 Okay, so today I want to start a new topic after contractions and we're going to talk about maps that are expanding or expansions. So I will explain what I mean in a second, but these maps will have very different kind of dynamical behavior and much more rich and in some sense also much more interesting than contractions. And we will use a completely different technique to study these maps and to study the topological conjugacy classes of these maps, the conjugacy class of these maps. And this technique is called symbolic dynamics. And I will start by just giving the basic idea because it's interesting to see it in general. So suppose you have your space, your usual map x to x, and suppose you have some partition of your space. So you divide your space into several sections, for example I0, I1, I2, I3, I4. And then you choose some arbitrary point x in your space and you look at the orbit of the point x and then you can describe this orbit in terms of its combinatorics in relation to this partition. So I can associate to each point a sequence, if x is in I1 I associate the symbol 1 and then I look at the image of x and suppose it's in I3 then I have I3 and then the next image may be still in I3 and I add another 3 and then the next image is in I2, I add 2 and I slowly build up a sequence that describes very roughly the orbit of the point x. It just describes where the image of the point x is in respect to this partition. Okay? So it seems like a very simple approach and also when you look at it like that it says okay what is this going to tell you nothing because you have no idea where this point is going to go or what it's going to do. But it turns out that it is really remarkable what kind of information we will be able to have from this, sorry? In some sense yes, in fact this is one of the keys approaches that will be very relevant to the ergodic theory course that we will do afterwards. But right now we will even also just from a topological and combinatorial point of view it will give some very interesting information. And we're going to apply this approach to first of all to the simplest possible setting which we can apply just to give an illustration. But even that we'll take at least a couple of lectures to describe it because we need to go carefully and do it step by step. First of all let me introduce some notation that we will need for this. So let me introduce the space of all possible sequences. So let's introduce the space sigma L plus is equal to the space, the set of sequences. I write this underline A and this is a sequence of the form A0, A1, A2, A3 and so on. Infinite sequence where AI belongs to the set 01 up to L minus 1. So there's L possible digits. So this corresponds to having L domains in your space, a partition into L domains. Okay, has anyone seen this space before, this set, no? From a topological point of view sometimes you can write this as a product space 01 to L minus 1 to the N because it's a product of sets in the sense that every element of this set has N coordinates if you want and each one is an element between 0 and L minus 1. But this is the space of all possible sequences. So if you choose a point and you look at the infinite image of this point you will associate to that image one such sequence because you will choose depending on X you will choose the first digit and then you look at f of X you will choose the second digit and so on. So I will also define so for A in sigma L plus, yes, thank you. So the meaning of the plus will not really appear in what we're doing but it's to do with this only infinite in one direction. You can also have spaces of sequence that are by infinite sequences. So the infinite, so we can also define set sigma L which is the sequence A minus 3, A minus 2, A minus 1, A0, A1, A2, okay. Also AI belongs to the set but it's a sequence that is infinite in both directions, okay. And in fact many properties of the dynamics you can also study some dynamics in this sense. If the map is invertible, okay, then you can also do this coding for the backward orbit of X and therefore you will get forward orbit will give you a sequence infinite in one direction the backward orbit infinite in the other direction and it is natural to associate to each point such a sequence like this, okay. But for the moment we will study not necessarily invertible maps and so we will mainly concentrate on the one sided infinite sequences. So for reasons that have to do with the construction that we need to do, we will rather than choose fixing a point and looking at the sequence associated to it, it is useful to fix a sequence and define the set of points that have that sequence as its coding. So let me define, let me fix one such sequence, okay, and let I of A bar be equal to the set of points X in X such that FI of X belongs to I, I, A, I for all I greater than equal to 0. Sorry, maybe this is okay. So I fix I, so let A underline be a sequence, okay. And now we let I, A, I, the definition, this is the set of points that have this property. So A, I is a digit between 0 and L minus 1, right, A. So if A belongs to this, then A is of this form, A0, A1, A2, A3, and so on, A, I, okay. So I fix a sequence, an infinite sequence, and now I take a set of points X that belongs to this set, if when I equals 0, okay, let me write it a little bit because we need to do this, if you're not familiar with this, it takes a little while. Soon it will be completely familiar with you. X in X, okay, such that X belongs to I, A0, F of X belongs to I, A1, F2 of X belongs to I, A2, and so on. I can be A0 can be 0, can be 1, can be 2, can be 3, these are the elements of the partition. So, sorry, maybe I should have formalized that, so I should have said suppose we have a map, we