 Good afternoon, everyone. So let's remember where we were. We have been studying the shift map. If you remember, we put a metric on the shift space that makes this into topological space. So I asked you to, as an exercise, to show that this is a canto set. And we saw last time that the sigma is a continuous map on this topological space, that the set of periodic points is dense. And then today we will start by showing that this dynamics also has some dense orbits. So lemma, sigma is transitive in sigma 2 plus, i.e. this means there exists some x such that omega of x bar equals the whole space. There exists a dense orbit. If you remember, we studied already one kind of system that has dense orbits. Do you remember? Circle rotations, that's right, irrational circle rotations. So the way you prove it is, well, the definition of dense orbit means that you approximate every orbit. So the proof has to prove that it's transitive by showing. So we need to show that. What do we need to show to show that an orbit is dense? An orbit is dense means that it comes arbitrarily close to every other point. Every point in sigma 2 plus is an accumulation point of the orbit. That's what it means. So how do we need to show that every point is accumulated by the orbit of x? So for all z in sigma 2 plus and for all epsilon greater than 0, there exists some n greater than or equal to 0 such that the distance between sigma n of x and z is less than epsilon. Do we agree? This is what it means to have a dense orbit. Is that right? Yes, sorry, we need to show that there exists an x. That's right. There exists an x bar such that this holds. Yes, thank you. Such that if you look long enough at the orbit of x, it comes as close as you want to any other point. What x is this going to be? Can a periodic point have a dense orbit? No, of course, because a periodic point has a finite orbit. It cannot be dense in this uncountable set. It needs to be a non-periodic orbit. It needs to be an infinite orbit. So how do we know that it comes close to the point z? Let's pick one z to begin with. And let's try to find an orbit that approximates that z. So fix some z in sigma 2 plus. What does it mean that the orbit of x approximates z? What is required on x? Well, x is whatever. If the beginning elements of x and z are the same, then the points x and z are very close. But all we need is that some iterate will shift that side. So it means that the beginning terms of z must appear somewhere in the sequence x. It doesn't matter where. After n iterates, we need to find the block of initial iterates of z. Let me write this. Let's write this down. So there exists, by the lemma that we proved before, there exists some n epsilon such that if the first n epsilon terms of sigma n of x and z are equal, then the distance between sigma n and x and z is less than epsilon. So it is sufficient that the finite sequence z0 z1 to zn epsilon appears somewhere in, for example, we could have x is equal to x0 x1 xn and then we have this block here and then x whatever it is here n plus n epsilon plus 1. So this block here appears after n steps. But what if we now change epsilon? So for all z, for every fixed z, this must be true for every epsilon. There exists an n such that this is true. So what does this tell us about the orbit of the sequence x? So now I take a smaller epsilon. So I get a bigger n epsilon and I take a bigger sequence xz n epsilon and so this longer sequence must appear somewhere in x where somewhere later. It doesn't have to be continuous with this. If I wait longer time, I will actually find here a longer sequence that coincides exactly with the first. So maybe n epsilon is 100 for a given epsilon. Now I take much smaller epsilon and the n epsilon that I need is 1000. And so I need to find somewhere inside the sequence x I need to find the first 1000 elements of the fixed sequence z. And maybe I will find them after 1 million n steps or whatever. I don't know where I will find them. But sometimes somewhere in the sequence I must find exactly this 1000 steps and so on. If I take epsilon even smaller, maybe n epsilon is 1 million. So I need to find somewhere inside the sequence x the first 1 million digits of the sequence z and I can find it. Is it possible to do this? Can I construct an x like this? Yes. Exactly. That's even more difficult. Let's try to do it for 1 z and then we'll see if we can do it for the other z's. You're right. Now I'm constructing an x that depends on z. But I will need exactly as you said I will need an x that does not depend on z because it needs to be true for every z. So I'm just trying to do it one step at a time. Let's see how we constructed depending on z. Is it possible to construct such an x that has somewhere the 1 million digits, you know, the first 1 million digits of z and then later on somewhere it has the first 1 billion digits of z and so on. How do I do that? So then z should be contained in x. No z. The whole infinite sequence of z is not... If the whole infinite sequence of z is contained in x this is possible but that means that after some iterates x falls on the point z. Right? And that is not exactly what we're looking for because even if it falls on the point z then the next iterated can leave the point z and it might never come back. So that is even not good enough that it falls on the point z. I really needed to come back closer and closer and closer and closer infinitely many times. Yes? You should construct all blocks. How do you construct all blocks? Okay, very good. Let me translate that. So let's see if we can... So I will leave the statement here because this is what we want to prove. So you're suggesting that I construct my sequence x like this. So I start with 0. Then I add 1. Then I put 0, 0. Then I put 0, 1. Then I put 1, 0. Then I put 1, 1. These are all the blocks of length 1. Then this is the blocks of length 2. Block of length 2. Block of length 2. Block of length 2. These are all the possible combinations with two digits. Right? And then I start with 3. 