 Hello, welcome to module 49 of NPTEL NOC, an introductory course on point set topology particle. So, we continue our study of ordinal topology. In a totally order set, every sequence has a monotone subsequence. This is an elementary result which goes under the name peak valley lemma. Maybe you have not seen it. On the other hand, this is actually important for us now. So, I would like to recall it completely. So, what is the meaning? See, you can imitate the real numbers. Real numbers, maybe you know this one, but do not use the full properties of real number. You have to use only that is totally order. No addition, no subtraction, no multiplication and so on. That is the whole idea. You can still prove this one just by using totally order. Every subsequence has a monotonically increasing sequence or a monotonically decreasing sequence. Some monotone subsequence should be there. That is the thing. So, fixing a sequence, consider the following two properties. Given any n inside n, you will have m bigger than n such that this s of m is you know less than or equal to s n follows s n. So, this is not to be confused with the order of real numbers. That is all. So, I have carefully written this one. s m is s n is s n. There is a larger m for which s m is smaller. Second one is given n inside n, there exists larger m such that s n is greater than s n. Rather, rather wrong. If s satisfies d, it follows that as such a decreasing subsequence. Similarly, if f satisfies i, it will have the increasing subsequence. That is why I have put d and i. So, decreasing sequence, increasing sequence will come forward by this property. So, there is nothing to prove. Therefore, let us assume that neither d nor i is true. Under this, let us see what happens to the sequence. So, if possible, let s be such that neither d nor i is true. Starting with s 1, for definiteness sake, we may assume that s 2 is bigger than 1. It is less than or equal to s 1. One of them has to be there. So, this is a symmetrical operation. So, we assume s 1 is less than or equal to s 2. Since s does not satisfy i, that means what? If you took s 2, I must be getting s 3 bigger than that. Some s and s k bigger than that. That is not possible. So, that is a positive. One, we always do not find. But is it true? It is not true indefinitely. It follows that there will be one n 1 belonging to n such that for all m bigger than n 1, s m will be smaller than s n 1. You cannot go further. So, pick up that n 1. What you may assume is that I have started the sequence at s n 1 itself. Just forget about the earlier part. Now, what happens to s n 1 plus 1? It has to be less than or equal to this one. m is bigger, h n plus 1 is bigger than m. So, it must be less than that. So, that is why I could have assumed. See in the beginning, I assume s 1 is s 2 is bigger than n 1. Now, I have another thing. Namely, this s n plus 1 is smaller than s n plus 2. The role is interchangeable. But now, can I keep going down and down and so on? d says it is not possible because I have assumed that d is not true. So, d is not possible. So, that means what? You see, this is what the picture says. I started with s n s 1 here. I keep going up. Maybe s 2 is there. But maybe s 5 is there. I have come somewhere here s n 1. After that, if you look at the whole thing, everything is smaller. There is nothing bigger than this. What is the meaning of this one? This is a peak. So, that is why p 1. So, now the next one is definitely smaller than that because everything is smaller now. So, you keep going down, down, down. If you have what? Subsequence which is decreasing. You keep coming down. In between, there may be something more. Don't worry. If you keep coming down, finally you will be stuck with something, namely this v 1. So, this is a valley. What is the meaning of this? Everything after that is bigger than p 1. So, they are above. But they are all smaller than p 1. See, because the p 1 has a property, this is bigger. So, repeat the process whatever you have done from here. The next one will be bigger. Keep going up. How far you can go somewhere wherein the same thing what happened to p 1 will happen to p 2. Namely, everything after that is smaller than p 2. Keep repeating. You get p 1, v 1, p 2, v 2 and so on. An interlaced, what is interlaced subsequence. What is the property of p 1? It is bigger than everything after that. What is the property of p 2? It is bigger than everything after that. Therefore, p 1, p 2, p 3 is a monotonically decreasing sequence. Indeed, you are getting two of them. v 1, v 2, v 3 will be monotonically decreasing sequence. So, either of them contradicts the assumption. Therefore, there must be a subsequence which is monotone. So, that is the proof. So, I have written down it completely. Started with n 1, it pick up n 2, etcetera. So, it is that n i is now bigger than s of n i plus 1, but the assumption does not allow you to