 Welcome to module 17 of point set topology course. This time we shall consider a few more interesting examples of topological spaces. The most interesting one that we are going to do today will be a subspace of r itself. Before that, let me consider something which is out of r or not, not the usual topology. Start with any infinite set, then I am denoting this topology COF, CO-finite, so you can read it as co-finite. So what is this? It is collection of all subsets A of x such that the complement of A is finite. Of course, I will have to allow A to be empty also because complement of empty set is the whole space, I do not want that, in this definition I do not have that one, x minus A is finite. If x is, if A is x, of course, it is allowed, A is empty is not allowed in this rule, so I have to allow it specifically, so either A is empty or x minus A is finite, then I put it inside this collection, co-finite topology. So I want to say that with this it becomes a topology, ok. So this is called the co-finite topology on the given set x. Once you consider such a topology, it is no point in considering x to be a finite set. Then what happens? This becomes just a discrete because all subsets will be there because their complements are also finite. So it is interesting only when you take x to be an infinite set, ok. How to verify that this is a topology that is not very difficult because what you have to do? Once you take two of them x minus A1 and x minus A2, right, their intersection, what is the complement of intersection? It is the union of the complements, right. So each of them is finite, so union is also finite. So intersection of two open sets is open or intersection of finitely many open sets is open. The union of arbitrary families is open because as soon as one of them complement is finite, if you enlarge it larger and larger, say it will be automatically finite. So that part is easier here. So that will verify that this co-finite becomes a, you know, a family of co-finites subsets, it becomes a topology. We will have many, many instances of, you know, usefulness of this one throughout the course, ok. So this is going to be kind of model for us. One of the interesting property, you can directly verify it right now, namely take any point in x, look at all the neighbourhoods of this point. Intersection is precisely the single written x, that is the only point x is common to all the neighbourhoods, ok. So verify the statement. Just to begin with, getting familiarity with this co-finite topology, we will have many instances of studying this one. So this is just an introduction right now. Similar to this one, but not so important is the following, ok. Take x to be an uncountable set, let COC is in a co-countable, denote the set of all subsets such that either it is empty, that empty set is allowed or the complement is countable. Now countable includes finiteness as well as infinite countable also, ok. So countable means that, alright. So once again verification that this topology is identical to the previous one, there is no problem, ok. Because countable, finite union of countable sets is countable, that is all I am using here, ok. So this is called the co-countable topologies, just similar to co-finite topology, ok. So it has similar properties. Once again, there is no point in taking this topology when x is a countable set. Then again it will be a discrete topology, therefore I take x to be starting with an uncountable set, ok. Let us go to the third example here, actually there are two of them. This time I come to the underlying set is R itself but topology is different. So I better write it in its different symbols. So here I have used LR respectively RR to denote collection of all intervals of the form minus infinity to A to denote that these are left-handed rays, LR, R for ray instead of arbitrary these are the intervals but to indicate that they are unbounded, unbounded intervals, right. Unbounded below, no. So LR, the other one is RR of the form A to infinity. A varies to minus infinity to A, this is LR, A to infinity there RR, right rays, ok. Where this A ray, A can be taken as all the way from minus infinity included here. So this one, what does it mean? Empty set. When it means minus infinity here, this will be empty set. When it is press infinity, open interval I am taking remember that. So that will be just the whole of R, ok. So this forms a topology because if you take intersection of minus infinity A with minus infinity B, look at whether A is B bigger than B or B is bigger than A, the smaller one intersection will be just smaller one. If you take some A alphas here and take the union, then the union will be what? Take the supreme of all these A alphas and take that open interval for minus infinity at one. This exactly similar thing will hold for this also. Here you may have to take infimum to show that arbitrary union of open rays is again an open ray. So this is even simpler than the topology in which all open intervals are allowed, right? The real line. So obviously in both of them there are fewer open sets than in the usual topology on R. So they are different topologies in any case. For example, the open ray here to the right will not be an open subset in the LR, see, and vice versa, all right? So these things are again important in analysis. So when an opportunity arises, I will again refer to this one. Now I come to what I told as the most important example. This time I have a subset of R itself with the usual topology, okay? So this is called Cantor set, which is a landmark result in topology, okay? A landmark result by