 Good afternoon, everyone. So if you remember, I introduced in the last lecture the notion of a full branch piecewise expanding map. So in short, the definition is that we have a finite number of intervals and a constant lambda greater than 1, such that the closure of the image of each interval is all of i. And this is a c1. And f, restricted to ii, is a c1 defer onto its image. And then two is that the derivative is uniformly bounded by lambda. And three is that the closure of the union is everything. So this is very similar to the kind of maps that we've studied before, except for the fact that the closures of the intervals intersect each other. i0, i1, i2. So the map globally does not have to be continuous. We might have, for example, it might look like this. It might or might not be continuous. So the theorem, so we're going to try to build a symbolic dynamics in the same way as we did in the other case. And the additional problem here comes from the fact that these intervals are not disjoint. So in points on the boundary, have an ambiguity as to whether we associate the symbol 0 or the symbol 1 to them. And because of this ambiguity, what we will get is just a subjection. So there exists h. There exists continuous surjective of sigma n plus 2i, such that, which is a semi-conjugation. Moreover, h fails to be injective. Only on a countable set. So we almost have a conjugacy. The fact that this is continuous and surjective and, in some sense, almost everywhere injective, so it's almost everywhere by injection, means that we can recover a lot of the dynamical properties of f from the dynamical properties of sigma plus l, of the shift map on sigma plus l, which is what we want to do. So the argument proceeds very similarly to the argument that we had before. So I'm not going to give completely all the details, but I will try to explain exactly the difference. I never will study quite carefully this issue about the injectivity because it can be a little bit confusing. So if you remember what we did is we introduced these maps on each branch. So in principle, the map f is defined on the open intervals, is differentiable on the open intervals. And on the boundaries, it might be defined here. On the boundary, it's either, strictly speaking, it's either defined on one side or the other side. And it's only differentiable on one side or on the other side. You have to be just a little bit careful about what exactly we mean by the boundaries. So to simplify that, we will extend each branch of the map on each interval to the closure by continuity because these are expanding and they're different morphisms. So we define branches. So for each i, l minus 1, we define fj is equal to f, restricted to ij, sorry, for each j, fj. And then we let fj bar to be equal to the extension, continuous extension of fj to the closure of ij. So this is well-defined. So we now have a family of maps. Each fj bar is defined on the closed interval, ij, and is well-defined on the boundaries of this closed interval, ij. And then we define a multivalued map. Define a multivalued map, f bar of x, to be equal to just fj of x, which of course is just equal to fj of x if x belongs to the interior of one of the open intervals, ij. And f i bar of x intersection fj bar of x if x belongs to the intersection between the closures of these two intervals. So of course, if I write ij, I could also write i and i plus 1. I mean, in general, we always draw these intervals as being ordered, 0, 1, 2, 3, 4. But they don't have to be like that. They can be labeled in any which way, OK? So of course, these points on the boundaries are just these points here on the boundaries between these intervals. I always draw three intervals, but you can imagine that there's 10,000 such intervals, right? So please ask if you have any questions or if there's anything that's confusing. So multivalued, of course, means that there are some points at which the image of this has more than one point and which are exactly these. Sorry, this should be union here. It might have two points. So I made the remark last time. This is all the review from the end of the last lecture. I made the remark that if these intervals are all disjoint, then this just gives the original map F. And using this map F, we define using this map F bar, we define the generalization of the intervals A i bar, which we had before. And we just write A i bar equals a set of x in i, such that F bar i of x intersection the closure of A i is non-empty. OK, so I think this is where we had got to last time. So the first proposition we want to prove is the analog of what we had before, which is that for each, so lemma, each A in sigma L plus i A bar is non-empty and consists of a single point. So notice, of course, that this does not imply injective. We're going to define later a map that goes from this shift space to the dynamical space exactly by associating to each sequence the corresponding set. So for each sequence, there's a well-defined single point. But there will be some situation in which there's two sequences go to the same point. This is just saying that if I fix the combinatorics, there's only one point that has exactly that combinatorics in this set here. So I fix this sequence here. And I want the point whose dynamics under this has exactly that sequence. And by this dynamics, I mean that if I get into this situation where this is multivalued, it is sufficient that one of the images is in here. One of the images is in here. OK, so really the only difficulty here is that it is a little bit confusing. Because formally, everything is exactly the same as before. So in fact, the only distinction we