 Welcome to module 52 of point set topology course part 1. So today we shall take up the study of normality again, characterization of normality, characterization in terms of continuous functions. In fact, we will have two different characterizations which are closely related to one another one is due to a horizon and then using that another one is due to T say. The central idea is that the set of all continuous real valued functions on a space must be able to reveal some properties of the space itself. One in algebraic geometry and sometimes in algebraic topology also, this is the central theme look at the set of real valued or complex valued functions or those which have some extra properties and so on declare them as what is known as coordinate space, coordinate ring and then the ring will dictate all the geometry and topology. So this is the theme they follow. We do not go into that depth here, but the idea of why we need such a thing like characterization only for that reason I am telling you so where it is much more important than to us. So we can ask a vague question how large is the set of all continuous real valued functions on a given space X. The study of this leads to a different kind of topological results altogether with applications to problems such as I have just mentioned a few of them here, matrization problem, embedding problem embedding problem is just means that take a space whether it can be embedded inside some Euclidean space of finite or infinite dimension and exploring the interrelationship between the ring structure, the algebraic structure of C XR with the topological properties of X. So these are the few things, for example when you go to function theory it is not all continuous functions they take, they will take analytic functions. So that is what is important for them in complex analysis for example. If you go to algebraic geometry they will only take polynomial functions ok, so depends upon for us if you want to study the entire topology you must better take continuous functions ok. In differential topology you will take differentiable functions, smooth functions and so on. So we shall not be able to discuss any of these topics I have mentioned three of them in this course more than what I have mentioned and I have mentioned it only because why the Euryjones characterization or discharacterization is important. But some of these problems and application will be taken up in part 2 of this course ok. So let us proceed with these characterizations. How one might have you know come up with this kind of thing I would like to tell I cannot read Euryjones mind nor I can go back in history and so on ok in being go back in time. So I am guessing that this may be the one which led him to consider such a thing. So again go back to metric spaces take any subset A consider distance function from A. So distance function is defined from x ok D underscore A x is equal to D x A you can write in two different notation which is nothing but infimum of all the distance between D comma x comma A where A raise over A x is 6 x. So that is the distance from A ok the infimum of all these systems. For the point x distance between x and A is defined like this ok it is easily checked that D A is continuous ok on the whole of x and clearly if A is inside A then D of you know we can put x equal to A so that will be 0 so infimum will be 0. So it will vanish right but it will vanish on A bar also. In fact A bar is precisely the set wherein this function will be identically 0 ok. So that is easy to check. Now let B be another set consider the function G from x to R defined by G x equal to D x A minus D x B. If for A and B we have defined D x A and D x B take the difference. I have no speciality here first I took A then I am taking B that is why I am rightly at otherwise I could have taken D x B minus D x A also as good as this one. This function is again continuous because it is sum of two continuous functions difference of two continuous functions and it is non positive on the closure of A because the first term will be 0 and on the closure of B it is non negative because the second term will be 0 ok. Now you take a special case when A and B are disjoint closed sets when A and B are disjoint closed sets at least one of them must be non 0 for all the points because A and B are disjoint. So if D x A and D x B are both 0 x will be inside A intersection B so there is no intersection right that is it. Therefore the sum function will never vanish D x A plus D x B what does that mean I can divide by that function ok. Sum and difference are both continuous sum is number 0 therefore this quotient function is also continuous I am calling it as F ok. So F X is the difference function divided by the sum function. Why when you think of this one is I itself say a more question but I think I have thought about this one so you really must have thought about this one. How I thought about it also this history I could come up to this one after long thinking oh this must be the reason. Then F from minus x x to minus 