 That's nice, I think that's something new. Okay, everyone. Welcome to the first talk of this semester of GSF. Like you know, I just be thrown here. And Andy would tell us about an extremely disconnected space. Yes, something about the power chain. I need to tell you that Andy just received the NSF post up award. So I'm going to tell you about extremely, not extremely, but extremal. So it has some extremal property. Extremely distanced spaces. It's actually kind of a lie because I'm going to talk way more about complete Boolean algebras. But we're going to see why those are not totally... So some preliminaries. I'm going to assume people know what a compact house door space is. We're going to assume, maybe a little bit less assume, but I'm not going to talk too much about what a Boolean algebra is. If you don't know what a Boolean algebra is, think about the collection of subsets of a given ambient set. That's a good example. And then take the sort of the finiteistic properties that that has. So you can union two things together. You can intersect two things. Union distributes over intersection. You can complement things. So that's roughly what a Boolean algebra is. And what are we going to do? I should tell you what an extremely disconnected space is. Define, uneds. Most of my extremely disconnected spaces are going to in fact be compact, extremely disconnected spaces. Everything is house door. We'll probably stop saying house door. Look at applications in topology. I'm running the talk. I may do some subset of this. I might get tired at 6.30 and say, let's go home. We'll see what happens. All right. You can do a second talk. A series. I don't think, I feel like I might get voted out after this one. So as a warm-up idea to extremely disconnected, there's a less extreme old notion called totally disconnected, or sometimes called zero-dimensional. Those are synonyms. So a compact house door space per-dimensional, a synonym per-dimensional is totally disconnected, means you have a non-trivial cloak-in-set. So totally disconnected. We have lots and lots and lots of cloak-in-sets, enough that they actually describe the topology. Some really simple examples. Cancer space is a great example of take. The compact space with two points, just the discrete topology and take a big product. That's probably a pretty familiar example. But compactness, I mean, you can take arbitrary products with a compact space. So I could put some bigger cardinal kappa up here. And that would also be totally disconnected. So just, these exist. Let me actually put this down here. So if zero-dimensional, let's say, I'm often going to even open it compact to, so like most of the spaces I talk about are going to be compact and house door. So if x is a compact house door zero-dimensional space, and it has lots of cloak-ins, let's give that a name. Cloth-x equals, I told you we'd be working with Julian. Well, here's our first one. Cloth-x is a Boolean algebra. Right, chocolate is a Boolean algebra. Even simpler, so to define a Boolean algebra, you need to define the meat operator. Meat is just going to be intersection. Union is just going for wedges with union. Compliment is just compliment in the space because things are clothing. That's also clothing. The zero is the empty set. And then since the operations are just the normal set theory, operation, there's nothing to verify. This is a Boolean algebra for free. All right, so we started with a compact zero-dimensional space and we get a Boolean algebra for free. Now, someone hands us a Boolean algebra. What's going to happen? Well, firstly, suppose it is a Boolean algebra. I might just write B-A for Boolean algebra. It would be nice if all Boolean algebras were of the form Cloth-x for some zero-dimensional compact space. That would be a really cool theorem. Maybe the Stone's theorem. We're going to prove Stone's representation theorem. We need to construct a compact zero-dimensional space that gives us, if you've seemingly given a GSS talk before, you might anticipate that I'm about to define an ultra-filter of some sort. So a filter, Boolean algebra, is a subset of Boolean algebra, two, three, and maybe not four. I'll explain what four is doing in a little bit. One are just basic non-trivialities. The whole thing is in there. The zero is not allowed to be in there. It's closed sections, which are finite applications of wedge. Closed, or not wedge, of meat. Closed under finite meats. So if you have one through three, you get a filter. Four is a bonus property. The bonus property four is that for every A in the Boolean algebra, either A is in, or the complement is in. So if four, what's called an ultra-filter, just take any set and upwards close it. That'll be a filter. It's not quite so clear that ultra-filters should exist, but they do once we observe the following. So a filter is ultra, if and only if, let's call it a filter, F. F is a maximal filter. So if you want a maximal filter, just start with any filter. It applies on its limit. You can get a maximal filter containing the filter you started with. So, ultra-filters exist. And let's give the collection of ultra-filters a name. So let's, this is ST for stone. The stone space of B. Stone