Cantor. First we shall define an operator. You can read it as Cantor. I don't know how to read this symbol. Here I read it as Cantor. On the class of all closed intervals A, B, A less than B, okay? I am not interested in singleton Bs. So A, B, A less than B, the closed intervals like that. Into the class of all closed and bounded subsets of R, okay? So what is this C? C is going to be an operator. It takes a closed interval and then produces a closed and bounded subset of the same interval A, B, okay? So prior to that we define another simpler operator free on the class of finite union of disjoint closed intervals, which you may call middle one-third radiators, okay? So this funny name, we will explain it right now. Start with any closed interval. It may denote it as J, okay? So A, B. Let us define phi of J to be the set obtained by deleting the open part of the one-third of the middle one-third, okay? Open part. So in notations what is it? The open one-third is nothing but A plus B minus A by 3 to A plus twice B minus A by 3. So delete that from J. So what you get is disjoint union of two closed intervals. So that is the definition of phi J. Take any closed interval, 0, 1, A to B, 5 to 15, whatever. So it will produce two disjoint closed interval, each of length one-third of the original one, right? From the starting point to that one-third distance and then two-third distance to the end, okay? Very simple operator. The middle one-third is deleted. That is why it is called middle one-third deleted, okay? Now you extend phi to the union of finitely many closed intervals, disjoint closed intervals, okay? So A is union of disjoint closed intervals. Define phi A to be union of phi of A, I, B, I. So here I have defined phi of A, B. So that is why you have defined phi of A, I, B, I. Take the union of this one. They will be again disjoint because each A, I, B, I is contained inside A. Each phi of A, I, B, I is contained inside A, I, B, I. And they are disjoint here, okay? So this is just extending from one interval to finitely many intervals. So phi A, definition of phi A is over. Now what we do? Now fix one interval A, B, okay? Let us call it I, naught. Inductively define, see I, one will be what? Phi of I, naught. That means the first I delete, the first one middle one-third of A, B. Now I have got two intervals there. Operate phi on that one, okay? So that is what I have to do. Okay, phi of I, one will have two intervals, right? Take that as phi I, two and so on. So I, n will be phi of I, n minus 1. So when you come to nth level, there will be how many intervals? Two power n intervals will be there, they are disjoint. All of them will be contained in the previous sets here. So this is a strictly monotonically decreasing sequence of closed and bounded sets, right? Strictly monotonically increasing sequence. What I am going to do, finally the canter of A, B is your operator. It is just take the intersection of all this. It is like taking the limit, limit of sets, but limits of sets is defined very easily. It just take intersection of, okay? This is definite. One thing is very clear that because all these are closed subsets, this will close. So this is a closed and bounded subset. Starting with an interval, I have produced a closed and bounded subset of A, B itself. I notice A, B and all these are subsets of that, okay? So here is a picture what I have done. Start with one interval, delete the middle one-third. So you have two intervals here. Now look at this one, delete the middle one-third. So this will give you two here, two here. So at the second level all day you have four of them. Now delete one-third here, delete one-third here and here also. So the third level you will have eight of them and so on. So keep going on so on. So you will get a lot of just, things looking just like dot, dot, dot, dots, okay? So that the limiting thing is the Cantor set, okay? The function C is called the Cantor's construction, okay? The set, now I am taking specialization. Instead of A, B put zero one, standard one. Take the closed interval zero one, apply the Cantor operation. Let us denote that by C, okay? Usually one can, one usually verifies Cantor set means that one, but there are more wider applicable meaning for this one, okay? Right now what is the meaning of Cantor set? When take A, B equal to zero one and apply C, okay? The sets C of A, B, so whenever there is arbitrary A, B I will write like this. When there is just zero one, I will just write C, that is all, okay? Have some wonderful properties, okay? So let me list a few of them. Not all these properties may be easy for you. Certain things may need some concepts which you do not know, but this is actually a topic analysis. So I will just list them right now. I may not explain all of them to you, okay? Quite a few of them I am doing it. They are easier, so or I will explain them right now, okay? The first thing is C of A, B is non-empty. I already told you it is closed and bounded. That is easy. Why it is non-empty? So this is Cantor's intersection theorem, okay? So the first thing is very important here. It is a non-trivial thing, but this is non-empty. If J is one of the intervals contained in I n. So what I mean by say, look at this picture. This is I naught, I 1, I 2, I 3. When one of I n, look at one of these intervals, the small intervals, okay? The intervals I am talking about, not arbitrary intervals. It is not subset of these intervals, but one of these intervals. You just take that. So if J is one of the intervals contained in I n for some n, then if you look at C of J, that would be also subset of C of