1 plus 1 because the numerator modulus of the numerator is never bigger than the modulus of the denominator ok. Denominatory is already non negative actually positive. So these always the value will be between minus 1 plus 1 ok. Moreover A and B are close subsets we know that F of A F of A x belong to A this is 0 this is 0 so it is minus of D divided by plus D which is minus 1. Similarly F of B equal to plus 1. So you see what we have produced is a continuous function from the whole of x which vanishes exactly sorry which is exactly minus 1 on A and plus 1 on B ok. Now you can take just a little small neighborhood of minus 1 and a small neighborhood of plus 1 I have taken last sufficiently are namely minus 1 to 0 and here 0 to 1 only thing is I have been careful that they are disjoint open subsets in minus 1 to plus 1. Take the inverse image call them as U and V they will be disjoint open subsets ok. Clearly they contain A and B respectively ok. So suddenly what we have proved is that disjoint closed subsets metric space can be separated by open subsets actually they can be separated by continuous functions. So in particular every metric space x is normal indeed since every metric space is normal a subspace of metric space is also metric space follows that every metric space is completely normal ok. It is completely normal also we have we have seen that singleton sets are closed in a metric space that means they are fresh spaces. Therefore complete normality implies regularity as well as house doorness also ok. Once they are fresh a regular implies house door complete normality implies regularity complete normality implies normality ok anyway. So all these things are true for a metric space. So why I am guessing is that perhaps this function F was the motivating example for the celebrated result known as Urizon's lemma of course it is due to Urizon there is no mistake in that. So this is the lemma a topological space x is normal if filled only if it satisfies the following Urizon's condition. So again and again I will be requiring this condition. So I will name it as Urizon's condition. What is this? For every pair of non-empty disjoint closed subsets A, B and X there exists a continuous function F from X to 01 such that F A is 0 and F B is 1. See if F A is singleton 0 just means that A has to be nonempty similarly B has to be nonempty. For that reason we have to assume that their nonempty disjoint closed sets ok. Otherwise in the statement of Urizon's you know definition of Urizon's normality you know you should take whatever definition you take three of them even if A or B is empty it is ok it is not so it does not cause any problem there. But if you want to have function theoretic characterization here then you have to take A and B are nonempty. Generally you can write F A is 0 and F B is 1. You can also do minus 1 to plus 1 here by changing the the interval the codomain interval by homomorphism ok. That is not so crucial getting a closed interval as a codomain ok and getting a function F from this one wherein the the two sets are going to two distinct points that is the crisp of the format. It could be F A equal to some A and F B equal to some B where A and B are distinct points of this ok. So, that is the two crucial matter here alright ok. So, one way is obvious which you have seen already in some sense. Assume that U C is satisfied by X take A and B any two nonempty disjoint closed sets ok and take a function F from X to 0, 1 continuous functions F is 0 and F B is 1. Then take 0 to half half open and half to 1 again half open take the inverse image call them as U and B they will contain A and B respectively. So, that is the condition for the normality. If one of them is empty you can always take suppose A is empty then you can take U empty and B equal to 0 of X. So, that is obviously satisfied there is no need to worry about that. So, normality is satisfied if U C is satisfied the converse is where we have to work hard out. Well hard work is done by your own we are doing hard work in a difference since we have to learn them properly right ok. Now, let X be normal and A and B be nonempty disjoint subsets. The second condition the definition in the set definition tells you if you take A and B complement A is contained in the B complement A is closed B is open there will be a open subset G such that A is contained inside G contained inside G bar contained inside B complement. So, I am using the second condition in the definition of normality. So, we have started a process here that is equal to be a iteration of this. So, I am denoting the first iteration by this G by G half ok. This notation will be clear in a moment so you have to wait. Right now it will be better if you write it as G 1 ok. So, but now what I want to do that is what I have to tell you apply the normality to both the situation. So, this is almost like you know this is one point and there is another point and I have chosen the half of the middle of them. If you have chosen middle from the first point to middle and the middle point