 Welcome to module 53 of Point Sector Biology course part 1. Today we will continue the study of normal spaces, characterization of normal spaces. Last time we did characterization due to the horizon, now we will do the characterization due to t-set. The t-set characterization is another landmark, but it actually uses the you know, the horizon's characterization, construction of continuous functions. But it is quite mysterious in a sense that here we have only one closest set. So it requires quite a ingenious, you know, mind to have explored this one. So it is not at all easy to come up with this kind of idea. So let us look at that statement here. A topological space X is normal if and only if it satisfies the following condition which I have put as t-set condition. So what is the condition? Given any close subset e of X and a continuous function g from a to minus 1 plus 1, there exists a continuous function f from X to minus 1 plus 1 such that f restricted to a is g. Obviously here e is a non-empty close subset. That is otherwise we do not have any continuous function g from e to minus 1 plus 1. So this is the hypothesis given a closest and a continuous function. So there is no need to worry about that e being close, e being our obtained so on. So every continuous function defined on a close set can be extended to a continuous function retaining the codomain as it is namely any close interval. So model we are using minus 1 to plus 1. I have already told you that you can always change the codomain to any close interval each time. So that is not very crucial but now instead of 0 1 as in the previous theorem, we will use minus 1 plus 1 which is more convenient for writing down the proof. Given that X satisfies the Tc, given any two disjoint close subsets a and b such that e put e equal to a union b and apply this condition to get a function g from minus 1 to plus 1 namely g of x equal to minus 1 and g of on x beyond a and g of x equal to plus 1 and x over 3 ok. G is continuous because a and b are disjoint close sets ok. So this makes sense. Only two values you have taken on a is minus 1 on b is plus 1. But now g will extend to a whole function from x to minus 1 plus 1. So there is an f here that f straight to e is g ok. So that is the condition for the urinals characterization you see. Therefore x is normal by the above theorem alright. So we have taken a shortcut here used the urinals characterization instead of trying to prove open subsets etcetera alright. But in the converse proof also we are going to use urinals theorem. So one way is all done. Now let us prove converse again the converse is taking up some time. Assume x is normal and e contained as x is closed and a map g is given ok. g is defined only on e and it is continuous there. We have to construct a map f from x to minus 1 plus 1 such that f restricted to e is g. So here we should use the fact that set of all continuous functions from x to minus 1 plus 1 ok. They form a Banach algebra Banach space remember that with the supremum norm. Remember that we had introduced the set of all x to r or c or whatever and then we took the bounded functions there on which we put the supremum norm. Now if you have continuous functions continuous functions on x ok they may not be bounded but here I am taking they are bounded by actually minus 1 plus 1. The values are inside minus 1 plus 1. So this will be a subspace of that Banach algebra alright. This norm makes sense because now all functions are taking values minus 1 plus 1 ok. We also saw that the continuous functions ok form a closed subspace of this Banach algebra and therefore this itself as a as a normal linear space it is a Banach algebra. That means Cauchy sequences will converge inside this one. It just means that if you take a Cauchy sequence of continuous functions from x to minus 1 plus 1 you can take the limit in the larger space but that limit is continuous. Therefore it is in the smaller space ok. So this also you have seen the convergence with respect to the supremum norm is nothing but automatically uniform convergence ok. So this is what we are going to use. Essentially if you do not want to use all these terminologies all that just means that if you have a sequence of continuous real world functions ok which converges uniformly now sequence of continuous uniformly to a function that function is continuous. So this is the fact that comes out of ordinary analysis so which you are going to use it now here ok. How do I going to use inductively we shall construct a sequence f n from x to minus 1 plus 1 such that summation f i is a partial sum that is a sequence not f n the partial sum that sequence converges uniformly to a function f from x to minus 1 plus 1. And it has this property that norm of g minus iron into 1 to f i converges to 0 this norm because g is only defined on e should be taken on e. So same norm restricted to e ok. So all these f i is restricted to e of course f i is make sense on the whole of it. But g makes only on e so you have to take c of e ok. So that is the meaning of this one. So this is what we want to do now so I will explain these things and more carefully in what comes here. Put R n equal to 2 power n minus 1 divided by 3 power n. Take g naught equal to g having defined the map g n from e to r for some n inductively we are going to construct the next g n plus 1. So for that put a n equal to g n inverse of minus 1 to r minus r n and b n equal to g n inverse of r n to 1. Then clearly a n and b n are disjoint closed subsets of e ok which itself is closed in x therefore a n and b n are disjoint closed subsets of x. Hence by Eugenius Lemma we get a continuous function chi a n b n from x 2 minus 1 plus 1. Remember this chi a n b n is a continuous function which has the property that on a n it is equal to minus 1 and on b n equal to plus 1 that is all we are going to use it here ok. After that we put f n equal to r n times chi of a n b n. Finally we define g n plus 1 as g n minus f n restricted to e. g n is defined on e already f n is defined on the whole of x but we take f n restricted to e and take g n plus 1 to be g n minus f n restricted to e on the substrate e. The induction inductive definition of the sequence f n and g n is so. Next task is to show that