 So, having seen a number of examples of CW complexes, we return to the more elaborate study of topological properties of CW complexes. To start with a CW complex X, then X power k is the kth skeleton, namely all, the union of all simplex is all cells of dimension less than or equal to k, right. So, this each skeleton Xk is a closed subset of X. This is the first statement. The second statement is each closer cell is also a closed subset of X. It is not just a closed cell, it is not I just name, but it refers to closeness of the underlying set inside X. A subset S of X is closed if it is only if S intersects each closed cell in E k, please call cell E k of X in a closed subset of E k. Topology on X is compactly generated. So, these are the four observations you want to make about the topology of a CW complex, okay. Remember a subset is closed in X if and only if it intersects each skeleton, okay, in a closed set. That is the definition of the topology on X, right. If X is actually a skeleton, then we do not need this definition because each skeleton Xk is obtained by attaching k cells to Xk minus 1. Therefore, the topology is well defined there as a quotient topology of a certain object, all right. So, this definition, this statement, statement A is needed only when X has infinitely many skeleton, the dimension of X is infinite. So, first of all, each X is a closed subset, how do you see that? Look at Xk intersect with Xl. If Xk, k is smaller than l, it is just Xk. If k is bigger than l, it is Xl. So, either it is Xl or Xk, they are closed subsets is what you have to show, right. Then only each Xk will be closed subset. If Xk is a closed subset, if Xk intersects in Xl, inside Xk is a closed subset. Now, if Xk is contained in Xl or l is bigger, then we know that Xl is obtained by attaching cells to Xk minus 1, Xk plus 1 and so on, Xl, Xl minus 1, etc. So, each Y, when you attach cells to Y to get X, we have seen that Y is a closed subset of X, okay. So, successfully Xk will be closed subset of Xk plus 1, closed in Xk plus 2, closed in Xk and so on, Xk will be closed in Xl. If l is bigger than l, so you can interchange the two roles. Therefore, intersection of Xk with every other thing is closed, therefore, Xl will be closed, okay. So, this much argument is needed here to say that Xk is closed. Next one, each closed cell is a closed subset of X. To see that, I have to show, again, take a closed cell, say let us call it as sigma, intersect it with each Xk, okay, then it must be closed inside Xk. If this cell, say cell itself is of dimension, say l, all right. If k is bigger than l, it will already contain the entire sigma, right. And each closed cell is, we have seen, when we attach a cell, they take a l, sigma is a dimension l, then it will be closed in Xl, right. Once it is closed in Xl, it will close in Xl plus 1, Xl plus 2 and so on, Xk also, all right. So, if l is smaller than k, it is fine. If l is bigger than k, what happens? Then the intersection, let us take a variable, sigma is of dimension k plus 1 and you are intersecting with Xk. What is the intersection? Intersection is the boundary and we have seen that the boundary is closed, okay. So, this way you will see that each cell is also a closed subset of X. The next thing is, take a subset S. If it is closed, we find only if, one part is obvious. If it is closed, then the intersections are all closed because we have shown that ek's are all closed. Intersection of two closed sets is closed. Now, we have to look at the converse. The converse, S intersection ek is closed in ek, right. But we want to show that S intersection Xk is closed in Xk for every k. Does that follow from this one? If this Xk, if k and this one are same, for example, this ek is closed in Xk and so S intersection ek is closed in ek and ek is closed in Xk, so it is closed in Xk also, okay. So, showing that X intersection, since S intersection Xk is closed, will follow from, I mean, very easy for it is closed in ek because this is ek's closed, okay. So, so, so C also follows. Finally, D also will follow for the following reason. What is this? What is the meaning of compactly generated? The same thing as instead of ek, I have to put a compact set for every compact set of X, original X. Take a set S such that S intersection k is closed in k. So, this should be true for every compact. Once this is true, S must be closed. Again, we have to show only this, only this part. The other part, if S is closed, S intersection k will be closed in k. That is obvious, okay. Take any compact set. We know that it is covered by what? Finitely many closed sets. Therefore, S intersection k will be the union of X intersection some sigma. The sigmas are finitely many of them. So, union of finitely many closed sets should be closed. So, S intersection sigmas are closed inside actually in the whole of space. Therefore, they will be closed inside k. So, S intersection k which is