need to make is that we have a finite version of this, just like in the previous case. So we let I a0 to an is equal to the set of x such that this is true up to time n. So f bar i x intersection i ai is non-empty for all i equals 0 up to n. OK, and then it is clear again that this is by definition. So by definition is equal to the nested intersection. Because this is just the infinite version of this. So you want to know if there's a point that follows the given combinatorics for all time. And so it is sufficient to show each a0 an for each finite for every, let me emphasize this, for every possible finite sequence a0 to an. The set i a0 to an is a non-empty closed interval. And that the length goes to 0 as n. You remember? This was the structure of the proof that we had before. So because these are nested by definition, all we need to show is that for any possible finite sequence you get a non-empty closed interval. That means that for any possible sequence that you choose here, this will always be a nested intersection of non-empty closed intervals. And therefore, by general results, the intersection is non-empty closed interval. And that gives the first part of the lemma. And the second part is just that these go to 0. So that means that this nested intersection has to be a single point. And this is exactly the same. We'll leave it as an exercise because it really is exactly the same. If you look at this definition, you define this use induction. For the first part, use induction on n just like previous section. So do you remember how the induction went? What was the first step of the induction? That's right. For n equals to 0, you just have these n intervals. Each one maps to the whole thing. And so inside each one there is n copies of the whole thing and so on. And you just assume that at the inductive step you have that. And each time you have that each of these intervals map to the whole thing after n plus 1 equals 3. And therefore, it gets further split up. It's exactly the same calculation, so I'll leave it. And the second part, the fact that this goes to 0, in the previous case, we used the fact that we had a specific derivative 3, remember, at each step. How do we know it in this case? How do we know that these go to 0? That's right. That's right. This depends on the fact that we have the assumption that the derivative is always bigger than lambda, bigger than 1, everywhere. And therefore, for the second part, second part, use that f n plus 1 from i a 0 to a n to all of i is c1 and f n plus 1 of x is greater than equal to lambda to the power n plus 1, because the derivative everywhere is greater than lambda. And so by the mean value theorem, we have that i a 0 a n plus 1 a n is less than or equal to lambda to the minus n plus 1, which tends to 0 as n. So these are the little generalizations that I asked you to think about already in the context in which the intervals are disjoint. We did it with a specific derivative 3, but I said that you could generalize it just to assume that the derivative was expanding and bigger than 1 everywhere and that you have some arbitrary finite number of intervals. And you could get exactly the same results. So this goes to the lemma. And this allows us to define the semiconducing. So we define a map h from sigma plus l to i by, this is now well defined, because for each sequence, there exists a unique point, which then we prove that it's continuous. So the fact that it's surjective is automatic inbuilt into the definition. Remember, because every point has some combinatorics, and therefore automatically there will be, for every point, if you choose that particular combinatorics of that point, you will get that point here. So the surjectivity is inbuilt into the definition. The continuity I will leave again, let me leave as an exercise, because again, it is just exactly the same argument that we proved that we used in the previous case. And injectivity. So the injectivity is a little bit more involved. So we're going to look at this a little bit more carefully. Which is injective, except countable set, which I will call I hat star, on which it is 2 to 1. So what set do you think this will be? So a basic picture, like the one we had before. So remember that this countable set, when we say that a map is non-injective on a certain set of points, what we're talking about is the points in the range, in the image, that have more than one pre-image. So what are the points where H is injective, and what are the points where H is not injective? What do you think? That's right. Those look like clearly points where it should be failed to be injective, because you don't know which interval you belong to. What are the points, any other points or just those? Exactly, pre-image. So this is really just that. It's all the points that, at some point, land into the intersection. At that point, they will fail to have a uniqueness of combinatorics. I will give, sorry? We cannot say that they're exactly those. Can we? Can we? Do you think, so your question is whether perhaps, at this point, H might be injective at that point? I didn't understand your question. Is it exactly these points? Let's see. Yes, the answer is yes. And let's analyze the situation quite carefully. So we make sure we understand exactly what's going on. So let's define, indeed, exactly the set. So let's define the set I star is just equal to all the intersection points. x in I such that x belongs to the intersection of some I and the closure of some j for ij in 0, l minus 1. So this adjusts the intersection points, the finite number of intersection points. And then we have the problem that there might be