A, B. You can just apply C of that because C is an operator operating on any intervals, closed intervals. So C of J is C of A, B because J is contained in A. Given any, the endpoints are always inside C of A, B. See, endpoints are never deleted. These two endpoints is endpoints point, okay? So beauty is to begin with I can say these are the endpoints. They are not deleted. But now this point and this point will never get deleted because from here I could have started and arrived at one. So this point, the endpoints are never deleted. The same thing is here. Endpoints, these endpoints are never deleted. So in the end, these endpoints are all there. They are never deleted. If something is somewhere in between, you never know whether it will get deleted. So that is the beauty of this. So there are all endpoints of successively cutting intervals. They are all there. That is the whole idea. Though I say it only for A, B, because it is true for all subintervals also, okay? Now take the map fx equal to A plus B minus A times x. It is a linear map, okay? What does it do? Put x equal to 0, you will get A. x equal to 1, you will get B. So that is a linear map, isomorphism, homeomorphism from 0, 1 to AB, right? It just expands. This one-third becomes B minus A by 3, right? So what happens is the cantor set C of AB and cantor set C of 01. This is C. They will be isomorphic under this continuous bijection. So this map preserves the cantor construction. It is, first of all, map defined from 0, 1, 3. Whatever middle one-third deleted, the middle one-third correspondingly will be deleted. This is not an isometry because 01 is expanded to AB, right? But it is a similarity. For now onwards, I will just concentrate on the cantor set C of 01. Each of the properties of C which we list below is carried over to an identical or similar property of C of AB, right? Because of similarity map. So that is why I do not have to say it for all of them. Once I say it for C, it will prove for C of AB for all of them. So here are some more properties of C. These are special in the sense that they are all points between 01. That is the speciality. Topologically, when you have, when you have similarity, they may not be inside AB, 01, but maybe it will be inside AB. That is the difference. The end points of every component of IN, n greater than or 0, they are inside this wall. We have, we have observed already. The set of all rationales of the form 1 to n, okay, the finite sum, ak divided by 3 power k, okay? What are ak? Ak is either 0 or 2. That is contained inside C. So why this happens? Look at the first time, the very first interval here, okay? Everything from here to here is one-third. From here to here, one-third plus something, okay? So that point from one-third plus something, that will never be there. From here to here, it is one-third to third plus something, that will be always there. So 2 will be there. From here to here, it is just within, it is in one-third. So the first will be 0. 0 plus something by 1 by 9, it may be. And plus something by something, I mean something by 1 by 27. So into some 27 and so on. So that is what this means. The ternary expansion is what is being done. I am writing every element as 1 by 3 or 0 by 3 plus something a by 9 plus something b by 27 and so on, right? So there is a sequence ak. These ak's will never be equal to 1, okay? They are either 0 or 2, okay? So this is an easy consequence of the construction here. The middle one-third is deleted, okay? All the rationales of this form, okay? They will be there, alright? Now C contains no open intervals. There is a first observation, okay, that I am making which is now serious. Take any open interval, okay? So first of all, it should be inside AB, inside 01, alright? So it is somewhere here. What happens? Go on making one-third, 1 by 9 and so on. Finally, some small portion will lie between the same interval because interval has positive length. And that one-third portion will get deleted, okay? So that is easy to see. So no open interval will be contained inside canter set in the final stage, okay? In between there, there are there. These are intervals. But at the end, there are no open intervals, okay? What is the meaning of that? Interior is empty. It is a closed set already. Therefore, it is no air dense, you see? So suddenly we have constructed no air dense set here, okay? So where are we? Here, okay? So C contains no open intervals. In particular, being a closed set, it is no air dense. Otherwise I have to take a closure and then look at the interior. So closure of C is C itself. Therefore, interior of that is empty. So this means it is no air dense. Every point of C, the limit point of C. Limit point means cluster point. Is that very difficult to see? Take a point of C, it is a limit point of C. Take an interval, okay? There will be some other point other than that point. That is all I have to see, right? So no point will be isolated, okay? So for this kind of thing, you can just keep glaring at this construction one by one. You can produce that. Or you can use your arithmetic and compute things and so on. Just look at every element of this form, okay? So near that, you can take any epsilon. You can produce one more element other than this one, okay? So that kind of arithmetic will produce proofs also. So it is not all this. So what is it? Every point is a cluster point. Every point is a limit point, okay? Such sets are called perfect sets. So L of C is equal to C. The fifth one here, the fifth one is C is uncountable. So this follows by looking at all these distinct, you know, suppose you have