 Hello, welcome to NPTEL NOC, an introductory course on point set topology part 2. So, as promised today we shall construct the ordinals. In this section, we shall construct the ordinals and study some of that topological properties. First of all, we have to make a definition and two well ordered sets x i less than equal to i i equal to 1 and 2 are said to be equivalent if there is a bijection f from x 1 to x 2 set theoretic function which preserves the order. What is the meaning of that a 1 less than equal to b a less than equal to b through this one first relation should imply f a is less than equal to b a second relation f a is less than equal to f b. See this order preserving which we have used earlier, there must be an order preserving bijection. Obviously, the inverse will be also ordered. An equivalence class of a well ordered set is called an ordinal order. You see you take the class the collection of all well ordered sets that is not a set. However, we can define the equivalence relation by this process. Nobody stops for that. Then you can verify that it is reflexive, asymmetric and transitive. So, you can look at equivalence class. This world class is very important now here. This class itself is not a set. That class is called an ordinal. So, in what follows you are going to construct a set of representatives for a class of ordinal, not all ordinal, but a large number of ordinals. So, we will construct that which itself has a well ordered one set of representatives and that will be a set. So, this is the whole idea and that will be a sufficiently large number of ordinal and that itself that set of collection itself will be a lot. The construction of this ordinal space. Of course, when you say when there is a well order then we can take the topology also then we can say ordinal space. The following construction works well with any uncountable set. Now, you must know that uncountable sets exist. So, that part is elementary or whatever you know set theory of cardinals. That part I am assuming that you know it or I will pretend that you know it that is all. So, I am not going to teach you the theory of cardinals here. So, be sure of that. So, indeed this is not the correct thing to do in the logical sequence, but since you are familiar with natural numbers in whatever quote unquote whatever way you know you can take the power set of the natural numbers and call it as x. That will do our job of understanding what are ordinals to begin with. Strictly speaking you should not do that even n will be constructed out of this thing. So, all that you have to know somehow that there is a uncountable set. Now, so fix an uncountable set x start with a well order. So, this is Jermelo's theorem that we have proved. Every set can be well order. Start with a well order next. Then you add one more point. So, once again I denoted by x infinity x star equal to x disjoint in x infinity and extend the order on a order on x to x star by declaring that all the elements of x are less than or equal to this in that way which will be strictly less than because they are different from infinity. That will be automatically a well order and a larger one. Clearly x star is also well order. That is what we have seen this extension. In fact below and above also we have already done that and we have used that one. Now let us put s equal to all points of x star such that l x is uncountable. Remember what is l x? It is a left raise. Take all those for l x is uncountable. Then s is non-empty. How do you prove that s is non-empty? Well you can take infinity prime here. x equal to infinity prime here. Then what is l x? l x is precisely x and we have started the x is uncountable. So, therefore this s contains infinity prime and so it is non-empty. That is the role of infinity prime here. If I do not put this one I will be hard to prove why this is non-empty. In fact it may not be true. That trick is there. Just put one extra thing. Then you have got a non-empty set. You are sure. Therefore once it is non-empty you can take intima mo base belong to x star because x star is the lord. Take the least element and call that as omega, large omega. So this makes sense because it is non-empty. That is all. So this omega will be automatically a member of this one. So it is s equal to infinity prime. We do not know whether this omega is less than infinity prime or for that matter it is equal to infinity prime. So this depends upon the x itself but we are not bothered about it. We have started with any uncountable set. Therefore we do not know what it is. If you take natural numbers and per set then maybe you can say that this has to be actually infinity. That is a different aspect. So we do not bother. This is an element of this x star. That is all we know. Also now let 0 denote the least element of x star. This time I am bold enough to use 0 itself. I better use 0 hat or 0 prime but that is too much of work. Whenever it is needed and later on I will do that. Right now let us just have this 0. 