 Welcome to NPTEL NOC course on point set ball V part 2, module 13, we shall continue our study of para-compactness. For subspaces 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 disk 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 the 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 over n. a may be any set you take the complement and from there a distance must be at least 1 by 2 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 over n minus 1 by 2 over n plus 1 which is 1 by 2 over n plus 1. So I again leave this elementary thing as an exercise clue. Note that for each a and each n what we have is this one this is 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 n. 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 condition bigger than 1 by 2 over n. So closure will be what bigger than equal to that is all. So that is the reason why you have these are open subsets and a n contains a n bar etc. 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 we have saved here namely with the use of matrix space we were able to write every open subset union of countable union union of a n with this property a n contained a n bar contained a n plus 1. So that is what I meant by you know writing every open set as a union of dislike 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. Okay 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 1 okay. 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 axiom of choice every set can be well order okay 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 that is no there is no change okay u n is just instead of a here put u okay u n is except all x belong right 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 you u is some element in this curly u it is well ordered okay so it is an ordering you take all the initial elements to u and take all their u v n plus 1 power only not the full v then take the union of all of them okay n is fixed here v is varying v is varying only this part initial segment okay take the interior of all this union and throw it away okay u n star is that so here is a picture of u and u 1 and v 1 okay u is the given one v occurring this v occurring before you okay 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 the before that so v 1 star will 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 okay so that is u 1 star only this part okay 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 okay 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 okay 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 you here you have to do this for all v okay so for all v before you which occur before you in this well order well order would be any order no matter just one well order you have to fix once for all okay then u n star is a closed subset of u n okay contained inside u so why is a closed subset what I have deleted some open subset that is interior of something whatever it is if you delete an open subset it is a closed subset of whatever wherever you have deleted closed subset of u n okay 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 okay v is before you or you is before me all right because if v is before you v n plus 1 will get subtracted from u n other or otherwise the u n plus 1 will subtract from this is all okay 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 okay so you don't 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 okay after that you have to just use whatever you have proved before all right so if you use property 2 this should 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 some 1 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 okay so first you chose u first okay u such that x is inside u and then since u is open and we have union of u n is equal to u okay so one of x must be inside one of the u n's okay we put now u n hat is see you can get board 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 less than 1 by 2 power n plus 3 okay and u n griddle is all those x such distance x and u n star is 1 by 2 power 1 more 2 power n plus 4 okay so both of them are here this is less than this is less than or 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 it is 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 okay 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 u n star is 1 divided by 2 n plus 1 okay 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 okay so distance between u n griddle and v n griddle is 1 by 2 power sorry u n hat or 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 a natural number put v n equal to this collection u n hat where u range is over all of you okay so each member is a open subset 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 okay so how to check x this 10th one this can be checked as follows as in the case of 6 what we what we have done here that each x belongs to some u n okay so each x will belong to some u n star also is what you have to say okay in as in 6 choose first u so that x is inside u then x will be inside some u n first okay but then it is also union star because it does not belong to any v all those v's which we 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 okay indeed it's also prove that x itself is in u n twiddle also because u n stars are contained in the u n twiddle also okay 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's and u n twiddle u n twiddles are close v itself is an open cover okay so that much we have done okay and it's a refinement of you 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 in the u n the u n hat which is actually all those x belong to it 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 okay and that is contained inside see u n was contained in u n plus 1 but u n hat is not contained in u n plus 1 but it contained in u n plus 2 okay so that is contained inside you okay so all these all these u n hats are also it is also an open cover for it is also refinement of you okay so finally I will have one more notation here let u n check equal to u n hat minus union of all v k twiddles v belonging to you and k is less than n so this is where the twiddles were used okay this is only case where we have to do what we are doing now u n check case I don't want even u n hat I am I am checking away some portion of that namely these are close subsets now take the union of all these v k hats where v is inside you but integer k is smaller than n now okay all we are taking inside you but k must be less than n so you throw away that part note that this v k twiddle where k is less than n there is a bound for under k v k