 Hello, welcome to NPTEL NOC, an introductory course on point set topology part 2, module 14. We continue our study with para-compactness. The following result which gives various characterization of para-compact spaces under additional condition of regularity is due to my care. Be warned about this condition regularity for some implication this may not be necessary, but there are other implications it will be necessary and there are counter examples we are not going that much deeper into it. The presentation that I have taken is close to what you can see in Willard's book. If you want a more elaborate description of para-compactness with more characterization then you can have a look at Kelly's book. Whatever we are doing today part of it we will be using in the final solution of metrizability problem. Okay, so we begin with a definition. A family of subsets of a topological space is called sigma locally finite if the family can be expressed as countable union of sub-families when each of them is locally finite. Obviously, if something is locally finite already then it is sigma locally finite. Okay, so here is the theorem that we want to go through today. Let X be a regular space. Then the following conditions are equivalent. X is para-compact. Every open cover for X has a sigma locally finite open refinement. Every open cover of X has locally finite refinement. The last one here is every open cover for X has locally finite closed refinement. Okay, so let us go through these proofs quickly. A implies B is very obvious because X is para-compact implies every open cover has a locally finite open refinement. Locally finite already implies sigma locally finite as we have pointed out earlier. So A implies B is obvious. A implies C, we start with every open cover having a sigma locally finite open refinement and then we want to produce a locally finite refinement. So this is an improvement but remember there is no, just notice that we have not put openness here. So we will not be achieving openness in this step. So we are not actually proving para-compactness but something weaker here, right? So B implies C, let us see how to prove that. Start with an open cover for X, say quality O, which is V be a countable unit of VN, be an open refinement of use such that each VN is locally finite. Okay, so sigma locally finite open refinement is there for you. So start with an open cover which admits a sigma locally finite open refinement. Okay, put WN equal to the union of all the elements in curly VN. Okay, then WN as N ranges over the natural number that will be an open cover for X because this curly V is an open cover. Okay, now we are defining AN equal to WN minus the union of all the previous WI, ranging from 1 to N minus 1. This is an open subset. I am subtracting an open subset from an open subset. So that will not, AN will not be an open subset in general. Okay, unless something discrete happens. These are also closed and so on. Then clearly what is true is that this curly A which is collection of AN as N ranges over the natural number is a refinement of WN because each AN is contained inside WN. Okay, and refinement is fine. But when you have a cover, WNs are cover. I have to prove that A is also a cover. Okay, for each X inside X, let NX denote the smallest number or you can say the first number such that X is inside WN. Okay, so X is inside WNX, smallest number N such that X is inside WNX. Then X will be inside ANX because it is not in any of these WIs. Right? So X will not get deleted. So it will be, it is inside WNX and it is not getting deleted. So it is there inside AN. And hence A is a cover. So now what we have got is A is a cover and it is refinement of WN. Alright, this cover W. Okay, also now you look at WNX is a neighborhood of X. Okay, WNX is they are all open and WNX will not meet any AM for M bigger than NX. Because in AM, M bigger than NX, this WNX gets subtracted. Okay, so those points will be deleted from WM. Okay, so AM will not intersect WNX for M bigger than NX. This means that this family is locally finite. Okay, so we have got a locally finite cover. Alright, but this is not a refinement of you or you want to get a refinement of you. Right, these are somewhat larger. So here is what we have to do. Put BN equal to AN cap V where V belongs to curly VN. So everything is now happening inside VN. Okay, so you take AN intersect with each member of V and put that family as BN. Then you take curly V to be the union of BN, N ranging over the integer with natural numbers. Given any X in NX, if NX is chosen as before, namely the first N for which X belongs to WN. Remember that. Then we have shown that X is inside ANX. Right, also X is inside WNX implies X is inside some V for some V where V is in BNX. Okay, it is already ANX. It must be in some V here. So it will be in the intersection. Okay, so X is VNX here. So X is inside ANX intersection V which is VNX. Therefore, B is a cover for X. So this is now written as countable union of this BN and this is a cover. Okay, still there is no openness here. Okay, so B is refinement of V because each element of B is BN intersection V. It will be inside BN. Okay, B is a refinement of V. B is a refinement of U. So B is a refinement of U. All that we have to see that is now this B is sigma locally finite. What we wanted to do here? So B is a sigma locally finite, sorry, open cover for X. It is a locally finite refinement. Okay, this B is locally finite is what we have to show now. Okay, once again we note that X inside N given X inside N, if NX equal to K is chosen as before. Once again I repeat, what is NX? NX is the smallest number N such that X is inside