 speaking today. So what are we doing today? So first off, I hope to explain what I even mean by a Ramsey-Like theorem. I'm certainly going to talk about Ramsey's theorem, that is sort of the prototypical Ramsey-Like theorem, but we're going to talk about a few others as well. And then we're going to take a detour to the abstract and sort of talk about some properties that the collection of sets might have. A collection of sets might be a filter or an ideal or an ultra-filter, etc. I'll define all these things. Three is where the connection between two and one becomes apparent. We're going to start talking about partition properties in terms of positivity on filters and in particular in terms of these ultra-filter things, which I will define. And then lastly, where the proofs and stuff? What sort of stuff you might ask? Well, we're going to... So a lot of you have probably seen proofs of Ramsey's theorem, especially the Torx people or the logic people probably seen proofs of Ramsey's theorem, but we're going to provide an ultra-filter proof of Ramsey's theorem and ultra-filter proofs of a few other things. And I'll explain what that means. And of course, the RFID is we're going to actually build an ultra-filter that do what we want. Let's start with number one. What are some examples of Ramsey-like theorem? These are all going to have the flavor of I have some set with or without structure. I split it into some finite number of pieces. I want one of those pieces to still contain some structure. So let's start easy. So if I take the natural numbers, let's just say my natural numbers contain zero for today, all it matters that it's countably infinite. So say that I take the natural numbers and split them up into finitely many pieces, but then some PI is infinite. Equivalently, I could say that if I take a finite union of finite sets, then their union is finite. And when the problems are this easy, that's sort of obviously equivalent. We're going to see later that those are sort of really dual, dual notions of attacking or anti-like problems. So you need to talk about partition regularity, the sets that are positive, or you talk about a collection of small sets being finite. So the small sets in this example are finite sets. Slightly harder. Save it. So this notation means the collection of unordered pairs of natural numbers. I'm going to take this and break it up into finitely many pieces. Infinite. So this, well, I've just written down as the first non-trivial instance of Ramsey's theorem. You've probably seen it described if I take a graph, a complete graph on countably many vertices, and I color the edges of this graph into a finite number of colors, then some color class contains an infinite click. Well, this is really the same thing. So these PI's are just what color are we assigning each pair. And then this set A here would correspond to the set of vertices in our infinite click. What we're going to see is that there's also a notion of, so this problem was sort of phrased in terms of a notion of largeness containing an infinite click like this. But it also, if you dig a little bit, there's a notion of smallness working here, namely collections of pairs which do not contain an infinite click are going to be small in some sense, and we're going to elaborate on this. Then let me, I'm going to just skip that example, and then let me bring up one kind of wackier example. So this is easy but different. So the thing that I'm going to say about trees, the trees are going to have roots. They're going to be branching. They're going to just keep going forever. They're going to be height omega, that's called. And every node has, what these things are that's not a problem to describe this example in more detail. If I get there at the very end, I might not. That's okay. Picture kind of looks like this. Here's a root. Here are a finite list of successors. Each of these nodes has a finite, but at least two successors, and then so on and so forth. I just do this forever. So that's what the tree is for me. To describe a Ramsey-like theorem for trees, if and some PI contains a, I'm going to put this in quotes for now, a strong subtree, fine strong subtree at the very end if I get there. I just want to put this example on the board now to illustrate that Ramsey-like theorems don't necessarily have to be Ramsey's theorem. They can look quite different, but they all, this flavor set with structure, split it up into some pieces, hope that some piece still has some structure. For those of you who are curious, this theorem is an instance of what's called the Halpern-Bloy-Clee theorem, which is a kind of fun partition result in that theory. So these are the sorts of problems we're going to be addressing today. So with these examples floating in the back of your mind, I want to sort of disembark on some abstract stuff, which I hope isn't too tedious, but they're technical things that we will need going forward, put x beyond the set section of subsets. I want some non-triviality conditions. These are going to look like this. So intuitively, as I'm writing down this definition, the filter is going to behave like a notion of largeness. So if I want to describe a notion of largeness that subsets of x might have, the whole thing had better be large. So I'm going to say that the whole thing isn't actually large. If I take nothing, the empty set, that had better be small. This should not be enough. If I take a as large and b