0, 0, 0. 0, 0. 1 and so on all the way to 1, 1, 1. And I put all the blocks of length 3. And I can continue doing that because the number of finite blocks is countable. Because for each length, there's only a finite number of blocks of that length. And therefore the union over all lengths is a countable set. And if it's countable, I can just put them one next to the other just like that. Do you believe this number exists? It exists, this particular sequence. Now does this sequence satisfy the properties that we need? Yes. Because it contains all finite blocks. So not only it will work for this Z, but it will work for any Z. Exactly the problem, right? So for any Z and for any epsilon, what we need is to find somewhere the n epsilon correct number of blocks corresponding to that Z. And if we wait long enough, we'll find it because X contains all possible blocks. So let me write this down. So construct X by enumerating all possible finite. Then omega of X bar is equal to sigma 2. That's it. We just needed to understand exactly what this approximation means. And then you can see what is the sequence that you need to achieve that. Can you tell me another point different from X that also has a dense orbit? That's right. I don't have to put them in this order. I can put them in any order I want. I can shuffle things around. I can also add, in between any of these blocks, I can add any random sequence I want. It will still work because it still contains all the sequence I want. So there's many, many possible distinct orbits that have dense... distinct points that have dense orbits. You can see that. Okay. So this is some interesting statements about the dynamics. As you can see, in the shift space, it is relatively easy to prove some fairly sophisticated properties of the dynamics like this one, right? Because it's very explicit working with the sequences. So finally, our objective was to prove the conjugacy with the tenth map. Remember that we had the tenth map. 0, 1, 3, 2, 3, 1. And we defined lambda to be equal to the set of points that stay inside. So this interval here is called I0. This interval here is called I1. fn of X is in I0, union I1 for all n greater than or equal to 0. And we already showed that sigma 2 is conjugate to lambda. And now what we will complete the proof by showing that f, restricted to lambda, is topologically conjugate to sigma. This is not really a restriction. I just like to clarify the space that we have. So the purpose of the topological conjugacy is to immediately transfer these topological results to f, restricted to lambda. So what we will prove, what is an immediate corollary of this is the fact that this set here, which we know now is a canto set, because we know that sigma 2 plus is a canto set. We know it has a dense set of periodic points. We know it has dense orbits. It's transitive. All those are topological properties that are preserved by the fact that we have a homeomorphism and a topological conjugacy between here. And we conclude that this set here, inside I0 and I1 is a canto set. It has a dense set of periodic points of all periods, just like sigma, which is not completely obvious here. And it has lots of dense orbits inside lambda, which is what we will prove from there. So a set of very non-trivial results for this that follow from this topological conjugacy. So the proof follows standard line. So first of all, we have already shown that h is sigma 2 plus 2 lambda given by h of a equals a is a conjugacy. So we just need to show, and it's a bijection, h is a bijection. So we just need to show that it's a homeomorphism. So just need to show that h and h minus 1 are continuous. So let's show first that h is continuous. What does it mean that h is continuous? It means that if we take two sequences that are nearby in this topology, so the sequences that have a very long number of terms, initial terms coincide, then the images in here are very close. Is that true? Can you see why that is true? If you take two sequences in here that are close, the images are close. This is in the metric of the shift of the symbolic space and this is just in the Euclidean metric. Why is that? What, what, what? No, in lambda we just use the Euclidean metric. Lambda is a subset of the interval. We just use the Euclidean metric. In sigma 2 plus, we use the metric on sigma 2 plus. Ah, but it's not a good idea to compare that metric with this metric or you will get, what? Think about it, okay? Think about it. You can't try to compare those two metrics, but... There's more. Just direct geometric approach. What is the relation between the sequence and the corresponding points, right? So if you have the sequence here, right, and this point here is h of a bar, this point x, what is the relation between the sequence and the point? This is the fundamental key of the construction is this relationship, right? Exactly. For each n, so remember this is exactly the definition, right? So