a point that is not in the intersection point. But after 200 iterates, it falls onto one of these. So let's define this set that I hat star is equal to the point x in I such that fn of x belongs to I star for some n greater than equal to 0. So this contains all the pre-images of I star. So we will show that this is exactly so I star. This is exactly the set in which we failed to have injectivity and on which we are exactly 2 to 1. So we have noticed that I star is a finite set. And so the set of pre-images here is a countable set. Now it's easy to see that if you do not belong to the set, then the map is injective. So if x belongs to I intersection minus I hat star, what does that mean? If it does not belong to I hat star, what does it mean about the combinatorics, the orbit of this point? It means that the full orbit of this point lies always in the interior of one of those elements. It never falls on the boundary. Or if it falls on the boundary, it has to be either 0 or 1. It falls in these boundaries, in these sets. But it never falls into any of those intersections. And so for these, the symbolic code in the combinatorics is always perfectly well-defined. So there exists a unique sequence A bar, such that fn of x belongs to I an, the closure of I an for all n greater than or equal to 0. And therefore, h of A bar equals x. H of b is not equal to x for all b different from A bar, because this is a different sequence. And if x does not have a combinatorics given by b, because x has a unique combinatorics, because it always falls uniquely inside one of the open intervals, so h has to be injected. So this shows that h is injective on the complement of I star. Now we need to study what happens on I star. So notice in particular that h is injective at the end points. This point here, we can assume this is the unit interval 0, 1. So what are the combinatorics? What is the symbolic sequences associated to these points, 0 and 1? What is the symbolic sequence associated to 0, the unique symbolic sequence associated to 0? 0, 0, 0, 0. It's a fixed point. What is the unique symbolic sequence associated to 1? 1, 0, 0, 0, because 1 maps to 0. Is it always like that? No, it depends on the shape of the graph. What are the other possibilities? Suppose that there's not that many possibilities. There's just a few possibilities, because it depends only on these two branches, the first branch and the last branch. So for example, if this branch was like this, was orientation preserving instead of reversing, what would be the sequence associated to 1? 1, 1, 1, 1, because it's a fixed point. And if this was instead of orientation reversing instead of orientation preserving, what would be the sequence associated to 0? 0, 1, 1, 1, if that's a fixed point. What if neither of them is a fixed point? What if we have this and this? Then what is the sequence associated to 0? Exactly. And what is the sequence associated to 1? The same. So we have a periodic orbit of 0 and 1. It's a periodic orbit of period 2. 0 gets mapped to 1, 1 gets mapped to 0, and so on. But they still have a perfectly well-defined unique sequence each of them. So this is an observation which is important to really analyze exactly. H is injective. So what are the possibilities? So the coding, the sequences associated to these end points depend on the orientation of the branches f0 and fl-1, the first branch and the last branch. So I won't write down all the possibilities, but let me just write down. So we may have, if both branches orientation preserving, then the end points are fixed points and we have that the point 0. So 0 corresponds to the sequence 0, 0, 0, 0, and 1 corresponds to the sequence 1, 1, 1, 1. Otherwise, that 0 corresponds to 0, 1, 1, and 1 corresponds. So in this case, 1 is a fixed point. So 1 may be just 1, 1, 1, 1, 1. Or we may have that 0 is a fixed point and 1 is not. So 1 goes 1, 0, 0, 0. Or we may have that 0 goes to 0, 1, 0, 1, 0, 1, and 1 goes to 1, 0, 1, 0. Sorry, not 1, not 1. This is L minus 1, OK? Sorry, this 0, 1 is on the case with two symbols. In the case with L symbols, this is L minus 1. And so I'm sorry, each of these should be L minus 1. L minus 1, L minus 1, L minus 1, and so on. And all of these 1s should be L minus 1s. And the same thing here. In this case, sorry? Ah, yes, thank you. No, because this is 1 in the interval here. No, these 0 and 1, they correspond to the end points on the interval. That's why it was a little bit confusing, right? So this is the point 1 in the interval. These are points inside the interval. And these are the corresponding sequences, OK? So the other, the last possibility, I think it's the last possibility, is that 0 goes to 0, L minus 1, 0, L minus 1, and so on. It's periodic. And 1 goes to L minus 1, 0, L minus 1, 0, L minus 1, OK? So this is not so important, the specific. I just wrote them out to make sure that it's clear that there's only a few possibilities and that they're all classified, and that we understand what they are. Yes? In this picture? Yes. Why not? OK, we'll see. We'll see. So you're saying maybe it's injective here because of the continuity? OK, it's possible. Let's see. Let's work through it. Yeah, so maybe there are some points. Maybe a first comment was like, maybe there are some points in eyesight with its injective, OK? So certainly, it is injective everywhere else. And the only points where it can go along is I star. Let's see. Let's go through the construction and we'll see exactly how it works, OK? So let X, so now let X in I star, OK? So what are the possible sequences? You're right. In this case, there is no ambiguity. Well, in this case, there is, yes, there