these things converge to whatever we wanted to namely first of all g n and f n are both continuous. Norm of g n is equal to r n times norm of chi n a n chi n a n b n that is 1 therefore it is r n. Which in summation r n 1 to infinity is actually 1 so it dominates this norm of r n just means that i r n to 1 to n f i converges uniformly to the function sum function continuous function of automatically because uniform convergent and it takes value between minus 1 plus 1 ok. So this function f is now continuous because it is continuous it is a uniform limit of finitely many f i is here and then n tends to infinity so these f i is finitely many f they are continuous n tends to infinity uniform convergent therefore f is uniform f is continuous. We now claim that the second part namely norm of g n on e that is supremum of all g n x modulus g n x are x range over x range is over e right that is the definition. We want to say that this is less than equal to twice r n minus 1. Again the proof will be by induction when n equal to 0 r minus 1 is nothing but 1 by 2 ok and twice r n is 1 and g naught is our function g which is taking value between minus 1 plus 1 therefore norm of g naught on e is less than equal to 1 ok. So assume that this statement is true for some n then we shall prove it for n plus 1 ok. So let us examine what happens to the function g n plus 1 suppose x is inside a n by the very definition of a n g n of x will be between minus 1 to minus r n so it is less than equal to minus r n. On the other hand the induction hypothesis says that minus twice r n minus 1 twice r n minus 1 is less than equal to g n x the modulus of this one is less than equal to twice r n minus 1 this is the definition this is the induction hypothesis. So if you put them together now you subtract f n subtracting f n f n is f n is now on a n it is minus r n right. So adding r n so what I get is minus 2 r n which you can verify is same thing as minus 2 r n minus 1 plus r n that is less than or equal to this is g n of x plus r n but this is the same thing as g n plus 1 of x now because plus r n is same thing as minus of f n ok. So g n is less than equal to minus r n here r n plus r n is less than equal to 0 and 0 is definitely less than twice r n plus 1 ok exactly similar reason not exactly same reason when you take x equal to x x is inside b n this becomes now minus f n becomes minus r n and we will have left and right and etc we have to correctly chose finally you will get a similar thing namely what you get is that minus 2 r n is less than equal to g n plus 1 plus 2 r n. So what remains now look at a point which is neither in a n or in b n ok take a point which is nor in a n nor in b n ok. Then both a n and g n are in the interval minus r n to r n by definition because if it is not in a n it is bigger than minus r n not in b n it is less than r n less than or equal to r n ok. So such a point automatically satisfies g n plus 1 of x ok less than equal to r n plus r n and bigger than equal to minus r n plus minus r n. So in modulus is less than equal to twice r n because both f n and g n are inside this one f n is always is minus r n times 1 minus 1 to plus 1 r n times minus 1 to plus 1 ok. So this completes the proof of inductive claim that namely g n of v is less than equal to twice r n minus 1 for the same reason because summation r n is convergent this is dominated by this twice this this thing it will follow that g n of x converges to 0 as n tends to infinity because this thing converges here ok. Now what is this g that mysterious thing why g n g n that is clear because g n is nothing but remainder after n terms ok. I could not say that one because before that I have to show that the series is convergent. So we have shown that g n converges uniformly to 0 inside e. But now if x is belong to e then g n plus 1 is g n x minus f n of x by definition. What is g n x? g n minus 1 x minus f n minus 1 of x. So you club them together it will be the g n minus 1 x minus f n minus 1 plus f n of x. But again you apply to g n minus 1 to g n minus 2 minus f n minus 2 of x plus all these. So go on doing that till you hit g naught g naught is g x and this term will become summation i rank to 1 to n f n first f n f n minus 1 etcetera up to f 1 all these terms will come. So g n plus 1 is nothing but g x minus i rank to 1 to n f i of x this is the second term there in the condition. We wanted to show that norm of this converges to 0 norm of this same thing as g n plus 1 of norm of this one and we have shown that this converges to 0 ok. So you see this is nothing but inside e this is just the partial sum ok and this is remaining after and that is the whole idea. Upon taking the limit this is 0 this will be become f now because i rank to 1 to infinity is our f right. So g x minus f x norm is 0 it just means that g x equal to f x. So this all happening for every x inside t. So that completes the proof of T says theorem. So let me make a few comments here since any two closed intervals I am making this comment again of positive length or homomorphic ok. We can use instead of minus 1 plus 1 or 0 1 we can use any interval a, b ok. But in the proof of the above theorem it was crucial that the codomain was a closed interval. You are going to take limits of certain for each point you are taking limits. So they are just point they are just values. So the limit point could be in the end points I mean that is possible right. So even if you start with the function suppose the function given g is taking values in the open interval extended function may not be taking values inside the open interval it may hit a or b that is possible ok yes or no. So because one if you take open intervals you know open interval is not complete so you cannot use all these completion results. So that is the whole idea ok. But you look at the statement of the T's theorem that will not suffer. In other words I want to have a theorem suppose g is a function from a closed interval to an open interval then I can extend it from x to open interval itself. So that is the statement ok. So that is not the part of the statement but it is extra statement. We have to work out and there is a trick which will help us to prove such a theorem. So let us do that. So this is the statement