space of B. Ultra-filters. We should give the stone space of B some sort of useful topology. We've given, we started with the Boolean algebra. We obtained the collection of points. We gave that collection of points some interesting topology. So I told you, typical basin of clopin. So why are these clopin? Well, an ultra-filter has to contain A or A complement. The complement of this is just replaced A with A complement here, and that's going to be of this form also. So these things are all clopin, either this topology. I told you that these are basic, so it's zero-dimensional. It actually ends up being compact. Why is it, let's, I'm not going to write down a proof, but let's think about how four is going to bias compactness. If I have, let's have an open cover of basic open sets, and I get, actually let me write this down. Just to see once how this sort of argument works. S-T-B is compact. The storefulness is clear. If I have two different ultra-filters, they're going to disagree on something, and that's a clopin containing this one, the complement clopin containing this one, those are disjoint open sets. So how's that? Why is it compact? Give this an A. So this will be use of A. That's the basic open set corresponding to A. Say that use of A, I, if there's no, the negations, all these sets, as I told you it wasn't finite sub-cover, so these negations, if, finitely, many of the negations failed to meet, that would mean that their compliments covered the whole space. So now why is that? That extends to some Q. And then this Q, it's, this Q cannot have been covered. Q is not covered. And that's a contradiction. Right, so what have we done? We have zero-dimensional compact spaces. Give us Boolean algebras. Boolean algebras give us zero-dimensional compact spaces. I should note that there are some nice, here's a quick remark. So I have the operator, if I start with X, I can hit it with the clop operator. It's really a functor, clop functor. And I can hit clop with the stone functor. What should I get? I should get what I started with. And similarly, if I start with a Boolean algebra, I hit it with the stone map, and then I hit that with the clop map, an isomorphic Boolean algebra. Both of these arguments, you need a little bit of compactness to make it work. Like they're sort of an obvious, for instance, there's an obvious inclusion of V into here. You need to argue that everything gets landed on, and that's a bit of a compactness argument. There's an obvious map from the stone space of the clop ends into X, and you need to argue it's objective, and that will get you a bit of a compactness. So these are totally disconnected. Let's make it extreme more. So here's the main definition. Space X is extremely closure. So this is going to sound like a really weird definition. And it is because it's a really weird property of spaces. If the closure wasn't set, so that's weird. That's another example that there are lots of... It would certainly appear that there are lots of clop insets. If I can obtain one just by closing off whatever open set I start with. Let me draw a picture to convince you that these are zero-dimensional. If X is compact, it should be disconnected. If you get zero-dimensional, it's failing me. And why is that? If I have a... Here's my space. Here's a point. Here's an open neighborhood. I want to argue that there's a base of clop insets. I need to put a clop in around this. Well, compact house-door spaces are normal, so certainly regular. I can put an open set here. V, whose closure is contained, finds V with V bar, so it's a few. That's just any compact space will do that. But then closures of open sets are clop in. So that's a quick argument that we actually get zero-dimensional spaces when we do this. Let's see, I'm trying to... So before you might ask for... Let me actually give a non-example of a... Oh, I don't know where to start. So I can't give you any examples quite yet because we need another characterization of what it means to be extremely disconnected, to know that they're going to exist, or to have an easy proof that something that we come across is extremely disconnected. But here's a non-existence proof. Sample, canter space, 2 to the omega is not... So I'm going to write down a... I wrote down a nice example somewhere. Here we go. So if I have a finite string, which goes to a basic neighborhood, and some s is the neighborhood in canter space with s, this is a typical basic clop in the neighborhood of the canter space. And I'm going to let... Let's consider the open set u, let me be the union, over k in the natural numbers, 0 to the 2k plus 1, 1. So what is this? So I'm looking at strings with a 1, an odd number of zeros, and then a 1, and the corresponding basic neighborhoods, and then just union them up, that's open. I'm going to show you that the closure is not... The closure, I claim that it's just going to be u, union, the string. So I mean, the only cluster point that's not in one of these things already is just if we kind of zoom off and let the zeros in the middle get longer and longer, we're going to get this string 1, 0, 0, 0, 0. But then this is closed. And now, but note, 1, 0, 0, 0, 0 is not in n sub 1. Oh, that's not enough. It's