is the ambiguity given by the 0, by n equals 0. So what are the codings associated to this point? So what are the sequences? Let's look at, let's choose one of these options for this picture, OK, so we don't get confused. So let me choose here the original picture like this. OK, so let's take this point here. Let's suppose X is in I0 intersection I1. So what are the possible sequences associated to this? We'll do it in this case and then we'll write out the general version, which is exactly the same. What is the symbolic coding? What is the combinatorics associated to this point? Well, we have an ambiguity right from the first moment, OK? It belongs both to I0 and to I1. So already that's an ambiguity about the sequence. It could start with 0 or it could start with 1. Well, what is the image of X? We have two possible images, OK? The image could be 0 or it could be 1. So is that another ambiguity? Does that mean that we have four possibilities? After it lands on 0 or on 1, it's OK. Because then once it lands on 0, it follows the same combinatorics of 0. And that's a unique combinatorics and there's no longer any problem. If it falls on 1, also from then on it's a unique combinatorics and there's no problem, OK? But here we have two possible sequences to begin with, 0 and 1. And then depending on where it lands, we have the sequence, whatever the sequence of 0 is, and whatever the sequence of 1 is, OK? So how do we decide if the image of this point is 0 or 1? How do we decide? Well, that's what we're proving is the case. But we actually need to make a decision in terms of the h. So our map h has already been defined. And we are trying to understand, given how many sequences can correspond to this point x. So are there two possible sequences that give rise? So do there exist two sequences? Do they exist? A, different from B, such that h of a equals x and h of b equals x. Or do they exist more than two sequences? We're trying to understand what these sequences are, OK? And we know that these sequences are given by a certain construction, by a certain definitions, because we have a precise definition here. Each sequence goes to i a bar, OK? So we need to look at that construction and see what sequence it gives. And if you look at that construction, if you remember, i a bar was exactly equal to the set of points such that f, so let me make sure I copy it exactly the way we had it before. Such that f bar i of x, intersection i bar a i, is non-empty for all i. So that construction actually corresponds, if you think carefully about what this corresponds, corresponds to the following fact. When we, if you remember, when we defined this f bar, we defined, first of all, I made a point of defining f bar as f bar defined on each closed interval, right? What I said is f is defined a priori on the open intervals here, but we can extend it by continuity to the closure. So we have f0 that's defined on this closed interval and f1 that is defined on the interval i1. So because this ambiguity of the point x, it may belong to i0 and may belong to i1, but if we decide that we think of ix as belonging to i0, then the correct map to apply is f0. If we think of it as belonging to i1, then correct map to apply is f1, right? So if we think of x in i0, we then apply f bar 0. And if we think of x belonging to i1, then we apply f bar 1 in this construction. And this decides exactly what the image of x is. If we apply f bar 0, which means the branch of f on i0, then what is the image of this point x under f bar 0 in this example, sorry? It's 1, right? Because f0 is the map defined on the open interval i0. Extending it by continuity to the closure means that the image of the endpoint is up here, which is 1. So x has two possible sequences. One of them is the one that starts with 0 and then follows whatever the combinatorics of the point 1 is. The other one is the combinatorics, the sequence that starts with 1. And then we apply f bar 1. And what is the image of this point under f bar 1? It's 0. And therefore then it follows the combinatorics of the point 0. So this is why I was very careful to define all these fj's and the closure, extension to the closure at the beginning because it is particularly crucial in this point to really understand what the images of these points are. Let's try doing the same thing for here. So let's take a point y here. What are the, how many sequences map to this point under this map 8? So how many sequences describe the combinatorics of this point? Well, again we have the same situation where the first symbol of the sequence could be 1 or it could be 2 because that's where the point is, okay? And now what is the image of this point? Now it's a little bit easier because we don't have the ambiguity. Either branch will map it to the point 1. That's okay. So it's a little bit simpler because whether you apply f1 or whether f bar 1 or f bar 2 in both cases you get to 1. So the possible sequences of this point is 1 followed by whatever the sequence corresponding to 1, to the point 1 is, or 2 followed by whatever the sequence corresponding to 2. So also in this case we have exactly two possible sequences corresponding to this point. So I will try to write this down now in the general setting. Write down this argument but it's important to keep this in mind. Let's try to write this down, okay? So for convenience let us denote by a are the sequences corresponding to the end points. Of the full interval I, okay? So 0 has the sequence B, okay? So H of A equals 0 and H of B equals 1. These are the sequences. As we've just seen, there's only a few choices for these sequences, right? They might be, this might