as bigger, b should be large also. And this is maybe a bit less intuitive. It depends on sort of what your mental picture is, but three I like to think of as saying that this particular notion of largeness is really, really large. All the sets that we're calling large are really large, although I mean by that. So if I have two sets, which I'm calling large, they are so large that if I intersect them, not only is their intersection not empty, but it is also large. It's worth noting, put this in a box over here, a little side, stronger looking prime, where instead of intersecting two sets, I can say a0, a k minus 1 are large, then their intersection is large. The filter is an ocean of largeness. The sets in there are really, really large. In particular, if I am a set and my complement is large, then I have better be really, really small. So there is actually, we can describe what we mean by smallness. And it is just going to be the dual of this notion of largeness, the dual to being a filter and being an ideal. But once again, I said I'm not going to set. Normally, it's going to be a collection of sets of x. And we're just, it's literally dual. So we're just going to flip all the things that we secure. How do you do that? The whole thing is small. Sub-sets of small sets are small. If I have a collection of small sets, their union is still small. That's the dual. All right, just for two sets, so five a and b are small. And it turns out this duality between filters and ideals, it's more than just superficial. We had this intuition that if I take, if I have a filter and a set in the filter, its complement looks really, really small. And we're actually going to be able to form an ideal out of that. So just to make this duality really explicit, good things to keep in mind. So if Mark is a, trying to be all of those subsets of x whose complements are large, this is an ideal. And that's not too surprising. This, this act of complementation is going to switch all the things that need it to be switched over here to get from one to the other. And similarly, every ideal gives rise to a filter in the exact same way. Similarly, it's going to be the exact same definition really though. Those subsets of x whose complements are small, but then this forms a filter. Just some abstract nonsense, but there's dual notions of large and small. And let's look at some examples. And these examples are going to be suspiciously related to the partition problems I brought up earlier. So very first really basic digital principle I wrote down. So if I take a finite collection of finite subsets of n, their union is still finite. That's all that this is saying. Two, this is going to be the notion of smallness related to Ramsey's theorem. So the ideal i given by, so I'm going to set i subsets of n squared, which don't contain it infinitely. I can say that so there does not exist an infinite so that there's going to be a similar notion for that third partition problem that I brought up. So if t the tree, the collection of if I set i the ideal of subsets of the tree so that there does not exist, I mean I'll use the letter a for my subsets. There does not exist a strong sub tree. Of course for two and three to really be examples of ideals, you have to know that the partition theorems I was telling you about are true or you have to take that in face value. So until we know that we don't really know that two and three are ideals yet. One is sort of obvious. We're going to take one in face value. Suggests let me decide how I want to proceed. I'm defining ultra filters so I think I'm going to talk about positivity. So filters give us notions of largeness. Sets and filters are huge and then if you're in the dual ideal you're really small. So what does mean? So there's a nice in-between notion which is called positivity and that's with respect to filter or an ideal and things of definition. I'll phrase it with respect to a filter. We say the subset of x is f positive. Currently you can form that dual ideal to the filter. I'll write this out explicitly. Equivalently a is not a member of positive sets are just those things which don't have to be small. If f is a filter and I'm a subset in the dual ideal, I am sort of certified as being a very small subset because my complement is in that filter. My complement was declared to be big. I can't possibly be a largen any sense. Let me expand on this intuition. Sort of to do is the f positive sets. So while positive sets aren't necessarily large, there are sets that could become large and I mean that in the following sense. a is a positive. I can take the filter and then I take all the sets of this form a and a second b for b in the filter if this is a filter. So this says that a is a positive, a might not be an f but there's some larger filter extending f which does contain a. So I can make a large by changing my notion of large. So we're about to get into the discussion of partition regularity and the key intuition is that collections of sets which are partition regular sort of behave like the collection of positive sets for some filter. So let me tell you how the collection of positive sets for a filter behaves. Let's give it some notation. So f plus I'm going to define that to be the positive sets. So here's a quick proposition. This looks eerily like those combinatorial problems we were thinking about at the very beginning. A set with some property we split it up. We want one of the pieces to