union of finitely many of them will be closed in k. And conversely, for each, for each ek you can show that actually that is what we wanted. For each ek is compact is what we have seen already. For every compact set is closed. So, in particular for each compact set also it will be closed. So, all these things are written down carefully here. So, let us go through this one once again. So, first we want to show that this k skeletons are closed. So, we have seen that if X is obtained by attaching kth l, then why is a closed set up? From here, it will follow that X naught is closed in X 1, X 1 is closed in X 2, X 2 is closed in X k and so on. Therefore, each X i is closed in X i plus n where n is any extra positive number. Therefore, if you reverse the rule either X k cross intersection X r is either X r where r is smaller or X k where k is smaller for the two closed sets. So, part b X naught being a discrete space is ausdorff. We have to take ausdorffness, the ausdorffness the ek's are not closed, need not be closed. We have you remember that by clock corollary whatever we have earlier and actively it follows that the skeleton X k is ausdorff. This was one of the exercises. If y is ausdorff by attaching a single dimension k cells, whatever you get is ausdorff. So, I am using that one here. That is why this will be these things will be drawn. So, each X k is ausdorff. Now, ek is a closed cell, then it is compact subset of it. A compact subset of a ausdorff space is closed. Since each X k is closed in X itself we are done. So, each ek is closed in X k. So, this is an easy way of saying the whole thing. But you have to use to show that a closed cell is actually closed. You have to use the ausdorffness of it. If a relative CW complex, then you should have started with y as a ausdorff space, then it is really true. The C part I will repeat this one namely I have to show that if S intersection E is closed for every E inside E then S itself is closed is what I will show. Inductively, we will show that S intersection X k is closed in X k for every k. How? S intersection S X naught, X naught being a discrete space every subspace is closed. So, S intersection X naught is also closed inside. So, there is nothing to prove. Suppose you have proved it for k, then we appeal to this Lemma and S intersection X k plus 1 is closed in X k plus 1. Remember, when you attach one single thing, what is the condition? It should be S intersection S, this set is closed if and only S intersection X k is closed and S intersection each k plus 1 cell must be closed and that is the condition here. So, step by step you will get S intersection X k plus 1 is closed in X k plus 1 and so on. So, that is part C. Now, above compactly generatedness. So, this is the meaning of co-induced topologies compactly generated. A subset is closed in X if and only if S intersection k is closed for every compact system. If you look at each cell E, it is compact. Therefore, by this condition S intersection E is closed in E. By C, it will be closed in the whole of A. The converse is obvious. So, now I come to this, why this finite is generated? There are only finite, limited, etcetera. Why I gave this argument? That was the argument used in olden days. Not only that, this was put as a condition in the definition itself by J H C Viter to give the first definition of CW complex. So, I will elaborate on that one little bit later. Right now, once you have watched this thing, if you have worked out from, for the attaching cell, the same exercise. If Y is host or regular or normal, then X is host or regular normal. That is what I have given an exercise. Now, inductively you can do that, that if A is a host or space or regular space or whatever it is, then so is X, where X is a relative CW complex. In particular, every CW complex, E is a T-force space. So, this is left as an exercise for you now. Having solved that exercise, this will be easier. The next one is, you should pay attention to this one. It is a very, very important result on CW complexes, all the time used. So, what is said? It just says that every CW complex is a disjoint union of its open cells. Don't be under the confusion that therefore it is disconnected, you may say. These open cells are just names. They are not open subsets of X, not necessarily. Some of them may be, some of them are not. Indeed, they are open cells if they are maximal, that no additional cell of higher dimension will intersect their interior. Then only they are open cells. Open. But they are open cells, they are fine. Open cells are by definition what? It is just the interior of Dn, whatever, under the image of V, that quotient map. So, that is the open part there. It is open inside the closer cell, that is all. So, what we are saying is, take all the interiors of these cells. That will be the union of the entire CW complex. This result, we have seen it in the Simplisher complexes. It is exactly same