be just 0, 0, 0, 1, 1, 1 or depending on the orientation there's a few other possibilities, few other combinations. But they're uniquely defined. So these points have uniquely defined sequences. So now let X belong to I, I intersection I, J. Intersection of these two closed intervals. They're exactly two possible symbolic sequences, two possible sequences associated to X. Yes. Intersection must be I, J and intersect I, J plus 1, 1. Excuse me, I didn't hear what you said. Intersection must be I, J plus 1, I, J plus 1, intersect I, J plus 1. Yes, so I made this comment before. In general, I prefer this just because maybe I don't want to assume that they're all ordered necessarily, right? They're all in increasing order. I could decide to call this I, 0 and call this I, 5 and call this I, 3. It doesn't really matter. I mean, in all the examples, generally, you put I, 0, I, 1, I, 2 all next to each other, in which case, of course, it would be I, J and I, J plus 1. But, you know, just a little bit more general. So this is just for, I'm assuming that this, there exists a point in this intersection. So since it's only a finite number, it needs to be, so this is for some I, J in 0, L minus 1. I guess if I want to be really precise, I should specify I different from J. So what are these two possible sequences? So depending on the orientation, so depending, let's say, how should we say that? So these two sequences, orientation, F bar, I, J sequences given by the symbol. So the first sequence, the sequence are given by the initial symbol I or J followed by the sequence associated to F bar I of X or F bar J of X respective. This says exactly what it needs to say. You just need to spend a little bit of time to think about what it says. So we have the point in the intersection. Because the point is in the intersection, then depending on whether we apply the map F bar I or the map F bar J, both of which have a well-defined image for the point X, like both of these map X are well-defined on X. They either map X to the end point, one end point of the interval, or to the other end point on the interval, depending on the orientation. But wherever they map them, this point here is a well-defined unique point. Because F bar is well-defined on the closed interval I, I. And therefore this point here, which is one of the end points, has a unique combinatorics. So the two possible symbols associated to X are I followed by the combinatorics of this, whatever it is, which are either or B. And the other possibility is the symbol J followed by the combinatorics of this, which may be either one. So we saw this exactly on this example. These are basically the only two possibilities. The general case is always exactly the same, even if you have 200 elements. If you look at the intersection of one, at this point you will have two possible branches that will map this point. One branch will map it to one of the end points, and the other one will map it also to one of the end points. The end points have uniquely defined itinerary. So the only ambiguity is given by the initial position of the point and not by what comes afterwards. OK, this is for the points in the intersection. What about the points that are in the pre-image of the intersection? So suppose now that X belongs to I hat star, and there exists some n. Let's say bigger than equal to 1, such that fn of x belongs to I star. OK, so in other words, let's suppose we have some point X here. And after 27 iterates, it falls on this point here. So what are the sequences that we're going to associate to the combinatorics of this point X here? Can you see that? That's right. That's right. So it's exactly the same, right? So until for the first n iterates, it's got a unique combinatorics, because it lies in the interior of these elements. And so it has a unique combinatorics. The first n terms of the sequence are given by its combinatorics inside the open intervals here. And then at some point, at time n, it falls inside here. And here, it has exactly the two possible combinations that this point here has. We already have decided which two sequences we will assign to each point in I star. So this point here will have two sequences associated to it. But the difference between these two sequences would not be in the first place, but in the nth place, 1 plus 1 place here. Whereas in this case here, it's in the nth place. In this case, it's in the 0th place. In this case, when you take a point in this intersection, then the 0th place is different. Sorry, there's two possibilities for the 0th place. And then the rest is all fixed. It's unique. So in this case, we have then the orbit of x has a unique itinerary until it lands in I star. At that point, there are two possibilities as before. Exactly the same as before. So what this lands in I star has exactly the same option. So this shows that it's exactly 2 to 1 on all the boundary points and all the pre-images of the boundary points. OK, so let's do a concrete example, which is very nice. And hopefully, it will help you to understand this construction better. So what are the easiest full branch maps? Example, let's consider this class of maps. It goes to lambda x, mod 1, where lambda greater than equals to 2 is an integer. So mod 1, of course, means that I just take the fractional part of 0, 1. If I take x and I multiply it by 2, it might become bigger than 1. So I only take the fractional part, the non-integer part. So in particular, choose first, consider first. OK, first of all, let's look at the structure of this. What is the structure of these maps? How many branches do these maps? Is this a full branch piecewise expanding