 Hello, welcome to NPTEL, NOCE, an introductory course on Point Setupology Part 2, module 27, today Nagata-Smirnov Metrization Theorem. As mentioned earlier, Euryzones Metrization Theorem does not give a complete answer to metrizability. This problem was solved by efforts of several people. There are various versions of metrizability theorem. We shall now concentrate on one such result which seems to be most satisfactory, namely Nagata-Smirnov Metrization Theorem. We shall present a proof due to Nagata. A proof due to Smirnov, you may look into Willard's book. Before that, we need to introduce another version of normality. So the Nagata's proof of metrization theorem is in spirit similar to Euryzones but since second countability is removed now, the embedding will be in a much larger product space and also we need a little more strengthening than normality. So that is what we want to introduce now. A topological space X is said to be perfectly normal if for every pair AB of disjoint closed sets inside X, there exists a map H from X to minus 1 plus 1 such that A is the precise inverse image of minus 1 and B is the precise inverse image of 1. So I have said AB disjoint closed subset. If one of them is empty, it does not make sense. So both of them have been non-empty also. That is necessary. That is said when you are talking about normality and so on. The precise inverse image, this is stronger thing. So even in a normal space, you will have A going to minus 1 and B going to 1, that is possible but there is more points may go to minus 1 and more points may go to plus 1. So that is where the perfect normality comes. This precise set of points where in this A is precise set of points where in H takes value minus 1. Similarly B is precise set of points where in H takes value plus 1. A normal space is perfectly normal if and only if each closed subset is G delta. So this is where the link between perfectly normal and normal is written. G delta is the key. Remember that under a continuous function, inverse image of any singleton, continuous function into a metric space, into a R for example. Inverse image of a singleton under a continuous function is a G delta. So if it is perfectly normal, A and B will be G delta sets. So what it says is each closed subset is G delta. Clearly perfectly normal is normal. There is no problem. So moving G delta for every A is every closed subset is G delta is what you have to show. That is all we want to say. So if only if part is obvious because once you have this, a closed subset must be G delta. If and only if is what I want to show. So to prove the if part, suppose A is intersection of Un and B is intersection of Vn for all Un and Vn. Un and Vn are all open subsets for all n and A intersection B is empty. Start with A and B are empty subsets. And write A as intersection of Un because it is G delta. B is intersection of G delta. Both Un and Vn for all of n, they are open subsets. By normality we can first show okay first of all assume that Un intersection Vm is empty. So you have A is contained inside Un, B is contained inside Vm but they may not be disjoint. A and B are closed subset, disjoint closed subset. Therefore inside these open subsets you can choose smaller open subsets. So set Un intersection Vm is okay. But it will contain A. So if we make some if you replace some Un by some Un prime, which still contains A and a smaller then this property will still be true. So one by one you can replace Un and Vn such that Un intersection Vm is empty for all n and N. Next we can also assume that Un is contained inside Un plus 1 bar contained inside Un plus 1 for all n. Once you have this Un, you can make it Un intersection U2 that will be open subset. Un intersection U2 intersection U3 will be open subset. Inside open subsets you can choose a smaller one such that Un plus 1 bar is contained inside that and so on. So you can make it decreasing. Similarly you can make Vn also decreasing. So that is what you can do by rearranging this redefining this Un. Now let fn from x to 01 is such that fn of x minus Un is singleton 1 and fn of Un plus 1 bar is 0. So this is by normality. I am here. I am not here. This is precise that this close set going to 1 and this complement is going to 0. This Un x minus Un, Un is larger. x minus Un does not intersect Un plus 1 bar. Un plus 1 bar and this bar disjoint subset. So you have got a sequence of function with this property. Take f to be sum of these fn's. Only thing is you should divide by some factor so that the whole thing is convergent. So I am dividing by 2 power n because I know that these fn's are bounded by 1. So this will be less than or equal to 1 by 2 power n summation. So this will also converge. So this dividing by some number is a general principle which you have to learn in analysis. Sometimes this may not be too far. Anything, if you have sequence of points, non-negative things, you can always divide them by some larger sequence of another sequence which is convergent so that you can make this one is convergent. It then follows that f from x to 0, 1 is continuous because each fn is fn divided by 2 power n is continuous. So this is uniformly convergent because 1 is dominated by 1 by 2 power n. So uniform limit therefore f is continuous and then now what happens? A is precise inverse of 0 of this one because if x belongs to A it must be inside every un. So there is no questions for every un. See this whole thing is 0 means