 Okay, so today is the last lecture of this first part of the course. And if you remember, in the previous lecture we talked about rotations of the circle, right? And how the rotations of the circle have a more interesting dynamics than the dynamics that we saw before, because every point is dense. Now we will study another example with a similar kind of dynamics. It's a very important example, also because it's an abstract setting. And I want to introduce an abstract metric space. This will be an example that is not on the real line or in Rn, but an abstract metric space. So we'll start by defining what I call symbolic spaces. So we start with an alphabet. So we let L greater than or equal to be an integer. And A is a kind of alphabet of L symbols, okay, 0, 1, 2, 3, up to L minus 1. So if L equals 2, then this is just a symbol 0 and 1. And we let sigma L plus is the space of all infinite sequences with AI in this alphabet. Words, infinite words or infinite sequences. Have you seen such spaces before, such sets? Has anyone encountered something like this, no? So let me just ask you a preliminary question. How many, how big is this set, sorry? How big, how big is this set? What is the cardinality of this set? How big is the set, sorry, sorry? 10 power L, where is the 10 coming from? First of all, it's infinite. Is it clear that it's infinite? It's clear that it's infinite because you can have all these different combinations, right? You can have an infinite number of possible signals. Is it countable or uncountable? Uncountable. Uncountable. Anybody thinks it's countable? Think it's countable? Good. Now we have two. We can vote. A is just a set of digits, right? 0, 1, 2, yes. So if you take L equals 10, then these are just the digits from 0 to 9, for example, right? So yes, so it can be useful to think of the decimal representation. In fact, often it can be very useful. When we're going to work with this space a lot, right? So when you try to visualize a space, there's mainly two examples that are most useful, either L equals 2 because then you have just two symbols and things become simpler, or L equals 10 because then you work with 0 to 9 and this is basically just, could be, you could think of this as the decimal representation of a number between 0 and 1, for example, okay? So it's the cardinality of R. It's uncountable. For those of you who know how to prove that the real numbers are uncountable, the diagonal argument, I don't know, then you can do this exactly like this and you get uncountable. You show that this is uncountable, okay? So this is uncountable. In fact, there's a close connection to R. So this is a set. We're now going to introduce a metric on this set. So we define a metric, or we define a function which we need to show as a metric. If you take two sequences a and b, 1 to infinity of ai minus bi. So I will put in the exercise, some simple exercises, such that, for example, to show that this is a metric in this space, to show that with this metric, this space, okay, first of all, let's discuss briefly what this metric looks like. So one of the conditions of the metric is that if the distance between two points is 0, then the points have to be the same. Is this the case? Yeah? Yes. Because if two sequences are the same, then ai equals bi for every i, and this is always 0, and you have the same, okay? Is it clear that this always converges? What is the maximum distance between two points, according to this metric? That's right. Just the sum of 1 over li here, right? Because the max, you can easily find two points that differ in every i, right? So wherever you have 0, for example, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, they differ, so you always get 1. So this is the sum of 1 over li, which is, yeah, 1 minus, or 1 over 1, minus m, okay, you have a specific, so that is the maximum, that is in some sense the diameter of the space, the maximum distance. When are two sequences close in this space? If I give you two sequences, you look at those sequences, how do you know if the distance between them is small, sorry, sorry? What do you mean by converge to each other? So look, wait, wait, wait, I think you're just kind of giving random answers here. So look at the sum, the explicit form of the sum, okay? What is the contribution of each term? This is just a sum, a countable sum of terms. The distance between two is given by this countable sum of terms, okay? Now, right, if the first term is different, okay, then the sum, these are all positive terms, so they all contribute to the metric. If the first is different, then you get 1 over l here, which is relatively big. If the second term is different, the contribution is 1 over l squared, right? If the third term is different, the contribution is 1 over l cubed. So if the first 