 Hello, welcome to NPTEL NOC and introductory course on point set apology part 2. Today module 59, a new topic classification of one dimensional manifolds. In every classification problem we must first of all have plenty of examples of okay of whatever you are looking for a likely representations of various objects which we want to classify. Suppose you want to classify a certain number of trees okay. So first of all you should have a number of trees, various number of trees and then you can say okay these are of this type, that is of this type and so on. That is the kind of thing right. So which are likely to represent all possible types we do not know yet. So we think that no we, this may be just exhaustive type okay. Only after that we can make a probable list of representatives which are mutually of different type okay. The final step is to draw a conclusion is to what is to show that every object that we wanted to classify belongs to precisely one of the types mentioned in the list. While doing that often what happens is somebody else or you yourself will find out another new object which was not listed in at all and which does not follow fall in any of these types. So you have to add that one to the next, that is the way classification keeps going on. When we were children the classification in a biological classification of species was not yet over. By now, by several years now people say that it is over now okay. So it is like that long, long back Mendel beef started classifying elements. So he predicted that this is what all the elements will be. So some elements were not even formed but he predicted them and so on right. So it is like that classification or scientific classification involves these two steps, fundamental steps here. In order to classify all manifolds clearly it suffices to consider only connected ones for any manifold is locally connected because they are locally Euclidean and hence its connected components are both open as well as closed. Therefore, every manifold can be written as disjoint union of its connected components even as a topological space. You see when you have disjoint union of topological spaces that is not the same thing as taking a topological space and writing it as disjoint union of closed sets okay. If it is closed sets as well as open sets then it will be disjoint union as a topological space okay. So all this justification you can only look at connected manifolds okay. So what are now, now we want to place one dimension manifold right. So what are the examples of connected one dimension manifold that you have so far, can you think of more of them. So this is the first step you have to do right. So first of all we look at inside R itself okay. Observe that any two closed intervals okay are homeomorphic you know non-linear single point you also may take a closed interval but take at least two points then they are all homeomorphic to each other. If it is single point, single point themselves are homeomorphic no problem. So what are there we have seen already via linear maps I mean affine linear maps. This homeomorphism can then be used to get homeomorphism between any two bounded open intervals as well or between any two bounded half open intervals okay. So suppose 0, 1 to A, B I have got then I can delete A and 0 I will get open 0 to 1 closed to open A to B closed and so on right that is what we have done. So we have we know that all open intervals are homeomorphic to each other half open intervals themselves will be homeomorphic to each other, closed intervals are themselves homeomorphic to each other. So there are three of them if you want to mention. Finally what we have if you look at x going to tan x or some other homeomorphism many other I will give you one example this defines homeomorphism of open interval minus pi by 2 to plus pi by 2 to the whole of r right. And if you restrict it to 0 to pi by 2 it will give you a homeomorphism from closed interval 0 to infinity. So these whole things unbounded things are also taken care under these open intervals half open intervals half closed intervals and so on. So thus as far as subsets of r are concerned we do not have too many of them what are they open intervals half open intervals or closed intervals. So there are only three classes when I say list whatever I mention here they are themselves homeomorphic to each other. So if you take one half interval 0 to half 0 to 1 1 open then there is no need to take no 0 open and 1 closed because they are already homeomorphic to each other. As soon as we go to subspaces of r2 we get some other types it seems like circles ellipses parabolas circles are definitely not there inside r right ellipses were not there parabolas are they different many more actually what are called as smooth curves you can talk about the smooth and non smooth this is something completely different game all together but we can take simultaneously you can study both of them also okay boundaries of a convex polygon like a triangle or a square or a rectangle or a pentagon you look at the boundary they are also one dimensional manifolds if we have one to one parameterization of any of these curves then clearly they will be homeomorphic to an interval or a circle and so on okay first of all a smooth curve open interval you know neighborhoods will be homeomorphic to intervals that is why they are manifolds this is the case with parabola a parabola can