corresponding fn is 0. Therefore you see the precise 0 set. Similarly we can construct a continuous function g from x to 0, 1 such that b is precisely g inverse of 0. All that you have to do is same construction do it with v n that is all. So you have two different functions. I want a single function to do that job. So this function f is such that its inverse image of 0 is A, g such that inverse image 0 is b. Now I want one single function such that inverse image under that is 0, inverse image of 0 is or minus 1 is A and plus 1 is b the other way. So for that I just take h equal to f minus 0 f plus. See f plus g makes sense because sum is never 0. See if f is 0 both of them are positive non-negative. If f is 0 that must be inside A which is disjoint from b. Outside b outside h is not 0 outside b g is not 0. So A and b are disjoint therefore f plus g is never 0. So I can divide by f plus g f minus g divided by f plus g always takes value between 0 and 1. So minus 1 plus 1 modulus of this one is less than 1 because this is always smaller than this one. Modulus of this is less than 4 to 1. So this h takes values with minus 1 plus 1. Finally suppose this h of x is 1 what does that mean these two are equal f minus g equal to f plus g equal to this is equal to 1 these two are equal what does that mean f minus g equal to f plus g. What does that mean? So such points are precisely g must be 0. So that will be a point of g. Similarly f minus g equal to f plus g equal to 0 means f equal to g. Suppose this is minus 1 this minus 1 means f minus g equal to minus f plus g. So that f must be 0. So h inverse of minus 1 is precisely A and h inverse of 1 is precisely b. So this is perfect normality given by normality plus every close of set being g divided by now let us come to Nagata-Smernau matrization theorem proof which will be much simpler now because we have perfect normality here. A t3 space is matrizable. See there is no second countability now. If and only it has a sigma locally finite basis sigma locally finite basis should ring a bell that is something to do with para compactness. So proof of the only if part if and only if is there only if it has sigma locally finite basis. So that all that you have to do is a matrizable space has sigma locally finite basis. So this is what we have to do it is actually a matrizable space we have seen that it is para compact. So for each n belonging to n let u n be a locally finite open refinement of the open cover b 1 by n x let us belong to x fix n take b 1 by n x as x y is over x this is a cover. Okay this is an open cover for each n so that take u n to be a locally finite open refinement of this one. So for each n I have got a locally finite open refinement okay all that I want to do is now if you take union of all these u ns call that as b this is a base for a topology on x okay this I have limited exercise because you have done so much of this kind of things okay. So for all n is involved that is all you have to use all right. So show that this b is a base now clearly this is sigma locally finite because by definition each now by choice u ns are locally finite so by definition this is sigma locally finite okay let us now prove the if part let x tau be a t3 space and b equal to union of this b n be a base for tau okay remember each base is also a cover for x open cover where each b n I am again writing down it indexing with b n alpha where alpha is in lambda n. So b ns are members of b ns are indexed by this set lambda n okay and each b n is locally finite family this is the starting hypothesis now now take lambda to be the union of all these lambda ns over n okay. So this lambda is going to be our indexing set for you know your last time we took i raise to i raise to this n natural numbers now I would like to take i raise to lambda but I will do slightly better job here namely the proof is now going to be similar to the proof of uterus naturalization thing the sense that we are going to produce an embedding of x inside the Hilbert space l2 lambda the l2 space okay with l2 norm so I recall what is this l2 of lambda on any set okay any set l2 of that makes sense okay that is what I am going to do now I am just recalling it the generalized Hilbert space okay for any function f from lambda to r let us denote sf we can reach at support of f to be all those lambda such that f lambda is not here okay take l2 of lambda collection of all functions from lambda to r okay such that sf is countable and when it is countable look at the sum of all the f lambda squares some total of that is defined so it is square summable function that is the terminology okay square summable functions form a Hilbert space okay I will just use the word Hilbert space because the inner product here is not being exactly used only the norm is used and what is the norm norm is square root of summation of f alpha square okay if you want to know what is the inner product is nothing but fg inner product is summation f alpha g alpha if you have complex value functions you take summation f alpha g alpha bar and so on we are not interested in that part we are just interested in this norm okay so this is the Hilbert space the first step is since every open cover has a refinement consisting of members of b because b is a base right every open cover for each point you can select a member of the b contained inside that that will be refinement so but b is locally finite therefore from 3.6 follows that x is per