 Hello, welcome to NPTEL NOC, an introductory course on points of topology part 2. So we will begin with a new chapter, Metrization. The problem of when an arbitrary topology comes from a metric, probably known as Metrization Theorem. Let us start from the theoretical importance of this question, it is solution play a solutions message, there are many solutions, play a very important role in our understanding of the logical space in general, in this chapter we shall focus on two such solutions, one is Eurydense Metrization, other one is Nagata Smirno Metrization, the same kind of result was proved separately by Nagata as well as Smirno. So here in the Nagata Smirno Metrization, Paracompactness enters into picture, namely with its property of admitting sigma locally finite refinements for every open cover. So we are going to present the proof due to Nagata, for Smirno's proof you can see in the book of Willard for example. So let us make a formal definition, take a topological space, then it is called Metrizable topological space, if there exists a metric on the underlying set X, with the family of all open balls forming a base for tau. In other words some open set is open, if in only for each point inside U, you have an open ball corresponding to the metric contained inside the open set U, contained inside given set U, then only you will get a U as a open set, that is the topology induced by the metric, if that coincides with tau, we will say tau is Metrizable, the difference between a metric space and a Metrizable space is that on a metric space, a metric has been chosen, whereas on a Metrizable space only the topology is chosen and it is possible to choose a metric which gives this particular topology, in fact there may be several topology, several metrics which use the same topology, but we will be interested only in study of the topology there, starting with a topological space, we naturally ask a question when it is Metrizable, any answer to this should be purely in terms of the topological properties, terms of that there are number of very useful answers to this question, rather than some useful characterization which may be used as characterization, often characterization can be merely topologies, that is why if this happens then I am sure kind of theorems are more valuable than if and only if theorems, if and only if part may be as difficult as finding the original you know satisfying the original condition, if not always the case several characterization always help you to find things also in a easy way, so that is not a global remark anyway, so we shall study only two such results here, one is Regions Metrization and the other one is Nagata Smirno as I have told you earlier, so today in module 26 let us concentrate on Regions Metrization theorem, to begin with we have this theorem about product which will be used in the proof of this theorem, Regions Metrization, what is it, take a countable family of metrizable spaces, this countable family is important finite is included of course, then the product is metrizable with the product biology here, whenever you take a product the first thing you do is take product biology, so how do we do that, to begin with you can choose any metric on Xn, there is one which gives you the corresponding tau n, whatever origin, but then I would like to choose it carefully by choosing it to be bounded by some number that you have that you know how to do that, namely if D is a metric you can define capital D as this minimum of distance xy or r for every xy, where r is any fixed number in particular for D and I will do the same thing with taking r equal to 1 by n square, maybe 1 by n will also do something some control for each n you know as the control should be becoming stronger and stronger that is all I need here, so you can take 2 power n also for example, 1 divided by 2 power n also, now once you have chosen that, now for each xn and yn, x and yn inside product space, x and yn are now sequences you can think of that one, we define delta of xy, so this is going to be a metric on the product by just taking the sum of Dn xn yn, so when I take this sum you should know that this is convergent otherwise this would not make sense as an element of r and that is precisely the role of this choosing these nth metric Dn is bounded by 1 by n square, so each element here Dn will be less than or equal to 1 by n square, therefore the sum total less than or equal to sum total 1 by n square which is convergent, so therefore this as a real number it makes sense no problem, moreover I want to claim that this delta is going to be a metric now, so first of all suppose Dx y is 0 that means you get this summation, this summation is sum of all non-negative numbers right, so if the total is 0 each of them must be 0, Dn of xn yn is 0, Dn's are metric therefore xn equal to yn for every n which means x and y are the same element, of course if you look at this one if you interchange xn yn this value does not change because Dn is symmetric therefore delta xy is same thing as delta of yx, the triangle inequality is also valid because each xn yn zn distance between Dn xn yn plus Dn xn yn zn is less than or equal to Dn xn zn okay, so it can take sum that will give you delta xy plus delta yz is less than delta xz is less than or equal to delta xy plus delta yz, so that is also easily valid okay, so what remains is why this metric gives you the topology that we have already chosen