20 terms are the same, then the maximum distance between those two points is the sum of the tail of that, of 1 over l to the 21 plus 1 over l to the 22 and so on. Whereas for i equal to 1 or 2, then this is a big contribution, right? So really these sequences are close, in some sense it's the opposite of the terms, the faraway terms they do not really matter, okay? The two sequences are close if the initial terms are the same. That's what you can see from the metric. So because this is a very important that we will use it, I will actually write it in a lemma formally, right? So let's write this as a lemma. So it's two parts to this lemma. One is for every epsilon greater than 0, there exists an n depending on epsilon with n epsilon going to infinity as epsilon goes to 0, such that if a and b are in sigma plus l and a i equals b i for all i less than equal to n epsilon, then the distance between a and b is less than epsilon. Do you agree with this? So what this is saying is that I want a criteria on the sequences that's going to guarantee that the distance between the sequences is less than epsilon. And what I'm saying is to guarantee that the distance is less than epsilon, all I need to know is that the first few terms coincide up to a certain n that depends only on epsilon, right? So this is fairly clear. You can calculate this explicitly, right? So there exists an n that depends only on epsilon. If two sequences coincide for the first n terms, then this will be less than equal to epsilon. I will let you do the calculation as an exercise, because it's fairly simple. And the other one is the converse. So if I want if the two are close, so what this is saying is that if they coincide for a large number of terms, then they're close. And now I'm going to say that if they're close, they must coincide for a large number of terms, right? So for all n greater than zero, there exists epsilon n greater than zero, with epsilon n tending to zero as n tends to infinity, such that if the distance between a and b is less than epsilon, then a i equals b i for all i up to n. Does it make sense, yeah? This is saying that the only way that these can be close is if they coincide for a large number of terms. Do you want me to write out the proof or are you happy to do the calculation yourselves? Do you want me to write it out? Yes? OK, I can do it. OK, let's write out the proof. But it's really very simple. So let's prove one. I will prove it because this intuition, once you understand that the intuition is crucial to understanding the topology and the properties of this as a metric space. So let epsilon greater than zero, and then let n epsilon be sufficiently large, such that 1 over l to the n epsilon is less than epsilon. And then I claim that this is good enough, right? So the statement says for every epsilon, there exists an n epsilon, OK, with this property satisfied. So I'm claiming that the n epsilon that works is exactly this one, right? The one, because since l is bigger than 1, 1 over ln epsilon, if n epsilon is big enough, this will be smaller than epsilon, whatever that epsilon is. So then I claim that this is it, and then we just have to check. Then the distance, the distance between a and b is equal to the sum i equals 1 to infinity of a i minus b i over l i. And this, what is this equal to? We assume that a i equals b i up to time n epsilon, right? So this is just equal to the sum i equal n epsilon plus 1 to infinity of a i minus b i over l i. You agree? Sorry, and this tends to zero, yeah, so it's clear. This is less than or equal to the sum of 1 over l i, i equals n epsilon plus 1 infinity, right? And in fact, the sum is, so this is actually equal to 1 over ln epsilon, and this is less than epsilon. So this is the tail, this is what I was saying about the tail of the sum. So the initial terms, they contribute a lot. If the initial terms coincide, then the distance becomes smaller and smaller because you just get this sum. And because this sum has l i, right, if you start from a large i, this will be very small, as small as you want this tail. Sorry, sorry, sorry, you're right, you're right, you're right. I was just, I was thinking of, of, you're right. So I need to change this a little bit, you're right. This is l, so this becomes l, ln epsilon minus 1, so I guess I want minus 1 here. Yes, thank you, that's it. I always think of the case 0 and 1 of the case, anyway. Okay, so the other direction is basically the same kind of calculation. So given n, let n greater than 0, and now define at n, and epsilon n, we define as equal to l to the minus n. Okay, so suppose the distance between a, b is less than epsilon n, okay, so this should be n. Then we need to show that it coincides, but it's obvious that this cannot be the case, right, because suppose by