 Good afternoon So let us remember what we were studying where we finished last week So we are looking at the family of tent maps and in particular at the tent map with slope 3 and We have studied already what happens outside the unit intervals 0 1 everything goes to minus infinity So we are concentrating on this interval 0 1 So this point here is the point one-third This point here is the point two-thirds We have these two closed intervals Zero one-third this interval here. We call I zero This closed interval here two-thirds one. We call I one and This interval here. We call Delta and we define the set lambda Equals the set of points x Let's stay inside this interval inside one of these two intervals for all time If we understand the dynamics on this set this set is invariant by definition number forward invariant in the sense that if x belongs to This set obviously the image f of x also belongs to this set So there is a dynamics on this set a Priority just looking at the map. We only know that there are these two fixed points in this set But the theorem we're going to show is that in fact lambda is a canto set In particular, it's a big set. It has a lot of points in here and the dynamics on lambda is non-trivial and the set of periodic points is Is Countable is dense What did I under this is dense and F restricted to lambda is transitive Which means the exists a point x in lambda? Such that The omega limits set of x equals lambda. This is the definition of what we mean by transitive So in particular not all points are periodic, so how are we going to prove this thing first of all we are going to Consider the set. Okay, so this is the statement In fact as in the course of the proof we will prove Additional information about this structure. So let's sigma 2 plus is equal to the space of all sequences of Zero and one so this is the sequences a Given by a zero a one a two and so on where each AI is Either zero or one So what we're doing is we're going to implement the strategy of symbolic coding that I described in the previous lecture In the previous lecture I described in abstract What you do if you just have a space a dynamical system on the space and you partition the space So here it's a slight generalization of that We don't really have a partition of the space But we know that all the dynamics we're interested occurs inside these two Elements and so these two elements are a cover of the space any element in the space in the said lambda Always for every it belongs to I see why one so we can use these two as the elements to code the dynamics So we introduce this space of all sequences of zeros and ones that correspond to these two and we define for a in the sequence let I a bar Equals a set of points x such that f n of x belongs to I a I Greater than equal to zero So I remember when I wrote it down I wrote it down in a very abstract setting So now we can think very concretely what this means so this means the following I You give me a sequence of zeros and ones an infinite sequence of zeros and ones and I say okay Let me consider the set of points whose combinatorics in terms of these two intervals has exactly that sequence So if your sequence is zero one one one one zero then I say is there a point that starts in I zero It's images and I one it's second images and I one it's third images and I one for however many ones I have in the sequence and then when there's a zero it's Kate images in I zero and so on so the full forward orbit Follows exactly the assa plea assigned combinatorics So as we said we know for example that if you just take the sequence zero zero zero zero zero Then which point will satisfy that The fixed point is always in I zero if I give you the fixed point one one one one one Then this other fixed point. There's another fixed point here What if I give you the sequence one zero zero zero zero zero So the orbit of one that's right the orbit of one If you look at the image of the point one is in I one and its image is in Isis and then it stays in Isis all the time So can we find some other points suppose? I give you the Sequence one one zero zero zero zero zero What two of a thing One of a thing would be what what's the sequence of this? Yes, but what's the sequence of this point? No, because it starts with the position of the point itself right when n is equal to zero when when I is equal to zero then this is the Interval in which x belongs All right, so this point here has a what sequence zero and then he maps to one and then zero zero zero Okay, so you can see you can start constructing some sequence this point here has a sequence one because it belongs to I one Then the next it that is one so it's one one zero zero zero zero Okay, here is the fixed point this fixed point has the sequence one one one one one which is the sequence zero one one one one one The pre-image of this fixed point right so there's a point here There's some point here and that has the starts with zero and then its image falls exactly here on the fixed point So this code is zero one one one one one Okay, so here we're taking specific points, but what I'm saying which is quite remarkable is that for any sequence Well, I'm not yet saying it I will say that for the moment I've defined this object as I said now We want to ask whether this object is non-empty. So is it true? that for any sequence of Zero ones I can give you twenty seven zeros followed by 200 ones followed by three million zeros followed by one one any sequence I give you there exists a point that does