compound okay so that is the the link between the condition sigma locally finite base and this one in particular x is normal what we want we want perfect normality or we will see why so first we have we have seen that it is normal space second space is we shall now prove that every open set u in x is f sigma so now we see g delta and f sigma are all both coming together here f sigma means what union of countable countably many closed subset right every open subset in x is f sigma okay then from theorem 6.7 it follows that fx is perfectly normal okay you want to recall theorem 382 normal space is perfectly normal if you know it is every closed set closed subset is a g delta you take the complements right that is all okay so so it is perfectly normal it is there next by regularity given x belonging to you there exists a basic element b which you may call it as b lambda b lambda x lambda x inside x such that x is contained inside b contained inside b bar contained inside u it follows that you put b n equal to union of all these b lambda x bar okay closures of all that as lambda rings over lambda n the union of closures of all locally finite of this is a locally finite family then is a closed subset okay because this is locally finite b curly b was sigma locally finite each b lambda b n square curly b n square locally finite and I am taking these things are inside these things are inside b lambda okay lambda n sorry here therefore this is locally finite and n's b n's are closed also know that b n is contained inside you because b lambda all of them b lambda bars are contained inside you therefore we have you itself is union of all these b n's for each point there is something when you take the whole of them it has to cover you itself so what we have shown u is countable union of closed subsets okay so that completes the proof that every open subset is f sigma third step is take each b lambda inside this base by step two we have a continuous function f lambda from x to i such that b lambda is precise non-zero points all lambda x all x such that f lambda of x is not equal to 0 okay the b lambdas are open subset every complement is a closed subset that will be precise set of 0 so it is not 0 that's all for each n inside n define g n from x to 0 infinity we are constructing these functions so that finally we can put it take all of them and put it them inside the Hilbert cube Hilbert space generalized Hilbert space so define g n to be from x x to 1 to infinity by the formula g n of x is this adding one is just for carefulness okay you could have added any epsilon no problem 1 plus f lambda x square where lambda is whole lambda so this is the countable family now okay this square summable so this makes sense so 1 plus that makes sense square root makes sense so that is g n of x by the local finiteness of the family number of b n's here okay this is actually finite now finite it follows that g n is well defined for the same reason it's continuous also because it's the finite sum this is continuous 1 plus that is continuous and this is never 0 that is why I put one error or some epsilon here I can take the square root the square root is also continuous note that for each lambda inside lambda it's unique n lambda inside n such that lambda is in lambda n lambda why look at this definition of our capital lambda is a disjoint union okay each lambda here belongs to exactly one of the lambda n that's all I'm using so that is important here to take the indexings at disjoint union of those countable disjoint union of arbitrary families whatever okay so where were we so so each n each lambda belongs to unique n lambda okay so I can talk about when lambda is in n lambda I can take h lambda of x to be f lambda of x divided by this number n lambda is some okay n lambda is some integer positive integer g n lambda also makes sense okay corresponding n lambda for each lambda is a unique n lambda okay g n lambda of x so divided by this one this is a non-zero function so divided by that so f x is f lambda is continuous this is continuous h lambda per continuous okay all these continuous functions taking values inside i now because f lambda is only one of you see if you take the square of this this square this square then this function is one of them here so the numerator is always smaller than the denominator and they are all positive so they take values inside i okay now I can define h from x to i raised to lambda itself okay in particular it will be instead r raised to lambda no problem by the formula hx of lambda equal to h lambda of x hx is a function from lambda to i so it's lambda value you can think of the function or you can think of it as a element in the product so it's lambda coordinate either way it is h lambda of x by the very definition if you take the product of all these should be continuous okay very first thing to do is take the h lambda is first of all you have defined it as an element of r raised to lambda but I want it to be inside l2 of lambda l2 of lambda is a subspace of r raised to lambda this is an element of r raised to lambda so this follows easily since first of all for each fixed x hx of lambda is not equal to 0 for finitely many lambda is inside lambda n for each lambda n is finite therefore set of points where in hx lambda is not equal to 0 is countable so s of this s of hx is the set of points where in it is not 0 this support that is countable now second thing is if you take the sum of all of