 Hello. Welcome to NPTEL NOC course on introductory point set topology part 2. This is module 6. We are beginning a new chapter back into down right point set topology. Recall that in part 1 we have introduced several topological concepts like compactness, lindelofness, reliability, first countability, second countability which may all be called as some kind of smallness properties. Then we also have introduced freshness which is T1, Hofstorzness which is T2, regularity, normality. These are called somewhat largeness properties. Now the idea is to mix the two of them. Of course, one has to be judicious, judicious mixture of such things is going to produce many, many interesting results which are also useful. So today we will concentrate upon just one of these things namely compactness on one part and Hofstorzness on the other side. So we are going to study compact Hofstorz space okay. We have already seen one such application of this one namely if you have a bijection, continuous bijection from a compact space to a Hofstorz space then it is a homeomorphism okay. So this is what we have already seen in the first part. A close examination of this theorem tells us that instead of having two different spaces so suppose you have the same set and the same set you may have different topologies okay. So that then you can compare them by using the inclusion maps from one to the other. If you have a Hofstorz space and then if you take another topology which is larger than that then we know that the inclusion map will be continuous right. Any topology to larger topology inclusion map is continuous. So it will be a continuous bijection. So you can use this theorem to control many things namely if the smaller one is compact and the larger one is Hofstorz then the inclusion map is a homeomorphism which means that the smaller one is also Hofstorz and the larger one is compact. So this is the way compactness and Hofstorzness you know one says there are lots of open sets. The compactness says there are not many open sets in some sense. So these come together very nicely okay. So obviously when you have compact order spaces they are going to dominate the scene in point set topology. So here we start with a very mild conclusion. Start with a Hofstorz space and a subspace which is compact. The same thing as saying that the subspace is both a compact and a Hofstorz. So that is the different way of saying it that is all. So start with a Hofstorz space right. Given a compact subset B and a point X not in B there exists disjoint open sets Q and B in X such that X is in U and B is contained inside V. So this disjointness you know immediately you must be remembering this kind of thing was studied in regularity okay. So I am not saying anything is regular here because something is assumed to be a subset is compact not the whole thing. We will apply that one later on right now. A point X outside this compact subset can be separated by open sets you know disjoint open sets lying in between X is inside U, B is inside V okay. So people who have studied regularity, although this will make immediate sense. So what I am going to do is just go through the proof directly now okay. So start with a point B find a disjoint open set U, B and V, B pair of disjoint open sets such that X is in U, B and Y is in sorry B is in V, B. Why this is possible because X is housed off that is all. Now we have done this one for each beam set V, B sorry each beam set B and B is compact this V under V suffix B they are open subsets and they contain little B. So union of all of them will be a cover for B and B is compact. So that means that I have finitely many V, B, 1, V, B, 2, V, B, N and so on which cover the set B okay. Now you see for each V, B, I I have a disjoint open set U, B, I. So I take U as intersection of these finitely many open sets that is an open subset which will be disjoint from all the V, B, I's therefore disjoint from the union okay. The union contains the entire of B the intersection contains the point X. So we are done okay. So this method will be repeated again and again. So just watch out this one again. So what we have done we have used the compactness of B after used in this separation by this horse darseness we are extracting a finite cover. The finite cover allows us to take intersection of certain other not from not from the cover itself for each for each member of the cover there are other things which are of interest. So there I take intersection here I take union okay. So we will we will keep playing this game again now all right. So as a corollary I told you now if you take B equal to the whole of X then what is the corollary? The corollary every compact subset of a horse darseness for a first of all B itself is taken as a compact subset of a horse darseness it will be closed. So why that is closed? Because B is compact as above for each point X belonging to the complement I have an open subset disjoint from B disjoint from open subset containing B. So it will disjoint from B also. So if you take union of all these X for each X in the complement of B that will be an open subset which will be precisely equal to what X minus B. So X minus B is over okay that is an easy corollary the next thing is what I am more interested in it namely a compact half star space is normal. What actually we have proved here already means that a compact half star space is regular okay. Why? Combining with this corollary take any closed set inside a compact space it will be compact. Therefore I can apply this theorem for every point outside B I will get this one means it is regular okay now we want to improve on that one namely compact half star space is normal recall normal means starting with the two disjoint closed subsets A and B I must produce open subsets containing them respectively say U containing A and V containing B and U and V must be disjoint okay. So what we do we start applying the proposition okay first for each point X inside A obviously it will be not in B right. So I can get a two disjoint open subsets which I will label with because they depend upon X U X and V X so X is in U X the whole of B is inside V X okay. So I am directly applying the proposition rather than just house doors mess here okay so whole B of container. Now let U X1, U X2, U XK be a finite sub cover operation okay why because this A being a closed subset of a compact space that is also compact. So there will be a finite cover again as why I indicated earlier. Now I take G to be the union of these XIs that is obviously an open set but on the other hand I take H S intersection of V XIs each V XI contains B. So the intersection contains being a finite intersection of open sets it is open okay. Look at this one each V