map? First of all, what is the derivative of this map? It's equal to lambda. That's right. It's equal to lambda. So for example, if lambda is 2, then what does the graph look like? If lambda is 2, you can see it has derivative 2. So there will be a point at the point 1, 1, 1, 1, 1 here. The graph will look like this. Because the mod 1 means that when you take something bigger than half, you multiply it by 2. You take it something bigger, and then you just take the non-integer part and you get the point that you get here. You shift it by an integer. So what if lambda equals 3? Then you can easily see that you have, so the derivative is 3 everywhere, and you have a map. Now these maps are, in some sense, the simplest possible example of piecewise expanding maps that defined very naturally. And we shall see that all this business about the symbolic coding is related in a very interesting way to some arithmetic properties, in particular to expansions in base 2, in base 3, in base 10. So let's look, for example, at the case lambda equals 10, which is the one we're most familiar with. So when lambda equals 10, we have 10 branches. What do the intervals look like here? The intervals of the branches. They go from 1 to 0.1, 0.2, and so on, right? So this is 0.8, 0.9, and the derivative is equal to 10 everywhere. We have a similar picture here, and then we have these branches. We have 10 branches. This is 0, this is 1. So according to our theorem that we've just proved, there is a semi-conjugacy from the space of sequences on 10 symbols into the interval that corresponds to the dynamics. So there is already some coding of these points on 10 symbols that has nothing to with dynamics. You know what it is? There's already, for each point, if I choose a point here, I already have a kind of coding using 10 symbols. What's the coding? The decimal expansion of this point, 0 to 9, right? So let's see what is the relation between this decimal expansion and the coding that we get dynamically. If there is any relation at all, is there any relation? So this is the map f of x equals 10x, mod 1. Multiply by 10 is good, because you've known this from when you were five years old, how to multiply by 10. So suppose you have a point x equals 0.371529 or whatever. This is the decimal expansion. What is f of x? Shift. I like that word. So we shift because we multiply by 10, and we throw away the 3. So we get 0.71529 and so on. What is the next digit? Same, OK? 0.1529 and so on. So you can see how you keep going. So it looks like a shift map on some kind of space of sequences, right? We don't think of this as a space of sequences. We think of this as the space on the line. But for each point, we associate a sequence. And in fact, the dynamics actually looks like the shift on the sequence, OK? What, moreover, how is this sequence related to this partition here? For example, where does this point lie? So let's give names to this partition. Let's call this i0. So let i0 equals, let me use this half. Let me call i0 equals 0 to 0.1. i1 equals 0.1 to 0.2, OK? All the way to i9, let me write as a 0.9 to 1. So where does this point here? Belongs to i3. How do you know that? Exactly. Where does this point belong to? i7. Where does this point belong to? Where's the next point going to belong to? Where's the next point going to belong to? So what is the relation between the dynamics? This is under the dynamics, right? Remember, we've introduced the dynamics. This is just the decimal expansion, OK? Now, of course, the orbit of the point is defined by the decimal expansion because it's just defined by the point. But what we're seeing here is that, in fact, if you look at the orbit of this point in terms of these partitions, the position of the point inside one of these elements is given exactly by the sequence, right? You can tell. If you look at the 20th term of the sequence and it's a 4, then you know that after 20 iterations, it's going to land in i4, OK? Which is exactly what we had before. That is how we had constructed the sequences. So in this case, really this example is just to give the observation that in this particular case, the decimal expansion gives you exactly the coding that we constructed in abstract before. In other words, if you take a sequence here and you look at the point that has that sequence as its decimal expansion, you will get exactly the point that we got with our abstract construction of the coding before. The decimal expansion is the coding that we constructed. Let me formalize that a little bit more. So in our setting, what did we do? We had some difficulty with these end points, which we also have here actually a little bit of difficulty between the end points, because it's not completely clear if these points map to 0 or map to 1. So the way we did it in our setting, we define open integrals. Let's call them j0 equals 0 to 0.1, j1 equals 0.1 to 0.2, and so on. Up to j9 equals 0.9 to 1. Oh, sorry, I wanted to call these j's and these i's, because this doesn't really matter. Because I'm trying to put this in the same setting that we had it before into open intervals here. So let's try to check what I just said. So let's check that this coding, if we try to apply our construction to this particular map, what we would get would be exactly the decimal expansion. And let's check it particularly with reference to these end points in which we have a little problem with the injectivity and see what happens. So first of all, if we apply our construction, if we apply construction, we get the following. So suppose x, suppose fn of x belongs to the interior, interior of some ij, OK, to the open set. So suppose it belongs