contradiction, suppose by contradiction that there exists some, so that there exists some j between 1 and n, okay, such that a, j is different from b, j. And then, of course, then a, j minus b, j is greater than or equal to 1, and so the distance between a bar and b bar, which is equal to the sum, i equals to 1 from infinity over a i minus bi over li, is greater than or equal to 1 over lj, right, is greater than or equal to a j minus bj over lj, which is greater than or equal to 1 over lj, which is greater than or equal to, which is greater than or equal to 1 over l to the n, which is equal to epsilon n, which is a contradiction. This induces a topology, this metric induces a topology in this space, and with this topology you can check, and I leave it as an exercise this, because we will not use it in a crucial way, but I think it's an important exercise. It shows that this set has certain topological properties, for example, it is compact, in the same way that the unit interval is compact, in the sense you take a sequence that converges in this metric, and it converges to some element, sequentially compact, right? It is, is it connected? You can check whether it's connected, but it's not connected. It is, in fact, totally disconnected, so the only connected components of this are points, and it is, it has no isolated points, so every point is accumulated by other points. In other words, it's a Cantor set. Have you done Cantor sets? No, Cantor sets, yes. So this is a Cantor set with this method. Exactly, so also in the exercise I think it's very interesting to try to understand what is the difference between the decimal expansion, between the Euclidean metric and this metric here. If you think of this as decimal expansion, and you think of these on the real line, and you think of these as the, this is not the metric that you have on the real line. It's a different kind of metric. It has the absolute value outside the sum, and not inside the sum. And I put this as an exercise, I don't want to spend time on this, too much time on this now, because it's something to think about at home. If you take these absolute values outside the sum, then you find that this is not a metric on the space. Why is it not a metric on the space? Because you can have two points, two sequences that have distance zero, and they are different sequences. For example, the sequence 0, 0, 0, 0, 0, the sequence 1, 0, 0, 0, 0, 0, and 0, 9, 9, 9, 9, 9, 9, 9. Because you get a sign here, right? So this can be positive or negative. So you get 1 over L minus the rest of the sum, which is exactly 1 over L, and you can get 0 here. And that is why the decimal representation of numbers is not unique. But as I said, I don't want to go into this. This is a different, there is a relation between this and the decimal representation, but at the moment it's important to think of them as distinct. It's a different topology. As you said, the topology you get induced by this is the same as the product topology on this alpha. But you can write this as a product A to the N, if you want, as a product space from a topological point of view. Excuse me? Can I open the topology? Exactly, yes, yes. So this, you can take the discrete topology on this set of discrete points. If you take the product topology, you get the same topology that is induced by this metric. Okay, just. But let's not spend too much time because what we want is to look at dynamics on this metric space. We have a metric space. We want to introduce a map. There are several maps you can introduce, but there are two that are particularly important. Okay, one of them I will introduce now and the other one I will introduce in the second part of the course. So, and they have very different properties. So the one that I want to introduce now is called the adding machine. So, define a map from sigma L plus to sigma L plus by add one and carry. And I will explain what this means. So, basically you take a sequence. It's like in the car, when you count the miles or the kilometers, you know, and you add the kilometers. And then when you get to the end of the cycle, you get to nine. It goes back to zero and you carry one to the other side. Okay, this is also sometimes called the odometer. This is called the adding machine. Okay, tau is called adding machine or sometimes it's called odometer. Odometer is the thing in the car that counts the kilometers that you've done. So, I will describe precisely what it is. So, you add one, you keep adding one here and then when you get to L minus one, then the next you add, this goes back to zero and you add one here and then you just keep going like that. So, let me define it more precisely, how it's defined. So, define in the following way. So, what we want to define is the image of some sequence x1, x2, x3. This is