0 be the least element of x star. Always least element exists anyway because this is a lord. We then take the order topology open intervals constituting base as we have seen earlier on 0 close and omega open as well as 0 close omega close. Both of them are subsets of now x star. And they are themselves well ordered. So you can take this topology order topology on both 0 omega open and 0 omega close. Clearly this capital omega will be in the closure of that. That much is clear. Apart from the properties we have listed above under total order except for one thing well order we did that but other things were generally total order. They are all valid here also. Now we will have more interesting properties of this space this omega 0 omega. So basic properties of this ordinal topology. Here we are not I told you I am not interested in teaching you cardinality theory. I hope you know. So we assume that you are familiar with this. So start with omega the least element of the set of all x such that cardinality of this left ray lx is not finite. For example if I take 0 lx is empty. If I take one successor successor of 0 immediate successor of 0 then it will have only one element and so on. So that is where actually Piano said you know start if you have studied it otherwise you can you know surmise here itself. So cardinality of lx is not finite. You take all that set and take the least element. The least element is also an element of this set. So this check denotes a cardinality. Then for every x inside 0 omega that means what x is less than omega right. Omega is the least element. Less than omega it is not. So it must be the cardinality of the left ray 0 to x 0 to open x is finite. We add one more point it is also finite. It follows that if you take this check as a function from 0 omega this interval to the natural number namely cardinality is now finite. It is a natural number. Take anything here it will at least have the 0 right lx will have at least 0. So it is non-empty. So natural number will be positive. So it is all 1, 2, 3 and so on you will get. You get a function here it is an order preserving bijection. You see it is an order preserving bijection. Here we have the order that is whatever we have started with. Here we are pretend we know the order of natural number. These order okay both of them are totally ordered sets. If you add both of them are now well ordered sets also. This is a order preserving bijection okay. So in our definition this equivalence class will itself denote an ordinal and that ordinal will be omega itself. That is the whole idea here okay. Indeed if you do not know what n is then 0 omega as above you can think of that as n okay and then construct the algebra out of that by using the ordinal the successor theory. That is not I am not going to do that. So here are few terms I want to recall. Every member of 0 omega, 0 close omega open is called a finite ordinal okay. Omega is called the first limit ordinal. There are other limit ordinal obviously okay. There are many more. This is called first limit ordinal. This is some name you can say things which are not finite. Amongst them omega is the least one. That is what what you are is the first one okay. So you can say not finite. Not finite ordinal omega is the least one. So that is called the first limit ordinal. Now cardinality of 0 omega is that of set of natural number because there is a bijection like that. Clearly elements of x, elements x of 0 omega close can be broadly classified by the cardinality of Lx okay. How many elements are there before that? Look at the cardinality. That is how many means okay. So it may be finite. It may be countably infinite like natural number 0 omega or it may be uncountable. I do not want to go on further classifying these what kind of countability is there. That is a you know that is for purely further logic which we are not interested in right in this space okay. We are only interested in broad classification. All elements x less than omega are finite. Omega is uncountable and all other elements are countably infinite. All finite elements bigger than 0 are immediate successor of the previous element. Omega is not an immediate successor. So all these things elementary things are intelligent. Note that there is a finer classification of ordinals and a very vast literature on them okay. You can study them elsewhere. We are not going to that. The elements of 0 omega close which are not immediate successor are called limit ordinals. I am just I am just repeating this one. Here we only define first limit ordinals but all of them which are not immediate successor are called limit ordinals okay. There are of course limit immediate successor for every element. So omega is not an immediate successor. Omega plus 1 is there. That will be immediate successor. Omega plus 1 plus 1 will be there and so on right. So again there will be another one which is not an immediate successor and then again after that only immediate successor will be there and so on. This is a wonderful space okay. Never ending. So that is why we have put this capital omega here. So that is the and this is the maximum element among all of them. There are several equivalent as well as slightly varying definitions of limit ordinals. Different names are also there. One of which we have chosen can be justified partially as follows. Why limit ordinals? Just ordinary name okay. So I am going to tell you that. Take any element in the open interval 0 omega. Suppose it is not a successor. I mean 0 is not a successor anyway. We do not know that omega will prove that soon. Immediate successor. Then there exists a strictly monotonically increasing sequence xn. Strictly monotonically increasing very important xn which converges to x. Every immediate successor has this property. Every element is not an immediate successor has this property. Because it is a limit in this then we are calling it limit ordinals. That is the justification. Soon we will see that it is not a very good justification either okay. First enumerate the countable set Lx, y1, y2 and so on. A countable set means just it can be enumerated. It can be put in the 1-1 correspondence with natural number okay. But these y1, y2 are not in the order in which you know there is an order in Lx already we have chosen. That order we do not know. This may not be the order okay. That does not matter. Just take a just an enumeration. Start with x0 equal to 0. I am constructing this sequence now. Start with x0 equal to 0. You can start with no