h of a bar, so recall that h of a bar equal i a bar is precisely equal to the set of points such that f i of x belongs to i a i for all i. So what's the way to remember that? The way to remember that is that this sequence is the combinatorics of the orbit of the point in terms of these two sets, i0 and i1. Very simply speaking, this sequence describes the fact that this point in its forward orbit visits i0 and i1 exactly in the order given by the sequence. It describes the visits to i0 and i1. And what we have proved in the process of proving that it's a conjugacy is that there's a unique point. So any two points will necessarily at some point visit b and belong to different elements. So there's only one point. So each sequence describes the combinatorics of the point. So now you take two sequence that are very, very close. This must correspond to two points that stay together for a very long time because if you have a and b, so you have another sequence b that maps to some h of b, right? And if I say that the first one million terms of these two sequences are the same, this means that these two points, x and y, must stay, have the same combinatorics for the first one million iterations. What does it mean that they have the same combinatorics for one million iterations? That they, for one million times, they stay together. So if one is the night zero, the other one is the night zero and so on, they stay together. Then after one million and one iterate, maybe they end up on opposite sides and one is here and one is there. So if you find two points that stay together for one million iterates, what can you tell me about how close these points are? Exactly. But exactly, we've proved it and it's clear. By now you should just see that it's true, right? Because the derivative is large. The nearby points are stretching. The distance, any small interval, is growing by a factor three each time. So these points are moving apart very quickly and at some point the distance between them will be so big that they land on opposite sides. So if they're going to stay always on the same side for a million iterates, they must be very, very, very close. Okay? And this proves the continuity of age. So let me, I will write down this calculation, it's really a calculation, you know, you have understood this argument when you really feel very natural to do this kind of calculation, right? So let a bar sigma plus two fix some epsilon greater than zero. We need to show that h is continuous. So we need to find a delta such that if two sequences are delta close then the images are epsilon close, right? Standard epsilon delta definition. So by the lemma that we proved before, so for all delta greater than zero there exists n delta greater than zero such that if the distance a, b is less than delta then this implies that a i equals b i for all i less than equal to n delta, okay? This is just the property of the metric, okay? So if this is true, this implies that h of a and h of b belong to the same interval. So h of a zero all the way to a n delta is equal to a b zero b n delta, right? Because these two sequences are the same and so h of a belongs to this interval, okay? And h of b belongs to this interval but these two intervals are the same so they belong to the same interval. Because these intervals describe the combinatorics up to this finite time. But then we know that these intervals are small, right? But an interval of this order a n delta is the size of this interval is less than three to the minus n delta plus one. So let delta be sufficiently small. So if delta is sufficiently small n delta is sufficiently large so that three to the minus n delta plus one is less than epsilon. And before and so x minus y is less than x. Any questions? Okay. So the intuition as I said is simply that if the sequences are small, the terms coincide for a long time. If the terms coincide for a long time by definition means that these points stay together for a long time. Staying together means they belong to the same finite interval of order n delta which is very small and therefore if these are sufficiently small then these two points will be sufficiently close and that's continuity. And continuity of h minus one is almost the same. It's very similar kind of argument. So let's think again why should h minus one be continuous? What is h minus one? Basically, basically, yes. I mean it's even simpler this, right? So h minus one means suppose x and y are very close then the corresponding sequences coincide for a large number of terms. Why is that so? That is almost obvious, right? So if they're very, very close they will stay together for a very long time. And if they stay together for a very long time their combinatorics is the same for a very long time and therefore the sequences are the same. Let me actually leave it as an exercise for you to formalize. It's basically the same kind of arguments as that. Okay, good. So this is our first theorem, right? So we have studied this set here. We've studied this map here. We've studied this set here. And we've shown quite a few things about the dynamics of this set here. So what is the natural question now? What is going to be my next... What do we want to ask ourselves next? Directly from this. Sorry? The structural stability. And structural stability means what? It means we perturb this. Are we still going to have a similar kind of dynamics? You know? It's not... Absolutely not to be taken for granted because now we have a very