X is disjoint from corresponding U X. So the intersection will be disjoint from all the U XIs the same kind of H will be disjoint from okay over. So what we have proved now is compact top star space is normal alright. Now little more a compact regular space itself is normal actually first what we proved is compact regular compact house door implies compact regular which I have not stated I indicated already right here from here if you if you see that I am using actually regularity here because regularity is now built in from our proposition compact top star already implied regularity okay. So I could have just stated here is regular there we have regular implies all. So this proof of this part here of this theorem gives you compact regular implies normal. Now why we make this first because you know that regularity does not imply house doorness even under compactness okay. Therefore the argument itself should be used okay do not rely upon house doorness directly whenever you have regular compact it is also normal okay. Proof is just exactly here yet another condition under which a space becomes normal okay is the following. Instead of compactness we just put lindelofness which is weaker than compactness recall lindelofness means every open cover has a countable sub cover okay that is enough this is a bit of a surprise because I cannot take intersection of countably many open sets and claim it countable yet this here works. So we have to sharpen our argument a little bit how let us see start with two disjoint close subsets of a regular space lindelof space is in the background first of all okay we will we will use it using the regularity for each point A inside A we get an open set U A such that A is contained inside U A contained inside U A bar contained inside the complement of B because to begin with A is contained inside complement of B which is open and A is closed. So this is another version of regularity likewise for each B inside B there exists open subset VB such that B belongs to VB contained inside VB closure contained inside AC here I am just reversing the two role of A and B that is all. Now we use the lindelof property we get a countable sub cover UN and countable sub cover VN for A and B respectively okay because A and B are closed subsets of a lindelof space just like closed subsets of a compact space is compact similarly closed subsets of a lindelof space is lindelof this we have seen earlier it is not difficult once I say this one you can verify it easily okay now we do process which is quite common in the study of measure theory okay there are lot of exchange of techniques from the study of measure theory as well as study of topology I cannot pinpoint whether these things were first used in measure theory or the other way okay there are many such things okay so what I do I start taking P1 as say U1 minus V1 bar okay V1 is V1 is open subset V1 bar is a closed set if you subtract a closed set from an open set it's still open it's just like the de Morgan law it is same thing as intersection with the complement here taking UN minus something inductively I do this PN is the UN part but now subtract K ring to 1 to N VK bar union all the sets up to K equal to 1 to N of the closure union of finitely many union of finitely many closed sets is closed so this PN is open okay for each N similarly on the other side QN is taken as VN minus K ring to 1 to N UK bar okay now PN and QN are open covers for A and B respectively I have just observed that these are open why they cover A point X take a point X inside A it is in one of the UNs okay but I am subtracting something here no but these things are all disjoint when we should start with how do we have taken this one B belongs to VB VB bar is contained inside a complement of A so when I subtract VB or VB bar points of A are not disturbed they are there that's why so these PNs will also cover take any X inside A it will be one of the UNs between the corresponding PN similarly B will these QNs will cover B okay now take P as union of all these PNs and Q as union of all these QNs okay then P and Q are open subset they contain A and B respectively all the circuits why you have done precisely to attain P intersection Q is empty section okay so let us be convinced why this intersection is empty what is the meaning of this is not empty take a point here in both P and Q okay this means X must be inside no because intersection of unions is union of all these intersections PN intersection QM okay N and M could be well it must be in one of them okay for some N and M without class of generality we may assume N is smaller than M otherwise we can interchange the rule no problem suppose N is smaller than M then Q is sorry X is inside QM implies that X is QM here means that X cannot be in any of this UK bar K range from 1 to M okay so X cannot be inside PK okay so for any one less than equal to K less than equal to M contradicting X is inside because N is smaller than M because you are subtracting those part if something is here fine but if same thing is in the subtract this part also then it will not be in here the fact that it is in QM so it is not in any of these PKs because PKs are what PK is UK minus something okay they are smaller subsets than if it is not in UK bar itself it cannot be in PK right so that is a contradiction okay so so this is the modification that we needed for countable beyond countability even this technique will fail okay so here is a remark in part one we have seen that regularity and normality do not imply each other seemingly normality is stronger because here we are expecting every close disjoint close subsets to be separated in the case of normality it is a point in a close subsets in the case of regularity so you know on a first thing you may think that normality is stronger than regularity under some conditions namely T1 as it is true that also you have seen so T4 implies T3 implies regularity so T4 is T1 plus normal okay but what I wanted to make these things are we have we have seen earlier now what is the what is the addition thing that I want to say what I want to say is really this regularity normality quite close that is why whenever we are discussing one of them we end up discussing the other one they are so close so this theorem for example does tell you that namely you know regular first compact hostile space is first we prove regularity and then we prove normality so compact hostile is normal once you prove normal of course it is regular because it is already hostile but to prove normality we went through regularity okay so that is the point I wanted to make here that is all okay so we will stop here today next time we will bring another new concept namely local compact thank you