have a space X, a map on the space, and suppose we have a partition, okay, sorry, let me write it more formally. So let X be a set, okay, and I0 to IL, a partition, let F from X to X be a map, okay. So you're right, I should have, I should have, okay. So then I can define this. So I fix a sequence, and then I say, take the set of points, whose combinatorics under this map corresponds exactly to the sequence. So I require, once I fix this sequence A, I have fixed an infinite sequence of digits, okay, a number. And so now there's no reason for which this sequence should be non-empty, okay, a priori. I'm not yet saying that I'm just making an abstract definition of an object. This is exactly the kind of question, but now I've defined these objects, and I'm going to start studying them in specific situations, okay. I'm going to ask myself if it's non-empty. So is Ibar non-empty? Is there a point that has that combinatorics? For example, if F is the identity map, then only very special sequences. This would be non-empty only for very special sequences, because if it's the identity map, every point will just stay inside the same partition element forever. So if you take a sequence that is the constant sequence, 111111, then there will be a point that is just inside that forever. But if you take a non-constant sequence for the identity map, there will be no such point, okay. But the identity map is very special. If it's non-empty, another question, of course, is a unique point. Or are there many points that have the same combinatorics? So is IA a single point? Again, if it's the identity map, in general, this will not be a single point, because all the points inside one of these partition elements will have the same sequence forever. If, for example, the map is a contraction, then in general, they're converging to the attracting fixed point. So if the attracting fixed point is inside one of these elements, then all the points that are in the neighborhood, they will all have the same. They will also spend all their time inside that element. So this approach is not very good for certain kinds of maps, like contractions or identity map and so on. But as we shall see, it is very good for expanding maps, which is how we will use it, okay. So notice one thing though, even in this level of abstractness, there's a very interesting observation we can make, which is think of this, so I use the word combinatorics, think of it as that. So by fixing this partition, what this gives you is the combinatorics of these points, in the sense, what I mean by combinatorics is that you have knowledge, only combinatorial knowledge of the orbit, right? So you don't know exactly where the orbit is. You don't know it from a medical or topological point of view. You just know where it is in relation to this partition. So that's why I call it the combinatorics of the orbit. Suppose I know that a point has this combinatorics here. What can you tell me about the combinatorics of the image of that point? So suppose x belongs to I a bar. What can you say? So a bar describes the combinatorics of the forward orbit of the point x. What can you say about the combinatorics of f of x? It's very easy, right? So then f of x belongs to I b, where b is equal to b0, b1, b2, b3. And bi is equal to a plus 1 for all i. So the new sequence, f of x, will have a combinatorics that is given just by shifting the whole sequence. What does it mean? It means that the new sequence corresponding to f of x, in its first digit, b0, it will have the digit a1. And in the second digit, it will have the digit a2. And here it will have the digit a3, and so on. So the combinatorics is just a shift, corresponds to a shift on the sequences. So this motivates us to define a dynamics on this space of sequences, which in fact corresponds exactly to the original dynamics in some sense. So we can define a map, and we take a sequence, and we map it to a sequence b. This is called the sigma, sigma of a. And how is this defined? This is defined precisely by this property here, where bi is equal to a i plus 1 greater than equal to 0. a0, a1, a2, a3, and so on. Maps 2, a1, a2, a3, and so on, which we write as b0, b1, b2. We shift to the left. So using this map, we can formalize this comment I made here. And we can just write that so. We have that so. Let's define it. So first, let me define. OK, yes. So if x belongs to i, a bar implies that f of x belongs to i, i what? Sigma a, OK? So maybe you can recognize the germs of a conjugacy already between these two dynamical systems, even though we still are working in a very abstract setting. I've said nothing precise at all. Where is the conjugacy between here? Well, it's saying that if you relate a point x with its combinatorics, applying the map f gives you a combinatorics that corresponds to applying the map sigma to the sequence. You see? We are, in some sense, conjugating the map f to the shift space already, OK? This is not quite a conjugacy, because we don't know exactly this. We don't know that i, a bar is well-defined or that it's a single point. But if we could show that, so let me write. So first, let me remind you this is called shift map. Suppose i, a bar is non-empty, and i, a bar is a single point for all a bar in sigma plus l. So suppose we can actually prove this. Then what? Then we can define a map h from sigma l plus to our space x, where we associate to each sequence the point i, a bar. If we can, if we have this. If we have this and this map is well-defined, for every sequence, this is non-empty, and it's a single point. So this becomes a well-defined surjective map, not necessarily injective. That would be an additional property. Would be to say that this is injective. Would be to say that two different