have the property. So this is sort of the abstract notion that's going to unify all these problems, the notion of positivity with respect to a filter. And in this level of abstraction, this proposition is just obvious. Why? Because the proof is just considerably dual ideal. If none of the PIs were f positive, that's exactly saying that all of the PIs are in the ideal, a finite union of small things are still small. But I told you that a wasn't small because a was f positive. So that can't happen. Positive sets for filters have this nice property. So what is this nice property? Let's abstract that out of this. So we're going to sort of abstract this partition regularity property. And then we're going to see that it's not much of an abstraction at all. I might actually modify something in my notes. So definition would be a subset of a first x, I'm not 97. We say that R, we say change what I'm doing. So if I make some mistakes, it's because I'm changing things right now. So I was going to define what partition regular means. I'm actually going to define weakly partition regular, because that's going to be a little bit more transparent for the applications we have. Weekly partition, then some contains a subset in, so what I mean explicitly, i.e., some a subset of PI. So this is abstracting away exactly what we want. We have these combinatorial problems. So we're just investigating, well, what types of collections of sets solve or for what sorts of collections of sets is the problem we're worried about, true list of examples. Once again, all very suspiciously related to the examples from the beginning. So one. And a lot of these are in fact going to be stronger than just weakly partition regular. The difference is I replace to go from weakly to just full partition regular, I replace x here with any member of R. So a small difference, but it's a little annoying to have to shrink down to a member of R and then prove the thing every single time. So one subset of, the collection of infinite sets is partition regular. If I take the natural numbers, split it into finitely many pieces, some piece has to be infinite. This is the partition regular family corresponding to Randy's theorem. Subset of power set of bracket square given by to be the collection of infinite clicks. Likewise, we have a subset of, let me actually preface this, if t is the tree, a subset of power set of t be the collection of strong subtrees. So partition regularity really captures exactly what we're looking for in all these problems. For some, yes, this is so, what we need to do now is, so we arrived at this notion of partition regularity by thinking about positivity. So now we need to go back, we need to describe some filter for which this is the case. Is that always the case? Ah, yes, this is what we're gonna do right now. Maybe I'm, maybe the, wait, I feel like the singleton containing only the empty set is weak. Singleton containing only, because if you decompose the set then the empty set is a subset. Well, let me, you're absolutely right. Empty is never in a weekly partition regularity. Okay, and does that fix it? Yes. Okay. Dang logicians. That's exactly our goal now. We want, we want to get back to the land of filters. And here's where it turns out some filters are better to think about than others. This is where the ultra filter comes in. So we need a tiny bit more abstract nonsense and then we're gonna do some cool things. Filter, filter subset of power set of some set x. x is a non-empty set. Maximal just literally means under set inclusion. So I'm gonna say that a filter f is a, might be a subset of a filter g in every set in f. Every set that is f large is also g large. And why do maximal filters exist? This is our friend Zorn. So Zorn's lemma tells us that these maximal filters, so we can do our hands on them. You do need a little bit of, you can't exhibit an ultra filter. You need some amount of choice. Not all of it though. So anything on the other side does not really get anywhere. Characterization of ultra filters. I mean, because I'm not going to grab a filter and ask, is this an ultra filter? Let's look at all the collections larger than it and see if any of those are filters. That's a really silly way of trying to verify that something's an ultra filter. Luckily, there's an easy way. So equivalently, set as either a, let's say that I write the whole thing as a disjoint union of two pieces. One of the pieces. And so we have those conditions one, two, and three for being a filter. So think of this as, by the number four, ultra filter. And this is going to be equivalent to, I'm going to write down a superficially stronger four prime, which is also always going to be true. So whenever a is a member, so there's two strength to things. I'm not going to start with the whole thing x. I'm going to start with just any member of you. I'm also going to allow not just two pieces, but finitely many pieces a zero. Then some characterization of this ultraness is also really useful to keep in mind. Equivalently, this is a pretty easy consequence of four and four prime. Equivalently, u is ultra. In fact, this is a direct consequence of four as written there. If u equals u plus. So there's no notion of medium for ultra filters. Every set with respect to an ultra filter is either large or small. And that's exactly what four sets here. If a is not in u, then b had to be in u. So there's no, no getting away with b and you have to be large