 Welcome to module 11 of NPTEL NOC course on point set topology part 2. So, today we shall pick up another important topic para compact spaces. Among several notions of compactness some of them we are going to study. It seems that para compactness is the best which captures certain features of compactness and yet encompasses a large class of interesting topological spaces. The feature that you are looking for is as usual having a large number of continuous real valued functions on a space. So, this is one of the interesting properties of para compactness that we will prove. So, let us begin with a formal definition or two. Take any set and two families of subsets of F1 and F2. We say F1 is a refinement of F2. If every member of F1 is contained in some member of F2, it may be contained in more than one member. Pay attention to this one. I am not saying that families of F1 is not a subfamily of F2, not necessarily. Sometimes it is possible. Actually if F1 is a subfamily of F2 then this condition is obvious because you can take the member here, the same member will be there in F2. You can take that. So, that member will contain itself. So, refinement is much better than something to do much better. Actually it comes from the practice of giving a subdivision of the interval 01 or interval AB. You are taking another subdivision which you call it as subdivision. Different divisions are there, but subdivision. That subdivision thing is refinement here. The word refinement takes place here. In the context of F2 is a cover for X, which just means that members of F2 together they contain all the points of X. Union of members of F2 is X. In that context, we will use the same word refinement for a slightly stronger meaning. Namely, F1 is a refinement of F2 only if F1 is also a cover of X and the old condition namely each member of F1 is contained in some member of F2. So, if it is a cover, this should be also a cover. F1, F2 is a cover then F1 must be also a cover. If X is a topological space, a refinement in which all members are open, respectively closed will be called an open refinement or a closed refinement accordingly. So far, we have actually made four different definitions here. Suppose now X is a topological space. We say a family F of subsets of X is locally finite if at each point X connects, there exists an open set U such that X is in U and U intersects only finitely many members of F. So, this is not the property of the space, it is just the property of the family of subsets inside a topological space. Similarly, we will have another definition later on what is called as point finite which just means that a given point can be found at most in finitely many members of F. Clearly locally finite implies point finite but not the converse. Anyway, I will recall the point finiteness when it is necessary. The following two simple results are the keys for usefulness of this local finiteness. So, these are the two things here namely take a family U of subsets of X which is locally finite. Then if you take union of all the closures that is another family U is inside U but I have taken closures now that will be also locally finite. The second thing is union of U bars. Remember if you have finite union then the closure is the union as a closure. But if it is infinite union then this may not hold but here is the case wherein it is possible namely union of U bars is also closed. So, it is in particular it is the closure of all the union. So, that is because local finite. So, let us have a look at it how this works. Given X the first part A I will not show that U bar is also locally finite. Let V be a neighborhood of X in which such that V intersects only finitely many members of U. If U is such that V intersection U is empty then V intersection U bar is also empty because V is an open subset. Therefore, V intersects only finitely many members U bar. It may we may intersect only those U i bars when U it intersects U i if it all if it intersects U i itself. Therefore, V is also finite that is all. Now, part B slightly more complicated actually this we have seen in part 1. Number 9 I will repeat it take a point in the closure of this set namely I have already taken U bar U inside U but I have taken the union then I do not know whether it is closed I want to show it is closed. So, take the closure that must be inside one of them is what I have to show right if it is inside the union of U bar it must be inside one of the U bar. So, first of all choose a open neighborhood W X of X such that W X intersects only finitely many members of U bar this U bar U inside U this is possible because of part A just now we have proved that. So, let us list these members U 1 bar U 2 bar U n bar we claim that X must be in one of these U 1 bar U 2 bar U n bar okay that will be enough after that it will follow that X is inside the union of U bar itself. Suppose this is not the case then what happens for each I we will get a neighborhood W I of X such that W I intersection U I bar must be empty because it is not in U I bar okay this is a closed subset already. So, I could have taken the compliment of this U W I bar for W I let us solve okay. Now, you take W equal to all these W I intersection along with the original W X that is a neighborhood of X now then this W will be a neighborhood of X okay and this W intersection all the U bars will be empty the entire union of U bar will be empty which just means that point X is not in the closure of this so that is absurd because you