rich dynamics. We have all these periodic points. We have all these dense orbits. You change it a little bit. You know, it's a very complicated structure. How do we know that it's going to survive this change? Okay. So really all we need to know that if you take a perturbation it will still be conjugate to the shift map because if two maps are conjugate both to the same map then they're conjugate to each other. So what we really just need to show is that this construction that we did is not depends on exactly the form of the map. This is a map that has a very special form. For example here, the derivative is always constant. You know, it's one-third, one-third. It's a very special map. I did that for simplicity. But if you look at the argument that we use where did we really use any of these specific characteristics of the map? So now I will write down a series of generalizations of what we have just proved and I will not give the proofs because the proofs are almost obvious generalizations of what we did, okay? But together they will show that we can perturb this and get some structural stability. So let me write down these generalizations. The first observation is that we can easily generalize this to more than two intervals, right? I mean, we introduced when we defined the shift map, I defined it on L numbers, L symbols, and clearly we can generalize very easily at least the shift dynamics to that. So we can define a metric, define, so on, sorry, on sigma L plus for L greater than or equal to 2, we can define the metric dA B is equal to I equals 0 to infinity of Ai minus Bi over Li. This is a direct generalization for the case L equals 2, right? And this is easy to check that this is also canto set and the shift map. So this is a proposition, sigma 2L is a canto set, sigma is continuous, periodic points, the set of periodic points of sigma on sigma plus L is dense, sigma is transitive, okay? All those conditions apply just immediately. You can generalize, you can see that the periodic orbits are dense, you can do the same construction for the dense, to find the dense orbits, right? You still have on L symbols, you still take all the finite blocks and you still do exactly the same construction. Then we have theorem, so let I0 to IL minus 1 closed this joint subintervals. Suppose there exists some lambda greater than 1, such that for each I from 0 to Li minus 1, we have the following properties, i e Fi i equals i and 2 f prime of x greater than equal to lambda. For all x, let lambda equals set, such that fn x belongs to I0 union i n minus 1 equal to 1. Then f, restricted to lambda, is topologically conjugate to sigma, restricted to sigma. So in particular, f restricted to lambda has this property. So in particular, lambda is a canto set, the periodic points are dense and it's transitive on lambda. So all the same conclusions that we had before. So this contains two generalizations of the theorem we have just proved in the last couple of lectures. What are these two generalizations? In what ways have we generalized the previous theorem? What's more general about this statement? One is obvious the number of intervals. We have several intervals, right? So what is this? So what we're saying here, what is the map we're describing here? This is an interval i. And we're saying that we have L minus 1 disjoint i0, i2. Here we'll just take L equals 3 for simplicity. Sorry? What is it? What is what? What is the function? Well, what is it? That's right. So I'm not giving you. I'm only saying the properties that I need, right? And I should say maybe f i equals i is a c1, sorry. I should say it is a c1 different office from ii to i. Okay? Sorry, I shouldn't have included that. And the derivative is bigger than lambda. So for example, I could have, so what I'm saying is that i0 maps to i1, maps to everything, i1 to everything. For example, like this. And inside here, I don't know how it does. I don't really care. It can do this. You see, I don't really care what happens outside i0, i1, i2. So this is a slightly more local version. So before we looked at the maps and I said everything that goes outside i0 and i1 goes to minus infinity. Okay, that's good. I'm going to give a global description of the dynamics because everything either goes to minus infinity or it belongs to lambda, right? Here, I'm generalizing a little bit. I'm saying, look, if I have some global information, that's good. But for the specific purpose of defining this set here, I really don't care what happens outside because I'm only looking at points that stay inside one of these intervals forever. Yes, so even though I can assume that f is globally c1, I don't even really need. It doesn't even need to be defined outside if I don't want it to be defined, okay? I don't really care. What I do care is that inside I have closed intervals on each. The map is c1 and it maps. It's a c1 different morphine from i0 onto all of i. So it goes from here to here, from here to here, from here to here. That's the only information I have. Plus, what information do I have about this? So this is the first generalization is that I have a finite number of intervals and I don't necessarily define the map outside these intervals, okay? If I want to, I can, but I don't have to. And the second generalization is that I'm not assuming that the derivative is constant inside each of these intervals. I'm only assuming that the derivative is bigger than equal to lambda, which is bigger than 1. And this