sequences must give different points, OK? That may or may not be true, but we don't need it yet for this moment. And what is this comment that we just gave here? What does that mean in terms of this map? What does this mean? This means that x in i of a means exactly that x equals h of a. So then if x equals h of a bar, then we have that h of sigma of a bar is equal to what? So h tells you for this sequence, which point corresponds to that sequence, OK? So I suppose that I have an x here, a bar maps to x. So now I take the shift of that sequence. What point am I going to get here? The image of x under f. So this is equal to the image of f under x, which is f of x, which is f of h of a bar. This is exactly i.e. We have an h composed with sigma equals f composed with h. Magic. Out of just like that, I've constructed a conjugacy of a dynamical system that you don't even know what dynamical system it is. This is not completely a conjugacy because we have not shown that h is a bijection, right? This is called a semi-conjugacy, OK? H is a semi-conjugacy, OK? If we also show that h is injective, then h is a conjugacy between and sigma, a conjugacy, OK? I haven't put any topology. We shall eventually also put a topology on the space so we'll have the possibility of asking for topological congex. So of course, we haven't actually done that, right? So in this construction, I have in some sense assumed that we're lucky and we have a system where we can carry out this construction and we get that i a bar is non-empty. It's a single point and possibly also two different sequences give rise to different points so this is also injective. That would be the perfect situation is that there is a one-to-one correspondence between the set of sequences and the set of points. Therefore, we have a bijection here and by construction automatically this bijection is a conjugacy. This is what I'm trying to show to you. So in various situations, sometimes it's easy to define bijections and then you have to struggle a lot to show that they actually conjugate the dynamics. In this case, the fact that it's a conjugacy, if you can prove this property, this is immediate. The difficulty is different. The difficulty in a specific situation would be to study these properties here. So before we have a break, let me just say and before we start actually studying a specific system, why do we care about such a conjugacy? Why should it be useful at all? It looks like when we apply this technique, we probably have a dynamical system or some map that we know something about. And we, assuming we can prove these things, we will conjugate it to a shift map, which is a very abstract dynamical system on the space of sequences. So what information do we have about the dynamics of the shift map on the space of sequences? This conjugacy is only really useful if it helps us because we know something about the dynamics of the shift and then helps us to say something about the dynamics of F, which is what we want to know. It turns out that the shift space is a very concrete space in which we can say a lot of things. And let's just briefly say, look briefly at the shift map. Actually, I think maybe we will just take a couple of minutes break before this, because I want to do have to take a little bit more time on this. So let's take two minutes break and then we'll come back. OK, so let's look a little bit at the shift map, because what I've described is an abstract strategy for constructing a conjugacy with the shift map on the abstract space of sequences. There's no guarantee just like that it will work for any system, but I'm telling you it will work for certain systems. So before we implement that, let's look a little bit at what is the advantage of that. So can we say something about the dynamics of the shift map? So we have here the space, we have the shift map. Are there any fixed points? Does the shift map have any fixed points? You tell me, does it have any fixed points? It may. Why not? Exactly, exactly. So how many shift points exactly does it have? Sorry? What? Fixed points. Fixed points. Come on. What is the sequence? Which sequence have fixed points? Hey. So we have the sequence 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, and so on up to L minus 1 equals L minus 1, L minus 1, and so on. These are all fixed points. You can see that they're fixed points. So it's going to have L fixed points. So that's a small bit of information, but it's some bit of information. If we construct the conjugacy with the map F, F is going to have to have L fixed points to be able to work. OK? Does it have periodic points of period 2? Give me a periodic point of period 2. OK, very good. I see we have a very bright class here. So for example, 1, 2, 1, 2, 1, 2, and so on. OK? So how many periodic points of period 2 does it have? Yes. If you don't repeat the fixed points, you don't count the fixed points as well. Yes? OK. So anyway, how many periodic points in general of any period? Countably many. You can calculate exactly for each period n. You can calculate exactly how many fixed points of period n. How many fixed points of period n minus 1? How many of them are plus 1? It's a finite number for every period n. There's an infinite number, countably infinite number. OK? So that is already some little bit of information. Is every point periodic? Or are there also some points that are not periodic? There's no periodic points. Just some sequence that's not periodic, and then the point will be not periodic. OK? Very good. So this is the first very important difference, like already from the systems of the systems that we have done so far. This system, as you remember, we've looked at systems that have a certain number of fixed points, and