 So, now we carry on with more examples of CW1 classes. For any sub family F of n cubes in P not necessarily finite, we have this notation WF equal to union of all the n cubes sigma which are present in F. We also have this notation F prime is all those sigma inside F which are maximal, what is the meaning of that? If sigma is contained inside tau, tau is also in F that means sigma is equal to tau, that is the meaning of maximal, only the maximal elements of F are taken in F prime. Then clearly the union of all the members is equal to union of all the maximal members, okay. So, this is where we use the fact that, you know remember I will show you that interior of sigma, intersection interior of sigma 2 is nonemptive, right? Either sigma is contained in sigma 2 or sigma 2 is contained in sigma 1. Either they are non-overlapping, okay, completely or one of them is contained in the other. So, you can take only the maximal one to take care of the union. Now, here is the extension of the earlier example for infinite case. So earlier we had this example for only on sigma. Now, I am taking it for finite members of P. Let F be a finite collection of members of P. Let F denote the collection of all k faces 0 less than k less than ornate of members of F prime. So earlier there is only one sigma. Now I am taking a finite collection. Then we can choose a function F on this large family F, okay, which is an integer value function satisfying condition 1 and 2 of example 1.7. Let me recall that. What are these conditions? This is F tau is taken to R for all tau whenever this is a member of P R, tau 1 and tau 2, one contained in the other implies F tau 1 is greater than or equal to F tau 2, okay. So this must be true for all now all members of this curly F, okay. So that is the condition 1 and 2. Such that the CW structure sigma F remember this is coming from 1.7 that example for F in F prime patch up to define a CW structure on WF so that the n cells are in 1-1 correspondence with members of F prime. There in that example let F be with this property. I did not say anything about whether there is one such and so on. There are many actually, okay. But here I say we can choose a function F. If some F is given with these two condition, I am not saying that this will happen, okay. Condition 1 and 2 are necessary even to have a CW structure on each sigma F, there is sigma F, okay. So once you have that each K phase here members of this one will have the sigma structure. Now the claim is that if F is chosen properly then these sigma structures, these sigma F structures on each of sigma will patch up to define a structure on WF itself. This is the claim not only that the second part says that the n cells of this structure are precisely the members of this P with which you started with which are maximal namely F prime, the members of F prime. So if you have understood the sigma F earlier, earlier example, then this will be an easy consequence of that, okay. So let me write down the formula. What is the F? Let F from curly F to n be the function defined as follows. Take A tau to be all those sigma prime in F prime, the maximal members here such that they will intersect the interior of tau, okay. There is at least one because each tau is a phase of some F prime, some F, some member of F, therefore it is a member of F prime, okay. So it is a phase of some member of F prime. So this set is non-empty but of course it is finite also. Anyway, all those sigma prime in F prime, sigma prime intersection interior of tau is non-empty. So A tau is a first of all, is contained inside P such a non-empty finite set, okay. We can take F tau to be the maximum of all T such that P T intersection A tau is non-empty. So each member of A tau belongs to some P T. If that P T is there, put that T in this set, okay. So P T intersection A tau must be non-empty. So take those set. This is anyway finite set. It is non-empty also. So take this maximum. It is some finite number. This defines F tau. As tau varies, this F tau will be different than matter. What you have to see is that it satisfies the condition 1 and 2. That is what we want, right. The condition 1 and 2 must be satisfied, okay. Note that if sigma is in F prime, intersection say P R, one of the members, sigma is one of the members in P R as well as it is a maximum one. Then F of sigma is R because this whole set itself will be just singleton here, namely R. So it is maximum, it is actually equal to R itself by the definition of F prime because F prime is maximal. It will not intersect any other sigma prime. Sorry, a sigma is maximal. It will not intersect any other sigma prime. It is maximal, right. The interior does not intersect. So this should be only sigma prime will intersect. So that is the only member. So that will be R, okay. You have to see that this satisfies the two conditions 1 and 2 in the above example, okay. The point here is, the interior I have put here, if the interior intersects, the sigma prime intersects the closure, the whole closure also. It will perhaps contain the whole closure, sigma prime first