not in the interior. There's no open ball I can put around the string to stay inside u closure. Because, for instance, if I... An open ball is like a string starting with 1 and some number of zeros. Extend that so that there's an even number of zeros. We're never going to be able to pick up things with an even number of zeros. That's just not... They're not there. I don't understand the notation. Here? Yeah. Okay, so cancer space is just strings of zeros and ones. I'm going off to the right. This is a finite string of zeros and ones. So these are the strings that start with the specified finite string. So I can take the finite string. So maybe like 1, 0, 1. And then maybe extend it to 1, 0, 1. That's my s. And then I extend it however I want. And I get an element of cancer space. Do you need the first one in these? I guess not yet. The key thing is the last thing. You're right. You don't need the first one. So we've isolated something that's definitely stronger than zero dimensional. We've exhibited a zero dimensional space. That is not extremely disconnected. So it's probably a good idea to convince you that this isn't a vacuous notion. So let's actually build an extremely disconnected space. And what we're going to do is we're going to use a random duality and isolate a property of the Boolean algebra that tells us when a space is extremely disconnected. There's a subset of the Boolean algebra and C is some other element. C is above. We still have a very, maybe a completeness labor because this is completeness. C is at least upper bound. So as it stands, there's no guarantee that every set has an upper bound one. The whole thing is an upper bound for every subset of the Boolean algebra. But there's no guarantee that least upper bounds exist. So we are going to define a property of Phoenician, a Boolean algebra, B is complete. If you have two elements which are least upper bounds for the same set, script A, well, each is below the other, so they are equal, with a giant wedge in front of the second question. So that's the least upper bound of the collection A. And there is a reason I define this. The proposition is let X be in a setting where we can use, we have access to Stone's theorem, we have access to the Boolean algebra, and we're going to characterize extremely disconnected in this. X is extremely disconnected and sort of the obvious thing, given that I just spent time defining this extra property of the Boolean algebra we have, we want it to be complete. And the proof actually isn't too bad. It's just a matter of writing down the things you sort of have to write down. So that one implies two. We want to define candidate for the wedge. Well, the wedge should better contain the union. Some clopin set that contains every clopin set in A better contain the union. That might not be, this is open, certainly, but it's not clopin. But wait, the closures of open sets are open. That's... And then it's really easy to check that this works. Two implies one. Let's see if co-op is complete. Let's say that U is an open subset of X. Well, we need to get our hands on some collection of clopins in the Boolean algebra. They're an obvious candidate. A, these will be the clopins, but A is a subset of U. At C, this is the direction that takes a tiny bit more work, not too much. So this is a subset of our Boolean algebra. It has a wedge. So let's, just for notational clarity, C will equal wedge A. C is a clopin. It seems like a decent candidate. We're trying to argue something about the closure of U. We want the closure of U to be open. So let's try to argue... We're going to argue we want that C is equal to U. C is equal to U closure. That would certainly work if we can do this. So certainly C contains U closure. C contains every clopin. So if I just take the union of A, that is U. It's just the union of all the basic clopin sets inside the open set. So if I take the union, I recover U. C has to contain U. C is closed. C will contain the closure. Anyway, round. If U bar were non-empty, then it's a non-empty open set. So let A non-empty a subset of C less U bar. C less A is an upper bound collection script A. But that's crazy. It's a smaller upper bound than the least one. So that's not going to work. So we get this equality. All right, so now we have a... We've used stone duality to give us a Boolean algebra characterization of extreme disconnectedness. So now this is where I kind of have to confess that talk might not be about extremely disconnected spaces as much. I'm going to start talking a lot about complete Boolean algebras. And in particular, we can use them to get some examples. So what's the simplest example? Finite discrete spaces are kind of boring. What's the simplest interesting example? So example, the power set of the natural numbers is a complete Boolean algebra. I often use just CVA for complete Boolean algebra. So the stone space is... Interesting, it's connected. But this has a much more common name. This is just beta omega. This is the space of ultra filters. I should say the space of ultra filters over omega. So the ultra filter on this Boolean algebra. So that's pretty easy. Here is a slightly more interesting one, maybe to bring in some analysis