one sequence. This is equal to y1, y2, y3, okay? And I want to say how I define the image of this sequence here. So, we'll consider several cases. So, first of all, suppose that xi equals L minus one for all i. So, this corresponds, if we think again in the decimal representation, which is more intuitive, that there are all nines. You have an infinite string of nines. What do you get when you add one to the infinite string of nines? You add one and you carry all the way down. You get zeros, okay? That's the natural thing, you just add zeros. Okay, then yi equals zero for all i greater than or equal to one. Okay, so I'm just saying, if these are all nines, all L minus ones, then by definition I'm defining this map now. The image is all zeros, okay? Now I'm going to, so if otherwise let i zero of x bar, so of this sequence here, be equal to the minimum i less than equal to one, greater than or equal to one, such that xi is different from L minus one. So, in this case, this is not defined, right? Because it is always L minus one. But if it's not, then there will be some time when it's not. So you have 9, 9, 9, 9, 5 at some point, okay? I look at the first time that it's not a 9, right? And then what am I going to do? How is it going to do? So what do I do? I add one to the nine and what happens? This becomes zero, right? And you carry one. You carry one, but that is a nine also. So I add one, it becomes zero, and I carry. And this becomes nine, it goes from nine, becomes zero, and I carry. And I keep carrying until I get to the five, and then I add one, and it becomes a six, and you stop there, okay? This is the system. So I'm going to write it formally. So if I of x, zero of x equals one, what does that mean? That means that this very first term is not a nine, okay? In words, I'm always going to use nine because it's easy to think of it. But it's not L minus one. And then what do I do? I just add one to this and finish. I just add one. I don't need to carry anything because it's not a nine. So then let y one equals xi plus x one plus one. And y i equals xi for all i greater than or equal to two, okay? So the image of this point, if this is not a nine, then the image of this point is just going to be add one kilometer here. And so you'll get y one is equal to x one plus one. But the others all stay the same because we didn't carry anything. If it's bigger, we just do the same thing. We carry until we hit this, right? So if i zero of x is greater than one, then we let y i equals zero for all i equals one all the way to i zero. All the way to i zero minus one, I guess this should be. Because at the moment i zero, you are no longer, you are not. So this should be i zero minus one, okay? And then we let y of i zero equals x of i zero plus one, okay? And y i equals xi for all i strictly greater than i zero. So this is the formal definition of what I, any questions? So it's not immediately, it takes a little bit of time to get used to it. But just think of the, of the adding. It's just, it's like an infinite kilometers. And instead of starting from the right, as we do here at least in Italy, you start the numbers from the left and these are the units and these are the tens and these are the hundreds or whatever, something like this. And you just keep adding one, and you just keep adding. This is the way to think about it. So this is a well-defined map because I have defined exactly what the image is. For any sequence, I have explained exactly how to define the image of this sequence and it turns out to be very consistent with this topology induced by this metric and in fact we can show that so lemma tau sigma plus l sigma plus l is a homomorphism. So okay, let me just sketch the steps of the proof here, okay? Even though again it's just a question of using these definitions to check. But so how do we show that h is injective? Why is h injective? Sorry, sorry, tau, tau, tau, tau is injective, yes. So it really just follows by looking properly at the definition, right? If you have two different points, okay, if x is different from x bar pi, what does it mean that two sequences are different? It means that there exists some point at which they're different, right? And then there exists some i such that xi is different from yi and so this implies. And so by the definition of tau, we have that the image, sorry, xi pi. And by the definition of pi, we have that yi is different from yi prime, where these are the images of these two sequences of course on the tau. It's just clear that you, if you have two points just from the definition that I equal up to some point and then here is different and then the images clearly will be different. Yes, yes, so in fact to show, yes, in fact to show the surjectivity that's what I say, we just defined the inverse map explicitly