to the second point again you can choose middle of them. So, that is the kind of thing that is going on here, but middle does not make sense something in between with the relation with the relation of what inclusion maps of sets that is what is being done. So, what I want to do is now between A and G half and G bar half and B complement ok. I introduced two more open subsets which we shall denote by G 1 4th and G 3 4th respectively such that from A to G half G half is open A is closed G G 1 4th will sit there contained in the G 1 4th closure. Similarly G half to ok G we have come up to G half right G half is contained in a G half bar that is already there we are using that one now is G bar is G half bar G half bar to B complement twice again open we will have one more open such that G 3 4 contained inside G 3 4 bar. See what I have done from 1 by 2 1 by 2 half of that is 1 by 4 between 1 by 2 and 1 I have 3 4th. So, keep on cutting down by half what are those numbers next time you will get 1 by 8 1 2 by 8 3 by 8 4 by 8 5 by 8 6 by 8 7 by 8 right. So, that those are numbers. So, what are they called? They are called derrick rational numbers rational numbers whose denominator is always a power of 2, but I want only all of them between 0 and 1. I am including 0 and 1 0 is taking the place of A and 1 is taken the B complement is taking the place of 1 there I do not need to write any symbols for them. In between I am going to put all these open sets such that they are contained inside their own closure and then the closure is contained inside the next one and so on. So, that is what I am going to do next step will be an open subset between A and G half in G 1 4 and one on this side also in everywhere in between you know G 4 is G 4 is closed and B C is open. So, between them I will put one more one more open set and so on. So, go on squeezing opens of sets in between A. So, let D denote the set of all derrick rationales in the interval 0, 1 namely integer divided 2 power n m and n are positive integers m less than 2 power n you can take m to be only odd numbers if you like. If there is some power you can cancel off, but I am on m to be less than 2 power n. Okay, carrying on with this process we obtain to each number here an open subset G T. So, all these new open subsets of X are indexed by this set. So, what is the property? All of them are neighborhoods of A, A is contained inside G T, G T is contained inside G T bar, G T bar is always contained inside B complement. So, this much is obvious, but more than that between T and G T and G S what is the relation as soon as T is smaller than S, okay, G T bar will be contained inside G S. So, this is the property. Okay, for example, it does not depend upon whether I have chosen in first or second or third and so on. It depends upon whether the corresponding index is a bigger or smaller. For example, the first one that I have chosen is G half. In the second stage I chose G 3 fourth. So, G half bar is contained in the G 3 fourth, okay. So, next I will make chosen you know 5 by 8 or 7 by 8 and so on. So, you have to compare the numbers first that can be compared. The corresponding sets also must be compared in a stronger way namely the closure of the smaller indexed one must be contained inside the other open set. So, G T bar is contained inside G S, okay. All that we have done is repetition of the normality condition inductively, okay. Now, we have produced a continuous function out of nowhere. The crux of the matter is that the added rationals are dense inside R, okay. So, once you have some way of nominating these indices one can cook up functions out of that. That is the whole idea. So, define f x equal to 1, okay. If all the x, x is all in G T complement for all t, it does not belong to any of G T's, okay. See all G T, G T's are inside what? G T bar itself is inside B, okay. So, B complement. So, if they are not in this itself x is in the G T complement then I am defining it as 1, okay. Otherwise, I will define it as infimum of all t such that x is inside G T. See at least this set is non-empty. Therefore, infimum makes sense, okay. Anyway, this set once it is infimum, once it is non-empty, it is bounded below. All these numbers are bigger than or equal to 1, actually bigger than 1, 0. So, it is bounded below by 0. So, infimum is a finite number. If it is empty then you would have a problem. So, whenever it is empty, it is defined as 1, that is all right. Take the infimum of all x such that all all t such that x is inside G T, okay. Since all the G T, all t's are between 0 and 1, infimum whatever is set is it has to be between 0 and 1 and this part is 1. So, 0 is less than or effect less than 1. This is obvious from this definition. Now, the first one says what all the G T's contain a, closures are contained inside B complement. So, from this it follows that f of a will be 0. The infimum of all the diatics inside 0, 1 is 0. And f of B is 1 because it is not in any of them then only it is 1. So, if it is in B, it cannot be in any of G T. So, it is in the