to some ij for all n, k turned n equal to 0. fn of x belongs to some ian, OK? So in that case, what is exactly the sequence an? Then the sequence is exactly the decimal expansion of the point. Then a equals a0, a1 on some is exactly. So these are the points that never fall on the intersection, never have a problem with this. Now in this case, what are the decimal expansions? What is the symbolic coding? Not the decimal expansion. We're constructing our symbolic coding, and we're checking that it coincides with the decimal expansion. What is the symbolic coding of the point 0? This corresponds with the decimal expansion of 0, OK? What's the symbolic coding of the point 1? No, no, no. I should apply the purpose exactly what we've been doing before. So coding of 0 is 0, 0, 0, 0, 0, OK? Coding of 1 is what element does it belong to? i9, OK? This is a fixed point. So what's the coding of this? 9, 9, 9, 9, 9. Does that make sense? Is this what you want it to be? What's the decimal expansion of the point 1? Exactly, OK? That's the sequence we have when we construct it dynamically, OK? Remember what I said. The exercise we're doing now is to use our construction to construct the symbolic coding and to check that what we get is exactly the decimal expansion of the points concerned, just to make sure that the two things give exactly the same thing, OK? So for 0 and for 1, it's OK, OK? Now, what if we take a point in the middle, like here? Now let x belong to the intersection of some i and some i plus 1 in this case, because in this case, we have them ordered. So sorry, let x, let x, OK, yes. i and i plus 1, OK? Then what is the symbolic coding that we're going to construct for this x here, for example? So this is 0.7. So we have one branch that does this and one branch that does this, OK? So now is the point in which you have to, hopefully will help you to understand all that discussion that we had previously, OK? So what are the two possibilities for this point here? So e.g., let's suppose it's in the intersection of i7 and i8, just to be more specific, OK? We have two options. Either we think of it as an element of i7, in which case the symbolic sequence is 7, followed by, now you look at the branch that is defined on i7 and you look at where it maps this endpoint. And in this case, it maps it to the point 1, because it's orientation preserved. And what is the symbolic sequence of 1 is 9, right? So the symbolic sequence we get is this, OK? Or we think of this point as belonging to the interval i8, in which case the first digit will be 8. And then we apply the map f bar 8, which is the map defined on this branch, which maps it to this, and it maps this point to 0. And what is the coding of 0? Exactly. So these are the two symbolic sequences associated to that point. Corresponds with the decimal expansion, OK? This is exactly the ambiguity in the decimal expansion. Did you ever ask yourself why there was no uniqueness in the decimal expansion of points? This is why. If you had unique decimal expansion, then you'd have canto set. You'd not have an interval. These are glued together exactly for this reason, OK? And you do exactly the same thing for all the others, OK? What if you have a point that is not on the intersection? So what if you have a point that after some time lands on the intersection? What is it going to look like? So suppose you have a point, so let x such that fn of x belongs to this, for example, to i7 intersection i8. Then what is the symbolic coding? And what is the decimal expansion of such a point? For some n greater than or equal to 1. For example, n equals 1. So suppose we have some point somewhere here, x. It belongs to the interior of one of these elements. But after one iterate, it's unlucky if you want, or lucky, depending on your point of view. And it falls exactly here. So what is its coding going to be, the dynamical coding? So suppose this belongs to i4, for example. It's inside i4. All 8, 0, 0, 0. It has two possibilities, OK? So this point here will have a certain number of sequences, a certain number of uniquely defined terms, which depend on its orbit up to time n, which is always in the interior. And then at time n, it lands in here. And in here, there's two possibilities. So this point itself has two possible symbolic codings given by both of these symbolic codings have the first n terms are the same. And then from then on, either it finishes 79999 or 8, 0, 0, 0. And those are exactly the two possibilities. And that is exactly the ambiguity. If you remember, of course, the ambiguity in the decimal expansion is basically all the digits that finish in 0, 0, 0, you can also write as finishing in 99999 like that, which is exactly what you get in our case. Sorry? Can I repeat it? So you mean what to do here in this case? OK, so in this case, then the associated sequence uniquely defined initial number of terms, then end either 79999 or 8 or other, both possibilities. If fn of x lands in this intersection, the symbolic sequence, there will be two possible symbolic sequences. One is given by, so the symbolic sequences will be given by some sequence a0 up to an and then 79999 or a0 up to an and then 8. OK, so just in the last few minutes, let me say first of all that, OK, so this is the same with the base. Remember that you can also, this is the decimal expansion of numbers. But as I'm sure you know, and I think this maybe helps to understand this, as the word says, the decimal expansion is just one specific way to represent the numbers. The numbers are not intrinsically given as in this form. The numbers are given as quantities. There are many different ways that we can use to