and I guess the fact that that is well defined gives the injectivity at the same time, yes. So to show that tau is surjective, we can define tau minus 1 explicitly, okay? And maybe I will leave that as an exercise to do that because that way you can think but it's very straightforward, it's just, okay, exercise. And you're right, I guess defining this inverse function also gives the injectivity, it gives the bijectivity at the same time. Continuity, okay, so continuity is a little bit, t is continuous, so okay, so what does it mean to, t is continuous it means that we take two nearby points that images are nearby, right? So we take x minus x prime close by less than delta, okay, then there exists some i depending on delta such that xi equals xi prime for all i less than equal to i delta, remember? This is just what we did before, right? The fact that two are close means that they must coincide for a certain number of iterations, okay? This is just a definition of close. And so letting tau i, tau x, so then if tau x bar equals y bar and tau x bar prime equals y prime, this implies that y i equals y i prime for all i less than equal to i delta, okay? By the definition of tau, if the two sequences are the same for a number of iterations, when you add you get also the images are the same for all these iterations. And therefore this implies that the distance between y bar and y bar prime is less than equal to some epsilon that you can make for any epsilon. As long as you choose i delta large enough then this will be less than epsilon and as long as you choose delta small enough here you will get i delta large enough here. So this is the continuity of h and continuity of h inverse is done in exactly the same way. So h minus one continuous same argument using the definition of h minus one, okay? So this proves that h is continuous, h minus one is continuous and this is a whole memo. Question? Sorry? H tau, what's the difference? So it's interesting that we have such an abstract metric space. We define a map in certain way fairly natural and this is actually a homomorphism of a space. So we have a homomorphism of this cantor set and we would like to study the dynamics of this homomorphism. So take a point and apply it, apply this over and over again iterated, right? This is what it means to study the dynamics of this set of this map. So what are we going to do? So I said that this was an example of a minimal dynamics, minimal homomorphism and indeed that is what we are going to show. So lemma, this is a proposition. So the dynamics of tau is minimal i.e. the omega limit of x bar is the whole space for all x bar. So every orbit is dense in this cantor set, in this metric space. So you do not get any, like all, this is an uncountable set of points. It's a cantor set and they somehow each point, now just a basic question to make sure you understand the subtle difference. Could it be that the orbit, the forward orbit touches every point in the space? When I say the omega limit set is equal to the space, what's the difference between saying this and saying that the orbit, wait, wait, let's, bar, is not the same? Why? Yes? Yes? Right. Okay. What did you want to say? Same thing? Yes. The definition is not the same. But would it be possible to have this? Could we hope to prove this maybe in some cases? Why not? It's countable, right, exactly, very important difference. This is a countable set, the orbit, whereas this is not a countable set, right? So there is this other very big difference. So when the orbit is dense, it's a dense subset of your space, so the omega limit is your whole space, but remember it's just a countable dense subset, right? So when you take a dense orbit, you cannot, it's very different from saying that the orbit touches every point. The same in the circle rotation. When we have the rotations of the circle, the points are dense, but they do not touch every point on the circle, because the orbit of a point is a countable set. It's just dense in the circle, okay? Okay. So how do we prove this? This is really just literally taking a point and iterating this point and seeing what happens. So proof, how do we show that the omega limit is everything? The same as we did for the circles, in the sense that you need to show that it comes epsilon close to any other point in your space for any epsilon, right? So let X bar, okay, let X bar be in sigma L plus. We need to show that for all Z bar in sigma plus L and for all epsilon greater than 0, there exists some n, there exists n greater than 0 to 1, such that the distance between tau n of X bar and Z is less than epsilon. This is a dense orbit because we fix the X bar. Now we look at the orbit, which is this Tn of X bar, and we show that given any epsilon and any other arbitrary point, at some moment we