problem. Having chosen xn, I want to define xn plus 1 inductively okay. So how do I do that? Once xn is there, look at a equal to maximum of xn and yn plus 1 here in this list. Go up to yn plus 1 okay. Take the maximum of them. Then clearly this a okay is still less than x because all these Lx are less than x okay. And this x is not equal to a plus 1 right because x is not an immediate successor. So if you even add one more you are not hitting the x. Therefore you can choose xn plus 1 such that which is a successor for xn such that which is bigger than xn such that this xn is less than xn plus 1. And yn is also less than xn plus 1 which means I am choosing something bigger than a bigger than both of them okay. That will be less than x. So what is this? This is maximum of these two. So I have given you what is xn plus 1 here okay. So why I can just do one more, one more and so on because x is never reached right. I can take one more, take x0 and one, two, three and so on. But you will never know whether you will reach x. So if you look at this list and I am going to exhaust this list by this method. Because once I am told x0 I am not interested in y1. I will go to y2. Bigger the bigger of the two I will take. Then having chosen something I will go to not y3, y4. I will go to something which is bigger than that one you see. Maximum of the two. So I will go to somewhere. This way I will be keeping on exhausting elements of lx. So that is the whole idea. That will imply that xn tends to x. So here is the caution. Do not be carried away by the above theorem. Omega is a limit ordinal. But it is not a limit of any sequence in 0 omega. If capital omega is not a successor. Yet there is no sequence which converges to this omega. The sequence being inside 0 omega. Of course you put omega, omega, omega that will converge. So we shall soon see why this is true. Observe that omega is a limit of the net. If you think of this 0 omega as the domain of a net. Because it is a directed set right. This is totally ordered set anyway. So 0 omega as the domain of a net. What is a net? Just the inclusion map. Take this net inside 0 omega close. Then this omega will be obviously a limit point of this. It is a limit of this set. So in this one single paragraph. I have both justified as well as cautioned you all. So I am giving you another justification for calling this one as limit ordinal. This time I cannot take a sequence but again take a net. That net has this as a limit. It is easy to see that the ordered topology on 0 omega is discreet. It is just like natural numbers. We have already seen that right. So ordered topology there is just discreet. However if you include omega it is not discreet. If you include omega something nice can be seen here. For singleton omega it is not open. Okay discreet means it is not open no. So to see that singleton omega you have to write it as maybe intersection of two left one left and right right right. That is the only way you can say this thing is open. So you cannot do that. But this one is a g delta set. So it is a countable intersection of open sets. Okay. What are they? Start with any n go to omega closed. That is an open subset in 0 omega. As n keeps increasing intersection over all n we will be left with only singleton omega. Okay. Indeed there is another way of looking at is 0 omega. There is a ordered reversing homeomorphism 0 omega to the space 1 by n n belonging to natural number include with 0. Send each x to 1 by n where n is its cardinality of Lx. Then this omega itself we send it to 0. So that will be a continuous function. That will be a bijection. It is an order reversing homeomorphism. You can take this also as the order of set. Okay. Order topology then you will have this. Okay. 0 is a limit point. So that is the way you have to think of this one. This omega is a limit point of 0 omega open. A subset A of 0 capital omega open is bounded in 0 omega. Bounded in 0 omega is very important here. Because everything is bounded in 0 omega if you include omega you know included. Okay. This is bounded above if and only if it is countable. If bounded below is obvious. So we are only interested in bounded means bounded above. Okay. Everything is bounded below in a well order set. Alright. So if it only if it is countable. Only countable sets and all countable subsets are bounded. Inside 0 omega itself. So this is the crucial thing I said to wait a minute. Wait a minute. Okay. Slowly we are building it up. So why? So in particular if this is the case then it happens that every compact subset of 0 omega is countable. Because compact subsets we have seen has to be bounded. That we have seen last time. Okay. So bounded subsets are countable is what we have to prove. Of course every subset A is bounded below. So we are only interested in things which are bounded whether it is bounded above or not. So let A be a countable subset of 0 omega open. To see that it is bounded above considered a subset of 0 omega it is bounded above. Right. What we have to show? Take the smallest bound namely least upper bound and show that it is inside omega open. Okay. That is what we have to show. See that it is bounded above whatever means inside 0 omega consider a subset of 0 it is bounded above and hence has a supremum S equal to sup of A. We claim that this S is less than omega. S is not omega. Then we are done. Got it? So how do we do that? Look at this set B union of all the left rays where X range is over A. So A is some scattered countable set. Right. So for each point you fill up all the things which are below that. So it is like a saturating this set. Everything below let them come inside. So that is B. So this is a very big set now. Okay. So for all X inside A you take LX and take the union. Then