started with the point which is in the closure okay now let us make a definition what we were aiming at a space is called para compact if each open cover for it has a locally finite open refinement okay I want to recall each of these terms once again you have open cover okay let us call this as F2 a refinement now means first of all each member of this refinement say F1 is contained a member of F2 union of members of F1 will be the whole of X because I started with F2 as a cover for X and then finally it must be locally finite at each point of X there must be an open neighborhood which will intersect at most finitely many members of this refinement for each open cover if we have locally finite open refinement such a space will be called para compact it is somewhat similar to the definition of compactness wherein each open cover has a finite sub cover we are not insisting on a finite sub cover first of all finiteness is replaced by local finiteness which is much much weaker not only that we are not insisting upon that this is a sub cover of the original cover it is a refinement so you can put more open sets here but they will be smaller so that is the beauty of this one since any finite families automatically locally finite it follows that a compact space is automatically para compact you start with any open cover first take a finite you know sub cover that will be refinement already now you take a refinement you do not have to because any any sub cover is a refinement you do not have to worry about local finiteness because it is already finite so automatically it is finite so compact spaces are para compact para compactness is very useful study very useful in the study of manifolds in the absence of compactness in fact it is also useful in more than you know manifolds which are called simplicial sets simplicial complexes and then what are called as CW complexes and so on okay so I will not much I cannot dwell much into that one those are the things which are studied in algebraic topology a locally compact lindelope space is para compact so we are now up to what are the kind of spaces which will become para compact what kind of conditions we can put to ensure para compactness of course definition is there finally but how to verify so suppose you know something is locally compact and lindelope that is para compact so this is the second kind of result here because easy one was compact implies para compact okay so this is very profound result actually the proof the proof itself is very very illuminating so you should pay attention to that okay so the first step I am going to prove them in separate steps so that you can just remember the step itself is such a kind of property first step is that there is a sequence of open sets un in x such that each un bar is compact un is contained inside un bar contained inside un plus 1 x in the union of un so every locally compact lindelope space can be written as union of open sets increasing union such that the closures are compact okay so that is what we are going to prove using local compactness and hence regularity for each x belonging to x we first find open sets x such that x is in x and ax bar is compact we do not need to assume that ax bar is contained inside some some some subset and so on right now using the lindelope property we find a countable sub cover from ax x belonging to x for each x I have put an open subset ax so this is an open cover for x every open cover of a lindelope space admits a countable sub cover let us rewrite this countable sub cover dropping out x here I will write this a1 a2 an and so on to be precise you may want to write it as ax1 ax2 and so on we do not worry about the point x we are only worried about this countable sequence of open sets whose closures are compact that is what we are worried about what is the extra property union of all a n c is the whole space x so you see part of that already come but now we have to arrange them so that the entire thing becomes increasing union okay so this is again quite useful this kind of argument is quite useful in measure theory you start with one of them any one of them so I have already enumerated them so take u1 equal to a1 since u1 bar is compact okay and all these a is together cover this u1 bar so a finitely many of them will cover it right so there is some n1 such that u1 bar is contained in the union of a1 a2 a1 okay take u2 to be the union of these finitely many of the first n1 of them carry u from 1 to n1 so each a k is open remember so u2 is an open set okay it contains this a1 this contains u1 and if you take the closure of u2 that will be equal to closure of you know union of the closures of a case therefore it is compact being a union of finitely many compact sets okay so what we have done is setting up an iteration process here u1 has been patterned to u1 bar which is patterned further to u2 is the open set and u2 bar is compact now we will repeat it repeat the same procedure for u2 bar again u2 bar is contained inside the union of a1 a2 a1 right so there will be some finite obviously that finite number okay can be taken bigger than equal to n2 you know in some funny cases it may happen that the new things which come they are already inside we have not arranged that one so it is possible that you do not get anything extra so but you can always take larger and larger the finite cover after all so you can take some more nobody stops you so you can always arrange n2 to be bigger than n1 and un bar is contained inside