exactly that Okay, this is the question and this is the answer So proposition for all a in a bar Sorry for all a in sigma two plus We have that I a bar is non-empty. Yes. Sorry. Sorry. Yes I and Consists of a single point. We will also then prove as a separate lemma That two different sequences correspond to do different points, right? So there's in the end we'll get a bijection between that. So this Because we've already shown that we get a conjugacy We're in the process soon of showing that there's a conjugacy between the shift map on two symbols and the set lambda So, how do we show this? Well, let's fix some set a So like a bar equals a zero a one a two and so on This is in sigma two plus and let us define finite sets a zero To a n minus one To be equal to the set of points x Such that if I Minus one, so what does this mean? It means we just look at finite block For example, if we just take One index here if we take n equals one Right, then a I a zero is the set of points that belongs to I a zero so if You take n equals one for the two different possibilities for the first digit. You just get I zero and I one So it follows obviously from the definition that these sets are nested All right notice that I a zero N minus one is contained in I a zero to a n minus two Okay, which is contained in I a zero which is contained in the unit interval I because This is the set of points in which this holds up to time n minus two This is the set of points in which the same thing holds for the same sequence here plus you've got an extra condition So these are clearly nested sequences and more over just from the definition. We clearly get that a I a bar is simply equal to the nested intersection all n greater than equal to one of I a zero because Just from the definition Because this is the set of points if the set belongs to I a it must belongs to this I a n minus one So we've got a nested sequence of sets Again, they could be empty right? It could be that there's a certain it could be that this is not empty But then if you look at the set of points in here whose image you add another term and here whose image belongs to And suddenly there's no points that actually have that combinatorics. So it could become empty All right, so we need to show that this is non-empty and However, so what is sufficient if we can show that each of these is a closed Non-empty set then this intersection is Non-empty because the nested intersection of closed non-empty sets is non-empty So so it is sufficient to show each and greater than equal to one to a minus one is closed non-empty Interval so there's a general results in topology probably you've done this that in general topological spaces nested intersection of closed Non-empty sets is closed and non-empty in the case of intervals. You can prove it Directly, okay I will leave it to you as an exercise if you have a closed sequence of non-empty intervals Then you can just take the end points on one side and the other end points on the other side and can never cross so you can show Easily this fact for intervals. Okay, but I will leave this an exercise So how are we going to show that this is true? We need to show that this is true for an arbitrary a right because we fixed a arbitrary There's no condition so by showing this for arbitrary a will show this that it will show that it's true for every a and Therefore we will have showed the first part of the result Yeah, so we will prove it by induction. So what is the first step of the induction? n equals 0 for n equals 0 Sorry for n equals 1 like this is not for n n equals 1 For n equals 1 what do we have? for n equals 1 We have I of a 0 is either so I of a 0 Equals I 0 if a 0 equals 0 or I of a 1 equals I 1 If a 0 equals 1 right from the definition because for n equals 0 This is just the set of points X such that X belongs to I of a 0 or sorry I 0 or I 1 depending on these values So this is clearly non-empty interval because it's I 0 or I 1 depending on what the value of a 0 is yes Sorry, sorry, but there's one crucial property. I mean so far We've we're not going to be able to base the induction on this because this is just the observation of what these are So we're going to use the extra ingredient in the induction is the following. What are the images of I a 0? Moreover We have that f from I 0 to I and F from I 1 to I Are bijections in fact homomorphism. Let's say in fact in this case. They're linear Okay, so I 0 maps to all of 0 1 I is equal to 0 1 Remember I is the whole interval So now we can state the induction assumption Active assumption But you can already see how the inductive assumption is going to go because the fact that I 0 Maps to all of I Means we can conclude something about the next step right about the intervals with two With two terms a 0 and I 1 because inside I 0 there is going to be one sub interval That maps under F2 I 0 and another sub interval that maps under F2 I 1 and inside I 1 There's one interval that maps to I 1 and another interval here that maps to I 0 So there's going to be exactly for the intervals I 0 0 I 0 1 I 1 1 and I 1 0 here so that shows that if you look at the the Elements of order 2 you have four possibilities and or each of those possibilities is a closed interval is a non-empty closed interval Which is what we're trying to show so let's suppose inductively suppose inductively for n Greater than equal to 1 There exists exactly 2 to the n Disjoint closed intervals I a 0 a n minus 1 Corresponding to all possible finite sequences a 0 a n minus 1 of