continue. So, you have to stop somewhere. So, you have to keep changing the role each time. So, finally, you get two monotones of sequences like this. One is increasing, one is decreasing. The next thing about now I come back to the ordinal 0 omega. Both 0 omega closed and 0 omega open, they are sequentially compact. Remember one sequentially compact, every sequence has subsequence which is cannot be changed. So, we are discussing this sequential compactness. So, this 0.9 here is done in that background. So, how do you prove that something is sequentially compact? Start with a n, any sequence in 0 omega. As seen above, just using total order. First of all, it is bounded. Any sequence is bounded that we have seen before and before that. In a well-ordered stack, we have seen that every countable set is bounded. It is bounded inside 0 omega itself. Therefore, a monotone subsequence will be also bounded above. Bounded above is what is important. Going down is always true here. But then it will convert to either supremum or supremum. According as the sequence is decreasing or increasing. Depending upon decreasing sequence, it will convert to the infimum. Increasing sequence will convert to supremum. Therefore, we have proved that every sequence has a convergent subsequence. So, 0.9 is used only to get a monotone subsequence. This monotone subsequence, I do not know whether it is increasing or decreasing. So, both of them you have to take the care. Both of them will converge in any case because it is bounded. This is also standard result in analysis. Namely, in real line, if we have a bounded monotone sequence, it is converted. There it is the property of the existence of least upper bound and the latest lower bound. So, we have more or less proved that theorem also here in this approach because all that you have to assume, all that you have to do is here is just a total order. That is all. And of course, assumption that least upper bound and latest lower bound exist, which we have proved separately for this 0 omega or well order sets. The next thing is 0 omega open is first countable and T1, actually T2 we have seen. And then the above result implies that it is limit point compact and countably compact also. So, I will try to recall these things. So, let me go through this one. Sequentially compactness implies limit point compactness under the axiom T1 and first countability. So, the converse is also true. Always sequential compactness implies these two, but under the T1-ness first countability then T1-ness, so they will imply limit point compactness implies sequential compactness. This is what we have seen. You can have a look at this picture. Remember, so this picture will tell you at this point. So, this is limit point compactness here. This is countable compactness here. Plus T1, limit point compact plus T1 will imply is countable compactness. T1 plus first countability limit point compact implies sequential compactness. From here to here you can come back. So, you can recall. I have just recalled that one for your ready reference. So, let us go back here now. So, 0 omega is first countable. 0 omega closure is not first countable. That is what we have seen. And of course, it is T1 because it is T2 also. Hence, the above result that it is sequential compactness implies that it is limit point compact and countably compact also. Of course, 0 omega, since it is not compact, this implies that 0 omega open is not Lindelof. Alright? Because once it is Lindelof and countably compact, it will be compact. Remember, Lindelof means what every open cover has a countable sub cover. Countably compact means what? Every countable sub cover has a countable cover has a finite sub cover. So, combining these two would have got compactness. So, important thing we have derived is 0 omega is not Lindelof. Okay? Next, if A and B are any two nonempty disjoint closed subsets in 0 omega, then at least one of them is countable and its closed ring 0 omega. So, we are slowly going towards normality here. So, take two disjoint closed subsets of 0 omega. Okay? Of course, nonempty are starting. At least one of them is countable. It is a very strong condition and that countable set is closed in 0 omega, omega closed. A closed subset here may not be closed here, right? How? Because this capital omega may be a limit point. So, this says that capital omega is not a limit point. This we know already because any countable subset of 0 omega is already bounded inside 0 omega. So, this part we know already, right? So, let us see how this one works. Part of it we have already seen. Let y bar denote the closure of y in 0 omega, omega closed in the larger space. Just for in this part. Otherwise, closure closure where 0 omega or 0 omega, I will have too many notations here. I do not want to have that. Okay? So, for any subset y of 0 omega, let the y bar, you know, the closure is 0 omega closure. Now, look at the