u3 here for n2 etc I have to say so continuously you can go on repeating this process such that un bar is contained inside carrying from 1 to this is u2 bar n2 ak okay that is that that is we will call it as u3 and so on and so on is there so then you can write this one so here it will be carrying from n to some nn so that is a funny notation so I do not want to write that one so the set properties are easily verified all that you have to do was property 2 and that is how we have selected this there is nothing more than that ultimately given any point x it will be in one of the un's right because x is union of un's right and the x is in one of the a n's here so therefore because this sequence is increasing strictly so some n a and k will be hit at some stage so it will be inside that un so here's a picture how things are going on you started with a1 equal to u1 the closure you can think of this one as it's a boundary it's a closure taking that one so this a2 a3 a4 is covering the closure of a1 okay that is u1 bar there may be other things here but you don't have to worry about that now you take the union of a1 a2 a3 a4 as your u2 now this u2 will have closure of that something like that that will be compact so again some u4 a4 a5 a6 etc will come which will cover u2 bar and go on okay ultimately a1 a2 a n a3 and so on these will cover x therefore u1 u2 u3s will cover okay so here is a remark before we go further a space which is a countable union of compact spaces is called a sigma compact space what we have proved above implies that every locally compact lindelow space is sigma compact in a strong sense what is the strong sense that these compact subsets are actually closures of open subsets the open subsets themselves cover the whole of x so that is the strongness here so we shall study sigma compactness a little more through some exercises later on okay this is also an important concept but mostly it is used in analysis now let us go to the step two towards proving that the space is para compact so in this step we claim that any space which satisfies the property stated in step one is para compact so now we are not going to use local compactness or lindelowness we will just use the property 1 2 3 here that there is such a family of open subsets in x that is all we are going to use to prove that the space is para compact okay so here is again another similar step which you have to learn start with w alpha any open cover we have to extract a locally finite open refinement for this okay for notational simplicity let us also put u naught and u minus 1 and u minus 2 see we have taken u 1 u 2 u 3 just for you know wherever you start you would like to have some left margin here so u naught u minus 1 u minus 2 as empty and then u onwards whatever we have okay so then you can write down inductive step very easily you can start induction at any of these points that is the whole idea so where u n's are chosen as the step one okay defined for each n greater than or equal to 1 and each alpha I am going to concentrate on w alpha okay so everything is intersected with w alpha what I am going to do for each n look at u n minus u n minus 3 bar throw away this compact hence closed subset and throw away a closed subset for no open subset this should be still open see I want from u n to n minus 3 that is why I have taken that u naught u minus 1 u minus 2 as also I have defined they are empty set in the case of n equal to 1 this sense this makes sense but it is actually u 1 because this part is empty does not matter what I insist as this is an open subset this is an open subset so this v and alpha is open subset okay so this is now defined for each alpha and each positive integer n clearly this is an open refinement of w alpha so what are the things that you have to verify every member is contained in some member here there w alpha that is fine it is open that is all fine these themselves cover the whole of whole of x y because whatever it is it is first of all in some w alpha after that it must be also in one of the u n's okay you take the first u n for me this happens because u n's are increasing sequence of open sets covering the whole of x so it is an open refinement we shall find a sub cover in fact a countable one which happens to be locally finite as well okay so this itself may not be locally finite so I will pass on to a sub cover sub cover means what it is a cover but there only a few members will be taken how many actually will take it countably many members from here all right yeah so this is what picture says here this rectangular thing represent w alpha this is the increasing union of our opens of sets which cover the whole of x these w alphas also cover the whole of x so what I am doing here is u n minus u n minus 3 closure so I am keeping one two three three steps away okay this whole u n minus the smallest u n minus 3 I I check it out I take u n minus 3 closure all these happening only inside w alpha so you look at only this portion and subtract this portion this heavily dotted bar is v n alpha heavily dotted v n alpha as alpha varies you will get a lot of things okay after that if you vary n you will get the entire of the spaces okay so that is how v n alphas have been constructed now some family sub family of this is going to be locally finite as well as a cover so that is what we have to find out for each fixed n the compact subset u n minus 1 bar minus u n minus 2 of this u n minus u n minus 3 bar okay I am not here I am not intersecting