itself, okay. So that is the trick here. So if you take sigma prime intersection interior tau non-empty and say tau prime is a phase of tau, this will be true for tau prime also. So all those t which are present in A tau will be there in A tau prime. Therefore, this will be a bigger set for tau prime than tau which means that maximum is bigger, okay. So tau 1 contain her tau 2 will imply f tau 1 is bigger than a quota of tau 2, etc. You have to verify this, okay. Hope you will be able to do it on your own now. Anyway, I have explained the second one. Condition 1 also you can see what happens, okay. So now the crucial thing in the lemma namely, we claim that the C-deb structures f sigma, as sigma varies over f prime, patch up to define a C-deb structure on w f prime which is same thing as w f, okay. So we want to do an inductive step here starting with suppose f prime has only one member. That is already covered in the example, okay. So therefore, we just want to do induction here, alright. So write f as sigma 1 sigma 2 sigma p. So this is an enumeration of members of f prime. It is a finite family. Let us check the notation w j equal to union of those sigma i for i less than equal to j. So w 1 is just sigma 1, w 2 is sigma 1 union sigma 2 and so on, okay. So we will induct on this number p. If this p is 1, we know it already. There is nothing to prove. Nothing to prove means what we have discussed this in the example 1.7. Inductively assume that the structure sigma f i have patched up to define a C-deb structure on w j. So i less than equal to j, suppose it is true for w j and that structure temporarily we should denote it by w j f. Is it coming from f, okay. Put next one a equal to w j plus 1, take the next member here. Intersect with w j, put a equal to that. If a is empty, you have two disjoint C-w structures, C-w complexes. Union is automatically C-w complex in which w j structure will be kept as it is and sigma j plus 1 structure will be kept as it is. Both of them will be sub complex. There is no problem, okay. The problem comes when a is non-empty, right. So suppose a is non-empty. The idea is to use our earlier lemma. So let us denote C-w, denote the C-w structure on a coming from w j 1 and coming from j plus 1, sigma j plus 1, okay. This intersection of these two things, right, respectively. Let us denote this one by a0, that one by a1, okay. What we wanted is that either a0 is finer than a1 or a1 is finer than a a0. Then we could have put them together. So that was our earlier lemma, right. So here we claim that a0 is actually equal to a1. So we are through. So we can patch it and then induction will take care of it, okay. So let us look at this. What is this a? a0, a1 are not structures. a is the subspace. a is the intersection of, a is the intersection of w a1. This intersection never contains any interior of sigma j plus 1. It is exactly some part of the boundary, okay. Because sigma j plus 1, these are all maximal simplex, this after all. So it is a boundary, part of boundary of j plus 1. A part of the boundary will be made up of k phases of sigma j plus 1 for k less than or equal to the dimension of sigma, namely n less than strictly less than n, okay. And hence the union of some proper phases of tau, phases tau of sigma j plus 1. On each of these tau, the CW structure coming from either side is precisely equal to that coming from p of f tau. By the very definition, f tau is what? the finest, you know, maximum of all those things. And that is what we have chosen, okay. Maximum, among us all the members, not only just sigma j plus 1, for all w j's, whatever sigma i's are there, if they intersect, they will give you the whatever number there. So this is the maximum. So this is finer than all these, okay, it is either side. The CW structure coming from either side, okay. This proves that A naught is A1 because it is actually equal to the p of f tau. So it just depends upon tau, it does not depend upon where this tau is, whether it is sigma j plus 1 or of w j or so on, okay. So we can put them together. That is it. So from them, it follows that w j plus 1 as a CW structure, which you denote by w j plus 1f, in which both the old one, the inductive step, w jf and sigma f are sub complexes, okay. So the structure keeps extending as you extend the the space itself, okay. So that completes the proof that wf prime has CW structure as proposed. And each time sigma, sigma, whatever sigma i have taken, sigma f has only one CW structure there because because f tau, f sigma itself is what we have observed, f sigma itself is r. Where was that? We have observed that one, no? f sigma itself is r, okay. So if sigma is in pr, there is no subdivision, it is just one single n cell, okay. So it should be noted, the final structure on wf, which is, let us call it as wp, does not depend upon the enumeration we have chosen on the members of f prime. And for j less than p, each wj is a sub complex of this wf, okay. So it all depends upon my f. Once f is chosen, no matter, this is only for our convenience that the enumeration was only for our convenience, okay, for