here. So I have stated a separate theorem that x p a normal space e contains or x closed then given any continuous function g from e to open interval minus 1 plus 1 there exists a continuous function f from x to open interval minus 1 plus 1 such that f free state is g ok. Directly from the theorem that we have proved this one does not follow that is what I made remark. What follows? You can think of this as closed interval by taking inclusion of this one in closed interval then you will get a extension also the closed interval right. So that much is true. Now how to come back to the open interval here is a trick. Applied T's extension theorem 4.39 to the map g treated as a function from e to minus 1 plus 1. We obtain a continuous function f 1 let us call it f 1 because this is not going to f from x to minus 1 plus 1 such that restricted to e it is g. What may happen is look at f 1 inverse of minus 1 and plus 1 just the two points put that as a that is a closed subset. This may be empty if a is empty then we are happy that means x is inside open interval minus 1 plus 1. If a is not empty then we have to work harder ok. In any case a is a closed subset of x disjoint from e why because to begin with we have started with a g such that g is taking values strictly inside minus 1 plus 1 and f restricted to e is g. So this a and e are disjoint subsets and both of them are closed. Hence you can apply your universe lemma here to get another function h from x to 0 1 such that h of a is 0 and h of a is 1. So you see here I am using again 0 1 instead of minus 1 plus 1. So that is as a it is deliberate it is not just it is not just arbitrary ok. So h of a is 0 and h of e is 1. So whole idea is you do not want to disturb the e part. So you have put it is 1 here you do not want this a part so you have killed it h of a is 0 that is the whole idea. Now you multiply the original f 1 with h x take f x equal to h x into f 1 of x ok. This is a product of two functions both of them taking values in minus 1 plus 1 minus 1 plus 1 and other one 0 1 anyway. So this will take values in minus 1 plus 1 there is no problem and it is continuous and when you put x equal to x inside e this h x is 1. So it is f 1 of x which is g the original g therefore that is also all that you have to show is that the new function f takes values inside minus 1 plus 1 open right then the theorem is over ok why it does not take the value minus 1 and plus 1 that is all there also all right. So suppose modulus of f x is equal to 1 but what is modulus of f x what is f x? f x is h x into f 1 of x ok. Suppose for some x it is equal to 1 either it is minus 1 or plus 1 then f 1 of x is less than equal to 1 modulus and h x is inside 0 1 for all x remember that we must have modulus of f x is actually equal to 1 even if any one of them is smaller than 1 the product will not be equal to modulus will not be equal to 1 ok. So both the modulus must be equal to 1 h x modulus does not make sense because this is already positive non negative so it is actually equal to x if h x is 1 after multiplying by h x what you get f 1 of x is equal to f x modulus less but modulus of f x f 1 of x is equal to 1 implies x is inside a remember that a was defined as f 1 inverse of minus 1 as well as plus 1 but then h is 0 h of a is 0 but just now we have h of a is 0 whether right hand side will be 0 this cannot be 1. So we have shown that h is also 1 so that contradiction proves that modulus of f x is never equal to 1 ok. So I have said that this is a trick this kind of tricks are used quite often in algebraic topology also. So in summary what we have proved is T section and theorem the core domain can be replaced by any bounded open interval instead of a closed interval but r is homeomorphic to any open interval therefore you can take the whole of r also. Finally if you carefully watch the proof of the above theorem you will notice that it can be easily adopted to half open intervals also if you do not want both of them you have to work only for one of them that a which you have taken f inverse of minus 1 plus 1 you take a includes only one of them ok inverse of one of them. So it will work for half open intervals also ok. Therefore all these statements can be collectively called as T's extension theorem for normal spaces. Given continuous function on a closed set taking values in open a, b, open r, closed a, b, closed a, open b whatever whatever you take the codome ok in the same codome there is a continuous extension of the function g to a function f on the whole of x. So that is the way you have to understand T section theorem at various places whichever form is necessary for you you can use that alright. Horizons characterization of normal space drew a lot of attention. Ticknauf came up with an idea of adopting a somewhat weaker version of the horizontal characterization but one is stronger than regularity. So another concept came in between what is called as complete regularity. A space x is called completely regular if it satisfies the following condition. Given a closed subset f of x and a point x belonging to x minus f remember this was the hypothesis for a regularity instead of two disjoint closed subset one closed subset and a point outside it ok. So then instead of open sets and so on now you have a function now there exists a continuous function f from x to 0, 1 such that f of x is 0 and f of capital F is 1. So this is perfect mixture of regularity condition and Euryzones criteria mixture such a thing is called complete regularity. As usual normality does not imply CR simply because singletons may not be closed. The complete this complete regularity is an important concept in matrizability problem. Even the Euryzones lemma I think Euryzone had finally to try to prove some you know embeddability results through which he wanted to prove matrization. That is why he he explored this. So he he came up with this what is so called a so called Euryzones lemma now ok. So this this complete regularity is important in matrizability problem that will be discussed and taken up in part 2. So I am just giving you a glimpse of that even if you forget it is ok. So I think you are now convinced that why Euryzones lemma is so important all right. So let us meet next time. Thank you.