 Hello, welcome to module 21 of NPTEL point set topology course part 2. Today's topic is stone check compactification. Having studied the minimal compactification namely the Alexander of Compactification, we shall now study the compactification that is maximal. The idea is to embed the given space in a large compact space and then take the closure of the image of this embedding. The large number of continuous maps of a completely regular space into the closure interval 0, 1 combined with the Ticknoff's theorem on product of compact spaces is the key in obtaining this maximal compactification which goes under the name stone check compactification. You should remember that all these we are doing only for austere spaces. So, our first lemma which you can name as Ticknoff's embedding lemma is the following. Start with a topological space, any topological space and any family of continuous functions which I will denote f from x to the the domain is always x, but co-domain may change that I will denote a yf. So, each member of G is a continuous function from x into some space that some space is yf that is all. The evaluation map is from x into the product space yf. You take the product of yf as f range is over this family. So, evaluation map is defined remember the fth coordinate of ex for each x inside x ex is a member of I have defined it as an element of this product space. So, I define it by taking its fth coordinate to be fx. So, statement is that this evaluation map is continuous. The second part is suppose the family G separates points of x, it just means that if you are given two distinct points x and x prime inside x, there exists a function f inside this family G such that fx is not equal to f of x prime that x and x prime have been separated by f that is the whole idea. So, if that condition is satisfied then the evaluation map is defined in part 1 is injective. This is the second part. The third statement is again some condition on the family G suppose the family G separates points and close sets that is just saying that given a close set and a point not in the close set there must be a function f inside G such that fx is not inside f of f closure. See f is a close set f of f may not be a close set you take the closure of that f of x little x must be outside this closure that is the meaning of this separation of a close set and a point which is not there. If this condition is satisfied then e is an open mapping of x on to its image which is e x that is what we have denoted e is a function e x in the image. So, third statement is precisely that this e is an open mapping an open mapping injective will be an embedding for that is why the name embedding in the lemma has come. So, there are three distinct things here. So, one by one let us have its proof which are all straightforward proofs. The first thing is continuity of a function namely e where is it it is some space x into the product space a function into a product space we know is continuous this is the more or less a definition or the property of product space is continuous a function is continuous it is well known if all its coordinate functions are continuous. So, starting with a g inside g let us take this pi g to be the projection map from the product y f to y g these are projection maps of coordinate projections. Then pi g of e operating upon any point x is nothing, but g x by definition g f coordinate is g x that is the definition of the point here e x of f is f x put f equal to g that is what we get here. What does that mean that pi g composite e is the function g that function g is continuous because we have taken all functions inside this curly g to be continuous function. So, this will just complete the proof that e itself is continuous. The second part start with two points x and y not equal to each other inside x. So, the statement x and x prime are there that is no problem x and y are different points of this capital X. As soon as you have that there will be some function f inside g such that f x is not equal to f y. So, that is the meaning of the property to that g separates points of x that is the condition that we have used. As soon as f x is not equal to f y remember f x is nothing, but the f th coordinate of e x and f y is the f th coordinate of e y we know that these two are different f x is not equal to f y. So, two coordinates of a same some point they are distinct which means the points are different that is all. So, property two is also proof. Now, the third statement. So, what we have to show we have to show that starting with an open subset of x take e u e u is open inside e x is what we have to show where is e x e x is a subset of subspace of the product space. So, how do one show that something is an open subset in the in the in the subspace only if you have an open subset of the whole space and then you intersect it with the subspace subset e x. So, we have to show that e u is open in e x it is enough to show that given any point x inside you there exists an open set v in the whole space product space namely y f f taking on inside capital G such that this way we has the property that e x belongs to this intersection v means it just belongs to e x always is inside e capital X intersection v this entire open subset must be concerned inside e u is the construct. Once you have that you can range this x for different x for entire over u for each u this should be true. So, that will show that this intersection is e u itself is open inside e x. So, I have to show that this e x e capital X intersection v that is contained inside e u this is what we have to show that e x belongs to v is obvious I mean to begin with we have taken such a open and we are we are constructing such an open set v. So, let us see how to construct this v for this all that I do is choose f such that f x is not in f of u closure then closure of that remember u is an open set. So, this is the compliment sorry this is not closure the compliment of u that is a closed set and x is not in the compliment of u because x is inside u. Therefore, we have such an f namely f x is not inside f of u compliment closure. So, such an f is there because of property 3. But now if you take v equal to pi f inverse of y f minus this subset remember f of u c etcetera where does this belong to this will be inside y f because f is a map from x to y f. So, these are subsets of y f take the compliment of f u compliment closure