 Welcome to NPTEL NOC course on point set biology part 2, module 13, we shall continue our study of para-compactness. For subspecies of RN, there is a result which says that every open cover has a subordinate partition of unity consisting of smooth function. The word smooth will not have any meaning when you are studying arbitrary topological spaces and para-compactness of that or partition of unity of that. Inside RN such a thing is possible and what is the additional thing that we have to do that namely on every disk you can have what is called as a bump function. For more you can see this reference here. The second thing is the local compactness in a very special way for RN namely the closed disks themselves are compact. So, if you carefully study the proof that we have given earlier, then we can get a similar result for any metric space with the help of decomposition of any open set into a countable union of increasing disk like open set. What I mean to say that is compactness is not all that necessary here the local compactness. So, something else will come to help namely the metric property and that is what we are going to do for more general results you may consult again Kelly's book. So, I will do the bare minimum here to expose you to the ideas behind this course is not something you can say concise or comprehensive and so on. Every pseudo metric space is para-compact. So, this is a big theorem now. So, slowly we have to have some patience slowly we will develop it and the development itself is quite educative. So, first of all let us have some notation here for every subset of this pseudo metric space for each integer n let us have a notation a n a is a subset and n is an integer a n is set of all x inside x such that distance of x from a complement is bigger than 1 by 2 power n. a may be any set you take the complement and from there a distance must be at least 1 by 2 power n of automatically these are subsets of a subsets of a. So, one thing very easy to see is that using triangle inequality distance between a n and a n plus 1 complement will be bigger than equal to 1 by 2 power n minus 1 by 2 power n plus 1 which is 1 by 2 power n plus 1. So, I again leave this element I think as an exercise clue. Note that for each a and each n what we have is this one this also easy a n is contained in a n bar that is obvious always, but a n bar is contained inside a n plus 1 and again a n plus 1 is contained inside a. So, already we have this is what we want it right only thing is these a i's are not compact or anything, but we have this increasing even increasing the phenomenon a n contained in a n bar contained a n plus 1 contained inside a and each a n is open. So, that is because this is this condition bigger than 1 by 2 power n ok. So, closure will be what bigger than or equal to that is all. So, that is the reason why you have these are open subsets and a n contained inside a n bar etcetera. Moreover if a itself is open then only a will be the union of a n in what we have proved earlier was when you have locally compact Lindelof space every open subset can be written like this with each a n bar compact and so on that is not that is not what we have here, but something which you have saved here namely with the use of matrix space we were able to write every open such that union of countable union union of a n with this property a n contained a n bar contained a n plus 1 ok. So, that is what I meant by you know writing every open set as a union of disk like open sets since a n minus 1 that is a n and the whole thing is that a slowly you will approach this a n ok. They are nested very strongly in the sense that the the previous thing is contained in a n plus 1 that is just nested, but the closure itself is contained in a n plus ok. So, these are just notations now you have to remember this one for the rest of the proof. Let now you be an open cover for x we want to extract a locally finite open refinement that is our purpose finally choose a well order on you. So, again here you have used a Ximov's choice or this one every set can be well order ok. For each n belonging to n and for each u inside u let us set up another set of notation now u n be the be defined as in one u n there is no there is no change ok u n is just instead of a here put u ok u n is except all x long time such that distance between x and u complement is bigger than 1 by 2 power n. But more definition I am going to give you namely u n star is a subset of u n wherein I have thrown away this is the sectaric complement sectaric minus interior of all the v n plus 1s where v occur before u u is some element in this curly u it is well ordered ok. So, it is an ordering you take all the initial elements to u and take all there u v n plus 1 part only not the full v then take the union of all of them ok n is fixed here v is varying v is varying only this part initial segment ok take the interior of all this union and throw it away ok u n star is that so here is a picture of u and u 1 and v 1 ok u is the given one v occurring this v occurring before u ok. So, if you look at v 1 star there is nothing suppose v 1 was v was the first one and u was the second one there is no throwing because there is there is nothing below before that. So, v 1 star will be be a full thing here but u 1 star what will happen I will have to throw away v 2 right v is there so v 2 has to be thrown away here ok. So, that is u 1 star only this part ok because of u n and v n definition if you take any element of u element of v 2 here the distance will be at least