or small with ultra filters. And now I believe let me just check something. Here's a remark that's going to be really useful. So it's related to this zorn thing again. So using zorn's lemma. So we don't just have to grab ultra filters out of the blue. We can be a little bit clever when choosing our ultra filters. So in particular, any filter can be extended to a maximum ultra filter or to an ultra filter can be extended. And so then just some terminology. So if an ideal filter u extends the dual filter of i, I'll say that u avoids i. So this will be really useful. Given any notion of smallness, I can cook up a maximal notion of largeness, which doesn't disagree with that previous notion of smallness. You don't want, I want, I do not want, I want to avoid placing members of i in the ultra filter. And I think we're ready for the key here, collection of subsets. Let's get that annoying empty set of case out of the way. Then the following. Sorry, weekly partition regular, I changed that. Weekly partition regular. And two. There is a filter that is actually non-part of purview filter. It applies two. So if r is partition regular, sub r be the collection those sets. e2, sorry this is the annoyance with weekly versus honest partition regular. So this is the i r prime collection function. Oh, this is weak pr. We put that there weak. We can partition regularity. This collection i sub r is an idea. That was exactly did I need the definition of weekly pr on the board? I did not. Let's think it through. I take my set x, I break it up or let me phrase it this way. Say that i r was not an ideal. What would have to go wrong? That would mean that there's some finite list of members of i r prime whose union is the whole thing, x. So I've written x as a finite union of pieces, supposedly none of which contain a member of r. r is partition regular, weekly partition regular, which means exactly the opposite. Whenever I take x, break it into finally many pieces, some piece contains a member of r. So it would help if I kept the definition of weak pr on the board, but this is this is just definition unfolding. Say I'm supposed not. This means pieces, no pi containing a member of r. We have no hope at building an ultra filter as a number two because of four prime. Any ultra filter on x contains one of these pieces and any one of those pieces is a certificate that you does not behave like this. You contains a subset or contains a set, no subsets of which r and r and that's a problem. So this was the key idea. This really connects weak partition regularity and ultra filtered. What it means for us, I have, I think I have enough time to cover the Ramsey theory example. I don't think I'll get to the tree example, but I will try my best. I want to, so remember goals, we had one two three, which were technical stuff, and then four was gruesome stuff or the gruesome stuff. So we can prove Ramsey like theorem by exhibiting ultra filters with certain properties. And let me show you what I mean by this by proving Ramsey's theorem. And to do that I need, so we need to, we need methods of constructing ultra filters not directly, of course, because ultra filters are these kind of ethereal objects in choice, et cetera. But if I have my hands on one ultra filter, can I cook up a more interesting ultra filter? That's the theme of these constructions. So the first, the construction that's going to come into play in proving Ramsey's theorem is the tensor product of two ultra filters. So let X and Y be sets. U, I didn't define this, beta X is the set of all ultra filters on X. So if I have U and beta X be in their tensor product, this is going to be an ultra filter on X cross Y is defined as a subset of the product. And when am I going to declare that it is large? And I do so if I'm going to use this ultra filter quantifier notation, I'll write it down and explain with a picture. So for U many little X's and X, there are V many little Y's and Y's little Y's and Y's so that X, Y is in. So what on earth does this mean? This is some new notation. So here's the product. The product looks like square is X is Y. And let's zoom in on a particular X0 over here. X0 is a member of X. And I'm going to look at this fiber above X. And A, A is a subset of the product. And it kind of looks like this. Here's A. So A sub X0, X0 looks up and thinks that these members of Y are good for getting into A. X1 over here, another member of X looks up and thinks that these members of Y are good for getting into A. So now some X's might be very sad. This one's very sad if they didn't see any members of A. But some of them might have seen so many members of A that we might be able to write something like this. We might be able to say that those Y and Y for which X0, Y is in A, it might be large. It might be in V. And so if this happens, I'm going to say that X0 is good for A. And then it might, we might just be so lucky that there are so many X0s or so many little X's that are good for A that the collection of X's that are good for A is a large sense of the X. It's a member of U. And so that is, that's exactly what this tensor product is describing. It says for lots and lots of little X's, there are lots and lots of little Y's that give us N. The other bay of compactification of a discrete space is the space of ultrafilters. For other spaces, it's weird. Any other questions, especially about the construction, we're going to use it to prove Granite's theorem. So what