complement of G T for all of them. So, the security property is already done very easily. The missing point, the important point is that f is continuous, okay. Producing continuous functions, you know very few such results are there. Even proving, after proving something you have to prove me more, they are easier here producing continuous function itself. So, this is something fantastic that you have done. So, first we observe that the collection of all intervals of the form 0, a and B open 1 for 0 less than a, B less than 1 forms a sub-base for the usual topology on the closed interval 0, 1. The usual topology and then I am taking restriction. So, I have to restrict the sub-base also. So, I do not want to take anything other than intervals which are sub-intervals of 0, 1. You have to take open there, but this is closed interval. So, 0 closed A open, this will be also an open subset. That is the difference. Similarly, P open and 1 closed, these are also open subsets of the closed interval 0, 1. If you take B plus safe, suppose you take minus epsilon to A and intersect it with 0, 1, what do you get? You will get 0 to A only. So, this is what it takes. Hence, once this is a base, sub-base, it is enough to prove that to continuity of, I want to prove continuity of this. F inverse of 0 A is open. Similarly, F inverse of B 1 is also open, where A and B are arbitrary points of 0 and 1, open. For that, I am explicitly proving that F inverse of 0 is union of all the gt's such that t is less than A. Each gt, remember, it is an open subset. So, union is also open. For all 0 less than A less than 1, I am going to prove this. Similarly, F inverse of B 1, I am going to prove its union of all gt bar compliment. You see, bar is closed, compliment is open. Again, it is open subset, but now this time t is strictly bigger than B. So, 3 and 4, if I prove, the theorem will be complete. Let us prove 3. Take a point x inside x such that it is on the left-hand side. What is the meaning of that? F of that x is strictly less than A. Of course, it is always bigger than equal to 0. This implies, remember, what is the definition of fx? Infimum of all t such that x is inside gt, this is fx. It is less than A. What is the meaning of infimum is less than some number. There must be something here which is less than that. So, there exists a gt such that x is in gt and this t is less than A, because infimum is taken over all fx t. So, I have not done anything other than just definition of infimum. Okay, not very serious also. So, a lectures of 3 is contained inside RHS. For each point x, I have found a gt here. All right? The other way inclusion is obvious. As soon as x belongs to gt, infimum will be smaller than that. Okay? And t is less than A. So, it will be here. f of fx will be less than A. Right? Even if one x, even if one t is such that x is in gt and that t is less than A, it will be, infimum will be smaller than A. So, this f of f of x will be less than equal to less than A. So, it is here. So, this way containment is obvious. All right? So, fourth one, two, four. So, let us take x which is that b is less than fx less than 1, less than or equal to 1. If fx is 1, then remember when it is 1, it is inside gt complement for all t contained in, you know, belongs to d. No gt will contain it. It is in the gt complement. So, fix some s belonging to, g, x belong to d, such that s is bigger than p. Take some s inside d because b is less than 1. Between b and 1, there must be some s. What is s? s is an element of t. So, this is where I am using the fact that the dyadic rationals are dense. So, we can find a t belonging to d such that this s is less than t. See, I have chosen b less than s less than 1. So, between s and t again, s and 1, I can choose another t. So, then it is also again density of this. So, I can choose actually b less than s less than t and t less than equal to 1. It follows that gs bar is contained inside gt. This was the choice property 2. And hence x is inside gs bar complement. X is in gt complement. Its gt complement will be contained in the gs bar complement. Now, consider the case. One case is over. Namely, if fx is 1, x belongs to one of them. If you remember what I have to show? I have to show that the point in the left hand side, namely point which looks like this one, b less than fx less than equal to 1 is in the union of all gt bar and so on. I have found one such gs bar complement. Now, it may happen that b is less than fx and fx is strictly less than 1. When fx is strictly less than 1, it is a second condition, namely fx is infimum of all t such that x is inside gt. This means that there is a t such that x is inside gt first of all. And then you have to take that fx that fx is bigger than b, but less than 1 is what I have. Also, it follows that x is not in gs for any b less than s less than fx. If there is a case, then fx would be infimum. It would be smaller than that one. So, if s is smaller than fx, x cannot be in gs because fx is the infimum. So, now you take b less than r less than s less than fx. Again