example. So let's consider unit interval, unit interval 01. Consider the measure algebra. This is the measure algebra, Borel. So certainly we know this is a Boolean algebra just because the Borels are a Boolean algebra. And the set theory operations respect the ideal. We know a few things just because of the null ideal. We know that, for instance, the measure algebra is countably complete. Any countable collection is going to have an upper bound just because I can take the union of those countably many Borels sets. And that's fine, that's an upper bound for a countable collection. How do we turn countable additivity into completeness? We need a little bit extra to actually get full completeness. So here is the quick fact is what's known as CCC, collection incomparable, called two elements of Boolean algebra incomparable if their beat is empty. Whereas incomparable elements is at most comparable. So why is this fact true in the measure algebra case? Well, I have this collection of Borels sets, pairwise intersections and null, but I know that everything in the collection is positive and it has a positive measure. So we're going to run into problems if I have uncountably many things of positive measure with pairwise null intersection. That's just ridiculous, that can't happen. So we get this property that Mt has. And then here we're going to tie it together with a quick proposition. If a Boolean algebra is countably Tc, then it is plenty sketch to prove. I don't want to spend too much time on this proof but I want to focus on my third and final example. What if I want to define the upper bound of a collection of size aleph 1? So proof sketch, collection, I go 1 many. This is the first uncountable ordinal. So I have uncountably many elements of this Boolean algebra. I want a least upper bound. What we're going to do is let... I'm going to truncate these collections. So A gamma, we're going to just be the A alphas for alpha less than gamma. We're at gamma. Well, each of these, well, now what's going on, these are all... these are countable ordinals. So these are countable collections. So by assumption, this wedge exists. So now we have an increasing... each of these is increasing because I'm throwing more and more sets from my original collection in. It's growing. I want to argue that it stops growing before the end. Well, if it kept growing, uncountably many sort of jumps in growth, but that sure looks like an uncountable anti-chain. That's a problem to our assumption that our anti-chains are all countable. So anyways, that provides in some sense another example of a compact, extremely disconnected space. You take the stone space in this. But in practice, people don't really do that. You just work with the measure algebra because this is pretty well understood. All right, the last example is taking a bit longer than I had hoped. So I might not do as much forcing as I said, but I don't think anyone is too heartbroken. Sorry, Jane. I want to talk about regular open algebras because this shows up into polity all the time. So I'm going to relax my convention that spaces are compact by default. So let's be any house dwarf space. I don't even know if you need a house dwarf. I just throw it in because I don't like thinking about non-house dwarf spaces. Any house dwarf space. In fact, I don't even... I don't think you need a house dwarf. But anyways, subset is regular open. U is the interior of its closure. So certainly you get the left-to-right inclusion for free. The reverse makes this an interesting definition. The fact that you recover exactly U and not some bonus stuff also. Or let RO be the collection of regular open sets. We are going to turn into a Boolean algebra. So I'm being... Notice I'm being kind of careful with how I'm phrasing this. I'm not saying that RO is a Boolean algebra. I'm saying we have to turn it into a Boolean algebra. Namely, the ordinary set theory operations are not going to work. For instance, the complement of my regular open set might not be open. The union of two regular open sets might not be regular open. So you have to tweak the ordinary operations. And then you have to argue that your definition actually works. If you get a Boolean algebra. So here are the definitions. The meet is just going to be the ordinary intersection. Which is a little trickier. So you want to write down union v. You want to write down this but it might not work. So you take its closure and you take the interior. Compliment. You want to take the complement, not be... I mean, that might not be open at all. So you take the interior of this. And then you argue... You have to argue that this works. So there's a few ways of seeing that. One is to just brute force your way through the axioms of what makes the Boolean algebra. Distributivity is really annoying if you do it that way. Maybe a slightly easier way of thinking about it is put all of the open sets into equivalence classes. And then each equivalence class based on having the same regular open representative. So if I start with any U and I form this, you get the regular open. Form equivalence classes based on having the same output. And then each of those is going to have a canonical regular open representative. And