length n with a I Belonging to 0 or 1 for I equals 0 Up to n minus 1 suppose moreover Fn from I 0 n minus 1 to I Is a homeomorphism Tell me if there's something you don't understand here of length n Because there's nth terms here right a 0 to n minus 1 so there's n Possibilities choices of zeros and ones here. So there's 2 to the n possibilities It's simply all 2 to the n possibilities of of zeros and ones of n of a sequence of n zeros and ones So we need to show So that the a that it we're proving the theme is an arbitrary a so even though it's fixed It can be any a right so we're going to show in the proof We're going to show all together that for any a this will work Right and so I am stating I'm proving something a little bit more general what I'm stating at this set of Set of at this stage of the induction is that for any finite sequence of length n There exists a closed interval that satisfies this condition In other words at stage n of this construction there exists exactly 2 to the n such intervals each one of these of this form But all of them appear for any Any combination of zeros and ones? Okay, and for each one we have the property that fn maps this interval homeomorphically In fact in this case it will be linearly, but homeomorphically to the whole interval zero one What I just wrote before was exactly the step zero of this induction right so what I wrote before was n equals One for n equals one There exists two this joint intervals I a of the form I a zero Corresponding to all possible sequences a zero a and one of length n so of length one with a zero in Zero one so I had I zero and I one those were the only two possibilities At the zero set and I showed that for n equals one I have I of a zero maps to I and I of a one maps to I that's exactly The zero step of the induction of this So what do I need to show now remember that I'm trying to show that given? One maybe this was a little question really just to remind ourselves What I'm trying to show is that given one infinite sequence if you take the nested finite terms they're all nested sequence of closed intervals, right and I am trying to show that for the for any sequence you have here It's a it's a closed interval the nest and this comes directly from the definition So I just need to show that each one of these is a non-empty closed interval. That's all I need to show So how do I show this? Can you tell me this is now the next step is kind of immediate almost I? Need to show that the same is true for n plus one So I need to show that this holds for n plus one So I need to show that there exists to n plus one this joint closed intervals of the form a zero a n minus one a n Corresponding to all possible finite sequence with the next term here, okay? So what do I need to try just need to show that for each one of these? I have two more possibilities one with the zero and one with the one here, and then I have all the possibilities Right, and this is obvious that this is true by this property here This maps homomorphically into I and I contains I zero and I one so that means like inside this interval here There are two subintervals that under these homomorphisms get mapped to I zero and to I one And that gives me the two subintervals that I need so I didn't understand your question Yes Yes, yes, it took two closed intervals The next step we subdivide each of these intervals into two subintervals. Yes Okay, so let me write down Let me write down the the the argument here in the general case And then we will check how it works in the in the first case Okay, just to to clarify but in the first case is just easy. So now we have Okay, so Since so we use the fact that f n I a zero to I n minus one to I Is a homomorphism then Clearly I a zero an minus one zero Which is equal to the set of points x in I a zero to I n minus one such that f n of x belongs to I zero and I a zero a n minus one one Which is equal to x in I a zero a n minus one such that f n x belongs to I one are non-empty closed intervals because I zero and I one Okay, because I zero and I one disjoint closed Subintervals So let's see how it works In the first case, right for n equals one We have I zero and I one Are the only two are the Two to the one the two disjoint intervals of the form I a zero Okay, they map homomorphically onto zero one But what is zero one zero one again? Remember this zero one on the vertical axis is only the presentation of the real zero one I say I repeat this every time right the dynamics is happening here on the horizontal axis The graph is just a way to help us look at the dynamics, right? What is really happening here is that this interval zero one third is mapped onto this interval zero one And this interval I one is mapped as you can see from the graph onto all of zero one Okay, so if you write a make a copy of I zero here And you make a copy of I one here You see that the homomorphism that maps I zero to all of I Means that there exists a smaller interval here That maps to I zero homomorphically and another smaller interval another small interval here That maps to I one homomorphically right from the dynamics like this And by definition, I will call this interval here I zero coming from the fact that this point in its I zero And zero coming from the fact that the image of these points lands in I zero And I will call this interval I Zero one because the first zero corresponds to n equals zero the fact that f of x belong that f of x belongs to From the definition that that x belongs to I zero and one here