them after squaring the square and then take some finite on on fixing lambda n this is a finite sum that is nothing but summation h lambda of x square lambda inside lambda n okay but by definition this is f lambda x square divided by n square because I have to take n lambda but lambda is in lambda n this n is this n itself here n square g n of x square I do not have to write g n lambda because n lambdas are all n here because lambda is in lambda n okay and this summation this this number is 1 plus this summation so this is smaller than this one so this is less than 1 by n square if for each n this is less than 1 by n square when you take summation of all of them that will less than summation of 1 by n square we succumb to convergent and we wanted that one convergent therefore hx is an element of l2 so we have got a function x2 l2 all right sir may you please repeat this step so we had a from starting with every element of x we wanted to associate a element of l2 of lambda that was from step 2 we found that every b lambda was open set so we can associate a function so that it is exactly non zero that was the first step first part of step that will come now in the that will come now in the proof of injectivity and continuity and all that right now the that this is opens up set precisely it is 0 etc is not used only local finiteness is used so that we have come you know this is countable the set of points s of s lambda right so to show that it is inside l2 of lambda only by n lambda right this n lambda term is got precisely for that not only that why in the definition of gn i have put one here for two different purposes first of all i should be sure that when i take square outs continuity etc i should not be getting into 0 don't bump into 0 secondly this term it is bigger than just this term so there are two purposes here okay so gn x is defined like this i could have i have told you any any positive constant would have done the job here okay so look at here now these are inside i it is not very crucial but this is but this numerator it divided by this this is so things less than n lambda that is important okay not just in i that is that won't give you anything it is less than 1 by n lambda 1 divided by n lambda that is important here okay so that will be when you fix lambda n lambda capital n capital lambda n what is n lambda n lambda is n so the same n is there so whole thing finite this finite collection is 1 by n square now you take some over n so that must be convergent that is convergent so we are inside l2 okay now comes you see the role of all this opens still be you will be there still you have to do that now check this h is injective given x1 not equal to x2 in x okay since x is hospital space is t3 space that we have assumed right you will have b lambda it is a basic element so say x1 is in b lambda and x2 is not in b lambda right so this immediately implies that corresponding f lambda of x1 is not 0 but f lambda of x2 is 0 so precise zeros have been used here now for f lambda but h lambda of x1 is not 0 whereas h lambda of x2 is not 0 because what is h lambda h lambda has f lambda's in the numerator and this is always non-zero this is 0 if it only if this is 0 right that is what it is so h lambda separate point okay therefore h hx1 of lambda will not be equal to hx2 of lambda so this lambda coordinates will be different as soon as x1 is not equal to x2 okay so this proves h is injective all right see countable family won't have been able to do in the local base we have used countable family of local base was possible if if x were second countable okay however some countability was necessary what is that sigma locally finite so that plays the role here next target to get l2 otherwise you won't be able to get l2 next we check that h from x to hx is a closer function okay injective closer continuity these three things we have to show right so continuity is at the last okay that a be any close subset of x you have to show the h a is close take a fee outside of h a it is inside hx of course okay everything is inside hx in the whole l2 we don't know we don't need that okay to show something is embedding you have to take the only image and work inside the image so fees h of some x for x inside x minus a because it is not in h a but then x belongs to b lambda contained inside x minus a okay a is starting with a close subset so you will have some b lambda so said x belongs to b lambda contains x minus for some lambda so that automatically implies h lambda of x is not zero whereas h lambda of a will be zero for all a inside a okay besides set of nonzero elements are inside b lambda so outside x minus outside a so for all a outside this inside x minus outside x minus a for all a it will be zero therefore if you take the norm of hx minus h a okay this is summation h lambda you know right of this minus h of that one there will be one term it's okay norm square is the summation squares and so on when you are taking square root there will be one term which is h lambda of x minus h lambda of a there will be many other terms they are all non negative terms this sum this therefore this thing is bigger than equal to this one okay one coordinate this is zero and this is not zero okay so that checks minus h lambda modulus is precisely equal to h lambda of x and that is positive this happens for all a one single number positive and this norm is bigger than that for all a that means the distance between hx and