here. In fact, much simpler. What is said? By very definition, you can write X as X naught union X1 minus X naught union X2 minus X1 and so on. So, Xk minus Xk minus 1. So, this union is patriotic union because Xk's are increasing subspaces of X. Here in this notation, I have taken X minus 1 is empty set. We start with X naught. There is nothing to subtract. X naught is the union of zero cells and zero cells are both open and closed. So, all these points belong to this disjoint union. So, we start with all the zero cells. They are open cells. Now, look at X1 minus X naught. What are they? They are open one cells and they are disjoint. Any two open one cells are always disjoint. So, Xk minus Xk minus 1 more generally is the disjoint union of open care cells. Take any K cell. Its boundary is contained inside Xk minus 1 and you are throwing away that what you left of it. So, it is interior. Okay. So, this is a totally obvious statement but this is very useful one. Now, let us see how it is useful. This lemma says take a subset that contains at most one point from each open cell. At most one point. Intersection of an open cell and S is a singleton or empty cell. So, this is the condition on S. Such a set is closed automatically and discrete subset of X. Okay. If you think a little bit, it is obvious but now let me write down the full proof of this one. Okay. There are books and books which give you long, long proofs of this one. So, even myself, I got this proof only little later and then there I found out that there are books which I have written like this also. To show that S is a closed set. What you have to do? Intercept with each Xk and show that it is closed inside Xk. Start with X0. X0 is a discrete space. Every subset is closed. That is fine. Now, assume that S intersection Xk is closed in Xk. I have to show that S intersection Xk plus 1 is closed in Xk plus 1. That is what I have to show. Then for any closed K plus 1 cell E, look at K plus 1 cell. E intersection S either is inside Xk or it is in the interior of E. If it is the interior of E, S intersection that part is at most one point. Therefore, if you look at E intersection S, it will have at most one extra point than E intersection S intersection Xk. In either case, this is a closed subset of Xk plus 1 because this already closed in Xk plus 1. What I get is one extra point. Where is the closed subset? Therefore, S intersection Xk plus 1 is closed in Xk plus 1 by our lemma because Xk plus 1 is attached is got by attaching K cells to K plus 1 cells to Xk. Intersection with say this you take this as Y. This intersection is closed and with each K plus 1 cell it is closed. Either it is empty in the interior or one extra point will come. Therefore, it is closed subset. Remember this is a hostile space. So, adding one extra point is a closed set because each point is a closed subset. Therefore, it will be closed in Xk plus 1. So, this way S intersection Xk will be closed for every K. Therefore, S is closed. So, what we have shown? S with that special property namely it contains only one extra one point or no points from each interior of interior of each cell such a property automatically is closed. Now, take any subset of S. It will have also same property. Therefore, every subset of S is closed. If every subset is closed, every subset is open also that is a discrete space. So, such a set is discrete as well as a closed set. We can use this one to prove certain properties that I was telling you that about finitely many cells and so on that comes now very easily. Take any compact set K of a CW complex. It is contained in the union of finitely many open cells. The funny thing is these open cells may not be an open cover. They do not form an open cover because they are not opens of sets not necessarily. Yet a compact set is contained in a finitely many open cells. This is an easy consequence of the previous lemma. X itself is a union of open cells. Therefore, every set inside it is contained in the union of open cells. You start with a compact set. Suppose you need infinitely many open cells to cover it. That means what? That means there will be infinitely many points, infinitely many cells which will intersect K on empty intersection. Pick up one point from each of these cells, open cells. Only one point you pick up, which is common with K. Then what will be the property of this K, this subset? That subset has the same property as previous lemma. But it is a subset of K. The previous lemma says that that subset is discrete. But it is a subset of K. K is compact. A discrete subset of a compact set is finite. So you could not have infinitely many points like that. So I repeat, if not, what happens? K will intersect infinitely many open cells, which we know are disjoint. Therefore, I can select one point in each such intersection. We will get a subset S of K, which is by lemma the discrete subset of a compact set. But by the