 Welcome to module 8 of NPTEL NOC course on point set apology part 2. So, we shall continue the study of local compactness. We have already observed that in our definition of local compactness regularity is built in under certain mild additional conditions like T1-ness, it will be a T3-space and then it will be a hostile space also ok. So, what we would like to see is that under this local compactness many stronger separation properties are possible. So, here is illustration let us begin with that. The first thing lemma here is take a compact set compact and close subset of a locally compact space that n be an open subset in x which contains k. Then there exist an open set u such that k is contained in u contained inside u bar contained inside n and u bar is compact. Moreover there is a stronger thing here that this kind of separation is actually coming from a function namely there is a continuous function f from x to 0 1 such that f of k is singleton 0 and f of the complement of u is singleton 1 ok. It started with the u is contained inside which contains sorry started with n actually n is a neighborhood ok. So, I could have stated is f of the complement of n is also singleton 1. So, this is also correct anyway. So, how do we how do you how do you perceive this one? Local compactness ensured for each point like this suppose if point in the neighborhood then there is a neighborhood u contained u bar contained inside n and u bar compact. From a point we are now extending it to a compact sector. So, that is the first thing and then not only that you can actually get functions like this like your results type results are true. So, proof is not all difficult let us go through the proof. For each point x inside k by local compactness we can find open sets u x such that x is inside u x u x bar contained inside n ok this is point wise. Since k is compact if you take all the u x x inside k that will be an open cover therefore, there are finite sub cover that means what k is contained inside some finite union of these numbers say I will call them as u 1 u 2 u n. Union of u n so, I will take it as u each u i is contained inside n. In fact, each u i bar is contained inside n therefore, u bar which is the union of u i bar because it is a finite union that is also contained inside n moreover it being a finite union of compact sets this u bar is compact also ok. So, what I want k contained inside u contained inside u bar contained inside n ok. So, first part is over now see u bar is compact and it is we are working inside a locally compact space ok. So, it is a closed subset this is locally compact also it is regular also regular and locally compact we have said it is normal ok u bar itself is normal therefore, you can apply the Uridon's characterization to this k and u bar minus u ok k is an open subset u bar minus u is k sir k is a closed subset u bar minus u is another closed subset they are disjoint therefore, you can get a function g from u bar to 0 1 that is the g of k is 0 g of u bar minus u is 1 ok. So, this is the standard Uridon's characterization which we have seen in the part one of any normal space any disjoint open closed subsets of a normal space can be separated by function itself like this that is what we have used there. Now, you define f from x to 0 1 finally, you want a function from x to 0 1 by the formula fx is gx inside u bar so far outside u bar I will extend it by identically 1. In fact, on entire of x minus u I can put it as 1 see g bar minus u is already 1 therefore, it will agree on these two closed subsets on the intersection of these two closed subsets and on this subset its g is continuous on this subset is 1 that is also continuous therefore, f is continuous from x to 0 1 itself restricted to u bar it will be g and it g has this property g of k is 0 g of u bar minus u is 1 right. So, whole function we have got this way ok. As an immediate corollary we obtain several interesting results the first one is quite useful in measure theory ok. In a locally compact host of space given a compact set k contained inside an open subset v there exists a continuous function g from x to 0 1 such that g of k is 1 g of v complement is 0 by interchanging g to 1 minus g you can always interchange the role of this k being 0 and g of this one being 1 that is always possible ok that is not that should not confuse you. So, what it says is in a locally compact host of space given an open set v containing a compact set we have g k equal to singleton 1 and g of complement of v equal to singleton 0 ok. All that I do is we know that every compact set in a host of space is a closed set. So, therefore, we can apply this lemma that is all ok. So, this is just a real statement of that part of a lemma in a special case that is all all right. So, this is what we get next by the way we will later on prove the same thing same result in a different way also ok different proof also we will give a locally compact space is completely regular is another corollary why I am calling this is a corollary because they are easy consequences of our lemma. A locally compact space is completely regular regularity we have already seen ok it is built in in our definition ok. So, that is again complete regularity means what given any point in x and a closed subset f outside you know x must be outside f this joint from x v k we have to produce a map continuous function g from x to 0 on such that g x is 1 g of f is singleton 0 ok. So, apply the previous theorem with k equal to singleton x and f equal to complement of n equal to complement of f ok previous lemma you can apply that then you have what it over ok. So, locally compact spaces are completely regular. So, this is what we meant by saying that there is more separation in locally compactness than which is obvious from the definition namely the built in built in property was regularity here now we have got complete regularity right of very different nature compact set and open subset of that one can be separated and so on. All right let us go ahead with