each LX is countable by the very definition. Okay. Of 0 omega. Okay. But X itself A is a countable, it is a countable union countable set. Therefore B is countable. Okay. But now if S is this omega. Okay. Then it follows that B must be the entire 0 omega because all of them will be there. Right. If S is supremum take any X that will they will satisfy this property. This less than that one. So B will be the whole of 0 omega. But 0 omega is uncountable. Right. So if we started with set X is uncountable then we constructed this omega. So set 0 omega is uncountable. In particular this implies that no sequence in 0 omega converges to omega. See because supremum will be inside omega. Inside 0 omega. Sequence is what countable set first of all. Right. So supremum. Supremum will be if at all if it is increasing you can take subsequence is increasing that it will be supremum. So 0 omega open. Okay. Inside that itself it will be the supremum will be there. So it cannot convert to this omega itself. Alright. So this fact is useful. So this is an extra thing I am telling already I have proved that if A is countable then it is bounded right bounded inside 0 omega. Now let us prove the converse. Suppose A is bounded in 0 omega. Okay. If X is in 0 omega. Okay. Is an upper bound for A that means bounded means what? Within 0 omega must have an upper bound. Then this A is contained inside LX plus 1. Right. In the way LX plus 1 because everything is less than X. Because since X plus 1 must be inside 0 omega because X is inside omega. Okay. But LX plus 1 is countable that is definition of 0 omega. Alright. So A is countable over. Okay. So now we come to little more serious business here. If X and Y and R they need to interlace increasing sequences in 0 omega. What is the meaning of interlace? You take X1 that will be less than equal to X2 or Y1 or X2 does not matter. X1 less than equal to Y1. X2 will be bigger than Y1. And then X2 will be less than equal to Y2 and so on. Okay. So they come interlaced. Interlacing can occur in different ways, slightly different ways also. Okay. This is one. If you change the labels there you will get the same thing. That is all. So I can define like this. Then they have the same supremum. The increasing sequences in 0 omega they are countable so they are bounded. This much we have seen. The supremum is the same. So this is similar to what happens in the real number. So I will leave it to you. Okay. 0 omega claw is first countable. Now we are talking about topology now. First countable at all points except omega. So how do we show that? Say X belongs to 0 omega open. If X is 0 then singleton 0 itself is open. Therefore it is because it is a discrete set. Singleton 0 is open. So that itself is a base. So it is a countable base. Otherwise look at open interval Y to X plus 1. Okay. Where Y is less than X. Look at all this. Now X is fixed. I am looking at all Y less than X. Okay. Then take interval Y to X plus 1. So it will cover X. These are open intervals. So this will form a countable set because you can't have any more than countable elements Y less than X. But this is a local base. Every open set must contain one of them. Okay. It may contain X plus 1, plus 1, plus 1, so on. But it can up to X plus 1 and we have to be there. And below it may be, you don't know how far you have to go. So you have to keep coming smaller and smaller. So there is always this. Alright. It's a local base at X. See, this X itself may not be a successor. That's why I have to do this trick. If X is a successor, it will be like 0. Okay. Lower part you don't have to take. And upper part you can take X plus 1. So singleton that will be an open set. So all immediate successors are singleton sensors are open. Now to see that there is no countable local base at omega. See, I said accept at that point. So why there is no countable local base at omega? Let us suppose there is one such. Then it follows that there is a countable set U. May Y n. I can denote them by Y n and inside natural number. Contains a 0 omega such that the half closed interval is Y n to omega closed because beyond that we don't have anything. This must be a local base. Each opens of set must contain such an open set. This part. So these themselves must be local base. Now you take supremum of this Y n because a countable set now, supremum of this set now. We have seen in 5 that S must be inside 0 omega. Now if you look at the open neutral S2 omega, no element will be there at all. None of them will be contained at this point. Because S is supremum. So this subset is not, this subset is an open neighborhood of omega which does not contain any member of U. Now that is a contradiction. This is more S1 actually. It is intersection of all of them. When you take supremum, you take intersection of Y n omega. This will be precisely S omega. So 0 omega, omega open is not separable. This is the next thing. For countability we have seen but separability is not possible. And hence of course the larger space is also not separable. Because separability is after all hereditary. So we shall prove this one is not separable. Automatically you follow that 0 omega close is not separable. So why this is true? Take any countable set. What happens? You will have supremum. If you take that supremum plus 1 that will be still inside omega. So that will not contain any of this. It will not be dense at all. So plus 1 is single 10 open. So it will not intersect this one at all. That is all. So these are the easy consequences. That 5 is the key here. Bounded subset is countable. And if it is not lived it is countable. Something is bounded inside 0 omega. So now it is countable. Well, so let us stop here. Next time we will see some more serious topology of this ordinal. Thank you.