the proof. But the final structure, so wf prime is nothing but wp here, okay, when the number a number of p is hit there, that structure is the same. So let us now go for infinite case. This was our aim finally. Let f be a family of members of p such that f prime is locally finite. So we are not into completely in the sea of all infinite family, that is not, that is not the aim, that is not possible either. So we allow f to be infinite but locally finite. Let f prime be the members of all k faces, one less than or two k less than or n, of all members of f prime. You take sub phases and so on for once you start with f prime, okay. Then there exists a function f on this f prime to end such that the cw structure sigma f patch up to define a cw structure on wf prime namely it is the incident of wf. So I am not saying here that f satisfies one and two, okay. So it should satisfy otherwise I do not talk about the structure here. The structure sigma f is defined only when f satisfies the condition one and two, okay. So here the finiteness is not there, you see. So we have to be little more careful but it is not at all all that difficult. We start off as in the proof of the previous lemma. So let f denote the collection of all k faces of all members of f prime, okay. Observe that each set a tau defined as in the above lemma, a tau equal to the set of all sigma prime and f prime such that sigma prime intersection integral tau is nonempty. This a tau is the same thing, okay. This will depend upon all those sigma prime which intersect integral tau and in particular they intersect tau. By local finiteness this family is finite, f prime may not be finite. So all this sigma prime which satisfies this one is finite. So each a tau is finite, okay. Therefore as before we can take f tau to be maximum of this thing, that will be finite. So this is the replacement for family not being finite. The local finite mistakes are obviously. Therefore, okay, we have the function f and it has satisfied the same properties one and no problem. The second thing is f prime is not just infinite but it is countable. Why? Because p itself is countable, okay. If it is finite then the proof is over by the previous lemma, okay. So assume that it is infinite and write f prime as again just like in the previous lemma, enumeration, sigma 1, sigma 2, sigma j and so on. The only difference is now this is infinite family. This take an enumeration, okay. As before put wj equal to the union of all sigma j i less than equal to j, okay. As in the example for each j, okay, the Cw structure sigma fi, i less than equal to j, patch up to define a Cw structure on wfj on wj which is a sub complex of wfj plus 1. So all these is there in the previous lemma because these are all finite now, okay. It is union of finite in many of them. It follows wfj themselves patch up define a Cw structure on wfj. So each finite stage you have and what is wf prime? It is union of these finite things. So one stereo structure is sub complex of the next one. Therefore, okay, look at any n cell or any k phase whatever is in wf prime. It would be in one some at finite stage and there things are nice. You can attach it, okay. You know attaching a k cell should make sense. How does it make sense provided its boundary is inside a k minus 1, k minus 1 skeleton. That is all you have to see. So that is automatically satisfied here, okay. Also note that since interior of any two members of a primary disjoint follows that for sigma belong to Pr, we have f sigma equal to r, same thing as in the previous case. Therefore, the n cells of wf are in one-one correspondence in member of f prime. There is no subdivision of the n cells, okay. In other words, the proof of previous lemma is due to words provided this once a tau is finite for all tau. And that is ensured by local finite. That is all the difference, okay. We now come to an important result which I have that is why we I have put it as theorem but this is also an example but this will be useful. This will be used elsewhere while studying manifolds we will be using this one. This was our aim of doing all this fundamental work here, okay. Take x to be any closed subset of an open set v inside Rn. So v is some open subset of Rn, x is a closed subset of that. I want to tell you right in the beginning that v could be the whole of Rn and on the other extreme x could be empty also. So that case is also allowed here, okay. And they are of some importance also, okay. They may be trivially true sometimes but they are of importance, okay. You can take just v to be any open subset and x to be empty and that is one important case, all right. So x is a closed subset of a open subset v of Rn. Let a equal to all the members of p. p, remember p is a union of all pk's such that this sigma is inside v. This is no more open set. This is some n cell which should be contained inside v and intersection of x with sigma must be empty. It should be away from x. So that is my a. Second one is b is all those p, all those sigma and p such that again sigma is contained inside v. So v is allowing