namely product topology, giving a metric is not at all difficult there are so many metric you know you can just take a discrete metric also and so on, so now the topology should be the same that is our thing okay, one thing is sure such things we have verified if this family is finite then we know that sum of the metrics is a metric okay, so that is the way we have become bold enough here to do this one and our boldness space here okay, so let epsilon positive be any nx belong to product of xn be any point, first choose a k such that summation 1 by n square after you know n after k bigger than k this is less than epsilon by 2, so this is the remainder term after n after k terms here okay that should be less than epsilon by 2, so I can choose given epsilon can choose k because summation n square is convergent, so now choose r positive such that this 1 by n square k1 to k the first k terms okay that sum into r is less than epsilon by 2, so this may be too large multiply by r such that it becomes less than epsilon by 2 r shall be told okay, so first k terms are controlled by this r, the remaining terms are already controlled here because we have chosen k in that way, so that is the whole idea. Now consider bn set of all yn belong into xn such that distance between xn and yn is less than r by corresponding n square, so this is an open ball inside xn centered around xn little xn okay, now take you to be product of inside product xn or y inside the product space such that yn is inside bn only for 1 less than or n less than or k the rest of the ends there is no condition, so that is a basic open subset see why each bn is an open subset in xn, so this will be a basic open subset in the product apology because conditions are only finitely many here the rest of them y could be anything right yn could be anything, so take you to be such an f after choosing bn like this take you to be this one clearly it follows that x is inside that one because the first x1, x2, xn are inside corresponding bn b1, b2, bn the rest of them there is no condition, so x must be inside u that is no clear that is clear and u is open also clear okay, also if y is inside u what happens the first n term yn and xn is less than less than r by n square right, so for first n to k equal to 1 to k each of term is correspondingly 1 by 1 square 1 by 2 square 1 by 1 by n square I have taken but r is common r times that one, so r is common it is summation 1 to k 1 by n square right the rest of them because k is bigger 1 by n square itself is less than epsilon by 2 this is less than epsilon by 2 so this whole thing okay r by n square has been chosen for that reason that means what distance between any point y and x is less than epsilon means the open ball of radius u is contained is a b epsilon of x for every x it happens right for every y axis inside u then for every y this happens so u is contains a b epsilon this means every open ball in the metric topology is open in the product ball okay, so therefore the metric topology is finer than the product topology so all the open balls here are contained inside that one so for each point so these are open subsets of open subsets in the product topology okay, now to show that an open set in the product topology is open in the metric topology we shall show that this curly m is the metric topology this is the product topology okay suppose I show this is continuous what does that mean take any open set in the product topology this identity map okay so it is open here okay just now we have shown metric topology metric is open here so I have to show that this identity map is continuous this way this way we have already shown that would mean that identity map is a homomorphism which is same here m is equal to tau okay, so how to show this is continuous any map into the product space is continuous if and only if all the coordinate projections of that map are continuous so take projection map pi n from xn to xn product of xn to xn composite with the identity it is just again product only what I get is I have to show that product xn with metric topology to xn with the usual topology whatever topology it comes this is continuous okay so that is what I have to show then this identity map will be continuous since both are metric space is now xn I have chosen metric right such that this topology is given by the metric here I have given already metric and I am looking at the metric topology so both are metric spaces this can be easily checked by sequential continuity suppose is x upper m is a sequence in the product space which converges to x upper 0 that is also an element in the product space what does that mean we must show that every projection map okay xmk or xmn you can put xmk to x0k they are you know convergent xmk is convergent to x0k that is what I would show sequential continuity is enough okay but what is the meaning of this one in the product suppose this sequence converges to this one what is the meaning of this delta xmx upper m to x0 0 can be made to x0 0 can be made less than some epsilon given epsilon all right the delta is here where I have defined this this function okay once that is the k this is sum of positive term non negative term each term will be less than epsilon right so fix one k here that will be less than epsilon something each term will be less than therefore sequence converges to a point in the product here implies each coordinate converges there that is because of our