case wherein a bar union b bar is contained inside 0 omega. It need not be, this is a case I am talking about. It need not happen this. Suppose this is true, then both are bounded in 0 omega, right? Because they are closed subset of now inside 0 omega. They are closed subset. Content in here does not matter. So, both are inside this one. Therefore, from our larger result, which we have just seen last time, both of them are commutable. Therefore, a bar is already a because a is closed, you know, inside this one, a is closed in 0 omega, b bar is already b and then both a and b are closed in 0 omega and commutable. Okay? It is a stronger thing we should have. So, we are through. So, now come to otherwise means what now? Suppose a bar in b bar will contain the this element capital omega. Otherwise, by symmetry, you may assume that capital omega is inside a bar. Okay? Interchange a and b if you want that is all. Suppose this capital omega is in b bar also. That can also happen, right? Then given b naught inside 0 omega open, we have this a intersection b naught omega closed is non-empty because all these open subsets must intersect a because omega is inside, capital omega is inside a bar. It is non-empty and then we get an element b naught. We get element l and b naught is there, b naught less than a1 inside a. Okay? Now, apply the same thing to a1 to omega. This is neighborhood of capital omega, intersection b that is non-empty. So, you get an element b1 inside b bigger than a1. Repeat this process. What you get? You will get an internalized sequence a1 less than b1 less than a2 less than b2 less than and so on, right? So, where are they going? Each time they are inside a smaller thing, the increasing sequence that is the whole idea. Okay? So, we observe that this internalized sequence, okay? They must have a same supremum. They have same limit. This is what we have seen earlier and that limit must be inside 0 omega. Okay? But then because it is any sequence which converges has to converge inside 0 omega only. The sequence is inside 0 omega. So, it does not converge to capital omega. No sequence converges to capital omega. But then s itself will be in a bar intersection 0 omega which is a and s is also in b bar intersection b. This means s is inside a intersection b and that is a contradiction because the intersection b is empty to begin with. So, they are disjoint closed subsets of 0 omega to begin with. So, therefore, if we assume a is inside a bar instead of omega is inside a bar, then omega is not inside b bar. But that means b itself is closed inside, b itself is closed inside 0 omega. Okay? Anything is closed inside 0 omega that is a compact. But now this compact subset is inside 0 omega open. Therefore, b is countable. This we have seen already. So, we wanted to prove one of them is countable. Right? Automatically, it will be closed. That part will be there. Countable and closed here. That is all. Now we will use this one in a meaningful way. Namely, to prove that 0 omega open is normal. We already know that 0 omega open is Hausdorff. Use this fact that 0 omega closed is normal and T2 actually is a compact and T2 space which is normal. Right? Therefore, starting with the two disjoint closed subsets here. Okay? You go to the closures of them inside 0 omega closure. Just now what we have shown is they are themselves closed. The closures are closed and disjoint. Because only omega bar is the question that belongs to only one of them. What does that mean? You can separate them inside 0 omega bar, 0 omega closure by two open subsets, disjoint open subsets. Intersect these disjoint open subsets with 0 omega open. So, that will give you disjoint closed subsets of these two closed subsets. Disjoint open subsets containing respectively these two open subsets within 0 omega open. Okay? That this space is normal. It is already T2. So, it is a T4. Okay? The next thing is the aim of all this. If you take the product of 0 omega open cross with 0 omega closed. Okay? Let us denote this one by T. This is going to be a wonderful example now. So, I am not taking 0 omega cross 0 omega. That will be compact cross compact. It will be compact. Okay? So, this is going to give you a number of counter examples. Okay? So, now we have some notational problem here. See, we have been using the ordered pair kind of A, B to denote intervals. Both inside real number systems as well as any ordered set. Right? But now the ordered pair we have to write here. So, I will not use that notation for ordered pairs. I will restrict it to open intervals only. So, ordered pairs will be denoted by X cross Y now. So, this is an element in the standard ordered pair notation. It would have been X comma Y brackets. Right? So, that notation will not be used for some time in this part now. Okay? So, with that convention, we shall show that T, this product is not normal. This is normal. This is actually compact