very choice S is infinite. That is not possible. So this is the consequence closure of every cell, which we know is compact now. Meets only finitely many open cells, because every compact set intersects only finitely many open cells. This property closure meeting finitely many open cells was called closure finite property. Okay. So this is the remark I wanted to make. I am going to elaborate it on. I was all the time referring to it. The property of CW complex is known as closure finiteness. Okay. That means what? Closure of every cell is will meet only finitely many open cells. Okay. So this explains the presence of the letter C, CW, CW complex. This C was corresponding to the closure finiteness in this mysterious nomenclature of CW complex. The CW complex was the name given by the creator J. H. C. Whitehead, creator of this concept. Historically, this property was an additional part of the definition itself. J. H. C. Whitehead definition of CW complex was very, very long. He had put many of the standard properties which we are discussing now as a consequence in the definition itself. Okay. So this was one of the property closure finiteness. We have just proved that we do not need to put this extra condition in the definition. By the way, there is W. What is this W corresponding to? In the letters CW. C corresponds to closure finiteness. What is W? W corresponded to the weak topology, the compactly generated topology. Okay. That is also called weak topology. Okay. Corresponds to weak topology on it. So it might have been appropriate to call this one. There is no need to W. There is no need for C we have seen. But W is not this weak, the compactly generated or co-industrial topology that part is necessary. Therefore, you could have just awarded C and kept only W. We can say W complexes. But nobody has made that kind of suggestion. So I am alone perhaps. So let us not stick to a new, let us not create a new nomenclature here. Let us, many people just call, they have given up W also but now they are calling it a cell complexes made up of cells. So that is a good name. Cell complexes is a good name. Okay. But I would say that you better stick to CW complex because there are many notions of cell complexes. Each author has his own definition. And sometimes they just differ slightly. Like somebody doesn't put house darkness. Somebody puts something else and so on. And the structure also somewhat different. So let us not bother about new terminology. Let us call this a CW complex. Okay. So is the closure finiteness property, if you read old papers, then what you will get is this kind of definition. The closure finiteness is a consequence. All right. Of actually the fundamental thing is the quotient space structure here. Okay. Attaching cells gives you X as a quotient space. And then from X1, X2, X3, one contains other. X infinity namely X itself has the topology co-induced by these inclusion maps. You could have taken directly that compactly generated. But before defining compactly generated, you have to have topology on it. So you can't say that compactly generated. So all these problems CW complex in different DLU complex, you know, though he had the ideas, things were not available to him. So many of these things will develop much later. So it was a great invention after all. So let me go ahead with the topology in the spirit of our earlier lemma of extending the neighbourhoods so that their deformation retracts and so on. Remember that lemma. Remember that proposition. Okay. So I am going to make another such lemma here which will be useful later in the study of CW complex says. What does it say is the following. Start with the CW complex. For each end, let Vn be a neighbourhood of Xn inside Xn plus 1. Okay. Let X be a point in the interior of say some k cell. And Wk be an open neighbourhood of X inside sigma. Okay. Choose some small epsilon neighbourhood for example, W is some neighbourhood of X inside sigma. So you can by the way, here also you can say open, open everywhere, open neighbourhood, no problem. Then there exists an open neighbourhood W of X in X such that W intersection Xk is Wk and W intersection Xn is contained in Vn for all n. So this Wk which is an open subset of open neighbourhood of X inside sigma gets extended to W. Extended means what? Intersected, Xk is Wk is exactly equal to Wk. And it is controlled in the sense that intersection with Xn is contained inside Vn. It is not going, it is not becoming too big. This W is a neighbourhood of X in X itself. A neighbourhood Wk which is inside Xn plus 1, sorry, where is it? Inside Wk, inside Xk may not be open inside X. Okay. It is like you have a point on the line and a neighbourhood of that point on the line like an interval. But now interval is contained in R2 that interval is not open inside R2. But you can extend it to a square, open square. And that is precisely what we have done in the first lemma. We took any