this one a locally compact half star space if you put half star space or just put t 1 s that is enough it is a tick now space. Why because we have already seen it is completely regular plus t 1 is by definition tick now space that is all ok. Now, we shall go to a different kind of result for local compact spaces. So, local compactness is sits in say you know in a juncture of so many different kind of concepts in topology. So, this is what we will try to see. So, next thing is the bare category type result for locally compactness ok. In other words the Bares theorem is true for locally compact spaces. So, every locally compact space is second category or you can say it is a bare space or you can say it is not first category and so on. Let me tell you what it is also. So, first of all the this this result was proved for complete metric spaces in part one. The proof is here is more or less similar in fact it is even simpler. Simplicity comes because of the local compactness you will see that. So, for proving this one in the in the case of metric spaces we have to do quite a bit of preparation and all that here everything is done now. So, we can immediately prove this one. So, what is the proof? Start with any countable family of nowhere dense subsets of x. We must prove that the union of all these f n's ok will not cover x actually we should prove stronger thing namely it is complement of union of all these f n's is itself dense in x. That is what we will prove and that is precisely the staying that the space x is bare space or it is second category ok. Start with a family countable family each of them must be nowhere dense even the complement will be dense. So, that much we will prove not only just it is non-empty is enough but we will show that it is dense. So, let u be any non-empty open set in x. We must show that this in the complement is dense none of these f n's we will intersect will take away a union of f n's will not take away this part is what we have shown right. So, choose any open set any point inside inside this open set ok. In fact, point is not necessary first of all we will see that take a neighborhood u naught of that only for that purpose I am taking a point here such that u naught bar is compact and u naught bar is contained inside u. So, what have we used here local compactness only for that part I want to start with a point inside inside this non-empty open set ok. Now u naught wise compact ok. Now there exists an open set non-empty open set u 1 bar contained inside u naught minus f 1 because f 1 is nowhere dense f 1 closure does not contain any open set non-empty open set that is the meaning of nowhere dense set. So, if closure does not contain f 1 also does not contain any non-empty open set that just means that complement of u naught minus f 1 is non-empty ok. So, there it is a non-empty open subset ok exactly same way take a point here then take u 1 and then closure of that and contain inside that. So, I am not writing all that here. So, I am taking an open subset u 1 of u naught such that even u 1 bar is contained inside u naught and this u 1 bar does not intersect f 1 repeat this process now u 2 bar contained inside u 1 minus f 2 u 3 bar contained inside u 2 minus f 3 and so on. So, keep doing that what you have what u n bar contained inside u n minus 1 minus f n ok that means u n bar does not intersect f n. So, what we have we have a decreasing sequence of open sets each contained inside the closure of each contained inside the next one inside the previous one and each of these is compact take the intersection ok. This w is clearly is in the complement of all the f n's why because even one of them does not intersect f n. So, this does not intersect the union of f n's ok and of course, w is contained inside u all right. u i bars are decreasing sequence of non-empty closed subsets of the original u naught bar and u naught bar is compact that is all I need ok once u naught bar is compact all u n bars are automatically compact anyway ok. So, a decreasing sequence of non-empty closed subsets in a compact set is intersection is non-empty. De Morgan law applied to compactness union of you know if finite union does not cover the whole space then the entire union also will not cover the whole space that is the property of compactness. So, we have produced a non-empty open subset inside this u outside all of u f n's. So, that is all what we wanted to prove ok. So, you see in three or four lines here yes I elaborated it fully we have complete proof of Bayer's theorem for locally compact spaces. Now, I am going to discuss something much deeper, but here I want full participation from you people. If for some chance you feel that this is too much for you you can you can you know you can take a leave I mean you can take ok I will I will think about it and so on you can you know sit on a side, but make it since an attempt to try this one ok. This is the way one becomes a mathematician or one starts doing research work. See you have observed two phenomenon namely a complete metric space is a bare space a locally compact space is a bare space. So, immediately you should ask what is common between them? What makes them both of them a bare space ok? Where is the hypothesis coming from? Apparent if you just look at local compactness of complete metric space they are nothing to do with each other. Indeed there are lots of complete metric spaces which are not locally compact. There are lot of locally compact spaces which are not matrixable either. So, what is this makes that both of them bare spaces? So, I have tried to explain this one in a set of exercises which are all doable. In any case you can try them and then your TAs will help you to solve them if you have difficulties only if you try otherwise you can take it easy. For example, question and so on we will not bother you with this one that is what I meant by you can take it easy ok. So, let us go through this exercise just I am not going to tell you what is it but they are all easily done. The first thing is