be parameterized by r completely in a one to one fashion what is that say for y square y could x square y square y x square that is the graph of that function itself will give you the parameterization right but if you look at hyperbola hyperbola is not connected so you have to take only one lap of hyperbola then again you can parameterize it so even if you go to r2 accept the circles and ellipses you do not get new things okay circles ellipses boundary of a triangle and so on what are they they are themselves homeomorphic to each other okay so this is what I want to tell you one can also easily see that any two circles are homeomorphic to each other first of all indeed you place a small circle inside an ellipse and then project it project the circle over to the ellipse you know radial projection you know from the center so that will be you homeomorphism of circle with ellipse okay you can write down formulas also but if you take a pentagon or you know any hexagon and so on arbitrary any convex polygon writing down a formula is a little more difficult but you know geometrically you can easily see that all of them are homeomorphic to a circle and any two circles themselves are homeomorphic to each other okay you try at least write down a formula from a circle from a triangle to a circle okay for each time what triangle is it an equilateral triangle wherever it is all this you have to bother about all right so do we get any other types of one-dimensional manifold if we go to r3 r4 and so on other higher dimensions you have to probe right why I am going inside only euclidean space is because our earlier theorem that any manifold is a closed subset of some r2n plus 1 therefore when I am hunting for one-dimensional manifolds you don't have to worry beyond r3 all one-dimensional manifolds copies of them will be there inside r3 right but in r3 there may be weird kind of embeddings of the circle very fact that they are embeddings of circle they are homeomorphic to circle are there different ones that is the point the embeddings may be very funny but they are all homeomorphic to circle so we are not bothered about weird embeddings okay so are there any other one-dimensional manifold the answer is no so that is the gist of whatever is going to come now okay maybe today and another one tomorrow so two more lectures we may have to take okay so it is the this is the theorem final theorem X be any connected one-dimensional manifold which just means remember it is Hausdorff and second countable abstract okay so topological manifold I have put a smooth in the bracket because the statement is true for smooth case also okay correspondingly instead of homeomorphism we will have diffeomorphism the final conclusions are the surprisingly they are the same here but for our purpose we will ignore this smooth path and diffeomorphism path we will be only proving the topological aspect okay so what is the statement take a connected one-dimensional manifold with or without boundaries specifically I have mentioned the boundary case also here then X is homeomorphic to one of the following one two three four oh what are they open interval half open interval closer interval or the circle look at this case the first one and the last one are manifolds that means manifolds without boundary the second one and third one are manifolds with boundary they are all connected of course otherwise I won't list them here this and this one these are non-compact these two are compact so if you want only compact one only these two you will get right if you want non-compact ones only these two you will get so don't put compactness you have all the four of them that's all now first thing what I will do is granting granting that we know the classification for manifolds without boundary namely boundary of X is empty okay what are the two then I will get only this one and this one first one and last one I will complete the classification for all of them that means when you allow boundary you will have two more that's what I will show you today okay granting this namely the classification theorem for when the boundary is empty okay we will prove the complete statement namely boundary of X non-empty also we will consider and complete the statement when boundary of X is non-empty I have to show that either it is this half open interval also closed interval right this one has no boundary this one has no boundary so that we assume that we know already as soon as boundary less it must be either open interval or the the self-cult alright okay so immediately it follows that start with a manifold with boundary X non-empty interior of X we know is a boundary less manifold it must be either the open interval or the all the circle but I have assumed a boundary of X is non-empty so it cannot be circle because if it is circle it's already closed okay so you can't have another extra boundary point boundary point must be in the closure of this one so so interior of X must be homeomorphic to an open interval okay the question is now how many boundary points can you put here that's all I want to say that you can put a boundary point around this end and and namely at zero you can put a boundary point here one of them or both of them and that's it if you put only one of them you will get this one or its carbon copy namely zero open and one closed that is homeomorphic this one so that is the same case if you put both of them you get