so then you just do the ordinary operations on any open set but then pass to the representative. And that's another way of... That simplifies some of the proofs. You still have to deal with complements. No way around dealing with complements. That's another way also. But it works. So what is nice about RO is it's a complete Boolean algebra proposition. It's actually really easy. The hard part is checking it's a Boolean algebra at the first place which I'm scooping under the rug. But if we assume it's a Boolean algebra, so we're going to let a be a subset of ROx. We set the interior of the closure of the union. So certainly it is. This is certainly an upper bound. It's certainly regular open because this is just the regular open operation. So what if we have another upper bound? So let's write... Let me call this union u. So if c is an upper bound... So c is some... Let's say c is ROx. It's a regular open set. u must be a subset of c. So let's just do the regular open operation. So close these off. You maintain the same inclusion interiors in front of that. You maintain the same inclusion. And that's the direction that you needed to show. So that this is a smaller upper bound than this one. So now we can... Now we have lots and lots of examples because there are lots and lots of Hausdorff topological spaces out there. So mainly we can revisit... It might be worthwhile to revisit the example from earlier when I showed you that canter space was not extremely disconnected. So what do we have? We know that if I look at canter space... Canter space has a Boolean algebra. Let's just let b be any Boolean algebra. This is a general statement. Oh no, let me do this more. So we plot... I can take the stone space of this i.e. 2 to the omega versus RO 2 to the omega. These seem like they should be pretty similar. This is complete. This is not. But how can we... What's the relation? So I'm going to plot every clopen. Every clopen is here. A clopen is a regular row. And that's going to turn out to be a very special map. So in general, if b is any Boolean algebra, we can embed some of the map from b into regular open of the stone space of b. Why is that? Well, every member of b is a... Describes a clopen subset of this. And so is regular open. So we're just going to include it in that sense. I b is an embedding with dense image. What does dense image mean in this context? It means given any member of the target Boolean algebra, there's something from the domain that maps below. So given, let's say, u, subset of RO of this regular open stone space of b, also u, no, u is an element of... So u is just some open subset of stone b. Let's find a clopen subset of u by stone duality. By stone duality, a has to come from our Boolean algebra. And it turns out that this is rather canonical. So we can call this the completion of the Boolean algebra b. And there's only one completion. Any two completions are going to be isomorphic. So that's something that's nice to know. If you are a Boolean algebra, you can make yourself a complete Boolean algebra without doing too much work. I have a few more minutes, so I'll take a vote. I have two applications I wanted to present. I thought of time for one of them. Or zero of them, that's an option. So I have a kind of dynamical interpretation or the set theoretic application or nothing. The set theory being forcing. It does. I'm going dynamic. Joe seems to be the only builder. So what is a good application? This application is called building more complicated dynamical systems from simple ones. Also because this is by far the shorter of the two applications to present. So we'll be done in five minutes. From old ones or from simple ones. I'll leave this as an example. Just an example of a simple dynamical system. Everyone's favorite irrational rotation. So you have a circle. You can identify with the unit circle of the complex plane. You have some irrational angle of alpha. And imagine, let's call this space x as my circle. z acts on the circle by each positive integer just does the irrational rotation. So this is an action of the group C by homeomorphisms. This is a classic system. Comes up a lot. But it's really simple. It's x is a compact metric space. It's a very simple compact metric space. It's isometric. There's loads of reasons to qualify that this is simple. Yet by abstract nonsense there exists really complicated z-system. Another reason it's simple or nice is that it's minimal. It's a minimal dynamical system meaning this is a... Maybe I'll define that in a second. But anyways, there's abstract reasons to believe that z admits tons of interesting dynamical systems. But I would like to build one using x and only x as my ingredient. Sort of a challenge. And we also want to preserve some dynamical information. So the dynamical information I want to preserve is the following. Let g act on x. x will be compact. This is an action by homeomorphisms. They are equivalent to this action being minimal. 