plays an important role. So if you look through the proof, there's almost no change, okay? Certainly the number of intervals is completely relevant. You will see that it really makes no difference whatsoever. This question about the derivative is a little bit. You have to worry about it a little bit. Where did we use? Do you remember where we used the derivative? Information about the derivative? Which part? Excuse me? That's right. We used it when we used the mean value theorem and said that each finite interval Ia0 to An is less than or equal to 3 to the minus n plus 1 or whatever. Here you will just still use the mean value theorem. You don't get an exact length, but you get an upper bound for the length because here you have the derivative is bigger than 1, which means that these small intervals are growing, at least by a factor lambda. So you'll get an upper bound of lambda to the minus n epsilon plus 1 or whatever that was, okay? So you will still get that those intervals are going to zero. So I'm going to give this as an exercise because it's really the same. There's no point doing this in class, but I am willing to mark this exercise. I think this exercise you should really do on your own at home and you can give it in and I will look at it, okay? Because it's a good way rather than just looking at going through the case we did in class and just copying everything, it makes a difference if you apply essentially the same proof to a situation that is different in a few respects and you can make sure that you check all the aspects of the proof, okay? So this proof is an exercise. I will not make it compulsory for credit, but I think it's in your advantage to do it. If you give it in, I will mark it and then you can make sure that you really understand all the steps. Okay? Any more questions on this? So in particular what is interesting about this is that we can... So now if we have two maps, so what's the immediate corollary of this? Okay, let me see if I can write it without writing everything down again. Suppose we have this and then we have another map, G, from some other interval J to R and we have J0 to Jl minus 1. I'll also close this joint subintervals of I. Suppose that there exists lambda greater than 1 such a full I. We have this is true and also G of Ii equals I is also C1 different, okay? And the same is true here, so f of xi for all... Sorry, this is J, J and J and here we have G prime of x greater than or equal to lambda for all x in J i and then we define lambda prime in the same way. Lambda prime is equal to the points y such that Gn of y belongs to J0 union Jl minus 1 greater than or equal to 0. Okay, and then what do we want here? f lambda is topologically conjugate to G. So it's immediate corollary. So notice, of course, kind of trivial observation. We say nothing about the actual lengths and sizes of these intervals, right? They can be completely different scales but it doesn't matter. Topology, as you know, homomorphism doesn't know about distances. It just knows about the topology, right? So these intervals J0 to J minus 1, I0, Il minus 1 might be 1,000 intervals here, 1,000 interval here. The sizes might be completely incompatible one to the other but it doesn't matter. In each case you get a canto set that is conjugate to the shift and therefore they're conjugate to each other. Okay? Tiam, you okay? Okay. Excuse me. If two shift maps are conjugate, you mean if sigma, so these are conjugate to the same shift map on the same space, right? Because this is the shift on sigma plus l. This is conjugate to sigma plus l and this is also conjugate to sigma plus l, right? So this is clear that these are conjugate to each other because they're conjugate to the same shift map, right? Same shift map because there's only one shift map on one space. You have sigma plus l, sigma plus l is one space and the shift map on sigma plus l is just the shift map on sigma plus l. So if this is conjugate to f from lambda to lambda and this is also conjugate to g from lambda prime to lambda prime, then these are also conjugate to each other and the transitivity of the equivalence relation because conjugate is an equivalence relation, right? So if two maps are conjugate to the same map, then they're conjugate to each other because you just take the composition of these two homomorphism and you get a homomorphism between these two. You mean on a different number of symbols? Right, that is a very good question, yes. So for example, if you have sigma 2 and sigma 3, right? So if you have, on the one hand you have sigma 2, on the one hand you have sigma 3, are they conjugate? This is actually a good question and it's not completely obvious. As it happens, they're not conjugate, they're not topologically conjugate, but it is not a trivial observation actually. And the problem of the complete classification, so in this course I'm not planning to go in depth into symbolic dynamics, but for example, this I was thinking actually of giving, I don't know how many of you are planning to do a project in dynamical systems, but I was planning to give a list of some possible projects that generalize some of the results we did here. But one possible project could be to look a little bit more into just the symbolic dynamics because here we've just looked at the, we've used the shift to study some specific maps using the topological conjugacy. But as you saw, even just the study of the dynamics of the shift map itself is also quite interesting. We