because f of x belongs to I one Right and similarly here this is one. I have One interval here What's the coding for this interval here? One one exactly and then I have one interval here Which is I one zero And this is really in this particular case just exactly what I've done here So I've used that's why I emphasize that this is the key ingredient in the induction really is this fact here Is the fact that we have Because this maps to this homomorphically then there is a sub interval of this which I denote like this Which is the set of points inside here which map into I zero Just like we did here and this is exactly By definition this is exactly the same definition as previously I had right because the fact that x is not here In here means that x belongs to I a zero f of x belongs to I a one and so on and here I'm adding the last term which is the fact that f of n of x belongs to I zero And I one so I get two Okay, so for each one of these two to the n disjoint intervals I get each one splits into two disjoint closed sub intervals which have zero and one here So if I do this for each one of these I get exactly two to the n plus one Disjoint sub intervals satisfying all the properties Okay, so this shows that for any finite sequence This interval here Is a non-empty closed interval and from what I thought before the infinite The interval given by the infinite sequence is just a nested intersection of all these finite intervals And therefore each of these is a closed interval and therefore the intersection is non-empty And that completes the prove that intersection Does that make sense? Yeah So now we need to show that it's a single point How do we know that they may not be They may be more than a single point Can is it possible that a nested intersection of closed intervals is more than a single point? How is that? Give me an example of an infinite nested intersection of closed intervals as more than a single point No closed closed intervals Compact intervals so you can take jn equals for example the interval zero one plus one over n Okay Or you can take even one over n One plus one over n anyway So it's not obvious. So how are we going to show that this nested sequence of closed intervals? Is a single point That's right. All we need to do is to estimate the diameter of these intervals Right because we know that this nested sequence is connected because this is a Sequence of nested intervals. So we know the nested intersection is just one connected piece. It's an interval itself If it's a single point that corresponds to being a trivial Interval so we just need to show that the length of these intervals goes to zero And why does it go to zero? So to show That a a bar equals nested intersection of a zero a n minus one Is a single point It is sufficient to show That the length of a zero a n minus one Goes to zero As n And how do we know the length of this interval here? Just what? What power series? Why is that? So i zero is one third i zero zero is one over nine Why is that? Because these are exactly One third one third one third and then this maps linearly to the whole interval And therefore these three bits are also of equal length because they map to paths that are one third So this divides the interval i zero into three equal paths So this is one third of a third one third of a third and more formally How can we state it in the general sense the general? In general well, we know that f n a zero a n minus one Is a homomorphism But we know much more Okay, because this is just the interval these are intervals that always stay either inside i zero or i one They're mapping it out. It's a small interval it maps either into i zero or i one and then it maps again By definition these intervals they always stay inside i zero and i one and what's the derivative inside i zero and i one It's always either three or minus three right So the absolute value of the derivative is always three for f n the absolute value of the derivative is three to the n Right and then we just have mean value theorem That says okay, so since The derivative is equal to three to the n for all x in i a zero i n minus one Okay, then we can just write that a okay, so i The length of i equals three to the n times the length of a i zero a minus one and so i a zero an n minus one Is equal to three to the minus n which goes to zero As and this completes the proof Of the proposition Okay, I want to make also a little remark which we will need later that follows from the construction so um You can start seeing the structure of these intervals right It's a little bit like the construction of the cantor set for those of you Who've done the construction of the cantor set because you see we have a central element here That's delta. That is the hole everything escapes right and then the intervals that stay inside zero one Inside these two intervals for two iterates are formed by these four closed intervals whose complement are these Three open intervals right and then each of these what happens to each of these well This interval here after one iterate it maps to i one and after two iterate it maps to i zero so We've just shown in the general case so f two of this maps to all of i zero So that means that it will be mapped and it maps in a a fine way in a linear way So that means that inside here there will be two intervals two small intervals fine and each of these after Three iterations will map to all of this because after two iterations This will map to either i zero i one and the other one will also