us, should not go out of v that is the first thing. But the second condition is x intersection sigma is non-empty, okay. So this is difference between these are quite different if x is non-empty then these are different. If x is empty set then this should be not there only. This is always true. So it would be just all those sigma adjacent, okay. So these two are quite different things which are both subspaces of v, okay. Then w a which is the union of all the members of a, w b which is the union of all members of v are both n dimensional Cw complexes which are countable, pure and locally finite, okay. So I am not including here when the case a and b may be empty and so on. So those things are obvious. You can just figure it out, okay. So the proof is though there are two statements here for a and b they are similar. So I am going to put them together that c denote either a or b as above. As observed before first of all w over c union of all members of c same thing as w over c prime which is c prime is what? a prime or b prime which is maximal numbers of c. In view of the lemma, okay all that we need to do is to verify that c prime is locally finite. That was our previous case if it is locally finite then we have the structure, okay. So that was the previous thing here. It is a locally finite family then there is an f then you have this sigma f's patch up to define such a structure. So in view of this lemma all that we need to do to verify c prime is locally finite. At all the points of w c this is your union this is the this is my x whatever or that whatever, okay. So at all points this family should be locally finite, okay. So here is pro take any point in w c but choose epsilon for epsilon positive let us have this notation standard notation of course u epsilon of x is the ball of radius epsilon open ball of radius epsilon centered at x norm of x minus norm of norm of x minus y is less than epsilon. So this is a ball in capital R n, right and this is an open ball strictly less than epsilon. So this is open ball centered at x and having radius epsilon. So this is notation. Now choose epsilon such that this ball is contained inside the open set v minus x if we are working namely c equal to a c equal to a x belongs to the v a which means that x is not a point of capital X and its x is a of course a point of v. So it seems a v minus x whenever v minus x is an open subset we can choose epsilon sufficiently small set u epsilon of x is contained inside that. So that is no problem. Similarly in the second case u epsilon x is just contained inside v, okay. In the second case this is easier. The first case you have to throw away x and then take that open set. Once you have taken u epsilon contains at v this u epsilon is going to ensure local finiteness. We claim that only finitely many members of c prime will intersect u epsilon fax. The emphasis is on c prime. If you ask for c members of c that is not true. Here it is definitely not true. May not be true is not correct not true at all, okay. Only for c prime it is true. Finitely many members of c prime will intersect u epsilon. Local finiteness is for c prime at every point, okay. Let us prove this one. Given any r, okay. Clearly there are only finitely many members of pr which intersect a given bounded set. Take a bounded set inside r. Members of pr will intersect it, okay. Because bounded set is inside some large square and inside the large square, large cube only finitely many members of pr will be there, okay. When r is fixed given any fixed r this is true, okay. Hence in particular how many will intersect u epsilon x, u epsilon x is the bounded set. Finitely many will intersect, okay. Therefore there are only finitely many members of if take all the s less than equal to r each of them will contribute finitely many. So how you have only finitely many for pr r union of pr where s is x quantity equal to r, okay. The problems comes with members of p s where s is bigger than r. That means these are smaller, smaller, smaller, smaller cubes. They can create problem. But this is the point is here. We shall see that none of them belong to c prime if r is chosen appropriately. U epsilon is already chosen. If I choose r sufficiently large after that no p s will intersect u epsilon x. I mean p s they will intersect but they are not members of c prime, sorry. They will not belong to c prime. So number of members of c prime which intersect u epsilon will be finite they will be coming from here only, okay. So once I have said this one it should be obvious to you but let me just make it more clear. To choose r let r belong to n be such that this number this is the real integer square root of n divided by 2 over r minus 1 is less than epsilon. This 2 over r minus 1 is side length. But what is square root of n? The square root of this denotes the diagonal of any n cube of side length 1 by 2 over r to 4 r minus 1. So that is why the square root of n times 2 over r minus 1 is there. So diagonal should be less than epsilon. Let f be the collection of all members of p r. Now r has been chosen correctly, okay. Which