studied the periodic orbits, the dense orbits and so on. And in particular, this question is a very interesting question, when are two shift maps topologically conjugate and even when are two sub-shifts? So these shift maps contain a lot of subsets that are invariant along the shift. For example, if you take the space of three symbols and you look inside the space as the subset, so this has three symbols, so this is all the sequences with the symbol zero, one, two. If you look inside this at all the sequences that never contain the digit two, what do you get? You get exactly the shift space on two symbols inside here. So the shift space of two symbols is a subset, you can think of it as a subset of the shift space on three symbols. And similarly you can define many other subsets, so you can define the subset that never contains the finite block zero, one, two. So it contains zero, it contains one, it contains all combinations, but it never contains that specific block of three letters. And that is a subset of sigma three, which is invariant, because if you take a sequence that does never contain this block and you look at all its image, also it will never contain any of its blocks. So you can study the dynamics on this subset and you can say does this still have an infinite number of periodic points and dense orbits and so on? So the general problem of which of these sub-shifts are topologically conjugate to each other is quite complicated, because it's not trivial, but in some sense it's not too difficult either to show that these two cannot be topologically conjugate, but it is still an open problem as to a complete classification. If I give you a sub-shift here of this shift and say, okay, look at all the sequences that never contain a certain block, maybe that is also topologically conjugate to sigma two, or maybe it's conjugate to the shift of five symbols when you remove all the sequences that don't contain various other blocks. Maybe they are. And this is not completely understood this problem. So I think this is also an interesting section. The problem of when things are not topologically conjugate is also a problem, as well as the problem of when things are topologically conjugate. It often happens in mathematics that you work very hard to prove something that's true, and then you have to work just as hard to show that in a different setting this is not true. Okay, but let me write down the question of structural stability. Then we'll take a couple of minutes break, because this immediately should give you a clue as to the structural stability. So we're going to not give a kind of full version of structural stability, but a kind of semi-local version like this, right? When we don't care what happens outside this interval. So let me write out theorem. So now let F to R. Now I'm going to assume that F is a C1 map, and i0 to iL minus 1 closed these joint subintervals, just like before. Okay, actually the same assumptions, but suppose there exists lambda greater than 1, such that 0 to L minus 1, we have F of ii equals i, that's a C1 differ from ii to i, and 2 F prime of X is greater than equal to lambda, all of X in ii. Okay, let lambda is always the same, lambda is equal to the set of X, such that fn of X belongs to i0 union i equal to 1. Then there exists epsilon greater than 0, such that if f i to R is C1 and is close in the C1 topology to G, there exists lambda prime in i invariant, so G of lambda prime equals lambda prime, such that f is stick to lambda is topologically conjugate to G is stick to lambda prime. So what is the proof of this? So I'm going to erase, but we're going to use this theorem. Okay, but I just want to do a picture. Can anyone see the proof? It's an immediate corollary of what we just did. Any ideas? So what does the graph of F look like? Well, we know that the graph of F must map i0, i1, i2 to itself. Actually, you know what? Let's just take a two-minute break now and then we'll come back and think about this a little bit carefully. Okay? Okay, so we need to show, what do we need to show? We have the theorem saying that as long as two maps satisfy these properties, they're both topologically conjugate. Yes? Well, this is true for, this is always true. I mean, yes, I haven't stated it like this before, G lambda prime equals lambda prime, but this is true in all these lambdas for F, right? All these that we've, what is the image of the shift map is the whole space, right? These are surjective, this map. Sigma is continuous and surjective. F is also continuous and surjective. In fact, I meant to make this comment, there's a little comment before. Remember that when we studied sigma, we thought that we saw that it's 2 to 1 in the example with sigma 2, right? In the example that we studied in detail, we have this map here, right? This is i0. This is i1. And lambda is contained inside here. It's a canto set in here, right? It's topologically conjugate to sigma. So sigma is 2 to 1, right? That means lambda is also 2 to 1. Can you see that it's 2 to 1 here? Where is lambda here? What does it look like? Lambda is a canto set here, right? It's a canto set in here, okay? So how can we know it's 2 to 1, right? So if you look at the image of points, so this is 0, this is 0, 1, this is 0, 1. So if you look at the copy of i0 and i1 here, right? And you look at a copy of the canto set here, right? You see that if this point maps to this point, then there is also another point inside lambda here, which also maps to