WN. Okay, so I am now writing it as K. Okay, this WK is a neighborhood of X because X belongs to WK and it does not make members of BN and bigger than K because this WK gets subtracted in them. That means that this WK may intersect only members up to B1, B2, BK. Members inside B1, B2, BK. If they are in BK plus 1 and so on, it won't intersect. Now for each 1, 2, 3 up to K, what happens? Each BI is locally finite because their refinements, they are sub-families of VI, okay. If VI is locally finite, at least. Okay, there exists neighborhood GI of X, which means only finitely many members of VI fix one I here, one, belong to one up to K. Then there will be only finitely many VI, okay, which will intersect GI. So I can choose GI in such a way because of local finiteness of VI. Now you look at intersection of all these GI, hiring one to K and WK also. Okay, WK will cut off things from K plus 1 onwards and this intersection will have the property that this will meet only finitely many members from each of B1, B2, BK and totally finitely many members of the curly B. So therefore this B is locally finite, alright. So that was some proof here. You see it needed from local finite, sigma local finiteness to local finiteness. But the price we have given is that we have not bothered to get the family, an open family. They are not members of, they are not members are not open. Now we have to improve on that, okay. So next step we will do that. Let now C implies D. So C implies D means what? Now starting with open, starting with arbitrary, locally finite refinement, we want to make it locally finite, closed refinement, okay. And from closed refinement, finally D implies A will be open refinement. That is the outcome factor, okay. So that is the idea. So now let us say C implies D. So let U be any open cover by regularity, it comes here now, okay. By regularity for each X belong to VX inside U, we get an open subset VX such that VX bar is inside UX. In particular, if you take all the VXs as X raise over X, this is a shrink of U, right. VX bars themselves are concerned inside UX. From hypothesis C, we get a locally finite refinement of A, refinement A of V. I am calling the refinement as curly A, okay. Then from lemma 3-3, I will tell you what is this lemma 3-3. The closure of A as A belongs to A is locally finite. So this precisely, if we have a locally finite family of subsets, then their closures is also locally finite. This was the lemma. In fact, that lemma had more things. In fact, union of arbitrary families of closed subsets here, the closures of here will be again closed. That is the second part here. So this we have proven, these are used elsewhere also. So these closures are locally finite. So clearly it is a refinement of U because all these C, all these VXs bar are contained inside UX, okay. So and A bars will be containing some VX bars if you can be there. So they are refinement of U as well, alright. So that complete C implies D. If A inside A is covering, A bar inside A also will cover. So there is no problem about that. So let us finally take this D implies A. From closed refinement, closed locally finite refinement, we are now proving that given any open cover, there is a locally finite open refinement. So that will be para-compiled. To start with any open cover, start with a locally finite closed refinement of U. We are going to put an open refinement of A now. So you can as well assume to begin with we have instead of U, you can just now take A, okay. A is a locally finite closed refinement of U, closed refinement having a cover itself. This means that for each X in X, there exists a neighborhood VX of X, which means only finitely many members of A, okay. Now there exists locally finite closed refinement of this cover, namely VX as X belongs to X. So starting with an arbitrary open cover, we passed on to a locally finite closed refinement. Using the local finiteness of that, we got a cover V, okay, with the property that members of VX are detecting the local finiteness of it. That itself will have a closed locally finite refinement and that is C, okay. So now you can just work with this A and C, you can forget about U and V. It follows that each C belonging to C means only finitely many members of A, okay, because they are refinements of VX and VX will mean finitely many members of A. For each A inside A, select a UA inside U such that A is contained inside UA. Remember, this U was an open cover and started with A as a refinement. A refinement means for each A there will be UA, okay. Right, select one such A in UA that A is contained inside A and put A star, this is just a notation, equal to UA, this is an open subset, minus all the elements or union of all the members of this C which intersect A and we do not intersect A. So what I am doing here is I am taking this element A and I am patterning it inside UA. Members of A are closed, they are locally finite and they are closed refinements. So, right, there exists an A, sorry, here locally finite, closed refinement, A is a locally closed refinement. Usually what you have done earlier in all earlier things starting with an open subset you got a shrink, a closed refinement you get. Now you are going to expand them, right. So how do I expand? I expand it inside UA. How do I do that? So UA is open, throw away a closed set. What is that closed set? It is a union of all the closed sets C inside B which do not meet A because when you are throwing away you do not want to throw away points of A, that is all. A intersection C is empty. Whatever you have thrown away it does not intersect A, so A part is kept as it is. Since this is locally finite family and