intersects u epsilon, okay. Look at all those they will intersect u epsilon. Then this u epsilon of x will be contained in the interior of all these w f. f is all of them which intersect. It should contain inside of them, okay. Like if I have single point for example, if I take all of them it will be a big bigger cube of side length double of that one 2 over r around that point. Similarly, if I take an edge also that will be the same thing. So this maybe may not be like look like that it may be bloated up thing. But it will be some open set that open set will contain u epsilon of x, okay. It will be union of cells after all, okay. So this is what you have to understand why this is happening. It is elementary geometry whatever topology which is inside r n, okay. Suppose now tau is in p s where s is bigger than r. And tau intersection u epsilon is non-empty, okay. So this is some member n cube in p s where s is bigger than r which is very tiny member s is bigger than r here, okay. So tau intersection interior is non-empty. This intersection tau intersection epsilon is non-empty would mean that interior of tau will mean the interior of this larger one because this is open subset contained inside this open set, okay. If some tau intersect interior tau intersects this open subset here it must intersect one of the interiors of members of this one, okay. So it will follow that interior of tau will mean the interior of WF and hence will mean the interior of some member sigma of s, okay. So this tau intersection interior tau intersection sigma that is non-empty that will imply that tau is contained in that sigma, okay. Just tau is smaller size sigma is of sigma is of size r where s is bigger than r. So it cannot be that sigma is inside tau but tau must be inside sigma. Both tau and sigma are inside c. Therefore, I mean that is what I want to say that we see how if both of them are inside c only one of them can be inside c prime. Sigma is already in c prime so tau is not in c prime. So you wanted to show that none of these tau are in c prime, right. So why both of them are in c? That is what I have to say, okay. So that is easy. The case a equal to c equal to a and b have to be done separately. In case c equal to a what does this imply? Both tau and sigma belong to c. Why? Because interior of tau, you see what is the definition of, definition of c here. Go back, okay c a definition of a. Sigma intersection x is non-empty. If tau intersection x is non-empty sigma intersection is also non-empty. And both of them are sigma itself is inside b tau will be also inside b. So both of them will be inside a, right. So that is the case when similarly the other one but this one is more since a both of c and tau are inside a. Therefore tau is not inside a prime. Okay. Now consider the case when c is b, okay. Now it is something different then it may happen that tau intersection x is empty, okay. But we want members of b to b that tau intersection x is non-empty. If this is empty that would mean tau is not in b already if it is not in b it cannot be in b prime. So that case is over. Now on the other hand tau intersection x is non-empty then tau sigma intersection x is also non-empty because sigma is larger. Both of them will be inside b. Once again tau cannot be inside b prime. So that completes the proof that this both a and b have a nice CW structures and dimensional CW complexes which are countable, pure and locally finite. You can further assume which I am not going to assume which I am not going to use but you can further assume that the n cells in each of them is equal to I mean they are coming from in 1-1 correspondence with members of a prime or members of a prime respectively. That can also be assured, okay. I want to make one more remark. W a's which I have made already I want to draw your attention. Note that W a is exactly equal to V minus x. Take any point in V minus x then there is a positive distance of that point to the the complement of V minus x, okay. After that you can choose a small neighborhood U epsilon which is completely contending side V minus V minus is open subset. Then if R is large enough then square of n divided by 2 power R minus 1 will be less than epsilon and so the smaller and smaller discs will intersect, okay. Therefore what we get is W a's actually equal to V minus x. That means V minus x gets a CW structure. If x is empty what we conclude? Every open subset of R n can be given a CW structure. Every open subset, okay. So that is the conclusion of this one. Further all these CW structures are so nice they are all made up of n cubes just to star them, okay. Each face of each of these n cubes the starring will take care of that it will become a simplicial complex. Therefore what we have proved here is every open subset of R n can be given a simplicial complex structure, okay. So we will use them in studying manifolds, okay. Here are some easy exercises for you. Try them, keep trying them. If we do not get them we are here to explain it to you. Our TAs will explain it to you and so on, okay. So that is it for today. Thank you.