2 is a little trickier to see, but it's really useful. Every non-empty open set, if I translate u by that many group elements, I get back to the whole space. So 2 implies 1 is pretty clear. I fix a u. Let me maybe draw a little diagram. So if I want to do 2, here's my dynamical system. Here's u. Here's a point inside my dynamical system. Let's assume that the orbit gets into the open set to have dense orbit. Well, I know if I'm assuming 2, finitely many translates cover the whole thing. In particular, if some translate moves you on top of x, then that the inverse of that translate will put x. We assume 1. Well, so every orbit is dense, so I start with a point. I know it lands into this open set somehow. So that means there is a group element moving you on to x. But how do you get down to finitely many? That's kind of the trick. Yeah, that's a compactness argument. So we have a group element g sub x for every point. g sub x, u unions up to x. This sure looks like an open cover of a compact space. And then you're going to use compactness. So it's useful to keep in mind. So now here is, as promised, a more complicated dynamical system from a simple one. I'm going to take where to go, here. So if g acts on x, g acting on the stone space of the regular open algebra of x by homomorphisms. So what's the action? A point inside here is just a collection of regular open sets of the original space. The group is moving around the original space. It'll move around these regular open sets. It'll move around the ultra-filter, which is a pretty straightforward fashion. It's going to push all the members that's open. So it's not hard to see. This is way more complicated. So even starting from irrational rotation, you get it's a massive space that has size. Oh gosh, I think you get up to two to the continuum in size. It's not matrizable. It's not equicontinuous. It's just a much more complicated dynamical system. But it's still going to be minimal. So that's the key property we wanted to preserve. So here's the proposition. I should say if g acting on, so is g's action on the stone space of the regular open. We use both 1 and 2. We're using them together. So fix str of x and a basic open neighborhood use of a of the form of ultra filters that contain the regular open set a. I want a group element sending p into use of a. Well, a, let's go back down to x. A is open. So this is open. So find, let's find those g's with their union. So now using part of the ultra, property of ultra filters, so there is some k. We covered the whole space with only many. X is regular open. These are regular open. Their join is the whole space. I have to find one in the ultra filter, g, i, a, and p. So g i inverse of p is in use of a. Has it not heard? That is building a more complicated minimal dynamical system from a simple minimal. All right, I think that's all I'll talk about today. So thank you. Can you guess what I'm going to ask you? About functors. Yes. Go on. I feel like you've been dangling so much junctions in functors. I think they're just, also stone and clock are both contravariant isomorphism. So I feel like they're junctions but for really simple, like the simplest possible reason. Oh, interesting. Yeah, they're fingers. They're fingers. Okay, okay. So if you take, so if you take like two ruling alphabets, I don't know what this is thinking of. And you take the stone spaces. Yes. You can add continuous math any other way. Continuous. Let's say the category of continuous surjections. Okay. And what about regular open? Oh, so regular open is a bit trickier. Right. Because that one, so you've got that map IB, go from B to row of stone of B. Yes. That looks a lot to me like you do there, but I don't care. IB from B to row of stone. It looks like stone should be left under row. Oh, that's possible. So I can tell you that extremely disconnected compact spaces. So you'll have to reverse all the errors, but extremely disconnected compact spaces are exactly the projective objects in the category of compact house surf spaces. Cool. So reverse the errors and then you get what completely, you know what they are. Yes. Um, so I don't really know what regular open sets are because I don't know topologies that aren't. So for like this example, this one. No, the Cantor space. Yes. Can you tell me like maybe some set which is on that side but not that side? Oh, sure. So for instance, when we talked about, remember when I showed that Cantor space was not extremely disconnected and I built that union of basic neighborhoods? Yeah. So that union was a regular open set which was not closed. An example of an open set that's not regular open, those are not to our team, we can be in an interval and I just have like here's an interval. Intervals are regular open. But then see I union it with another interval that is just, its start is exactly the endpoint of the other. So then when I take the closure when I get the whole thing then the interior is just going to forget that there was a gap there. So that's where you are open. So your application, why did you use the, did you use any completion of the way out? Um, the, you're bringing up a good point. I didn't need the, the only thing I needed about it is that it's an extremely disconnected and disgusting space. So, so starting with this space, you can do many of the same dynamical things with an extremely disconnected space. And those are interesting to me at least. Any questions? Alright, let's see.