every compact regular space is a bare space. See what we have done is locally compact space is bare space. But when I say compact a compact in number definition may not be locally compact remember that. So, I have to put regularity here also. But the proof is if you have understood locally compact space bare space just now you can prove this one also. The next thing is once you prove this one you can prove this one namely every compact horse door space is a bare space. This is the consequences of exercise 1 if you think properly. The third thing is every open subset of a compact space is a bare space. If you prove 3 you have already proved 2 but why I have given it like this first you prove this one 2 and then deduce that every open subset of a compact space is bare space that will take a little more effort ok. Now all these three is going to tell you that there is more bare spaces than just locally compact spaces or complete metric spacing only for that reason I have to prove this one. Now comes the crucial thing every g delta in a compact horse door space is a bare space ok. So, every g delta concept comes here g delta remember just means that intersection of a countably many open sets. Open set is a in a compact horse door space is exercise 3 from open set you have come to g delta here. So, this is set of the first 4 exercises ok. Once you have done that now we will turn our attention to complete metric spaces. Ok, indeed the metric space is not completely is all that necessary you can just work all these things in a pseudo metric space ok. Pseudo metric space is just a metric space all other properties are true but property 1 d x y equal to 0 need not imply x equal to y that is all I am just recalling that ok. So, take a pseudo metric space it will also give you a topology show that every closed subset in it is a g delta. So, now we have got one single property common to both these families pseudo metric spaces every closed subset g delta and here you know every g delta set in a compact horse door space is a bare space. So, that must be something that is the key for these two things to have bare space ok. So, let x d be a bounded and complete metric space the boundedness can be always attained because every metric space every metric is equivalent to a bounded metric by just taking d divided by 1 plus d that will be a bounded metric always it will give you same topology. So, this is a mild restriction but it is needed in this exercise start with a bounded and complete metric space ok. Now, wide tau is a compact horse door space ok let tau d denote the topology induced by this metric on x deem induce a metric ok. So, that I am taking here let e from x tau d to y tau be a topological embedding see here you have metric here you do not have metric but both of them have topology in the topological sense this is an embedding embedding means what continuous mapping such that onto the image that mapping is a homeomorphism. Suppose for every bounded continuous function f from x to r that is defined on x you have an extension of that to the whole of y by continuous extension. Now, why can I can talk about continuous action but this is embedded here. So, you can think of this as subspace here. More precisely all that I want is a continuous function f fact such f fact composite is f ok. So, suppose this property is also true all these I am hypothesis here for each x belonging to x let fx from x to r denote the function the distance function fx of x prime is distance between x and x prime this distance is the metric coming from x here ok. With this function I want to define a map d hat from y cross y to r by the formula d hat of y 1 y 2 is the infimum of f fact of x on y 1 minus f fact of x on y 2 take the modulus. So, these are all non-negative real numbers take the infimum. So, there you will get a non-negative function here ok on y cross y what is this this is has all the characters of metric except it is not exactly metric it is pseudo metric on on e on x x thought of a subspace of y where e x and e x prime distance d hat the same thing has the original distance. So, what it has happened is d hat is an extension of d on x ok to the space y. So, d hat is not a metric perhaps it may fail to be a metric, but it is a pseudo metric d hat is continuous on y cross y remember y has already a topology on it right d hat will introduce a topology we have to wait for that yet. So, but this d hat is continuous on y cross y and hence identity map from the original y tau to y the metric topology tau d twiddle this is continuous ok. The fourth part is the image of x under e is a closed subset of y tau d twiddle in the new topology metric topology pseudo metric topology is a closed subset all that y I have done I have put an arbitrary metric space inside a y tau which is a compact metric space ok now e x becomes a g delta set in y tau ok. So, this will really complete the picture why a complete metric space and locally compact spaces have this beautifully one property namely being bare space they share this property ok. So, that is the conclusion of this last exercise which you can read by itself, but you will have to wait to see the full thing ok you will have to wait till chapter 5 when we will do compactifications of various things. Namely a locally compact half dog space can be compactified what is called as one point compactification whereas a pseudo metric space can be compacted where as metric space can be compactified what is called as stone check compactification and so on ok that will fully complete the picture, but for preparing for before you can take this as granted and then you would have understood the difference here ok. So, you have plenty of time to understand this one not necessarily by the by the time we come to chapter 5 or finish this chapter 5 you can again come back to these exercises part of these exercise you might have solved before by that time and then you will be able to see the full picture ok. So, that is it for today. So, thank you so we will meet again next time.