one fourth in particular distance between any element here and any element here will be at least one fourth ok. So, this is the picture that is showing. So, why we are making this kind of you know adding or subtracting some portions of earlier elements only that is what we are doing here ok u n star is this one. So, only u 1 v 1 picture I have shown. So, you have to do if you selecting one element u here you have to do this for all v ok. So, for all v before u which occur before you in this well order well order would be any order does not matter just one well order you have to fix once for all ok. Then u n star is a close subset of u n ok contained inside u. So, why is a close subset what I have just deleted some open subset that is interior of something whatever it is if you delete an open subset it is a close subset of whatever wherever you have deleted close subset of u n ok. So, that is what it is and it is a subset u n is contained inside u. If u and v are not equal they are different distinct elements of u then u n star is inside x minus v n plus 1 or v n star is in x minus v n plus 1 depending on whether v is first or u is first ok v is before you or u is before v all right because if v is before you v n plus 1 will get subtracted from u n or otherwise u n plus 1 will subtract from this is all ok. So, that is the definition therefore, in either case what happens is from this general remark here distance between a n and a n plus 1 complement is bigger than 1 by 2 power n plus 1 which I showed you in this picture what happens is distance between u n star and v n star is always bigger than equal to 1 by 2 power n plus 1. So, in this picture it was n equal to 1. So, it is one fourth ok. So, you do not have to do any pictures at all if you follow the logic one by step each step is a very small picture in your mind ok. After that you have to just use whatever you have proved before all right. So, if you use property 2 this will be obvious. The next thing is each x belongs to u n for some n and some u right first of all it belongs to some u because u is a cover, but once it belongs to some u from the complement of u its distance will be positive. So, there will be someone by n for which it is be smaller than that bigger than that. So, that is all you have to show then it will be inside u n ok. So, first you chose choose u first u such that x is inside u and then since u is open and we have union of u n is equal to u ok. So, one of x must be inside one of the u n ok. We put now u n hat is see you can get both. So, first you have u n star here now u n hat is set of all x such that distance between x and u n star is less than 1 by 2 power n plus 3 ok. And u n griddle is all those x set distance x and u n star is 1 by 2 power 1 more 2 power n plus 4 ok. So, both of them are here this is less than equal to that you have to pay attention each u n is open once again and u n griddle is closed because there is equality here and because you see 2 power n plus 4 u n griddle will be contained inside u n hat. Once again the same property 2 will tell you the distance between u n hat and v n hat is bigger than or equal to 1 divided by 2 power n plus 2 ok for every u for every v inside u. So, this time you can directly use the fifth property here that distance with u n star union star u n divided by 2 n plus 4 ok. So, u n hat v n hat similarly you can talk about u n griddle this u n griddle really will not be needed in the final proof, but it will play some auxiliary role. So, I have kept it ok. So, distance between u n griddle and v n griddle is 1 by 2 power sorry u n hat 1 by 2 power n plus 2 and they are open subsets no matter whether u occurs first or v occurs first. Next for each n inside natural number put v n equal to this collection u n hat where u range is over all of u ok. So, each member is a open subsets. So, this is an open family of open sets all right. Take v to be union of v n's. So, we have written v as a countable union of these families. What is the property of these v n's and v? v is an open cover for x not v n's, but when you take all of them v that is an open cover for x and v is a refinement of u ok. So, how to check x this 10th one this can be checked as follows as in the case of 6 what we have what we have done here that each x belongs to some u n ok. So, each x will belong to some u n star also is what you have to think ok in as in 6 choose first u. So, that x is inside u then x will be inside some u n first ok, but then it is also in u n star because it does not belong to any v all those v's which you have subtracted it will not be inside v. So, that will be in u n star all right, but why it is in u n hat since u n star is contained inside u n hat we are done ok. Indeed it is also proved that x itself is in u n twiddle also because u n stars are contained in the u n twiddle also ok. So, remember these are less than less than. So, u n hat and u n twiddle actually fatten the u n star right all those which are of distance smaller than that one. So, you have to understand this inequality he has in the beginning we are bigger than bigger than right. So, both of them are used left and right you are cutting down things all right. So, we are we know that this u n star and u n twiddle u n twiddles are closed v itself is an open cover ok. So, that much we have done ok. And it is a refinement of u because all these elements are for each u n is subset of u corresponding u right. So, they are inside all right. Now, look at u n star is contained as a u n. So, u n hat which is actually all those x