this one so these are the only two different cases so that is what we have to prove so in other words I want to show that the boundary of zero one okay zero one is sitting inside interior of X okay X is a manifold with boundary and that boundary non-empty will consist of one point two point that's it and in those cases it must be precise so this is what we have to show all right so I repeat this one interior of X is homeomorphic should be zero one and hence without loss of generative we may assume zero one itself is interior by taking the homeomorphism copy of this one by the very definition if X is in the boundary there is a homeomorphism psi from zero epsilon to U this is the remember this is an open subset in the half H space H one right zero infinity I can assume but I can just take zero epsilon also to U where U is an open subset of X so it's a homeomorphism such that the zero goes to X so this is parameterization it's the inverse of a coordinate chart around the point X clearly if you look at psi of open interval zero epsilon that is homeomorph to U minus X being a connected open subset of zero one contained inside X okay is a b for some zero less than a less than b less because we know all the connected subsets of an open interval okay when you throw away one point what you have what is U minus X is homeomorphic zero epsilon okay also the psi X is in its closure right if we have chosen a and b between zero and one then it follows that psi X X is a well chosen point in the interior we are we are in the boundary we are looking at psi X and a or psi X and b psi X does not hit a is no open troll is there only a is not in the image of psi X psi X comma a this is a ordered just two two element set similarly psi X or b you take will be a pair of distinct points of X which violate the host or condition because every neighborhood of this a see a is now a point you know inside our interior of X okay every neighborhood of a and every neighborhood of psi X they will intersect each other some a plus epsilon some part will come on this side on the other side I do not know okay a plus something because a to b is this psi of zero epsilon right so so that is what happens this means that host darkness is violated host darkness is violated so this a and b cannot be strictly between zero and one okay so they must be a must be zero or b must be one okay so which implies that zero one contained inside X because I started with zero one itself is contained inside X I did not want to distinguish between the homeomorphism under this homeomorphism you take this zero one this just facilitates this to think of this points a and b are also inside or it may be the one is in the neighborhood so zero one is inside X so I always in one of them by symmetry I could have changed the psi so zero one is inside X okay with this zero being one of the boundary points the X that is I will start at it okay for each point I have got something like this one single point I have completed I have no contradiction but it has to be like this of course X may be equal to zero one then the case is over there is no more to bother about this is allowed then we are done so consider the case when X minus zero one is non empty okay there is some more point that Y be another boundary point of X arguing exactly as before it now implies that Y has to be equal to one now because it cannot be equal to again another zero then the zero and this Y will be violating the host darkness it cannot be in the other in the other end so it has to be this end and then this Y and one should be the same unless if they are not same then host darkness condition will be violated so they must be the same that means now I have got zero one is contained inside X okay now zero one is compact so it is closed in X it follows that they cannot hide cannot be any may any you know more boundary points of X at all thus we have completed X equal to zero one okay the moment zero one is a closed subset it is automatically open subset of the bound is the entire X because the boundary points are there okay this zero comma zero closed comma one open this in open subset inside X okay so it is both open and closed and X is connected so it is the whole space so you are saying something yeah X can be equal to close interval zero so we have only noted zero is the boundary point that's precisely what we are doing here now we are we are not assuming we are not assuming the rest of the real number system has anything to do with this this zero this one is a copy of the open interval zero one inside X that is subspace of this one open interval this we started with that okay this zero one using the homomorphism I have identified this notation here now this zero is a point of this one okay this zero one is a carbon copy of open interval so there is an order X has no order okay that order is being used here okay so suppose now it were something like some a comma b after removing this see see there is a there is a map from zero to zero close to this one okay you don't know where is this zero is going this zero is going where it's going to X and X is not a point of the interior okay it X is inside this open subset and X minus this open set in the complement of that okay so I am showing that if if you have this open interval sitting here okay U minus X you throw with U is the whole thing U minus X means what I am throwing away X okay so where does it come from if