choice of the product metric here so what we have proved here is that countable product of metric spaces is matrizable countable product matrizable space is matrizable okay now every subspace of metric space is matrizable there is no problem right you have seen this one in the in first part itself okay you can take the restricted metric that will give you restricted ball that is from the above theorem it follows that if you take 0 1 raise to a countable product 0 1 raise to this natural number a countable product of copies of the closed interval 0 1 that is matrizable okay in general I can call it Hilbert cube when I am talking about a set topological space there is no problem as a metric space you will have to see what metric you give right here I am taking only product topology it can be given the metric quite often people call it Hilbert cube only after choosing a metric so that is that is why I have said it is matrizable thus we would like to embed a given metric space in a Hilbert cube then it will follow that metric space is also that topological space is also matrizable okay thus it remains to find out which are the spaces that can be embedded in a Hilbert cube okay a topological space is embeddable in a Hilbert cube if and only if it is a second countable T3 space in particular a second countable T3 space is matrizable so this is the final theorem of Eurydense matrization okay what we are going to do we take Hilbert we take second countable T3 space regular and T1 okay we will show that it can be embedded in the Hilbert cube in fact what it says is if you have a subspace of Hilbert cube it has to be second countable T3 space if and only there is no other choice so Eurydense matrization theorem is only if part in the sense that there may be many other metric spaces which are not necessarily second countable right it does not answer those things right so it is only a partial answer but it is a very useful theorem okay proof is very easy now see i power n is compact and every subspace you know of a compact metric space is second countable therefore the necessary condition necessity of the condition in the statement theorem follows okay so subspace compact metric space is T4 also T in fact T5 also so it will be automatic subspace is T3 so T3 has to be there also now conversely let xp a second countable T3 space we have proved earlier that a regular Lindelow space is normal okay second countable implies Lindelow T3 includes regularity therefore our space is automatically normal to show that it is embeddable in i power n we will use Ticknoff's embedding theorem okay so 5.19 it is sufficient to find a countable family of continuous functions from x to 0, 1 which separates points and closed sets if it separates points the corresponding embedding will be corresponding function that we are showing it will be injective if separates closed sets and points it will be open mapping or a closed mapping that is why it is it is embeddable that is how we have proved it so we will only prove this part now that there is a countable family of continuous functions which separates points as well as closed sets automatically separate points because points are closed in our x because x is also the host of space also T1 space also so begin with a countable base B for x it is countable family is now coming because of the second countability of x so start with a countable base for the topology of x okay put f equal to curly f equal to ordered these pairs u comma v inside B cross B such that u bar is contained inside B okay so this kind of things will make sense because of the regularity otherwise such things may be empty that should not happen then this is there are plenty of this capital curly f there are plenty of them nevertheless it is a countable family because B cross B is countable so it is a sub family so it is countable family now for each member u comma v of f let f u comma v from x to 0 1 be such that this f u v suffix here operating upon u bar is singleton 0 operating upon x minus v is singleton 1 so this is where we have used normality you see u bar is closed x minus v is closed they are disjoint because u bar is contained inside B okay for each pair u v you have an f u v which take u bar to 0 and x minus v to 1 this family f u v as u v varies over f is required family that we have to show that it separates points and close subsets given a closed set f and a point outside it we can first find a v belonging to B in the base right such that x is inside v contained inside x minus f x belongs to x minus f and x minus f is open okay f is closed therefore you can find a basic element v x belongs to be contained in x minus f okay then we can find u again inside B such that x is inside u contained inside u bar contained inside B because regularity of x now what we have what is this u comma v is a member of here it follows that corresponding f u v separates x from f over right because the whole of u bar goes to 0 for x goes to 0 and they are complement x minus v goes to 1 and x minus v contains f okay so so our theorem Urizon's matrization theorem is over okay basically because we have already done Tikhonov's theorem and then we just finished the product theorem product of matrizable spaces is countable product of matrizable spaces matrizable so this is the way to remember okay Tikhonov's immunity theorem and this product metric for countable product metric Urizon's theorem so let us do next time thank you.