belong to x. So, the distance between x and u n is 1 by 2 power n plus 3 I am just recalling this definition is contained inside u n plus 2 ok. And that is contained inside u n was contained in u n plus 1, but u n hat is not contained in u n plus 1, but in contained in u n plus 2 ok. So, that is contained inside u ok. So, all these all these u n hats are also it is also an open cover for it is also a refinement of u ok. So, finally, I will have one more notation here let u n check equal to u n hat minus union of all vk twiddles v belonging to u and k is less than n. So, this is where the twiddles were used ok this is only case where we have to do what we are doing now u n check case I do not want even u n hat I am I am checking away some portion of that namely these are closed subsets now take the union of all these vk hats where v is inside u, but integer k is smaller than n now all v I am taking inside u, but k must be less than n. So, you you throw away that part note that this vk twiddles where k is less than n there is a bound for under k vk twiddles, but v serviced in the entire of you this is a locally finite family of closed sets ok. Therefore, it follows that each u n check is open when you have locally if I remember this thing about locally finiteness locally finite family of closed sets when you take arbitrary union it is still a closed set ok. So, this whole thing is closed set the complement will be open subset now u n hats were open subset this u n twiddles were not they were closed subset. So, u n check these are open subset ok. I have a question why this family is locally finite because you have taken only finitely many of them ok. So, what is vk if you take any two of them different from same v the same k they are actually disjoint. So, that is what this property may have to use u n hat v n hat do not worry about u and v I mean what they are they are two different elements of u of capital U or curly u ok their distance between them is bigger than 1 by 2 power n plus and 2 power n plus 2 ok. Sir, I have one more question here. So, starting with the open cover first we made u n stars the collection of which is also a cover for x, but that was not open that is why you consider u n hat. So, they are not they are closed subset. Yeah and so, this u n hat collection was open refinement, but that may not be locally finite. So, that is why you are coming to u n check right. Yes that this precise is subtracting these things you know vk hat that will make it u n check locally finite we will see that ok yeah. So, first of all u n check is open now you see we we did not even stop at u n hats also ok. So, first of all these are open itself you you have to look for that these these things are locally finite that is fine vk hat, but they are closed things right. So, now these are open subset first thing. So, what we want to do is that this is now take w to be the collection of all subsets of the form u n hat u range over u n range over n that is like all u n's first n fixed and then taking the union over n you can say double that that is a cover of x it is an open refinement this is locally finite itself all right there is no sigma locally finiteness here this w is actually locally finite. So, we have to prove this 13, 14 and 15 all right. So, say how do you prove 13? 13 is that that w covers the whole thing ok given any x again let n be the first inch is as x belongs to u n twiddle for some u ok once it is in some u there will be some n for which it belongs to ok. So, let us take x belong to some the first u n twiddle for some u it follows that it is not occurring here at all right it is always before that I have k less than n I have taken, but n is the first one to which it belongs to it will not be inside some other u k k less than n. So, this part it is not there so it must be inside u n check once it is here it will be inside u n check. So, that is the trick here. So, these things cover ok 14th is what it is an open refinement we have we have told this open and these are refinements they are all subsets of you that is clear 15 is why it is locally finite ok to see the 15th one notice that u n twiddle chosen as above is a neighborhood of x and does not intersect any v m check for m bigger than n because these n twiddle u n twiddle would have been subtracted from v m right. So, it does not intersect v m twiddle at all therefore, if we choose 0 less than r less than 1 divided by 2 power n plus 3 ok then this ball v r x it will be contained inside u n twiddle ok I sorry what I want to say r must be less than this it may be even further smaller so that v r x is contained u n twiddle because u n twiddle are open subsets neighborhoods ok they are open subsets also then v r x will intersect at most one member of v m ok because m less than equal to n n less than equal to m both of them it cannot be as soon as m m it will not intersect anything smaller or bigger 161 of them it will intersect ok so that is why this 1 by 2 power n plus 3 I have to show once there is an r such that v r of x is in u n twiddle you can make it smaller than any further also you can take it smaller than 1 by 2 power n plus 3 also ok. So, if such a choice is possible then it happens that it will intersect only one of them ok. So, this completes the proof of a theorem there is a remark here which which is a bit deep it takes you a little deeper. So, I do not mind even if you do not understand in the first reading ok so, but I will make this remark a family a of subsets of the topological space is called sigma discrete family