will come within epsilon of that point. And that clearly shows that this point Z is in the omega limit, right? Because you can take smaller epsilon and then I can take just a bigger n and eventually I will come within that small epsilon of Z bar and so on. That's exactly what we're doing, right? Because you can take a sequence of epsilon is going to 0 and then you take a corresponding sequence of n's going to infinity, such that the orbit, so you have a subsequence of iterates that is converging to the point Z, right? So that's exactly what we're doing here, we're showing that there's a subsequence of iterates that converges. That is the definition of omega limit, right? But that's exactly what we're doing because we're taking epsilon arbitrary. So how do we look at, how do we study the iterates of the point X? How do we show this? It's not that easy because it doesn't look that easy because this can be any point, okay? And we don't know anything about this point, this also can be any point, okay? What we do know, however, is what it means to be close. This is what we have to use in a crucial way. I knew. What does it mean to be close for these two sequences to be smaller than epsilon? Yes. That is the definition in a metric, but in terms of the sequences, so these are specific sequences. We have a specific metric which we've studied in this sequence. So if I tell you what Z is, if I give you the sequence and I give you this sequence here, how do you know if these sequences are smaller than epsilon? Exactly. Right? So what you want from these sequences is the first 2,000 digits are the same or whatever. A number that depends on epsilon. The smaller epsilon, the more number of digits are the same. Remember we proved this lemma, this is the fundamental thing that you must keep in mind. Two sequences are close if the digits, the first digits coincide, right? Because when you look at the metric, you sum the difference between the digits over L to the I and if a lot of them are the same, then they're close. The two sequences are close. So this is the only element we have to go on to try to show this, right? So we need to somehow show Z is arbitrary. We're iterating this. So what we're looking for is an n that will make this close. So once you fix an epsilon, you also fix the number of digits that have to coincide for this to be true. So if you fix epsilon, you say, okay, the only way this is true is if the first 2,000,000 digits are the same. So the question is, can you find an n such that the first 2,000,000 digits of this sequence are the same as the first 2,000,000 digits of this? This is the question, okay? So let's try to do that. Suppose first, wait, let me write it. Okay, so first let me say, write down what I just said, which is that to show this, it is sufficient. So need to show this. So it is sufficient to show that first n epsilon digits of tau n x bar coincide with the first n epsilon digits of z bar, okay, where n epsilon is sufficiently large, depending only on epsilon. So once you fix the epsilon, there exists an n epsilon such that if this is true, if the first n epsilon digits coincide, then you have this property. Now, what looks difficult is that z bar is arbitrary, right? This has to be true for all z bar. So what we're saying is that you must allow for the first n epsilon digits of z bar to be anything, which means that we must be able to find an n that will give that the first n epsilon digits of this are any possible combination that we want, okay? Any combination that you give me, I must find an n so that the first n epsilon digits are those, okay? So in some sense, we're saying that the units of this contain as an initial block of digits any possible combination, okay? That's what we need to show. So let's suppose first that, let's prove it first, suppose first that x bar is equal to zero, the sequence zero. Just as an example, now we will prove it for x bar equals zero, and then we will prove it for every other points. So what is the image of zero under tau? So we have 0, 0, 0, 0, 0, 0. We apply tau, what do we get? 1, 0, 0, 0, okay? We apply tau again, what do we get? 2, 2, 0, 0, 0, okay? We apply tau several times until we get L minus 1, 0, 0, 0, 0, right? Then we apply tau again, what do we get? L minus 1 has completed a cycle, it's like a 9, it goes to 0, and here we get 1, right? So 0, 1, 0, 0, 0, then we apply tau again, what do we get now? Okay, you start to see the pattern. We get 1, 1, 0, 0, 0, right? And then we apply tau again, we get 2, 1, 0, 0, 0, and then we keep applying it until we get L minus 1, 1, 0, 0, 0. Then, one more cycle, so we now apply add 1 again and we get 0, 2, 0, 0, 0, okay? You start seeing now how it proceeds, then we apply tau again. We will get 1, 2, 0, 0, 0, 2, 2, 0, 0, 0. All the way to L minus 1, L minus 1, 0, 0, 0. And then, okay, now here we need to carry twice, right? Because we add 1, this becomes 0. We carry, we add 1, this becomes 0, and we carry. So this becomes 0, 0, 1, 0, 0, 0. 