order set. Yet the product is not normal. So, that is a strong example for all. You know, product is not normal was promised in part one but we did not prove any examples there. So, here is an example. Beautiful example. So, let us have this standard notation. Delta denotes the diagonal X cross X where X varies over 0 omega closed. But I do not want to go to 0 omega cross 0 omega but I want to be inside T. Take A to be delta intersection T. Okay? So, last point omega cross omega is not there. That is the problem. And B equal to 0 omega open cross singleton omega. So, second factor has omega here. Right? I can take that. So, this A and B are subsets of T. Obviously, they are disjoint. Alright? This part omega cross omega is not there but here there is no omega cross omega either. So, anyway, these are all. So, this is like a top line and this is a diagonal course all the way near but the intersection point is not there. That would have been omega cross omega. So, these two are disjoint closed subsets. Since 0 omega bar with this closure is half door, this delta is closed. Therefore, A intersection that delta intersection T is A. A is closed in T. Okay? Similarly, singleton is closed. So, this opens up that is closed in 0, T. So, this is closed in T. Alright? Also, A intersection B is empty. Is that clear? So, now we claim that there are no disjoint open subsets U and B inside T such that A is inside U and B is inside V. So, that will complete the proof that T is not normal. So, you have to do a little bit of work here. Assume on the contrary. That means what? Start with two open subsets, pretend they are non, you know, they contain those two closed subset A and B respectively and their intersection is non, intersection is empty. You will arrive at a contrary. That is the whole idea. Okay? Assume the contrary. Fix each, for each x belonging to 0 omega, you just fix an element 0 omega. Look at the subspace, the vertical subspace mx equal to singleton x cross 0 omega. That is an element of T, that is a subset of T, right? And it is homomorphic 0 omega because I first fact prime singleton set. Then if you take V, U and V are open subsets, right? V intersection mx, that will be a neighborhood of x cross omega. X cross omega is one of the points inside B, right? So, V contain B. So, x cross omega is inside V. So, it is also inside mx. So, this is a neighborhood of x cross omega inside mx. Okay? Therefore, V intersection mx minus singleton x cross omega is a punctured neighborhood. That is non-empty. Okay? It is an open subset and this is a singleton. Okay? This is non-empty subset. If you put rx equal to all of second coordinate y such that this x is less than y. Okay? This is our old notation, sorry. This is just the right ray. It follows that this is this non-empty set. What is this? V intersection mx minus this one. This is non-empty set. But this is actually contained inside x cross rx minus U. Okay? So, let me show all these things in a picture just to see that you are not completely lost. So, this is my T. This is 0 cross, you know, this is just 0 cross. This whole thing is omega open and this way I put 0 comma omega closed. So, that is a cross. This is the second factor. This is the first factor. Okay? So, I have taken U here, this A here. This is the diagonal minus this point. This point omega cross omega is not there. Okay? Now, B is here. This is just this singleton x, singleton 0 cross or this B here except this point is not there. So, these two disjoints are set and I have taken this dot dot dot here U is an open neighborhood of this A and V is an open neighborhood of V and I am pretending they are disjoint. What is happening when you keep coming closer to closer? I do not know. Okay? So, that is where the mystery is there. Finally, we will get a contradiction there. So, what I have done? Fixy x here on this axis, similarly x cross 0. Look at this vertical line. Okay? That is my Mx. This V intersection Mx, this portion is the neighborhood of this x cross omega. If you throw away this point, this part is non-empty. Okay? This U is disjoint from that whether I throw it away or not, this will still be non-empty. That is all I am telling you. Okay? So, Rx is what? Take any x here, x has come here somewhere. Okay? x is here, x comma x. So, take everything above that, that is Rx. So, this is from here to here, it is x cross Rx. Throw away U, this is U part. Okay? So, this part will be still non-empty all the time. And it contains V minus, sorry, V intersection Mx, this line minus this point. So, that is all so far we have done. This V intersection Mx minus x cross omega, it contains the x cross Rx minus 0. Okay? So, that is non-empty set. So, we can define a function here. Namely, take the minimum of all elements y such that x cross y is inside this non-empty set, x cross Rx minus 0. So, what is this in this picture? This is the minimum alpha x. Okay? And all