subset of the boundary and then we extend it to a neighbourhood of that inside the disk. Right. And this fundamental elementary construction is going to help us in all this. So this is the lemma that we had. Okay. Start with attaching cells. Okay. Start with a subset A, where is it? B inside Y. The B gets thickened to n epsilon B. Okay. This epsilon could be chosen at your wheel and that is a neighbourhood. That is an open subset of X. Okay. And if you intersect it with the Y, then it is just B itself. So it is extended. Q, B, P, Q is in it. It is a strong differentiator. This part is extra thing. We do not need this one right now. Okay. This is also true we can do. But all that I want is extension from not from Y to X, not from Xk to Xk plus 1, but Xk to directly to X we want now. So we have to do inductive steps. So this theorem is what we are going to use. Okay. So now I have to go back, but this is not where I am to go. So by this proposition for each k plus 1 cell tau, okay, tau intersection vk plus 1 is open subset in tau contained in the entire boundary of tau because what are v ns? v ns are neighbourhoods of x n. So they contain the boundary of any simple n plus 1 simplex. Boundary of any n plus 1 is x n. So it is contained in v n. So when you take the inverse image, it will contain the entire boundary, the boundary sphere. So I am not going above at all. I am just working on in X itself all the time. So the notations etcetera are simpler here. Okay. So for each k plus 1 cell tau, tau intersection vk plus 1 is an open subset of tau containing the entire boundary which is actually a subset of xk and contained inside vk plus 1. Therefore we can choose epsilon to epsilon tau such that once the entire sphere is there which is compact, there will be an epsilon. If it is non-compact then there would have been a problem. So a tau is equal to n epsilon of tau dot. Remember this is what is the meaning of tau dot. Tau dot is a subset, it is a sphere itself actually n epsilon of tau dot is contained inside tau intersection vk plus 1. This happens for every tau. So there I have to have you know sigma alpha and so on. I have reduced the notation here that is all. Okay. Therefore if you take beta which is equal to a smaller set namely n epsilon tau dot intersection wk, that will be a subset of tau intersection vk plus 1. Okay. So this is the intersection tau, this is the service in tau intersection vk plus 1. I am taking a subset here by taking the intersection of this one. Okay. This is going to be an open subset. Therefore n epsilon of that will be open subset. This is the import of this lemma. Okay. So take now, take wk plus 1 to be the union of all these beta's as done in the proposition. If you take union of all these beta's that is going to be an open neighborhood of this x because x is there in xk plus. That will be an open subset of xk plus 1. So it is an open neighborhood of xk plus 1. Okay. And its intersection with xk is precisely wk because all these neighborhoods had this one. B remember B is equal to QB is equal to what? It is intersection of n epsilon B intersection with y. So that is the meaning of wk plus 1 xk plus wk. And wk plus 1 is the union of all these beta's. It contains a vk plus 1. So inductive step is over. Okay. Now what you do? You just take w equal to, so we have got this inductive step for all of them. Now you just take w equal to union of all the wx. You start with n greater or equal to k here. You start with wk because x happens to be inside the k cell in the indeed of k cell. If x is a vertex, if x is 0 cell, you would have taken all these things n greater or equal to 0. Okay. Why this is open? Intersection with each xn is a neighborhood open subset. Wn minus. Intersection with xn is this one. Why it is contained vn plus 1? But if you have w and intersection it is precisely, so it is containing vn. That is how we have chosen. So this is controlled by these vn's but it is a neighborhood of x inside x itself. Okay. So this I call it as controlled extension. So we will use this one much later of course but we will use that. All right. So we have done quite a bit of topology but there are many more things to be done. One of the most important thing is that the cw complexes, the weak topology is very, very useful in constructing continuous functions on cw complexes. So this was the motivation for taking that or verifying, suppose you have constructed something verifying that it is a continuous defining a continuous function as well as if there is a function to verifying continuous function, the cw complex structure is very useful. Whereas if you take just an arbitrary space and so on like r infinity and so on it would be very unwieldy to do that kind of things on that. In particular we will show that there are lots of continuous functions namely what is called as partition of infinity. So that we will discuss next time. So today this is over. Thank you.