1, 0, 1, 0, 0, 0, 2, 0, 1, 0, 0, 0, 0, and so on, and so forth, okay? So the observation is the following. First of all, we get all the possible values. Sorry, 2, no, no, sorry, I jumped some step here because here you always have 2. So here I have 2, well, I just went ahead, right? So this really should be 2, 0, 0, 0, 0, sorry. And then you do the same thing many times, right? Until you end up with, so this is L minus 1, 3, and so on until you get L minus 1, L minus 1, I had jumped a few steps here, right? And then you do, and then you get the situation where I, that I described before. So you get 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, and so on, and so forth. So now the simple observation is that we get every possible finite block in the beginning if we wait long enough, right? Because look, at the beginning for the first, if you just look at the, if you suppose that n epsilon is equal to 1, okay? So you just look at the first term, you get all possibilities. You get 0, you get 1, you get 2, up to L minus 1. Now look at the blocks of length 2. What are the possibilities? Well, you have 0, 1, 1, 1, 2, 1, all the way to L minus 1, 1, 0, 2, 1, 2, 2, 2, L minus 1, 2, and so on, up to L minus 1, L minus 1. It's just like the decimal digits, right? You have a 10, 11, 12, and so on, all the way to 99. You have a 0, 1, you have all possible blocks of two digits. And then you can see that I'm going to continue, and I'm going to get, now for many iterates, nothing is going to happen. Everything is going to happen only to the first three iterates, right? All the way, and then at some point I'm going to get L minus 1, L minus 1, L minus 1, 0, 0, 0, right? But before I get to this stage, I have to cycle through all possible combinations. On this slide, I'm going to go through 0, 0, 1, 1, 0, 1, 2, 0, 1, and so on. Then when I carry, I will start again with 0, 1, 1, and so I will carry. And then I will keep going, and I cycle through them, and I end up with all the possible blocks with 3. And then when I carry this time, I will get 0, 0, 0, 1. And then I will go on for a long time, where I will go through all the possible combinations of 4, of 4 digits, and so on and so forth, right? So if I say, okay, I fix a Z, and I fix a small epsilon. So I want some, such that the iterate of 0 coincides with Z for the first 1,000 steps is not a problem. I can see that if I wait long enough, I'm just going to cycle through all the combinations. Eventually, I will start cycling through the 1 million, 1 million digits. And if I wait long enough at some point, I will clearly get exactly that combination, okay? This is really just the observation. As soon as you observe that this is what you get, then you have your result. So in this case, so yes, that's a good question, yes. It's related to the comment I made before. You tell me, will it be the whole space or not? Is there any sequence that is left out from the orbit? That's a good question, actually. What are the sequences that would not be included? For example, yes, so these sequences always finish in zeros. All the sequences you get here all terminate in an infinite number of zeros. So really, you are literally approximating by countable set. This is a little bit like just a collection of all the finite words. So the set of all finite words is countable. And what you've got here in this sequence of orbits is you've got just every single finite word followed by all infinite zeros. And that is what this orbit is. It's a countable set of points, finished with all zeros. So how do we generalize this to a generic point? So now, so do I need to say anything more? So clearly, so the orbit, let me just say the orbit of zero and the tau cycles through all possible finite words as the initial digits. Thus, we have that omega of zero is equal to the whole space, okay? Because of what we just said here. Now, how do we do it for general x bar? So we have x bar equals x1, x2. There's a rotation of what? Rotation of zero? Yes, yes, I think that would work, that would work. That's a very good observation. So you need to add the fact that tau is an isometry which we have not discussed. So in fact, tau preserves the distance between points. So if you take two points at a certain distance and you apply tau to them both, the distance stays the same in this case. Because if the terms are the same up to a certain moment, then they're the same. Yes, that's right. So this would be a very good proof. I will give a slightly different argument here, but I think that would be just fine. So what Sahel is saying is that since this orbit here