represent all the numbers on the real line. The real numbers are not defined by a specific representation. So the decimal representation of the numbers is just one way that we can write them. And we can also write them in base 2, for example. So we can also write in 0, 1 in base lambda, where lambda is greater than or equal to 2, and lambda is an integer, by writing in two ways. If you want to just look at it algebraically, then we write x equals a0 over lambda plus a1 over lambda squared plus a2 over lambda cubed and so on. So when lambda equals 10, this sequence is the decimal expansion. So when lambda equals 10, this is the tenths. This is a0 over 10, which gives you the first digit in decimal expansion. This is over 100, which gives you the second digit and so on. But you can do it exactly in the same way for any other lambda. And you get this is the base lambda. OK, so base. So then we can just write it in some sense a0, a1, a2, basically base lambda, expansion less. Where ai belongs to 0 to lambda minus 1. So if lambda equals 2, this is just 0s or 1s, of course. And you get the well-known binary expansion that is used in computer science, for example, the binary expansion in 0, 1s. And you can write every number in 0, 1. And you can do exactly the same thing. What is the dynamics? If you have a number f lambda, what is f of x equals lambda of x equals lambda of a0 over lambda plus a1 over lambda squared plus a2 over lambda cubed. What do you get when you multiply by lambda? You get the shift again. Because here you get a0 plus a1 over lambda plus a2 over lambda squared and so on. But if we take this mod 1, that means we only want the fractional part, a0 is an integer. So we disregard it. And this gives exactly mod 1. It gives, OK, we write it as a1, a2, a3 is the expansion of the image. And if you do exactly the same thing that we had before, you get exactly the same. Everything is exactly the same. So you can see this base lambda expansion as dynamically defined by using the map 2x mod 1 or lambda x mod 1. And here we have the two intervals i0, i1. And here we have the lack of injectivity exactly here. This is the points that has two possible sequences, either 1, 0, 0, 0, 0, 0 or 0, 1, 1, 1, 1. And everything is exactly the same. You get exactly. And if you put it into here, you will see also if you try to write this as an infinite series, you will get exactly that those two, the distance between these two points is 0 or these give exactly the same sum. That's the same thing that you do in the decimal expansion. If you write lambda equals 10 and you put these two in there and you compute the infinite series, you get the same number. So there's many different ways to see this that I'm writing here. So we're not going to go any deeper into this, but we will actually come back to this map in the algorithm theory course. This will be one of the basic examples and very interesting examples because they will be interested in studying the statistics of points, not just the existence of periodic points and so on, but the statistical properties of points. So this is all this construction. We will not need the construction, but these kind of maps and these kind of ideas will play a big role in the algorithm theory course. It's one of the reasons I'm spending quite a bit of time on this. So let me finish with just one question or remark or just a suggestion for something for you to think about is the question is the following. Well, in all of this that we've done, do we use the fact that this map is full-branch, that these maps are full-branch? So for example, if I take lambda not to be an integer, so the fact that lambda is an integer gives a full-branch map, okay? But if lambda was not an integer, we would still get something very similar and we would still get something like this. If lambda was not an integer, maybe if it's between two and three, then we would get something like this. We still have three intervals, I zero, I two. We still could hope to, there still is some kind of symbolic coding because I can still take a point and look, I can still define the objects that I have there. I can still take an infinite sequence of zero ones and twos and I can define the set, the set of points that has exactly that sequence and it's not difficult for you with the techniques that we've done before. It's quite easy in fact for you to show that you cannot have the two, that in general two points, but you can recover some of the results that we had before, okay? But the question is what fails? That's correct and in this case, we don't have a homomorphism between this and the whole interval, but so what? What exactly does that, what problems does that cause? You know, what fails? Suppose I still wanted to prove a topological conjugacy between this and this. Why not? Okay, so I think it's an interesting, it's a bit of a start question in the sense that I, it's not really, it's a question that I think it's interesting for you to think about to understand better the limits of the construction because always of course when you do a big construction under certain assumptions, you should always ask yourself, are those assumptions really necessary, right? I think this is the way mathematics progresses, one of the way mathematics progresses is by saying wait, we can actually do the same construction in more general situations. So the question is you should always explore the boundaries of the techniques that you have and what exactly you can achieve, okay? And in fact in this case, you cannot, of course, there's problems, but I encourage you to think about this, okay? So I think that's good for today. Thank you.