it is like this there will be problem okay so what is in other words what I am saying you can see case take a sequence of points okay converging to this zero okay converging to the zero means what that point is not there in X at all that is why I cannot do that converging to zero and so on so if it converges to some other point inside this interval say a or b on the other side then you have problem that's what you have to see see this is similar to our earlier problem wherein compactification of open interval zero one can you have one point compactification two point compactification three point compactification and so on the one point compactification of zero one is the circle open interval okay the two point compactification is the closed interval zero one can you have a three point compactification four point compactification so this was a question I think We have left it to you as an exercise, maybe by now you have solved it. I will tell you the answer now, okay, what is the answer? If you don't put the word horse darseness, then it is possible. If you put the word horse darseness, the same argument, argument here or you can do it separately, without many manifolds at all now, just horse darseness, okay, you cannot put a third point at all over. Remember the compactification means what of the original space X, this X must be dense inside the larger space X bar, right. Just use that, use the property that X bar is house door. You cannot have three extra points from 0, 1, open, two points, fine, one point, fine, okay, so that is the answer. Just use house door, so that is what I have done here, but here I have used even stronger property, so it is easier here, namely that the whole space, I am not assuming that it is compactification, I am just assuming it is manifold with boundary, manifold, one-dimensional manifold with boundary, okay. Since it is boundary, I also have that namely interior of X, closure is the whole of X, all that I am using is that, okay. So what we have done is, we have reduced the proof of the original theorem to the case when boundary of X is empty, that means it is a manifold in our original definition with which we shall proceed now, okay. So let me do a little bit of it. Starting with a countable open cover U for X, consisting of charts, we need to understand how any two members Ui, Uj of U intersect each other, then only we will be able to assemble these various open intervals and produce the new objects, how they look like. Of course, they may not intersect each other, that is well and good, no problem, okay. The moment all of them do not intersect each other at all, then it will be disconnected, okay. So some of them have to intersect in some other way because the whole thing is connected. Assume that they intersect, guess two of them at a time, don't jump to the whole thing, okay. One, we know already, it is homomorphic to an open interval, the other one is also homomorphic to open interval. We want to say how they intersect, they are not subspace of R now, you see, if they are inside R there is nothing to prove, all right. Each copy is hanging somewhere, I don't want to use even in R2, R3 or anything. We can use the picture inside R3 because we have another theorem that whole thing is inside R3. So if you want, you can drop pictures, no problem. Beyond that, you can't do anything, okay. So what we have to understand first is how two opens of sets Ui, Uj which are coordinate neighborhoods, namely they are homomorphic to open interval, how they intersect, how they intersect means what nobody has told you. So you have to understand all possible ways of them, they may have intersect. So that is the precisely the meaning of classification here, okay. So what we will do is assuming that they intersect, we shall first take two cases which we wish to happen and examine what best we can do in those cases, nice cases, that's what you may say. But then finally we should say that no bad case has occurred. So that is the whole idea, okay. So today we shall see these nice cases and then later on we shall do the other case. So start with a manifold X, one-dimensional manifold. U1 and U2 are any two non-empty open subsets in X, neither of them contained in the other. This is an obvious thing that if one is contained in the other, there is nothing. You can take the bigger one that is open interval. So you can go ahead to the third one, right. So don't get into that kind of cases. Take the case wherein U1 and U2 are proper, you know, they are not contained in the one container, they intersect across and choose open interval a, i, b, i and homeomorphisms from psi i to ui, this is the homeomorphism. Suppose further that intersection is non-empty and connected. This connectedness is the biggest type of this here, right. Non-empty you would like to have already, anyway. Next thing is look at the intersection. Under psi 1 inverse, it is some open interval. Why? Because psi 1 inverse is an open interval here and this will be a connected subset, a connected open subset of an interval is again an interval, okay. C1, B1. It is not the whole of a1, B1. Why? Because if it is the whole of a1, B1, that means U1 is contained inside, U1 intersection U2 is contained inside U2. That should not happen. So this is a proper open interval of a1, B1, okay. Similarly psi 2 inverse is a proper open subset of a2, B2. So what I have done? I have actually assumed that it