence' in a metric space. It would be nice to have a similar definition of a cauchy filter.
[14:33] (Kit) F_x_n essentially encapsulates the convergence properties of x_n, so in order for this to make sense we want x_n cauchy iff F_x_n is cauchy
[14:33] #mathematics: mode change '+v mattjf' by Kit!
[14:34] (Kit) Exercise: x_n is cauchy iff F_x_n contains sets of arbitrarily small diameter.
[14:34] (Kit) In proving this it may help you to note that sets of the form { x_n : n >= m } form a filter base for F_x_n
[14:35] (Kit) So, we define a cauchy filter to be one which contains sets of arbitrarily small diameter.
[14:36] (Kit) It would be nice if completeness was equivalent to every cauchy filter converging. And it is, which is good.
[14:36] (Kit) First we need to show the following:
[14:36] (Kit) Let F be cauchy, F <= G and G -> x. Then F -> x
[14:36] (Kit) Proof:
[14:37] (Kit) We know that x \in cl(U) for any U \in F
[14:37] (Kit) By roughly the same arguments we made back in the discussion of compactness
[14:37] (Kit) So pick some neighbourhood V of x.
[14:38] (Kit) Say this contains B(x, e).
[14:39] (Kit) Now. F contains sets of arbitrarily small diameter. In particular it contains a set U of diameter <= e/2
[14:40] (Kit) It also contains cl(U), which has diameter <= e/2, so we may assume U is closed and thus x \in U.
[14:41] (Kit) So in particular U <= cl(B(x, e/2)) <= B(x, e)
[14:41] (Kit) So because F is a filter, B(x, e) \in G and thus V \in F
[14:41] (Kit) So F -> x
[14:41] (Kit) That proof always takes longer than I expect.
[14:41] (Kit) For use later, lets analyse exactly what I've used about the metric.
[14:42] (Kit) I needed the following two things:
[14:42] (Kit) Firstly, that there was a neighbourhood V' of x with cl(V') <= V
[14:43] (Kit) Secondly, that a neighbourhood of x contains all sets of some sufficiently small diameter containing x.
[14:43] (Kit) And that's all I've used the metric for.
[14:43] (Kit) (Except of course for defining what 'sufficiently small diameter' actually means)
[14:43] (Kit) We'll need this later.
[14:44] (Kit) Now to prove that completeness = every cauchy filter converges.
[14:44] (Kit) Suppose every cauchy filter converges. Let x_n be a sequence. Then F_x_n is cauchy, so it converges, so x_n converges.
[14:44] (Kit) The converse is slightly harder.
[14:44] (Kit) (But only slightly)
[14:44] (Kit) Suppose every cauchy sequence converges.
[14:45] (Kit) Let F be a cauchy filter.
[14:45] (Kit) We can find a sequence of elements V_i of F with the following properties:
[14:45] (Kit) If j > i then V_j <= V_i
[14:45] (Kit) diam(V_n) <= 1/2^n
[14:45] (Kit) Pick x_n \in V_n
[14:46] (Kit) Then by construction the tail of x_n after m has diameter <= 1/2^m, so x_n is cauch, so it converges.
[14:46] (Kit) cauchy
[14:46] (Kit) Oh. Lets make the V_n closed.
[14:47] (Kit) Say x_n -> x
[14:47] (Kit) Then we have x \in V_n for all n, because the V_n are closed.
[14:48] (Kit) So given e > 0, pick n such that 1/2^n < e. Then V_n <= B(x, e), so B(x, e) \in F.
[14:48] (Kit) Thus F -> x
[14:48] (Kit) (Sorry, I changed directions halfway through that proof, as I realised a better way of proving it than I'd originally intended. :)
[14:48] (Kit) e.g. it wasn't wrong. :)
[14:49] (Kit) ok. So we've got a characterisation of completeness in terms of filters.
[14:50] (Kit) In particular note that it's obvious from this that every compact space is complete, because you can extend every cauchy filter to a convergent filter, so it must have been convergent in the first place.
[14:50] (Kit) (Although this is fairly obvious anyway)
[14:50] (Kit) Another nice characterisation:
[14:50] (Kit) Let X be totally bounded. Then every ultrafilter is cauchy.
[14:50] (Moduli) !
[14:52] (Kit) Moduli: Yes?
[14:52] (Moduli) nevermind, planetmath helped.