with with with w alpha yet okay so I am in this in this concentric ellipses here look at u n minus u n minus 3 inside that I have another open subset namely u n minus 1 minus u n minus 2 that is what I am looking at only I am looking at the other way around now u n minus 1 bar minus u n minus 2 so closure minus the open set so this is a compact subset of this open subset okay all of them are covered by v n alpha therefore we will have a finitely many of them which will cover this part call them f n those members finitely many members which you select to cover this one put them instead f n so I have taken only finitely many members from here they are members here but they are here now what is the purpose of this one this family okay each member is covering this part what does it mean they are substrates of u n minus 3 already so they will not intersect the u n minus 3 closure at all so they are contained somewhere here like that okay when you intersect with w alpha of course for each alpha you have to take that because when you have taken alpha that is where I have taken f n you know here alpha also varies and also varies but f n there are only finitely many members the construction is over all that I do is now is put f equal to union of f n's each f n has finitely many members so the union will have countably many members okay and they are all members of v n alpha so this is a sub family a sub family of a refinement is also a refinement okay since x is a increasing union of u n bar given any x let k be the smallest n such that x is in u n bar which means that x is in u k bar but not in u n minus k bar so that is the smallest k okay then x will be inside u k bar minus u k minus 1 therefore belongs to some member of f k plus 1 for n it will be f n oh this must be in f k plus 1 therefore every member of x is inside some f k plus 1 union of all the members namely this f will be a cover for x okay finally y f is locally finite given any x we take n such that x is inside u n it is clear that u n does not intersect any member of f n plus 3 right so in this picture it is rather around if you take something here in u n minus 3 then members of f n will not intersect this one so when you index high add three of them so it will never come so it may be inside f say this is f 1 it may be inside f 2 f 3 but it will not be inside f 4 therefore it just means that this u n will not intersect any member of f n plus 3 up to f n plus 2 how many members are there finally many members so u n itself will serve the purpose of local finite therefore f is locally finite so that completes the proof of this theorem we now consider an important result that says that every para compact half star space is normal similar to the result that every compact half star space is normal on the way we shall prove that it is regular also so proof is somewhat similar in that same indeed that seems to be the key step to prove it is so normal we first have to prove regularity here just like in the proof of the fact that compact half star space is normal which we have proved a couple of days before right so let us go through the proof of this one every regular para compact space is normal start with any two disjoint closed sets pick up a point x in a pick up a point pick up an open neighborhood u is a set a belongs to u a contained in your a bar contained in the complement of b which is an open set okay then the family u a such that a belong is to a union ac is an open cover for x so we have just used regularity here a and b are disjoint sets now we have got an open cover now we use para compactness so let f be a locally finite open refinement of this family and f prime is a sub family with a collection of all members of f which are not contained in ac being a refinement every member of f will be contained either in here or in some member here i don't want those which are contained in here i want to concentrate on this part where is f prime okay in particular each member of f prime is disjoint from b okay it is not contained in a complement it might contain somewhere so it is disjoint from b put u equal to union of all d such that okay d belongs to f prime okay so all these open sets which are in f prime none of them intersects b so u will not intersect b and of course a belong to a they are all covered by this one so a belongs to u a is contained in u and u intersection b is empty okay i repeat here look at members of f prime they are contained in this part okay and u a is in the complement of bc so there that's why they don't intersect b that's all okay they are open cover a so u intersection b is empty okay so we have got one part namely an open subset around a which does not intersect b all right now look at f prime is locally finite because it's a sub family sub cover of it is a sub cover of it's a sub family of locally finite family which is f so f prime is locally finite therefore for each b inside b there is a neighborhood open neighborhood vb of b such that only finitely many members let us say d1, d2, dn belonging to this f prime will intersect vb so look at these di's they will have to be contained in one of these members now because members of f prime are coming from here okay so let us call them ua y ua 1 ua 2 ua n ua is now you look at wb which is intersection of this vb along with the closures of closures of ai bar complement okay i am taking complement of ui that will be close subset but if you take complement of the closures that those still the open subsets finitely