everything converges to the fixed points. Or we have looked at circle rotations in which maybe every point is periodic, or every or no point is periodic. OK? We've just looked at contractions in which everything converges to a unique fixed point. Already we can see just from this that there is a rich structure in here, because we have an infinite number of periodic points of all different periods, not all of the same period. We also have lots of non-periodic points. OK? We also have pre-periodic points. So we have points that are not periodic, but after a while they become periodic. Right? So for example, if you take a point 3, 5, 2, 1, 3, 1, 3, 1, 3, and it continues like that, it's not periodic, but then it falls into a periodic point of period 2. Does that? OK? So at the moment we don't have a topology on this, but later on tomorrow probably we'll define a topology on this, and we will be able to say that the periodic points are dense and that there's dense orbits, and we will be able to study the topological property. At the moment we're just looking at the combinatorial, from some sense, combinatorial point of view. OK, so this is just the first glimpse of how useful it is to study a symbolic system like this, because there's quite a few things that you can just immediately tell, like the number of periodic points and so on. So let's look at the kind of systems that we're going to try to apply this strategy to. I'm going to introduce a family of maps called the tent maps. So I'm going to define a family of maps from R to R, lambda is going to be a parameter, and I'm going to define it like this. Half lambda over x is equal to lambda x. If x is less than or equal to one-half, and minus lambda x plus lambda is greater than or equal to one-half. So what does this look like? What's the graph of this family? OK, so for x less than or equal to half, it's easy, it's just lambda x. So it's a constant slope. It's a straight line, slope lambda. When x equals 0, this is 0. When x equals a half, this is a half of lambda, OK? So we fix some lambda, and it looks like this. So this point here is lambda over 2. And what does it look like on the other side? This side is symmetric, it's got minus lambda as the slope. You see that when x is one-half, when x equals one-half, this gives minus one-half of lambda plus lambda equals half of lambda. So it's a continuous map. And then when x is equal to 1, this is minus lambda plus lambda is equal to 0. So this just goes exactly. So it's very easy. Lambda parameterizes this family. You just have these two lines. As you shift lambda, you just shift this fixed point. When lambda is small, it will go like this. When lambda is large, it will go like that. So you can see the whole family. So for the construction that I want to do, let us fix a value of lambda equal to 3. So the following construction for all lambda is strictly greater than 2, what I'm going to do. But for simplicity, let us just choose a specific value of lambda. So let's just choose lambda equals 3. So which means that unfortunately I need to redraw this picture because I didn't draw it very well. So for lambda equals 3, I have 0, 1. This is 1, 3 halves. This is 1 half. What is the special nature of this lambda 2 here? What am I saying? What happens? I haven't given you the construction yet, but you can immediately see that 2 is a special parameter here. What happens at the parameter 2? If lambda is equal to 2, then the image of 1 half is 1. So for any lambda less than 2, this graph will be contained inside this box. Whereas for lambda bigger than 2, it stretches outside this box. And we will see that this plays a key difference in what we're going to analyze. So let's first of all try to study the dynamics of this map. Several of you drew this picture in the exam. And I realized then that there was an impression that the dynamics was very simple of this map. And I would like to show you now that instead hidden inside this map, there's a very interesting dynamical structure. So some things are easy to study about this map. So what happens if x, so first of all, fixed points. What are the fixed points of this map? There's two fixed points. There's one fixed point here and one fixed point here. Those you can see. Can you see any periodic points? It's actually very difficult to see periodic points just by looking at the graph of the map. I mean, if you try it very hard, maybe you might be able to find periodic points of period 2. But anything else would be very difficult just by looking at the map. Because you really need to draw the graph of the ethers of the map and so on. It's very difficult. So let's analyze this systematically for different points. So suppose you take x negative. So let x negative. What can we say about the dynamics of this point x here? Exactly. Can you see that? What happens to the point x? Make sure you are, sorry? So if you look, I mean, we can use the graph to get some idea. You should always be able to formalize this in a formal argument, right? But it's clear from the graph. And if you want to prove it, you can look at the formula as well. Because the graph is below the diagonal here, if you remember, what this means is that the image of x is here. This is f of x, right? And because it's below the diagonal, this means that f of x is quite a bit less than x. And in fact, you can see by reflecting the diagonal, you see that this is f of x. If you start with x here, you see that this is f of x, which is this point here. But the dynamics is happening on the horizontal axis. This is a tricky point that I've mentioned before, I think, but always keep in mind that what we're doing when we're looking at a one dimensional map like this, we're drawing here the graph of the map. But