[14:52] (Kit) ok
[14:52] (Kit) I do apologise for the analysis by the way. I promise I am leading on to something topological with this. :)
[14:54] (Kit) Proof: Suppose not. Let F be an ultrafilter which isn't cauchy. Then for some e > 0 F contains no set of diameter <= 2e
[14:54] (Kit) Let x_1, ... , x_n be an e-net.
[14:54] (Kit) i.e. X = Union_{i=1}^n B(x, e)
[14:55] (Kit) B(x_i, e)
[14:55] (Kit) Then these sets all have diameter <= 2e, so none of them are in F.
[14:55] (Kit) So their complements B(x_i, e)^c are in F, as F is an ultrafilter.
[14:55] (Kit) But then the intersection of these is in F, as F is closed under finite intersections.
[14:55] (Kit) But the intersection is empty. Contradiction.
[14:56] (Kit) So let X be a complete and totally bounded metric space. Then every ultrafilter is cauchy, and so convergent. Thus every ultrafilter converges, so X is compact.
[14:56] (Kit) (And it's an easy consequence of the open cover formulation of compactness that every compact metric space is totally bounded)
[14:57] (Kit) So we've got a nice proof of the usual equivalence compactnes iff complete and totally bounded.
[14:57] (Kit) One final thing on metric spaces before we move on to topological groups:
[14:58] (Kit) Theorem: Let X, Y be metric spaces with Y complete. Let A <= X be dense in X, and f : A -> Y. There is a unique extension of f to X.
[14:58] (Kit) This is normally proved by sequence picking.
[14:58] (Kit) Oh.
[14:58] (Kit) Sorry. f is uniformly continuous.
[14:59] (Kit) Proof:
[14:59] (Kit) First an easy exercise: If F is cauchy and f is uniformly continuous, then f(F) is cauchy.
[15:00] (Kit) Let x \not\in A. We want to define a value of f(x).
[15:00] (Kit) We take the only possible definition:
[15:00] (Kit) We've got a filter N_{A,x} on A given by { U n A : U \in N_x }
[15:00] (Kit) This is a filter because x is a limit point of A, so it doesn't contain the empty set.
[15:00] (Kit) This filter is cauchy.
[15:00] (Kit) Thus f(F) is cauchy.
[15:01] (Kit) But Y is complete, so f(F) converges to a unique point y.
[15:01] (Kit) Define f(x) = y
[15:01] (Kit) Actually, that works for all x. Not just x \not\in A.
[15:01] (Kit) Hold on. Let me make that more coherent. :)
[15:02] (Kit) Let x \in X. N_{x, A} is a filter because x \in cl(A). Define g(x) = lim f(N_{x, A})
[15:02] (Kit) I claim g is a continuous extension of f.
[15:02] (Kit) It's clear from the limit definition of continuity that this agrees with f on A.
[15:03] (Kit) As for x \in A N_{x, A} = N_x regarding A as a metric space in its own right.
[15:04] (Kit) Further, I claim g is continuous.
[15:04] (Kit) And I've just realised I've fudged the details in my proof of that. So I'm going to have to stop and think for a minute. :)
[15:05] (Kit) Any questions while I do?
[15:06] (Kit) No. I'm really sorry and I'm sure I'm being a moron but I don't see how to fix it offhand. :)
[15:07] (Kit) So I'm going to carry on for now. Lets just pretend I've proved it. I'll write up a proof later and ask for it to be posted as an addendum to the log when it goes onto the website.
[15:07] (Kit) (Which is a pity, as this result was part of one of the reasons that what I'm about to do for topological groups is interesting).
[15:08] (Kit) Ok. That's the end of the metric space stuff.
[15:08] (Kit) Now I'm going to show you how to do the same with a topological group.
[15:08] (Kit) First I'm going to remind you what a topological group is: It's a group which is also a topological space, such that all the operations are continuous and points are closed.
[15:09] (Kit) I'm going to quote the following result from the theory of topological groups (it's not hard to prove, but it would take up too much time and not really be relevant):
[15:09] (Kit) Every topological group is regular as a topological space. That is to say for a point x and a closed set F, there are disjoint open sets U, V such that x \in U, F <= V.