I am using density of d here. Then x cannot be in gs implies that x cannot be in gr closure because all the closure of corresponding things are contained here. Which is the same thing as saying that x is in gr bar complement. So, we have found out another element which contains on the left. Hence, lx is contained inside rh. So, you see the proof of 4 required us to use the density of d at least 3 times here. So, proof of 3 was easier. The proof of 4 took some time. We have not yet completed a proof. What we have to do? rhs is contained inside lhs. Even that is also not. In the third case, it was very easy. This part little bit more we have to say. Suppose fx belongs to rhs. rhs is what union of all g bar sc namely g bar complement. So, x is in one of them for some s which is bigger than b. That is the definition of the right hand side. This implies that x cannot be in gt for rs bigger than t. That is bigger than b. So, between s and b if you take another t here, x will not be inside gt because x is not in gs bar itself. x is inside gs bar complement. On the other hand, suppose x is not in the lhs of 4. I took it is in rhs. I want to show that it is in the lhs. lhs means what? b is bigger than fx. fx is of course bigger than 1. So, that is the meaning of x. So, if that is not the case, that is what I am telling you. Hence, so suppose on the other hand x is not in lhs. Then fx must be less than or equal to b. fx is less than or equal to b means what? infimum is less than or equal to b. Then x must be in gt for some for all t bigger than b because the infimum is less than or equal to b. If something is bigger than b, x will be already inside some gt. Bigger than that one, it will be definitely inside that because gt is contained inside gs for all t less than s. So, but that is absurd because we have just shown here that you know just here we have shown that x is not in gt as soon as sc is t is bigger between s and b. But here it says that for all t bigger than that it should happen. So, that is absurd. So, lhs will be, rhs will be contained inside lhs. You start with a point in rhs, it is contained inside lhs. This is what it is. So, this proves for and hence continuity of a phase established. Therefore, the completion of the proof is done. So, I make some remarks here. In uc, I have already done this one, I will repeat it. We can freely use any closer interval a, b, ls and b in place of 0, 1 by merely composing which is the linear homeomorphism t going to b minus a times t plus a. Of one it is convenient to use the interval minus 1 plus 1 instead of 0, 1. Like the metric that we considered you know d of x a minus d of x b divided by d of x a plus d of x b it was between minus 1 plus 1. So, that is what we are going to do next time. But sometimes you may have to choose some other numbers also. So, any closer interval you can take no problem. That is the comment here. There is no assertion about the uniqueness of the continuous maps in the uc there exists some function continuous function. There may be plenty of them in that there are lots and lots of function. The difficulty was showing that there exists one. After that you can cook up many, many of them. So, one of the special function you know very specific one I want to have and I want to go to I am going to use that. So, I am going to introduce a notation here. Temporary notation like in measure theory you have this the characteristic function. Let us introduce a temporary notation which we will use in the proof of next theorem that we are coming. Given any to disjoint closed subset a and b of x let us denote chi of a b. So, it depends upon both a and b a function a continuous function for x 2 minus 1 plus 1 such that on a it is minus 1 and on b it is plus 1. There are many of them. Any of them I will just call this chi of a b depends upon the context all that I need is it has this property and it is defined on the whole of x. So, it is continuous. That is why I am just writing the chi of a b. The only thing that we need here is that if x is normal then such functions exist such continuous functions exist ok. So, choose any one of them and temporarily write it as chi of a b. So, this is what I am going to do next time. I am going to use it next time. So, this comment also I have made but I will repeat it. We have mentioned that normality is not hereditary. We have not proved it however it is weakly hereditary namely every closed subspace of a normal space is normal. That is very easy to find out easy to prove because closed subspace of a closed subspace is closed. So, start with a start with y as a closed subspace of x if a and b are closed subspace of y then they will be closer in x itself. Therefore, normality of x will produce opens of sets in x which contain a and b and disjoint. Now, we will intersect it with y ok. So, y will be normal. So, closed subspace of a normal space is normal ok and that information is important and we are going to use that also in the next theorem ok. So, let us stop here for the next theorem tomorrow. Thank you.