remember what the graph of the map means. This is just the graph of the map. The dynamics is happening on the horizontal axis, right? So when you take a point here, this graph tells you, gives you a graphical representation of the image of this point, but the real image of this point is on the corresponding point on the horizontal axis, right? That's where all the dynamics is happening. So that's why what we do here is we just use the graph to understand where the point goes. So the image of the graph in the vertical axis gives you the value of the image, which you then have to represent on the horizontal axis. So you can see that this f of x is less than x. Then f of x is less than x. And the same thing when you apply the same thing again, you will see that f2 of x is less than f of x. And it's not difficult to see that fn of x tends to minus infinity as n tends to infinity, right? So this point here is moving. This is f2 of x, and so on, is moving towards minus infinity. Stop me if you don't understand or you're confused, OK? Because we're now entering kind of the heart of the course, and I want to make sure you don't get lost. Otherwise it will just get worse and worse as we go along. So now we're doing a fairly simple analysis of what we know how to do here. So what happens if x is bigger than 1? So let x bigger than 1. Then what? Yes? What's f of x? This is x. What is f of x? Where is f of x? Well, it's impossible if you go up. That's because you need to go down, right? Because the graph is below the horizontal axis because the value of f of x is negative, right? That's what it means to have the graph below. The value is negative. So this is x. Let me draw it here. Maybe it's better. x. So this is f of x. OK? What does it mean that this is f of x? This is just a graph. This is a representation of the dynamics. So what this graph tells you is that f of x, the value of f of x is here, which is a negative value. But this means it's a negative value on the axis where the dynamics is happening, which is somewhere here. OK? Project along the diagonal. So if you start with x bigger than 1, then f of x is less than 0. And so what's going to happen afterwards? If f of x is less than 0, then we already saw that every negative point, as you iterated, will converge to minus infinity. So f of x, if we now iterate this, will converge to minus infinity. So the images of x will converge to minus infinity. OK? So f will converge to minus infinity. That's it. So we've already dealt with a lot of points. OK? All the points to the left of 0, all the points to the right of 1 disappear into minus infinity. Maybe everything will end up in minus infinity. Not everything, because we know there's at least two fixed points. Maybe everything, except for these fixed points, will go to minus infinity. Let's see. What happens to the point 1 half? So let x equals 1 half. What happens exactly? And so f of x is 3 over 2, and it's greater than 1. And therefore, we already know what happens. OK? So x is a half, then f of x equals 3 over 2, which is greater than 1. And therefore, we're in the situation that we had previously, and so converges to minus infinity as x, as n. What are the points? Are the points close to half? They would also do that. So we have an interval here. OK, it should be symmetric here, but it's not. Let me just decide that I have something else. What is this interval, in fact? Because we know if lambda is equal to 3, what is this point here? This is exactly 1 third, exactly 1 third. And this is exactly 2 thirds. But as I said, for what we will do, it's not important. We use 3 just because it's easier. But if we had anything lambda bigger than 2, you will still have a little interval around the half that has this property. OK? So now let x belong to the open interval 1 third, 2 thirds. OK? Then f of x is greater than 1. Everything inside here will map to here, which is bigger than 1. And therefore, the next z that will be negative, and it just goes to minus infinity. So we're narrowing things down. OK? So fn of x tends to minus infinity as n tends to. Situation is not looking good. Everything is escaping. Right? Well, these points here, they don't escape for the moment. The first iterate, what happens? They stay inside the interval 0, 1. Can we say anything about these points? Well, they all stay inside the interval 0, 1. But some of them will now fall inside here. And those, we know that they will eventually escape. Right? Because look, this one, this point maps to 0. This point maps to 1. So in fact, the image of this interval is exactly the whole interval 0, 1. So inside this interval, there is a small interval that maps exactly to this. So here there will be an interval of points that maps to here and then escapes. Exactly. Exactly. Three parts. Two that stay inside. So how do we do this systematically? We need to start. OK, so now is where we need to be a little bit careful. Right? So it looks like more points are escaping. Maybe, as I said, everything eventually will fall inside this hole. This is like a hole now. That everything, once it falls in this hole, then we know the next theta is bigger than 1. The next theta is negative, and it's gone. It's like you fall in the hole, you fall to minus infinity. Even these points here that have these points that fall inside a hole. If you take this point, after one iterated, it falls into here. And then after one more iterated, it falls onto everything else. So there's another piece in here that will fall inside the hole and so on. So how are we going to study this? So first of all, let's define the set. So let me call i0 is equal to 0 1 third. i1 is equal to 2 thirds 1, and delta is equal to the open interval 1 third 2 thirds. And then the interval