this point. And these two points, what will be the distance, the difference between the sequences of these two points? The corresponding sequences in the shift map. These are two points that have the same image, and the rest is the same, of course, right? This is exactly the 2 to 1. If you remember in the shift map, you'll notice that every sequence has exactly two pre-images, right? This point here has exactly two pre-images. And if you look at the corresponding sequences of this point and of these two points, these two points, after one iterate, they map on the same point, so their sequence is exactly the same except for the two initial terms. And of course, it's surjective, right? Sigma is surjective. Maybe I should have remarked that. Sigma is surjective because given any sequence, it has pre-images, right? So it's surjective. There's sequences that map to every sequence. So the same is true for lambda. It's surjective. The image of this cantor set maps exactly onto the whole cantor set. It's very fine, right? Even though it's a cantor set, so it's a lot of holes, it's just designed so that if you look at the image of this cantor set, all the points will map exactly the points on the cantor set. None of them will fall inside the holes, okay? Just by construction. Right, so now we need to show. We start with some f. And the information we have is just about the behavior of f inside i0 and i1. Okay? The derivative is bigger than lambda everywhere. So notice that this means that it comes out the other side somehow. Here it goes like this. Here maybe it also goes like this. Okay? We don't know what happens in between, but here I'm assuming for simplicity, because I'm going to take g that is c1 close. It's easier to define it as c1 function globally. So in the simplest situation, it could just be like this. And this could be whatever. It could be something like this. Okay? It doesn't really matter. Outside it can do anything we want. But what we're mostly interested is in these bits here. So what can we say about c1, small c1 perturbation of this map? A small c1 perturbation of course can map, does not need to map i0 onto the whole interval, right? But it will map some other small interval near i0 onto the whole interval. Because the c1 perturbation means in particular it's a c0. So it's inside, the graph is inside some small neighborhood of the graph here, right? And because it's c1 and the derivative is bigger than lambda, bigger, strictly bigger than 1 everywhere, then the c1 perturbation will mean that inside close enough to this interval, the derivative will still be bigger than some lambda minus epsilon, let's say it's still bigger than 1. So there will be some other smaller interval here, i0 prime, which will map, where the map g will map the small interval i0 prime exactly onto all of i. And the same will be true for all the other intervals. If you have a small enough perturbation, you will necessarily have some small intervals here in which the same condition is satisfied and here also the same condition is satisfied. So proof, if epsilon is sufficiently small, there exists intervals j0 to jl minus 1 close to i0 il minus 1, such that g satisfies same conditions in 2. You agree with this? Yes? And therefore by the theorem that we had before, the invariant set for these intervals j0 to jl minus 1, it will be of course the same number of intervals, so the invariant set lambda prime defined in a similar way to this with respect to these intervals here is conjugate to the shift map on l minus 1, but lambda is also conjugate to the shift map and therefore they're conjugate to each other. So letting lambda prime equal to be the set of y, such that g and y belongs to j0, jl minus 1, like this equal to 1. We have that f is connected to lambda is topologically conjugate to g. So it's fair enough, you're justified in not seeing all this immediately because even though these are all fairly straightforward consequences of the constructions, there's not something, you know, you still need to absorb this construction and the generalizations of this construction. So when you finish the exercise to generalize the construction to l intervals, you should also fill in the proofs of these additional corollaries because once you really understand that construction, these corollaries are completely immediate like this one here. So notice that we're not saying that f and g are topologically conjugate as maps globally, right? They're just topologically conjugate on these sets because I don't know exactly what happens outside and how it affects the conjugacy class and so on. That's correct, that's correct. But remember that the dynamics was very simple in those cases because when you have only one interval, it's a monotone map and everything is asymptotic to a fixed point, right? So this is a very different kind of dynamical structure. Excuse me? It's not exactly the same conditions, no. Those were just the condition that you had the diffeomorphism from an interval to itself depending on the number of fixed points you had the conjugacy class was parameterized by the number of hyperbolic fixed points, yes. But there there was no periodic points, there was no dense orbits, right? So it was very different because it was invertible. So the non-invertibility of here is exactly what allows this very rich structure to occur. The non-invertibility has two intervals or more than two intervals all mapping to the whole space and it allows all this very rich