is C1, B1. So the other end is fully taken. This part has to be bigger than a1. Similarly a2, C2 I have assumed. What does that mean? I have taken this end already. The other end B2, the B2 must be strictly bigger than C2. So that is what I have written. a1 less than or C1 less than B1, a2 less than C2 less than B2, okay. So the second assumption is very much stronger. I do not know why it should happen but I would like this one to happen, okay. The third condition is that is not very strong. Psi 2 inverse psi 1 from C1, B1 to a2, C2, right. C1, B1 goes into U1 intersection U2 and comes back to a2, C2. This must be order preserving. Look at this one. If I am working inside x, I do not have any order. Therefore I take two homeomorphisms here, take the intersection, take on the intersection, go to 1 and come back. So this way I am getting a homeomorphism from interval to interval. Here I can talk about whether it is order preserving or order reversing. It is very important but it can be, either of them can be handled. So I am assuming order preserving. With all this, the conclusion is very nice. Namely, the union is homeomorphism between open interval, okay. So what has happened is you have an open interval like this and another like this, okay. So what you have assumed there is they are intersecting like this. The union will be again an open interval. So this portion is there fully, this portion is there. It is not something like this open interval, that open interval intersecting like that, okay. It is not of that nature, okay. Let us go ahead, see whether we can do something. So this is, we want to prove this. Union is homeomorphism between interval. Very easy proof. Pick up any point D1 between C1 and B1, okay. Let D2 be the unique point inside A2 to C2 such that if Psi1 is defined here, Psi1 of D1 is equal to Psi2 of D2. Remember C1 to B1, C1 to B1 is intersection of this one. So image of Psi1 and image of Psi2, they are same here in intersection in U1 intersection U2. So if I take a point here D1, Psi1 of that, okay, that will be in the intersection. It will be Psi2 of some point. Well that point must be between A2 and C2. That is all I have written down. This is nothing very, very special. You pick up one point and choose the other point so that Psi2 of D2 is equal to Psi1 of D1. Now you take this number B1 prime as D1 plus B2 minus D2, okay. So add to D1 you add this B2 minus D2. Now I am going to define a map P on the interval A1 to B1 prime, okay. A1 to B1 what I have? I have Psi1. But I do not want to take the whole of Psi1. I will take Psi1 of T up to D1. After that I use Psi2 of T plus D2 minus D1. I can use Psi2 but I have to shift the origin because at D1 these two, these common thing they should agree, right. Psi1 of Psi2 should agree. So Psi1 of D1 exists by D2. So it should become D2. So if you take T times T plus D2 minus D1, T equal to D1, what happens? D1 and D1 cancel. This is Psi1 of D2. So Psi1 of D1 here, Psi1 of D2. So they coincide. So this map is well defined. This function is well defined. This part is continuous. This part is continuous. So phi T is a continuous function from this interval A1 to B1 prime inside X. So how it goes inside X? Infide actually U1 union U2. It does not go out at that. So all the points of U1 union U2 are taken care. If they are inside U1, then this Psi1 takes care of up to D1. Beyond D1, Psi2 will take care of. All the points of this one are taken care of here. So phi is subjective on to what? U1 union U2. I should say that one here. So subjective means U1, U2, U1. Because of 3, what is this 3? That depends. It is order preserving. What happens is, it is very easy. This function becomes injective. If two points are in this interval, Psi1 will take care of that injective. If they are in this interval, Psi2 will take care of it. The problem is suppose you have a point here and a point here, then you have two different formula. Why they are not equal? Yes, they are equal only when D1 is equal to this one, the end point. That is all. For that, you have to use that it is order preserving. Just write down. Suppose T is in this interval and S is in this interval. Phi of T equal to Phi of S, which just means Psi1 of T equal to Psi2 of S. See that use that order preserving thing. You can easily prove that this is injective. To see that phi inverse is injective or is continuous, this is bijection. Now look at phi inverse. Phi inverse injective, you have to observe that on Psi1 of A1 D1, A1 to D1 here, we have what is the inverse of this one? It is Psi1 inverse. On Psi2 of D2 to B2, which happens to be in this interval D1 to B1 prime, it is, this phi inverse is, it is not Psi2 inverse directly because there is a shift here. What is that shift? It is a translation by this T. T is, what is this translation? It is X plus D1 minus D2. D2 minus D1 is added. The inverse will be X plus D1 minus D2. After taking Psi2 inverse, you translate this one, then what you get is the inverse of this map, phi. So both of them are continuous and they agree at the point D1. Therefore, the entire theorem is continuous. So this proves that the union is homomorphic to an interval. That was the statement here. Next time, we shall do another thing but we will wait for that. Another wishful thinking is, is fulfilled. The wish is fulfilled. So that will be next, we will do that. And after that, we will vote the full classification. Thank you.