if each x belong to x has a neighborhood which means at most one member of a. So, one of you ask this question that is why this remark will be even more relevant here now. So, why is sigma locally what happens is you see this condition right in the beginning distance between these two is bigger than 1 by 2 power n plus 1. So, this u n star and v n star same n, but u and v are different elements of the same cover what happens to these sets they are disjoint right not only that this is stronger than being disjoint namely I can take small open subsets around them right for all of them around all of them simultaneously. So, that all these neighborhoods are disjoint. So, because of this metric property we are able to do that one such a thing is you can make that as an axiom in the general case then it is called sigma discrete that is what I am trying to say here family is discrete family if each point x belongs to x has a neighborhood which means at most one member of a. You see if the distance between as something positive each point x I can take the ball of radius half of that distance whatever positive I have say half the half that radius then what I get is that that open ball cannot intersect both of them that is all. So, that is what we have achieved here it is called sigma discrete if it is union of countable union of a n's where each a n is discrete. So, that is why that n has come when you fix n it is a discrete family you take the union it becomes a cover and so on. Only countable unionality such a thing is called sigma discrete this family a itself is not discrete that is one thing you have to understand is sigma discrete. Properties is 8 to 9 say with that this v which we have constructed is sigma discrete open refinement of view. So, in general what one does it without assuming metric space you would like to prove this one out of some other properties by making this definition sigma discrete. We have not used this concept anywhere in the core and so that is what I want to say if you prefer you can simply ignore it for the time being if you are interested more then you can look into Kelly's book ok. So, coming back to R n we have remarked earlier that due to the existence of some smooth functions we get politicians affinity subordinate to an open cover. Moreover, since step one of the proof of 3.6 is valid for all open subsets of R n it follows that every open subset of R n is paracompact indeed it is also true that every subspace of R n is normal because it is a metric space, but what is important here is that given an open cover for any subspace there is a smooth partition affinity subordinate to that open cover but the functions are all defined on the entire of R n ok not on just that open subset thinking a little further along this line ok you will be able to prove the following theorem ok this is this I will get it as a corollary to whatever we have done so far. All this remark is for getting such a motivation from R n let x be a second countable locally compact half dot space then every closed subset of x is the precise zero set of a continuous real valued function you can choose the the codement to be 0 1 the closed interval 0 1 once you have this it follows that such a thing is a g delta set also because the precise set it is a zero set of a continuous function is a g delta set ok you can just write it as intersection of inverse image alpha inverse of 0 closed 1 by n o 1 ok so how do we prove this one it is not difficult start with any f contained in the x closed subset for each x in the complement choose a function alpha x from x to 0 1 such that alpha x at x is 1 and alpha x of f is 0 ok consider the open cover u is alpha x inverse of open 0 1 closed x belonging to this fc I have such function as alpha x so take u to be this open set this will be an open cover for fc ok fc being a closed subset of a second countable say is second countable locally compact and t2 ok therefore it is para compact so any open cover so this open cover u there is a locally finite open refinement further by second countability any open cover you can get a countable sub cover it will be again a locally finite refinement ok so we get a countable locally finite open refinement u n n belonging to n I can write like this of u ok these u n's are not necessarily subsets of u I mean they are not members of u but they are refinements they are contained in some members of u since each u n is contained in some alpha x inverse of 0 1 that is the way it of course right for some x we can select one alpha x such that and relabel it as alpha n instead of alpha x alpha x n I am cutting it short to alpha n that is all ok so define alpha now from x to 0 infinity no index here right equal to sum of all alpha n's but divided by 2 power n each of them after dividing 2 power n take the sum these 2 power n's are there obviously to make the whole sum convergent these alpha n's are bounded by you know they are 0 to 1 so any number 0 to 1 divided by 2 power n summation is convergent ok not only convergent uniformly convergent so alpha will be automatically continuous function now it is elementary thing to check that this alpha x is 0 exactly on f that is all we wanted to prove ok so go through this this prove carefully again and again maybe three times does not matter ok the each step each way why you are doing all this somewhat circus kind of thing they have meaning there next time we will do some general rate result which comes from nowhere but the motivation is here if you know this one you know where it is coming ok thank you