[15:10] (Kit) Or, equivalently, for a point x and a neighbourhood V of x, there is a neighbourhood U of x such that cl(U) <= V
[15:10] (Kit) (Those of you paying attention may be thinking 'ah ha' and remembering what I did when I showed that a cauchy filter with a convergent extension was in fact convergent)
[15:11] (Kit) So. What we need is a notion of 'arbitrarily small' for a topological group.
[15:11] (Kit) The following proves to be right one:
[15:12] (Kit) Let G be a topological group, and U be a neighbourhood of the identity. A subset of G of size <= U is one contained in the translate gU for some g \in G
[15:12] (Kit) We may think of this as saying that left translation acts as an isometry, but that's not really what we're doing because there are non-metrizable topological groups.
[15:13] (Kit) So we now have a notion of what it means for F to contain arbitrarily small sets. For every U a nbhd of the identity, gU \in F for some g.
[15:13] (Kit) So we've got a notion of cauchy filters for topological groups!
[15:14] (Kit) The analogue of total boundedness is clear: X is totally bounded iff for every neighbourhood U of e, we have elements g_1, ... , g_n \in G such that U g_i U = G
[15:15] (Kit) Oh. Sorry, I forgot to mention. If you use what I said about the regularity of G, and go back and check our proof for metric spaces, you'll see that exactly the same proof now shows that if F is cauchy and F <= H with H with H -> x then F -> x in a topological group.
[15:15] (Kit) So. Total boundedness.
[15:15] (Kit) The same proof as before shows that Total boundedness => every ultrafilter is cauchy.
[15:16] (Kit) So once again we have Compactness = Completeness + Total Boundedness
[15:16] (Kit) (Where once again completeness = every cauchy filter converges)
[15:17] (Kit) For a function f : G -> H between two topological groups, it's reasonably clear what we mean by 'f is uniformly continuous'
[15:17] (Kit) For every neighbourhood V of e_h we have some neighbourhood U of e_G such that if X has size <= U then f(X) has size <= V
[15:18] Nick change: mickske -> mkiskce
[15:18] (Kit) (Stare at this for a moment or two and you'll find yourself convinced that it's really the same thing as the metric space version)
[15:19] (Kit) Given this definition, if I'd actually shown you a proof you'd be able to see that the extension theorem also works for topological groups. i.e. if you've got a function defined on a dense subset into a complete group, you can extend it to a unique function on the entire group.
[15:19] (Kit) (Oh, I should note: Regular clearly implies hausdorff, so we do have unique limits in topological groups)
[15:20] (Kit) This is nice, because for topological groups we have a rather nice selection of uniformly continuous functions:
[15:20] (Kit) Lemma: Let G, H be topological groups and f : G -> H be a continuous group homomorphism. Then f is uniformly continuous.
[15:21] (Kit) Proof: Let V be a neighbourhood of e_H. Then there is a neighbourhood U of e_G by continuity.
[15:21] (Kit) But then f(gU) = f(g)f(U) <= f(g) V
[15:21] (Kit) So a set of size <= U is mapped into a set of size <= V
[15:23] (Kit) Corollary: Let G, H be topological groups with H complete and let M <= G be a dense subgroup. Let f : M -> H be a continuous homomorphism. We can extend f to a continuous homomorphism f : G -> H
[15:23] (Kit) Which is nice, and not something it would be entirely obvious how to prove without the mechanisms of filters that we've set up.
[15:23] (Kit) What I've said for topological groups also works for topological rings, vector spaces, etc.
[15:24] (Kit) Complete topological vector spaces are particularly of interest to us analysts, because they're the natural generalisation of banach spaces: A lot of cool results that we use in functional analysis work on complete TVS, even when the TVS isn't metrizable.
[15:25] (Kit) I should mention, we do tend to use nets for such things. However you can use the theory developed for filters to develop the theory for nets as an almost trivial consequence, so this isn't wasted effort. :)
[15:25] (Kit) And the theory for filters is prettier.
[15:25] (Kit) So. That's done then. :)
[15:25] (Kit) Sorry it took so long. It ran on longer than I expected.
[15:25] (Kit) Any questions?
[15:26] (Kit) Or have I put you all to sleep? :)
[15:26] (lsmsrbls) I'm sure I'll have questions for you in the future, as I do the exercises.
[15:27] Action: Moduli applauds
[15:27] (Moduli) do I have your permission to upload the log?
[15:27] (Kit) Of course.