they are closed, arbitrary union of such closed sets is closed. This again used, this lemma 3.3 is used here. So UA minus this set is a open subset. So each A star is a patterning of A to an open subset. Moreover they are inside UA. That means they are refinements of UA. This family A star as A belongs to this curly A it is a locally finite lower A. It is a cover for A because A is a cover for A it is a refinement of UA. So there is lot of hope with this family. In fact we will show that A star is locally finite also and that will complete the proof that D implies A. Remember I just recall what we have done starting with this A which is a locally finite closed refinement. You select one single member UA which contains you, contains A which is inside you. From UA you subtract a closed subset. That closed subset does not intersect A therefore A is contained inside A star. That is the patterning of A. Since you have subtracted a closed subset this is an open subset. A covers the entire A. So A star will cover the entire A. They are refinements. This much we have done. We have to show that A star is locally finite. A star is a refinement, it is locally finite. This is what remains. Now again use the fact that C is locally finite. Given any X in X we can find a neighborhood WS of X which intersects finitely many members C1, C2, CK of C. Since C is a cover it follows that this neighborhood WS must be contained in whatever members which intersect namely I range from 1 to K C I. Other members do not intersect so they are not needed to cover it. But all of them together have to cover WS. Therefore WS is contained inside I range from 1 to K C I. This implies now suppose WS intersects some A star. We want to finally prove that A star is locally finite. Take a member of A star say A star. WS suppose intersects A star. WS intersects A star is non-empty. But WS is contained inside this union. Therefore one of the C I intersection A star must be non-empty. If all C I intersection A is non-empty then this will not happen. So for some I it may happen for several of them. It does not matter at least one I for which C I intersection A star is non-empty must exist. But C I intersection A star look at this definition. C I intersection A star is non-empty means C I is not here. Not here means what A intersection C I itself is non-empty. So C I intersection A star is non-empty implies that C I intersection A itself is non-empty. So C I meets only finitely many members of A. So that is the choice of this C right from the beginning. Using local finiteness of A we have got this V and then C I was C was the refinement of that. For close refinement. So these mean finitely many members of A. So with all of that W X intersection A star is empty for all but finitely many members of A star. Fix an A star look at W X intersection A star. If it is non-empty then they must be inside this one of the C I. And each C I will contribute only finitely many of the members. Therefore there will be only finitely many members of A star for which W X intersection A star is empty. So this completes the proof that A star is locally finite and the theorem of my care is completed. As an immediate consequence we get a stronger version of this theorem 3.6 that we have proved. Finally if you have locally compact of the space which is Lindelof then it is para compact. Locally compactness automatically implies regularity. Therefore regular Lindelof space is para compact is a stronger theorem now. How do you get this one? Every open cover has a countable sub cover which is automatically sigma locally finite. A countable family can write it as singleton union of singleton countable union of singleton. So singleton opens at singleton families are automatically finite locally finite. So using this criterion once you have regularity sigma locally finite refinement or the first thing here A and B every sigma locally finite open refinement that will give you X is para compact. So easy proof of this one. You can directly prove this one also. This won't be all that easy. I mean much easier than what we have done already. It's a ready made example of sigma locally finite refinement. We have defined but we haven't given an example. So I want to give an example. Consider a family V which we met in the proof of theorem 3.1, 3.21. Namely while proving that a pseudo matrix space is para compact. There we have actually proved that that family V is sigma discrete. Remember sigma discrete means what? It is a countable union of sub family. Each sub family for each point there will be a neighborhood which will intersect only one of them. One of the members to address sigma discreteness. That is a discreteness. Sigma discreteness is such a union of such countable union of such families. That's what we have done. So sigma discrete is automatically sigma locally finite. So we have such ready made examples there. So sigma locally finite is slightly more general family than sigma discreteness. Sigma discreteness is a very strong company. Therefore, when you come up to several steps there, the step 11th step. If we use theorem 3.26 of Michael, you don't have to go further at all. Because immediately you can conclude that it's para compact. Whereas there we have to work harder and completed. Because we wanted to have an independent proof for matrix space, pseudo matrix space. That's all. So what we intended the finally results of Michael were based on experience which we have felt in the case of matrix space. So you dig deeper into them, you get better theorem. That's all. So let us stop here and we shall meet next time with other notions of compactness, a new chapter. Thank you.