map to i zero i one And what you've got is a complement this little bit here that falls inside delta and that then escapes Right, so from the construction We have that Two new intervals Two new intervals Let's say these ones i a zero a n minus one zero And i a zero a n minus one one Right the two that we construct by using the fact that you have at some scale a small interval That a time n maps on to zero one you get two they are separated So they have length Have length these are of order n plus one. So they have length three to the minus n plus one, right? and Are separated by A pre-image Of delta Which also is length three the minus n plus one Right because these gaps also have the same length of the intervals that we are creating So in particular, so we have that if x and y belong to lambda and x minus y Less than three to the n minus m plus one They're closer Then this distance here Then x and y belong to the same interval a zero to a n because any two intervals of this level Are separated by a pre-image of delta Which is of at least of length three to the minus n plus one or bigger and similarly We have that Moreover conversely x and y are in lambda And x minus y is greater and three to the minus n plus one they necessarily belong to different To distinct intervals Of the form Well, that's right. Then it depends because you could have They could be the end points of the same interval or they could be the end points of intervals that are next to each other So both cases can occur But you know for sure that if you have these stick the qualities then it gives you some Condition about what okay, let's have a couple of minutes break and then we'll come back Okay, so this proves That the sets i a are well defined And this allows us to define the map which associates To each sequence the point i a Because it's a single point There exists a point and we've got the single point Yes Is a set but it's a set made of a single point the this this Is a single point is a single point It's a one point set. Okay Let's agree to identify that one point set with the point that it contains Yes So the last thing we would like to show is injectivity of this then we will get a nice bijection and a nice conjugacy All right, so the injectivity we've essentially already proved it, but let me give a little argument here, so h is Injective So what does it mean that h is injective? That if you fix two sequences Cannot correspond to the same point But this yes two different sequences So injecting yes injecting means that if you take two different sequences So wait, uh, no, this is not what I want to show. I want to show that So what i've shown is that for every a We have a non-empty Single point here. Yeah, so now I want to show that if I have two different sequences You cannot have the same point, but this is kind of Obvious right? What is the reason for that? I mean there's several there's several ways to see it What's what's the argument? Yes, there's many ways one of them is first of all this that if two points Are at any distance apart then for sufficiently large n they will satisfy this condition and therefore they will belong to distinct Intervals here and so they cannot belong if they belong to distinct intervals here Then obviously they will belong to distinct intervals Of this form All right, or if two Sorry, it's the other way around if two sequences are different then at some point They must have a different okay, let's put it in a slightly more Simple way if two sequences are different all the points in here have this sequence as the combinatorics Right, so you cannot have a single point having two different combinatorics because that would mean that this point at time n Is both in i zero and i one actually so the injectivity is much more basic It's kind of included in this In this definition right because all these points here if you look at the definition Okay, so let me write it here i a bar is equal to the set of points x such that f i x belongs to a i for all i greater than equals either this is the definition of i a bar So if you just look at the definition you see that it's obvious that if you have If a bar is different from b bar Okay, then there exists some i greater than equal to zero such that a i is different from bi And so we cannot have obviously And so i a bar is different. Yeah, so this is really Injectivity is almost intrinsic to the definition that What we're doing is we're coding the combinatorics of the point. So What we have here before is a conjugacy h is a conjugacy between f on lambda and sigma on sigma two plus So as an immediate corollary of this conjugacy, what do we have? So we have that lambda Um is uncountable Because f is a h is a bijection and we know that sigma two plus is uncountable We have something about periodic points Right, what do we know remember about the dynamics of sigma? We studied it last time. We know that it has infinitely many periodic points of all periods, right? And f is stick to lambda has infinitely Okay, so now to go a little bit deeper into this Conjugacy, we would like to show that h is a topological conjugacy So we would like to have a topology on sigma two plus And show that h is a homomorphism between this topology here and the topology on lambda induced by the euclidean metric here And also we will want to study the topological structure of the dynamics here So that's how we're going to get our final result that the periodic points are dense here or that the there's a dense orbit here We will just show that this