i equals 0 1 is precisely equal to i0 union delta union i1. So I've made a natural partition here of our space. This is i0, this is delta, and this is i1. So delta is the set of points which escape. So for all x in delta, we have that fn of x tends to minus infinity. We've just shown that. And the same also for any point which at any moment falls in delta. So now we define lambda equal the set of points x in i such that fn of x is in i0 union i1. So what happens to the points that do not belong to lambda? What can we say about the points that do not belong to lambda? What's going to happen to them? Because the only way it cannot belong to lambda, if at some point it falls in delta. And if it falls in delta, it goes to minus infinity. So really we've divided all the points on the real line into two possibilities, points in lambda, which are the points that stay bounded in here forever, and all the other points that just go to minus infinity. So they're lost. What we are interested in is this set here. We know that this set is non-empty. How do we know that this set is non-empty? Exactly. It contains at least these two fixed points. Maybe it contains only these two fixed points. And this is going to be our theorem. So theorem lambda is a Cantor set. You know what the Cantor set is? OK. One of the most important, in particular, it's an infinite set. It's more than an infinite set. It's an uncountable set. It's a big set, in other words. The set of periodic points of f, this is the set of periodic points of f. The set of periodic points is dense in lambda. In other words, there's a lot of periodic points, but not all points are periodic. And f is transitive, which means that there exists some x in lambda whose orbit is dense, such that omega x equals lambda. So what we're saying is that, in fact, there is a large set of points that never fall in delta. There's a large set of points that stay here. So by definition, you can see that from the very definition, lambda is invariant by the map. Because if x belongs to lambda, then obviously the image of x also belongs to lambda, from the very definition of lambda. That means that lambda is a set in there on which you can look at the dynamics restricted to the set lambda. In other words, you can say, OK, you can almost take lambda as a metric space in itself. It's got a metric induced from the Euclidean metric here. And it's got a topology induced by the Euclidean metric. And so you can say, OK, what's the dynamics on lambda? And I'm saying that the dynamics of lambda has a lot of periodic points of all orders we shall see, because we shall construct the conjugacy with a shift map. And moreover, these periodic points are dense in lambda. And also, there are many points in particular. There's one orbit that is dense where the whole orbit is dense in lambda. So the orbit approximates every point. So there's a very rich dynamical structure. So should we start the proof? Or are you tired? I think maybe we might stop a little bit early today, because it's all very new. So I want to give you a little bit of time to think about that. So you have a couple of days to review this whole notion of the symbolic dynamics and everything we've done. And it would be very useful, because what we're going to do now, the way we're going to prove this theorem, the strategy is very simple. We're going to show. We're going to define that map. We're going to look at those sets I, A. We're going to look at the sequences of 0's and 1's. I0 and I1 is not so much a partition of lambda, but it's a cover of lambda. And it works just as well. So lambda will be contained in I0 and I1. So for every point, we can define the symbolic coding just with 0 and 1. And then we will be able to show that all those things that we wanted to be true are true, that for every symbolic sequence of 0's and 1's, there exists a point that has that combinatorics. There exists a unique point that has that combinatorics. And the map H is injective, so that if you take two different sequences of 0's and 1's, they will correspond to two different points. There cannot be two points that have the same combinatorics. So everything will be perfect. We will get a one-to-one correspondence between the points of lambda and the infinite sequences of 0's and 1's. And that will give a conjugacy between the dynamics on lambda and this, which in particular will give the fact that we have an infinite number of periodic points. But most of these statements are topological in nature. So we will also, after we've constructed this combinatorial conjugacy, we will define a topology on sigma. And we will show that, in fact, we have a topological conjugacy between the two. And then we will use that. We will derive this from the analogous statements about sigma, which are quite easy. Once we have the topology on the symbolic space, it's quite easy to see that all those periodic points that we had before are dense in sigma. And that sigma has dense orbits. And therefore, we will recover these properties about the set lambda. So this is the general strategy. So as you see, it's a fairly naive strategy. But it's really remarkable. When you look at a map like this and you just look at it, you cannot see any way that, by elementary arguments, you can say that you have a canto set that's invariant and has infinitely many periodic orbits and so on. Instead, we'll be able to give quite a good description of that. So to get the most out of this construction, it's really worth your while making sure that you understand not so much that you understand, but that you start to familiarize yourself with this idea of the shift map and what we did today. It will make it much easier for you to understand what we do afterwards. OK, thank you.