structure to occur. So in non-invertible systems, it's much easier to get some very complicated rich structure. Okay, so now I want to just a couple more observations here but first let me say that we're going to continue this topic in the next couple of lectures and we're going to continue by looking at what happens if the intervals are no longer disjoint. You can imagine you do this construction but now suppose the gap between the intervals is very small and then suppose the intervals are actually next to each other so the end points coincide. And then there is some difficulty with the conjugacy because when the point lands on the boundary you don't know if it belongs to one or to the other. It's a little bit more complicated. Okay, but that's what we're going to do. So I'm saying that this is an important topic not just for what we've completed this section where you have disjoint intervals and canto sets. When you close the gaps we will also get that the set lambda is no longer a canto set because the canto set comes from the gaps. The gaps create lots of holes and you create a canto set. Okay, but before that let me make a final couple of observations in the last 10 minutes on a property of these maps which is not quite topological. It's metric and it's sometimes used as one of the definition of chaos in dynamical systems. So let me write the definition here. So definition if x is a metric space and f of x then we say that f exhibits sensitive dependence on initial conditions if there exists epsilon such that for all x, x and for all delta greater than 0 there exists y in x with distance y over x less than delta and there exists n greater than 0 such that the distance between fn of x and fn of y is greater than epsilon. For example, let's think about some system that do not exhibit sensitive dependence on initial conditions. Suppose you have a circular rotation. Does it satisfy sensitive dependence on initial conditions? What does this mean? Sensitive dependence on initial conditions. So what you want is two points that at some point are further away from each other than epsilon in your space. So what this is saying is that any two points however small, not any two points but let's say a neighborhood arbitrarily small neighborhood of the point x contains a point that at some point moves far away from x by distance epsilon. So epsilon is a kind of large scale. You think of delta as being much smaller than epsilon. This is the idea. So however small you take delta there exist two points that are really, really close to each other that eventually move apart from each other. So if you have a circle rotation examples, suppose we have f s1 to s1. Does this exhibit sensitive dependence on initial conditions? What do two nearby points do? x, y, x minus y less than delta. Distance, okay, x minus y less than delta. What's the distance between the images of x and y? It's the same. So what's the image of the next images? It's always the same. So will they ever, if delta is very small if the two points are very close will they ever grow? The distance between them ever grow to size epsilon. Ah, I like this conjecture. You think continuity is enough? Excuse me, what? You're going to f n. Exactly, exactly. The family f n is not uniformly continuous in general. That's correct. But we agree that this is not sensitive dependence on initial. This does not exhibit, right? Because if two points x and y are close they will stay the same distance apart every time. So if you take delta less than epsilon this will never hold. The two points will always stay a distance delta apart. What about if f is a contraction so that it has a unique fixed point? That is a unique fixed point. Everything converges to the fixed point somehow with this exhibit sensitive dependence on initial condition. No, because all the points are converging to each other. They're not moving apart from each other, right? So even though at some, you know they might start far away if they start close, I mean they may or may not for a while move apart but if you take them close enough and at some point they will come close to that fixed point and they will just be staying closer and closer together the whole time. So it cannot exhibit sensitive dependence on initial condition. So we've run out of time but at the beginning of the next lecture we'll start with this and we will show that the shift map and the map, the tenth map that we studied they both exhibit sensitive dependence on initial condition. It's very easy to show. You can try to do it. It's very trivial when you think about it. And it's natural that the sensitive dependence I remember I have emphasized the contraction as a general philosophy. Maps are contracting or expanding and this is in fact an example of neutral. This is neither contracting nor expanding. The distance is like an isometry. The distance is the same. It's a kind of intermediate step, right? These are the basic categories that distinguish these classes of maps. If the map is expanding then this is likely to happen because you're saying that no matter how close two points are if the map is expanding that the distance between them is increasing at each iteration. We use that argument already in the proof of our theorem. The distance is increasing so sooner or later these points will be far away from each other. And this is seen as a kind of property of chaotic systems. I will discuss it a little bit more maybe next time. Okay, thank you very much for today.