y, second coordinate, first coordinate x, x cross y such that, you know, they are inside this Rx minus U. Okay? So, that is the definition, x cross Rx minus 0. Take the minimum. So, this is the function now because these are non-empty subsets for each x. Okay? Alpha x is some point of 0 omega for each x. Okay? Starting with any x one times at 0 omega, inductively you define xn equal to alpha of xn minus 1. So, you see now the induction process starts. Maybe I started here. Okay? Now alpha of this one will be some point here. Okay? So, that point will be somewhere here. Then you have to take alpha of that will be some point here. That point will be somewhere here. Now alpha of that point will be somewhere here. You take alpha of that and so on. Okay? So, this is the induction process here. Namely, starting with x1 belongs to 0 omega, anything wherever you want to start. Inductively define x2 equal to alpha of x1, x3 equal to alpha of x2 and so on. So, obtain a sequence xn inside 0 omega which is monotonically increasing. Why? Look at this one. This alpha x is bigger than x because x, x is inside u. So, no chance that this alpha x will be the minimum of this one will be less than that. It is actually away from that. It is bigger than that. Okay? So, this is a strictly monotonically increasing sequence, converging to some point in 0 omega. Any sequence in 0 omega which is monotonic has to converge inside 0 omega. That is all. But then the sequence xn cross alpha xn, where does it converge to? xn is going to, why? Alpha xn is bigger than that. So, it must go to bigger than that one. But alpha xn is dominated by again xn plus 1 and so on. So, alpha xn is nothing but the same sequence xn, except the indexing is changed. So, both of them converge to same point. So, their same sequence is except indexing is changed. That is all. So, it will converge to y cross y inside a which is contained inside u by definition. Whereas, none of the terms of this sequence is inside u. Right? Look at this point, x comma alpha x is this point. Right? When you take xn plus 1, xn plus 1, alpha xn plus 1 will be in this point like that. Though they are not inside u. So, that is a contradiction. So, this proves that t is not normal. So, finally, why this notation t? Because this is called Ticknoff's Planck. This example is called Ticknoff's Planck. In conclusion, we say normality is not finite productive. As a conclusion, we can say that Ticknoff's Planck is not para-compact. Para-compact, Hoss-Dorff's space is para-compact. Regular space will be normal. That is what we have seen already. It follows that 0 omega is not para-compact because para-compact, cross para-compact would have been para-compact. So, 0 omega itself is not para-compact. So, in one single example, you have so many contra-examples here. There will be more to come. This is another wonderful property of 0 omega. Every continuous function 0 omega to R is eventually a constant. Real value continuous functions are all eventually constant. See, the role is reversed here. Usually, if you have a connected space and a continuous function R into a disconnected space, a connected space into a disconnected space or a discrete space, then it is a constant. That is the way you have been all the time doing. So, this is highly connected. This is a nice connected space. This is the one which has problems in a disconnected space. But this is the other way around now. This is a eventually constant. So, what is the meaning of that? There are some x inside 0 omega such that f of the right ray Rx, x2 omega, that is a single term. By composing with a homeomorphism g from R to 0 1, because any homeomorphism, you can compose with a 0 1 because you can take a tannin or whatever. There are many. You may assume that the function is bounded first of all. It is actually taking value 0 1. If I show this is eventually constant, then f will be also eventually constant. So, let us assume f is bounded. Put Rx equal to all y inside 0 omega. This is the old notation. I am changing here. Now, I am taking a closure here instead of R bar of x. We shall claim that there exists a monotonically increasing sequence xn in 0 omega such that the diameter of f of Rxn, f of Rxn is instead of R. So, diameter of that makes sense because this is a matrix space. That is less than 2 third rays to n. Some number raised to n, that number is less than 1 is important for r. Then, if you take x as a limit of xn, it will follow that f of Rx is a singleton. This kind of term, you know balls of radius smaller and smaller converging to 0, it will be in the intersection f of Rxn. So, it must be singleton. So, that is the way we are going to prove this theorem. Just by proving that, you know a construction of monotonically increasing sequence in 0 omega, which will automatically converge and that converging point will be our x. Then, f of Rx contained inside each of them being the smallest