a sequence here, okay? And take infinite sequence also. See, this I have not said here. Look at all infinite sequences. The entries are either 0 or 2. Take a sequence and then conform this sum. This will be a convergent sequence, okay? Right? If you truncate it as finite level, all these points are there. The limit is also there because it is a closed set, right? So infinite sequences of this form 0, 2, 0, 2, whatever give you a number here in the cantile set. But these sequences are themselves uncountable, okay? Now all that you have to see is that if you take two different sequences, they always give you different numbers, okay? This is similar to what you do with the static expansions. So here we are doing ternary expansion. You could have done it with any number here, okay? With the decimal expansion also you could have done, okay? If it is one-third, there is a charm here which is not there with some other number. So we will see what come to that one later. Okay? So where were we? Uncountable. C is of length 0. Now you may not know what is the meaning of length of this arbitrary subsets here, right? If you do not know that, don't worry about that. I will tell you what it is whenever it is needed. The third, the last point here I am going to make is given any two points inside C, okay? So assume X is less than Y. There exists disjoint clause of states A, B of C such that C is A union B, A and B are disjoint clause. X is inside A and Y is inside B. So such a thing is called a separation. Whenever any two points can be separated like this, okay? Such a set is called, such a topological space is called totally disconnected, okay? So don't worry about this one. This is a very profound topological property which we will discuss only in part two. But this property I have very nicely defined and this can be proved right directly. There is no problem for the canter set, okay? For the canter set, we will prove this one, all right? So that is not very difficult. So in any case, I have put all these proofs here also. Let me do the last one first and then come back to all of them. So look at this one. Take any interval. This interval is not contained inside C, which we have observed earlier. So take a point x, say z between x and y, strictly between x and y, which is not in C. If it is not in C, what happens? Take the closed interval 0 z, intersect with C. See, this is a closed subset of R. This is a closed subset of R. Intersection is a closed subset. Similarly, B equal to z to 1 and intersect with C, okay? z to 1 contains y, right? So B will be inside y, sorry, y is inside B. Similarly, x will be inside A. And by the very definition, the intersection of this will be just z, but z is not there. So intersection is empty, okay? So both A, B are closed in C, x is inside A, y inside B. So this is a proof for last k, okay? Now this is what I am telling you. What is the meaning of length is 0, okay? What I am subtracting? For the first time, for the first time, I am doing 1 third, right? Then I am again subtracting 1 by 9, right? So this j, I am trying to approve this j. C is of length 0, okay? You sum up all the lengths, okay? That summation is equal to 1. Whatever your lengths, you know, you sum up all. At a finite state, you see that it is less than, this minus this minus is less than epsilon. So if some number is less than epsilon for every epsilon, that must be 0. So that is the meaning of this. This totality of total length of all the intervals that I am going to subtract. First, this is 1 third. Then from whatever remaining, I am taking 1 third here and 1 third here. 1 third here, 1 third here. 1 third, 1 third, 1 third, 1 third. 1 third of the original thing, right? So I keep taking. So this is 1 third. This will be 1 by 9, 1 by 9, 2 by 9. This will be 1 by 27, 1 by 27, like 4 by 27. So you see nothing but 2 power n divided by 3 power n. Take the summation. That is what it is here. 2 power n minus 1 divided by 3 power n. Some total is 1. So whatever remaining is 1 minus 1. So length will be 0. So I think I will turn down all these proofs here. More carefully, you can go through that. You will get the nodes in a case, okay? So y interior is empty. I have explained it completely here. So this canter set is going to play a lot of interesting role in analysis and also in topology, okay? So once again, let me give you some exercises here. All right? Start with a equal to 1 by n. n belong to n. Include 0. So compute all these things under the... This is when old XI remain under topology R, under topology is different topology. Co-finite topology, co-countable topology and LR topology. The original topology you have done it was... Now you do it for these three topologies. Once you do it for LR, it is similar to RR. So it won't be much difficult for you. That is similar. Okay? So do that exercise. Here there is another exercise. Convergent sequences and dense subsets. You have described in co-finite topology and co-countable topology. Okay? All this doesn't need many more education. You have been given the definition. Definition of what is a convergent sequence. Definition of what is the meaning of dense subsets and so on. You have to work out for them for here. Okay? A sequence converges to a point if given any neighborhood, there must be some n such that k bigger than n means Xk is inside that neighborhood. So that is the kind of definition. But now you apply to cleverly chosen neighborhoods of points inside co-finite topology and see what happens. Similarly for this one. See what happens. Okay? So that will be the material for today. Let us stop here. Thank you.