we're going to do, so goal is construct an ultrafilter, the set of pairs of numbers, so that every, so every instead contains an infinite click. If I can do this, then by our theorem, we're done. We've proved Granite's theorem. And we're going to do this by, by looking at tensor products. So step one, on principle, I'll explain what on principle means in a bit, U in beta M. This is going to get us started. On principle means just in terms of that vocabulary of avoidance of ideals, it just means that you avoid the finite ideal. You contains no finite sets. So every, every large set is infinite. Step two, we're going to form a tensor product. Form U. You can, you can form the tensor product of two filters. That's, that's certainly something that's good to do. But what's nice about doing it with ultrafilters is exactly that we get an ultrafilter when we do it. Something might look a little fishy, because this is, this is really, this is not really on bracket N squared. This really, if we're being honest, it lives in beta N cross N, a superficially different object. The first thing we're going to do is let's convince ourselves that this really lives here. So lemma collection less than N is in U cross U. It's really easy. The proof is I'm just going to draw, I'm going to draw a rectangle. So here's, I need the diagonal. So how many, so M, this is exactly the top. The second question is all the stuff up here. I want to show that this is large, whether it's that T cross U. Well, we know that U is non-principled. So if I look at any point here, and I look at the line above it, how many, so this is some little M. How many M to get little M, little N into the set? Well, almost all of them. Whatever N is bigger than M, it's going to work. And we know that U is non-principled, so it doesn't contain this finite initial segment. So it has to have contained the rest of everything. So it turns out every point down here works, and ultra filters contain all N. So that's proof by picture. And so now we can make, this allows us to make an identification. I'm going to identify a pair of natural numbers with this ordered pair here. So just the pair MN with M less than N is going to play the role of the pair, the ordered pair M. We're ready to actually show that this works. And I'm going to do this kind of carefully because I definitely don't have time for the tree example, so I'm just going to do this example really carefully. So let's, so we need to show that every U tensor U large set contains infinite click. So let it be, we need to, the first thing we kind of have to do, all these, these proofs have a similar play, we need to unfold this definition that's a pretty bundled up definition. So I'm going to let B sub little U equal those, so that the collection, so drawing that, drawing that square again, little B sub U is all the points on the bottom whose fiber above that point hits B lots of times. And by definition of tensor product, so definition U tensor U, B sub U is a U, because that's just, that's just what it means for membership in U tensor U. And this B sub U, the engine is going to keep our construction. So I'm going to start, I'm going to pick n zero in right, so B sub n zero to be those n, so that n zero n is in, so what does it mean to be in B sub U? It means that this is in U, so this is in U. So this is in U, this is in U, what's their intersection? Their intersection is in U. So B sub n zero intersect B sub U, is it in U? This first of all, the first thing we should know, this is a pair, n zero and one, is in B. Why? Because n sub one is in B, n sub zero. And that means exactly this, n sub zero, at stage K, I'm going to be a little of the ideas clear at this point, at stage K, what do we need to do? We need to pick n sub K. Membership in B sub U is what allows us to keep going. It had better n zero, nK had better B in B, and one nK had better B in B, all the way up to n sub K minus one, n sub K minus one, nK had better B. Why is this possible? Well this is just a finite intersection of large sets, so we can do that. Finite intersections of large sets are still U large. And so then, then election n sub K, that's all I have time for today, so any questions? Joe? This is, this is quite amazing because it feels like you're proving infinite runcy with only a single pass through the vertices. Exactly, the ultrafilter tells you, the ultrafilter navigates for you. Right, so I have to ask, is there a known proof of finite runcy using some finite reequivalent of? Oh, um, I'm going to say no, we really needed the non-principality to do anything. So the moment we're talking about finite sets, ultrafilters on finite sets are really boring, they're all principle. So, so this really is some, some extra machinery that we're using. That is, that is what I was working on all winter break. Um, so I have, I have something for dimension two, I can't get higher. The one for dimension one is pretty fun, though. I'll make a quick aside, so Ramsey's theorem, Ramsey-Kruti's theorem for not just coloring the pairs, you can color the triples, you can color the k-tuples, what have you, what do you need to do to generalize this sort of machinery? You just form the, let's say for three, you form a three-fold tensor product of your favorite non-principal ultrafilter, same idea. For, for most x-coordinates, there are most y-coordinates, so that for most z-coordinates, you're large, that's an ultrafilter, that's exactly what you need.