the set a is positive okay therefore hx cannot be in the closure of h a bar h a bar closure distance between hx and h of a is positive okay this says point wise but this is true for all a therefore distance between hx and h capital a is positive that means hx is not in the closure of h a okay we want to show that h a is closed so first of all hx is not in the closure since this is true for all points x minus a hx minus h a this implies that hx minus h a itself is contained inside h a minus h a bar because it take any point here it just now we have shown that it is not in h a bar that is all but by demorgan law this just means that h a bar is contained inside h a bar is contained inside a set itself that means that that set is closed okay maybe you can directly prove that from here here it is a open set I think close through in closeness is easier that is all all right now let us check h is continuous okay now you see the full force of all these things will be used how we have constructed given x belonging to x choose epsilon let epsilon be positive we have to produce some open subset u around x such that x belongs to x prime belongs to implies norm of hx prime minus hx is less than epsilon so this is the continuity I am using the the norm on the core domain okay x is just a topological space so I have to use only open subsets around that one which has for every x for every epsilon I have to prove this as soon as epsilon is given choose some n n such that n to infinity I can put it k also earlier I have to k also we have capital no problem n plus 1 to infinity 1 by n square is less than epsilon by 4 maybe epsilon by 8 if you want you can choose none matter okay by local finiteness of all these b n we get an open set v around x such that this open set v meets only finitely many members of b n for all n less than equal to capital N for each of them you will get a say b 1 you will get v 1 v 2 you will get v 2 up to b capital N you get v capital N but then you can take the intersection of v 1 v 2 with capital N call that at v that is all okay this v is the neighborhood of x which will make finitely many members of b n n ranging from 1 to capital N all right let us denote these members by b lambda i say b lambda i belong to all they will all belong to b 1 b 2 b 3 or b up to n where I do not care so totally they are in number k finite now choose a neighborhood view of x such that u is inside this v smaller neighborhood I am going to choose and these finitely many come continuous functions no h lambda i of x prime minus h lambda i of x is less than some epsilon I have chosen carefully some epsilon for this epsilon square root of epsilon by square root of 2k choose neighborhood okay for each of them choose neighborhood and then take the intersection take intersection with v you will get a subset u contained inside u for which all these inequalities will be true for all points x prime inside u okay that is by continuity of each h lambda i finitely many of them you can do this is for all x inside u if you now take the summation n bring to 1 to n and all points lambda inside lambda n h modulus of h x lambda minus h lambda of x square square if I take here that will be epsilon divided by 2k how many are there k of them so it is k times epsilon divided by 2k okay and that is nothing but epsilon bit alright all the sums up to capital n here is less than epsilon bit okay now on the other end what happens the second part namely n plus 1 to infinity n plus 1 to infinity take summation lambda inside capital lambda n is a double summation h x h lambda of x prime minus h lambda of x square okay what happens to this summation this is n to infinity i as it is lambda i as it is h lambda of x prime square plus h lambda of x square okay so squares or difference is less than equal to square you know square of each individual term so that is a very rough estimate actually but that is enough because this summation these are positive things okay though this is maybe infinite when lambda equal to lambda and this is finite when actually okay then we might we can separate them out if you if you take the first summation this n to infinity is to be less than 1 by n square by choice this also less than 1 by n square for each x prime x this is the property that we have chosen okay so that is twice summation and n plus 1 to infinity 1 by n square each of them is less than epsilon by 4 so twice that is less than epsilon by 2 okay thus it follows that u is the required neighborhood of x so guess if you take sum together this sum square is less than epsilon by 2 plus this sum sort of square is less than epsilon so we have instead of showing the epsilon square we have shown that it does not matter it should have been square root of epsilon start at it okay so this proves the continuity of h and thereby the theorem is low okay so here are a few exercise sorry one exercise here it is not very difficult at all every matrix space has a sigma locally finite space okay I have put already as exercise and I have put it as separately in the proof somewhere I should prove this one as an exercise remember so I have separately mentioned it that is off maybe in the earlier theorem never mind so there are many problems in vectorization like you may talk about when can a topological space be a given a metric which is complete such things are called complete materialization we shall not be able to discuss such things which are too special okay thank you