many intersection will be finitely intersection is also finitely many intersection so it is open so wb is a neighborhood of what b for each b then wb is a neighborhood of b and disjoint from you right so let me see why it disjoint from you because of this one let me prove this one why this disjoint from you once you have done that we are finished roughly okay u and v will be open subsets which are disjoint so this wb for each b we are taking okay so let us see why wb intersection u is empty wb intersection u is wb intersection d1 union d2 union dn okay that is contained in wb intersection ua1 etc uan each di is contained inside uai okay so here is containment but this is nothing but vb intersection ua1 bar ua1 bar complement ua2 bar complement and intersection of this and then this union intersection of this union see the brackets are important here I am taking unions here this intersection here here I am taking intersection this is also intersection the intersection of one two three sets but this itself is the intersection this is a union but now this is then we have vb intersection intersection of unions we keep one this one single but in this union this this union this this you have take the union of all intersections now this intersection ua1 this intersection ua2 this intersection uan and so on but each time uai comes the complement of uai by closures are there they are disjoint therefore each member here is empty it is a union finite union of empty sets so vb intersection empty okay therefore vbs are each vbs disjoint from you you now you take v to be the union of all the vbs b running over b so we have found out two disjoint open subsets u and v respectively containing a and b that completes the proof that a regular paracompact space is normal the next step will be proved to prove a half-dar paracompact space is regular okay so that is the next step a half-dar paracompact is normal what we have to do you have just prove regularity and use the previous theorem right so let x belong to x u be an open subset such that x is inside u for every y in the complement of complement of u that is a close subset choose disjoint open sets ui and vy such that x is inside ui and y is inside vy okay x and y are distinct and then I can do this one because it is a half-dar space now as you vary y over the complement of u okay this vy will cover the complement of u along with u that will be an open cover for x so this admits a locally finite open refinement that is where paracompactness is used again as in the previous step take only those members of f which are not contained in this singleton u here okay they may be coming from here they have to come if it is not coming from here they have to be subsets of some members here obviously this being a subfamily it is also locally finite take w be an open set such that x is inside w and w means only finitely many members of f prime which we will write as w 1 w 2 wi each wi will be contained inside vy i vy i's are in w is view complement okay see these steps are identical almost identical as in the previous proof of the previous theorem okay idea is identical at least now take n to be intersection of w with u y 1 u y n where u y i's are chosen v i's are this y i's are chosen in this way I am passing to u y 1's which are disjoint from v y 1's okay I do not have to take the complement and closures and complements and so on here they are ready made open subsets which are disjoint from that one clearly this is an open set containing x because u y's are all neighbors of x here x is inside u y for all y okay we claim that the closure of n is inside u that will complete the regularity x contained inside u to find x contained in n contained in n bar contained in u that is the regularity okay so let v which is union of all a a inside f prime so this is an open these are members of open this is an open subset then v is an open set and u complement will be contained inside v because together they cover the whole thing right so members after all we started f as a covering for this one right so this u and members of here they will cover the whole thing right so u c is contained inside the union of all these v therefore it is enough to show that n intersection v itself is empty is n is open subset then n intersection v bar will be authentic okay or n bar intersection sorry n bar intersection v v will be empty so n bar will be contained inside u okay so look at n intersection v same argument okay w intersection u i one intersection intersection u i n intersection with v but now expression for v i have i have written w one union w two w n okay the u i ones are as it is so now you look at this w one you can replace them by v y ones v y to each w i is contained some v i right so v y one v y to v y n but now when you distribute it because this is intersection u one union to un intersection with this one so each of them intersection with one member here whichever member it will be empty so it is w intersection empty set that will be empty so all this part proof is identical to the previous steps so in conclusion para compact tors dos space is normal plus t one is already there so it is t four all right so we have done some major topological properties of para compact spaces the next thing is functional properties namely what are called a partition unity partition of unity is one of the major purpose of introducing para compactness by the great mathematician dironi so thank you let us do this para partition of unity next time