is a topological conjugacy and then we just need to show The properties the corresponding properties here and there are topological properties that are preserved by topological conjugacy So the first thing we do it let's State let's lie down what the metric Is on sigma two which we're going to use to define the topology so define a distance function between two sequences as the sum From i equals zero to infinity of ai minus bi over two to the i is a metric sigma two plus So I will leave this for you as an exercise Okay, remember they recall the definitions of a metric and just check Basically the triangle inequality Okay, it's not difficult to check So what does this do? Let's uh to work with this metric we need a little bit of intuition about what this metric does As you can see what this metric sums up So what's the maximum distance between two points here? Maybe two for i equals zero For i equals zero this is one And the rest of the sum is at most one. Okay, so the maximum distance is two Okay And when are two sequences close to each other when the sum is small? Yes But when is the sum small The first part should be equal exactly right because If for example for i equals zero the first terms are different already the distance is at least one So they're far apart even if everything else is the same Right the distance is one But if the first two million terms are the same then the tail is very very small Okay So the way to think of this metric is that two sequences are closed if the first Large number of terms is close That's the intuition you need to have about this metric In fact, just as a remark, this is the topology if you did in the in the I don't know in the cosine topology I don't know if you did product topologies, but if you think of the space sigma two plus has the product The countable product like this which is the space of sequences And you take the discrete topology on zero one and you take the product of these topologies You get the same topology basically so you get a topology in which sequences converge if they have larger than larger initial Finite blocks converge. Okay, that was just Common there are many metrics as usually metrics spaces There are many metrics that induce the same topology and there's several if you look at the books You might find some slightly different definitions, but they all are essentially equivalent so Uh, let's formalize because we will use this a lot Let's formalize a little bit these properties of the metrics that we just said So let me write this as a lemma So for all epsilon There exists n epsilon greater than zero With n epsilon going to infinity As epsilon goes to zero such that If a i equals b i for all i equals zero to n epsilon Then the distance between a and b Is less than epsilon. This is just what we said if We'll do the calculation in a second, but I think it's clear That this corresponds to what we said, right? So if The relevant thing in some sense the reason why I'm writing this as a calculation Is just to make sure we see that this n epsilon depends only on epsilon and not on the specific sequence, right? For every epsilon there exists an n that depends only on epsilon Such that if two sequences are the same for the first n epsilon terms Then they are close by epsilon and the n depends only on epsilon, right and conversely For all n Greater than zero there exists epsilon n greater than zero with epsilon n Going to zero as n tends to infinity such that if the distance Between a and b is less than epsilon n Then a i equals b i for all i equals zero to n This also if you think about it a bit is obvious what this is saying is that If two sequences are close in this metric then they must be close The first few the first initial terms must be the same Okay, so let me do the calculation which is very simple I could leave it as an exercise, but I will do it because Because we will use this a lot actually in in in all the estimates the calculation, so it's good to really Fix it in your mind. Okay, so proof so Let epsilon greater than zero And n epsilon Sufficiently large such that two to the minus n epsilon is less than epsilon Okay, in fact, this is a Fairly explicit way of defining n epsilon in terms of epsilon by this condition Then if a i equals b i for all i equals zero To an epsilon we have that the distance between a and b Which is equal to the sum a i minus b i Over two to the i Is equal to the sum i equals n epsilon Yes, an epsilon plus one to infinity of a i minus b i Over two to the i Which is less than or equal to the sum i equals n epsilon plus one to infinity of one over two to the i Which is less than or equal to one over two to the n epsilon Which is less than epsilon so completely state forward calculation, but just to Clarify this now let n greater than zero And let epsilon n equals two to the minus n then If the distance between a bar and b bar is less than epsilon n supposed by contradiction that there exists sum j in zero n such that a j is different from b j Then we would have that the distance Yeah Then we would just have a contradiction the distance between a and b which is equal to the sum a i minus b i over two to the i Would be greater than or equal one over two to the j Which is greater than or equal to one over two to the n Which is uh equal to epsilon Which is a contradiction because we assume that they're less than epsilon So just a little remark Exercise which I think is quite interesting exercise Consider this metric this function What's the difference between this one