subset of that. Therefore, it is contained in the intersection. Therefore, it is happening right now. So, we will have to construct this sequence. Given, so we are going to prove slightly different statement here, inductive statement here and then apply it. So, this statement is followed. This s given x belong to 0 omega, let g from this right ray to AB be a continuous function. Right now, recall this is I have taken as a closed ray. Suppose you have continuous function, which is bounded. That is the whole A to B. Then, there exists a y bigger than x. That is an element of Rx such that g of R y is contained inside the first two-third or the second two-third. The first two-third means what? A to A plus two-third of B minus A. B minus A is the length of this one. I am adding that. Or it is A plus one-third of B minus A to all that A B. So, that is again a two-third length. The first two-third or second two-third. They are not disjoint by the way. Remember that. That is fine. I want to say it is either here or here. It may be in the intersection also. That is good enough for me. No problem. It is contained in one of them is important. That is what happens. Then, we construct a sequence x n as follows. See, first I had A B. Then, if this A B is zero one, then what happens? R of g of R y will be the maximum radius of the diameter of that one will be two-third. Apply this g. Apply the same thing to this function. Repeat this process. You will get another y two such that diameter of the new thing will be two-third of that. That is what we are going to do. So, then we construct a sequence x n as follows. Start with x naught belongs to zero omega. Take g equal to s. Take x 1 equal to y as given in s. Having defined x n, repeat the step to the function g which is restricted to f x n. You get x n equal to 1. This s is a general statement. I am starting with this hypothesis. f is from zero one to zero one to r, but I have converted the g from r. This f is zero one to zero omega to zero one. So, begin with radius is one. This is the second stage. The first stage I am getting the diameter two-third. Next, I will get two-third into two-third and so on. So, we have to prove this statement. Once we prove this one, this inductive step is over. Then the proof is over. So, let us prove this statement in s now. So, let us prove. Look at the two disjoint rows of sets A plus A comma A plus one-third of B minus A and g inverse of A plus two-third of B minus A to B. This means these are one-third one-third now. Like cantor set, okay, in the construction of cantor set. A to one-third length and then one-third length to entire B. They are disjoint. If you take g inverse of that, they will be disjoint subsets of zero comma omega. What do you know about the disjoint rows subsets of zero comma omega? We have done something, you see. So, that can be used now. As seen before, one of them is countable, okay, and so as an upper bound, say the first one is bounded by y. Which one? This one, that one, does not matter symmetrically. So, let us say, assume this is bounded by y. This y is an element of what? Element of g inverse of, you know, sorry, element of the bound of the main leaf AB. g is a function, sorry, in rx. So, y is bigger than x, that is all, okay. So, we get a y. That is all as seen above. One of them is countable and hence we have this. This means that g of r y is contained inside this part. Likewise, if the second one is bounded by y, then you will get g of r y is contained inside this part. So, that is all, okay. So, why I separated out this one, the normality, directly instead of proving a and b are disjoint rows subsets, find open subsets and so on. Disjoint rows subsets have themselves some interesting property. Namely, one of them has to be countable and that is here. So, next thing is a strong conclusion, okay. The first thing is 0 omega closure is the Alexander compact, one point compactification of 0 omega open. This is very easy to see by the very definition of this, there is only one extra point, right. What are the neighborhoods of this omega? Take any neighborhood system, something like alpha and x open to omega capital. Compliment is what? Compliment is a bounded subset, it is countable, right. So, it is compact also. That is the definition of neighborhoods for Alexander of one point compactification. The complements must be compact subsets of the given space, okay. In particular, if you want Alexander's compactification, you should see that this is a horse door space to begin with 0 omega and it is locally compact also, okay. Namely, finally it is a subspace of a compact horse door space. So, it is locally compact also. So, Alexander's one point compactification makes sense. If the extra point here is denoted by infinity, that infinity can send it to this capital omega to get a homeomorphism, that is all. So, whichever way you want to, so it is easy to see that this is