is dense, okay? Then for any epsilon, you can wait long enough until it comes close to the point x. And then they will stay together for the whole time. So the density of this basically implies the density of that. Because for arbitrarily small epsilon, when you come close, you take it with you and you start wrapping it around. So that is absolutely fine, okay? That would be a good argument. Another argument which is in the same style of this, which is kind of just as simple, is that you can see that if you just iterate this directly, which is basically the same thing, you will end up, you can end up with a whole bunch of zeros at the beginning, right? So iterating, so for any n, for any n epsilon, let me write this for any k, for any k greater than or equal to 1, there exists some n sufficiently large such that tau n of x bar equals y1, y2, y3 and so on. And yi equals zero for all i less than or equal to k. So what I'm saying is that if you start with this, you can make these zeros, right? Because what do you do? You start, so you start with x1, x2, x3, x4 and so on, right? And then you apply tau, and after some point this will become zero. So you apply tau a certain number of times, and then you will get zero, and here you will get x2 plus one. So before that, at some point this will become l minus one because you keep adding here, right? So at some point, this will become l minus one, x2, x3 and so on, right? Then at the next iterate, this will become zero, and this will become x2 plus one, x3, x4 and so on, right? Then now you keep cycling. Then as you add one, this cycles through another cycle here, and this will become x plus two, and then you cycle through another cycle, and this will become x plus three, right? And at some point, this will become zero, l minus one, x3, x4, x5 and so on. And then you iterate one more time and you will get zero, zero, and here you get x3 plus one, x4 and so on. And then you keep going, right? And then now you cycle through 100 of these steps, and then you add one to here. And then you cycle through more, and then you get zero, zero, and then you add one to here. And then at some point, this will reach l minus one. And then you add one more, and you end up, you get l minus one, l minus one. Sorry, this should be, yeah, you get this. And then at some, maybe there's one more intermediate step. So at some point, this will become l minus one, l minus one, x3, x4, x5. And then you add one, and then this becomes zero, zero, x3 plus one, x4, and so on. And then at some point, this one also becomes zero, and so on. So if you wait long enough, you can get that the first one million digits is zero. And if one million digits is all you need, then you have exactly this situation here. If the first one million digits is zero, then you start from this point, and you finish, and you get the sequence that you want from these zeros, okay? I think your proof was probably more elegant. This proof that I used here is the famous mathematician's proof about how you boil water. You know these mathematicians how they boil water in a pot? Because in Italy, we make pasta all the time, right? So if you want to boil water, what do you got? You go in the kitchen, and you open your cupboard, and you take out the pot, and you fill it with water, and then you put it on the stove, and you put it on the fire, right? So the next day, mathematician comes in and finds the pot already on the table full of water, and they say, so what do I do now? Or I know, I throw the water out, I put it back in the cupboard, and now I'm in the same situation as I was yesterday, and I know what to do. So this is a danger here. But anyway, this is a similar strategy, right? So we go a little bit, we take this general system and we reduce this general point, and we reduce it to one that has sufficiently many zeros, so that then we can apply exactly the same argument, and we get any possible sequence, okay? So this also shows, this completes the proof, because it shows that this point also has a dense orbit, and so every single point has a dense orbit, okay? So, I think this is all I wanted to say about the adding machine. This is not the only map that occurs on this symbolic space. That in the next term, we will look at a different map that has completely different properties on this symbolic space. In fact, in the next term, we will look at some other dynamical systems. We will continue our study of topological conjugacy, but we will look at some other dynamical systems that have different properties, including a mixture of having fixed and periodic orbits and having dense orbits in the same system, okay? These are systems in which you have a more rich and interesting dynamical behavior, even on the one. Okay, so I think we can finish for today. Thank you very much.