twiddle but v is servicing the entire of you this is a locally finite family of closed sets okay 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 okay so this whole thing is closed set the complement will be an open subset now u n hats are open subset this u n twiddles were not they were closed subset so u n check these are open subset okay I have question why this family is locally fine because you have taken only finite limiting of them okay so what is v k 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 don't worry about u and v I mean what they are they are two different elements of u v of capital u or curly u okay their distance between them is bigger than 1 by 2 power n plus and 2 power n plus 2 okay 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's why you consider right right so they are not they are closed subset yeah yeah and so this u n hat collection was open refinement but that may not be locally finite so that's why you are coming to u n check right yes that this precise is subtracting these things you know v k hats that will make it in u n check locally finite we will see that okay yeah so first of all u n check is open now you see we we didn't even stop it to u n hats also okay so first of all these are open itself you you have to look for that these these things are locally finite that is fine we can't but they are closed things right so now these are opens are set 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 raise over u n raise over n okay that is like all u n's first n fix it and then taking the union over n you can say double that that is a cover of x it's 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 x that w covers the whole thing okay given any x again let n be the first inches as x belongs to u n twiddle for some u okay once it's in some u there will be some n for which it belongs to okay so let's 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's 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 uk 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 okay 14th is what it's an open refinement we have we have told this open and these are refinement they are all subsets of you that is clear 15 is why it's locally finite okay 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 twiddles 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 okay then this ball v r x which will be contained inside u n twiddle okay 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 twiddles are open subsets neighborhoods okay they are open subsets also then v r x will intersect at most one member of v m okay 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 one six one of them it will intersect okay so that is why this one by two power n plus three 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 one by two power n plus three also okay so such a choice is possible then it happens that it will intersect only one of them okay so this completes the proof of a theorem there is a remark here which which is a bit takes you a little deeper so I don't mind even if you don't understand in the first reading okay so but I will make this remark a family a of subsets of topological space is called sigma discrete family if each x belong to x has a neighborhood which makes 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 it my lovely what happens is see this condition right in the beginning distance between these two is bigger than my two power n plus one so this u n star and v star same n but you and we 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 okay then it be it's called sigma discrete or that's 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 okay 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's all okay so that is what we have achieved here okay it is called sigma discrete if it is a union of countable union of a n's where each a n is discrete okay so that's 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 okay only countable union you need to take such a thing is called sigma discrete this family a itself is not discrete okay that's 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 u so in general what one does it without assuming matrix space you would like to prove this one out of some other properties by making this high this definition sigma discrete okay we have not used this concept anywhere in the core and so that's 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 okay so coming back to Rn we have remarked earlier that due to the existence of some smooth functions we get partitions of unity subordinate to an open cover moreover since step one of the proof of 3.6 is valid for all open subsets of Rn it follows that every open subset of Rn is paracompact indeed it's also true that every subspace of Rn is normal because it's a matrix space but what is important here is that given an open cover for any subspace there is a smooth partition of unity subordinate to that open cover but the functions are all defined on the entire of Rn okay not on just to that open subset thinking a little further along this line okay you will be able to prove the following theorem okay 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 Rn let X be a second countable locally compact half-dark 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 is zero set of a continuous function is a g delta set okay you can just write it as intersection of inverse image alpha inverse of 0 closed 1 by n open okay so how do we prove this one it's 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 okay 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 okay fc being a closed subset of a second countable say is second countable locally compact and t2 okay 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 okay so we get a countable locally finite open refinement U n and belonging to n I can write like this of U okay these Un'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 Un 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 if we alpha X n I am cutting it short to alpha n that is all okay 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 divided to 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 to power n summation is convergent okay not only convergent uniformly convergent so alpha will be automatically continuous function now it's elementary thing to check that this alpha X is 0 exactly on F that's all we wanted to prove okay so go through this this proof carefully again and again maybe three times it doesn't matter okay 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 okay thank you