and the other one? Sorry This is not a metric That's right. So exercise show this is not a metric and why is it not a metric? It's to do with the axioms of metric which axiom does it not satisfy What is the test? Which is what? Exactly Okay, you can find two sequences that are equal to zero Here No symmetric is true Because a minus bi we've got the norm outside But so I'm not going to spend more time on this But this will be relevant for what we're going to do next week actually So I think it's interesting to think about it a little bit. This is called a pseudometric Because it satisfies in fact the other assumptions of the metric But it doesn't satisfy this one That two points at zero distance are the same But pseudometric you can always turn it into a metric on a quotient space by identifying all the points That have zero distance if you identify them they become the same point and then you have a metric Okay, so exercise or something for you to think about is what kind of space do we get When we identify these points, what kind of metric is it? What kind of space is it? And which sequences do we identify you can you can work out quite explicitly which sequence need to identify Okay, I will just leave it for you to think about a little bit And the second exercise Also is to show That our space Sigma two plus With this metric the real metric that we've defined is a canto set So a canto set is a topological notion. It means it's a totally disconnected Close set with no isolated points Okay, so those are all topological notions that can just be checked from the main Okay, so we'll leave again. So these are all very good exercise to really Consolidate the properties of this metric in your mind. Okay, even though it's fairly simple It's good to consolidate it because we use it a lot So before we finish, let's just prove a couple of things or maybe just one thing depending on how much time we've got On using the fact that we now have a metric on sigma Let's show for example That the map sigma is continuous in this metric So lemma sigma From sigma two to sigma two Is continuous Before we prove this, can you tell me if it's invertible? Not invertible Why not? Why not? And so Okay, so remember that sigma of a zero a one a two Is equal to the shift to the left. So you lose this so you get b zero b one b two and so on Where bi is equal to ai plus one All right, this is the this is the definition of the map So why is it not invertible? sorry Why not? What is the pre-image? of this sequence here There's two possible pre-image is exactly you could have a zero here or the one here and both of them go to the same image So it's two to one at every point It's not injective, but it's exactly two to one at every point Okay, nevertheless, we can ask if it's continuous This is what we're going to ask so map may be not invertible But it may be continuous which means that nearby points map to nearby points So how do we show that it's continuous? How do we show that it's continuous? Sorry Basically, yes Yes, basically exactly exactly. So Let epsilon so fix epsilon greater than zero and we want to find delta so Okay, let epsilon greater than zero then the distance between sigma Of a and sigma of b is less than epsilon if The initial n terms if the first if the initial n epsilon terms of sigma a And sigma b are equal right by that lemma there So So as long as the initial n epsilon plus one terms of a Of a and b are equal then We will have d sigma of a Sigma b less than epsilon Because if the n epsilon plus one terms of this are equal Then the first n epsilon terms of these two will be equal and then this will hold Okay, and so we can just let Choose delta So now let Delta Delta equals okay delta sufficiently small delta Is sufficiently Small then if The distance between a and b is less than delta implies Implies this implies that the first n epsilon plus one terms Equal sigma is lip sheets With lip sheets constant two that's two The image of two nearby points Is multiplied by a factor two Okay, let's To one more lemma. We just have a few more minutes. So Let's start studying the dynamics Of sigma now that we have the topology now that we have a continuous map We studied topological properties of the dynamics before we had the topology We looked at all the periodic points And we decided that we had an infinite number countable number of periodic points Now we can say a little bit more So the set of periodic points of sigma is dense in sigma two plus Again, this is just a straightforward application of this Of the properties of the matrix. So how do we of the metric? How do we prove this? Yeah Exactly. So Yes, right. So for any epsilon so let um z bar We need to show that arbitrarily close to this point. There is a periodic Point that it's obvious, right? What's the periodic point that is arbitrarily close? So for any epsilon There exists A periodic point Let's call it p p bar equals p 0 p 1 p 2 and so on such that Distance between p and z is less than epsilon Because indeed How do we choose the periodic point? Exactly Let p bar Is equal to z 0 Z n Okay for some sufficiently large and depending on epsilon In fact, it's exactly the n epsilon of the previous Lemma, okay So next week we will proceed to study the further Topology the fact that it has a dense orbit And to show that h now that we have a metric on this and a topology h is actually topological conjugacy So the topological properties of the dynamics they map Topological properties of the dynamics of f Okay, thank you very much