the Alexander of one point compactification of the open 0 omega. It is also the stone-checked compactification because of this property that every continuous function 0 omega to r is eventually a constant. Therefore, you can extend it, so you can take 0 omega open, 0 omega, instead of taking the whole of r, take just the real rad functions taking value inside this closed interval, okay. It must be eventually constant, right. Once it is eventually constant, you can extend it continuously to take f of capital omega to that constant. So, every continuous function from 0 omega to 0 1 can be extended to 0 omega closed. That is the characterization of stone-checked compactification. So, that is what we have to remember now, okay. Any space which contains this one and compact, okay, saving this property of extension of continuous functions into any, into 0 1, closed interval 0 1 has the property that it must be a homeon, you know equivalent, equivalent to the stone-checked compactification. That is the meaning of characterization of stone-checked compactification. Therefore, the closed ordinal 0 to omega closed is a stone-checked compactification of 0 to omega also, okay. There is a final thing here. Since 0 omega is a T4 space, okay, T1 and regular that is enough. So, it is T4 actually. Its wallman compactification exists. And being normal, we have seen that it was an exercise to you. The, if you have T4 space, the wallman compactification is house door. So, we have got this one as a house door space. Then we have made a remark that whenever the wallman compactification is a house door space, it is the same as stone-checked compactification, okay. So, that is another remark which we have studied all these things. So, we conclude that 0, omega closed is three different compactifications of 0, omega open. I do not know any other space having this such a property. So, here is some more concluding remark. Though 0 omega is compact house door, it has a subspace Y which is not compactly generated. We wanted to have an example of a space which is, you know, compactly generated. The subspace is not compactly generated. This seems to be a simplest example. I have studied all these. This now is easy example for us. The original space is compact house door. Obviously, it is compactly generated. But the subspace, we have a subspace here which is not compactly generated. So, what is that subspace? Let us see. Take Y to be obtained by 0 omega, you know, as a subspace of this, by deleting all the limit ordinars except the top one, capital omega. Like for the first thing I will omit is little omega. Then I will omit twice omega. Then I will omit three times omega. Then I will omit omega square and all those limit ordinars are omitted. I will deal with the last one I will keep. We first claim that every compact subset of subset K of Y is finite now. Every compact subset of this one which does not contain omega, we have seen that it is countable. Now we are claiming something more stronger. Every compact subset of this subset Y, automatically it is compact in 0 omega but it is subset of Y. Then it is finite. For K will then be compact as a subspace of 0 omega. If K were infinite, first of all it is countable. If it is infinite countable, we can then extract a strictly increasing sequence in K which will converge to 0, converge to some point in 0 omega. And the limit point that will be limit ordinal. The limit being the limit point, it will be inside K but K is inside Y but Y does not contain any limit ordinal if inside 0 omega open, so that is the countable. So K has to be finite. Now once you have that, it follows that if U equal to Y minus omega, you throw a omega also. This will make every compact subset of Y in a closed subset because it is just finite. Yet U is not closed. I am taking U equal to Y minus just omega. Yet U is not closed inside Y obviously because Y is the limit point of this. Remember compactly generated means what? For every compact subset, U intersection K is a closed subset inside K then U itself must be closed inside the ordinal space. And that is violated here because take any subset K which is compact inside Y that is called finite obviously. Then U intersection that will be a finite subset and finite subsets are closed inside the host of space. So they are all closed inside K. That means for every K this is true. That means U should have been closed inside Y minus omega inside Y. But this U contains this capital omega as a limit point. So that is a contradiction. Therefore this space Y is not compactly generated. I think you must have been satisfied with so many properties, wonderful properties, topological properties of this zero omega. So that justifies the efforts in studying these ordinals. So next time we will do one more example using the ordinal but we will construct something more, some more interesting things from the topology point of view. Thank you.