that is an open subset. So, pi f inverse of that will be an open subset and that is what I am taking v. So, this v will be an open subset of now the entire product space because pi f is after all a projection map from the entire of this product space into y f. So, this v is open obviously, e x belongs to v because pi f of pi f of v h which is something the e x of f is f x. So, that is already there because f x is not in this one means f x in the compliment that is all. Also if x prime is any other point such that e x prime is also inside v. So, that is the point some other point of this intersection then what happens f of x prime will be inside pi f of e x prime apply f to both sides it will be pi f of e x prime which is nothing but f of this one. So, this will be inside y f minus f of compliment of c bar. So, that is just the meaning that x is not inside u c bar which is same thing as saying that u c this is what this is just compliment. So, this same thing as saying x is inside u therefore, e x intersection v is contained inside e u. So, that will show that e u is open inside e x. So, the lemma is proved now we can read some important conclusion here. So, that I am calling it as Tyknoff's embedding theorem. So, for this I start with a Tyknoff's space. Tyknoff's space means what completely regular and a Hausdorff or T1 space. So, start with a Tyknoff's space and let f x be the set of all continuous maps f from x to 0 1. So, here I have a liberal choice of this curly g in the lemma. So, curly g is the entire space of all continuous functions from x to 0 1. Then the evaluation map e from x to this closed interval 0 1 taken product taken f x times indexed over f x for each continuous function you take a copy earlier we had arbitrary y f. Now, each y f is equal to the closed interval 0 1. So, that is the special case of this previous lemma. So, take that map that is an embedding that is a statement. This statement is an immediate consequence all that you have to show you that the condition 2 and 3 are automatically satisfied if x is a Tyknoff's space. Tyknoff's space is Hausdorff's space then we know that this is true and both each point is closed and any disjoint two disjoint closed point and a closed subset are separated is the complete regularity. So, 2 and 3 will be automatically satisfied if y if x is for Tyknoff's space. Of course, always the evaluation map is continuous there is no no extra assumption on this family g. So, 2 and 3 come only because we have assumed x is a Tyknoff's space. So, what we have what is final conclusion is every Tyknoff's space the evaluation map from x to the product of 0 1 taken f x times namely the index over the family of continuous functions from x to 0 1 that itself is an embedding. Now, we make the definition of the Stonchuk compactification. Look at any Tyknoff's space take f x and the evaluation map etc as we have just discussed earlier then the pair e the map e comma e x closure this closure is inside this product space this is called the Stonchuk compactification of the Tyknoff's space x. Of course, the product space is compact every closed subset will be compact that is why e x bar will be compact. Each factor here is 0 1 that is host of space the product is host of space. So, the subspace is also host of space. So, these are all host of compactification and the image e x by the very definition is dense inside this whole space namely e x bar. So, all that is automatically satisfied. So, this is the compactification of x. x can be identified with e x here through the embedding a then this x bar is just you know you can think of this as a extension of the space x itself that is the way to think about a compactification. However, we have its elaborate definition wherein we just do not forget the embedding the embedding e if this is different then we do not call it a compactification. It is necessary that this is an embedding if you change the embedding it may be called a different in a compactification that is all. We shall now establish a certain canonical property of stone check compactification. I will explain the word canonical a little bit later. As a preparatory result we have this lemma this itself is a some kind of you know canonical word already comes here not the word, but the explanation comes here. Pay attention to this lemma. Let z be any topological space and theta from a to b any set map this is a set theoretic map there is no topology here z is any topological space then the function theta star from z power b z power b is what z cross z cross z etcetera taken b number of times indexed over this set b to z power a same thing z taken product taken index over a here. So, what is this theta star theta star is look at any map from any element in z power b that is a function from b to z. So, composite theta b to z you have the function alpha composite theta alpha composite theta is a function inside is an element inside a to z that means z power a. So, this map itself is a continuous function where this z power b z power a are given product topology under the product topology is theta star theta is any map there is no continuity condition there is no topology here, but theta star there is product topology on both domain and core domain this theta star is continuous. This statement looks somewhat strange, but proof is entirely one line here because what we have to do is to prove the continuity of a function into a product space we have to check its continuity after composing with each coordinate function that is all. So, that we will have to. So, here is a diagram a representation of theta star theta is this any function here each continuous function each sorry each functions of b to z these also set theta functions right element they are elements of z power we mean just functions here set theta function you compose this then you get a theta star of alpha is alpha going to theta star of alpha the theta star is continuous is what we want to show for that all that I have to do is for each point a inside a I have to show the pi a composite theta star is continuous these are the projection coordinate projections of theta star. So, operate it on any alpha by the very definition it is nothing but pi a of alpha composite theta because theta star of alpha is a continuous but what is this one this is nothing but alpha of theta a. So, theta a is an element of b alpha composite theta a is nothing but the projection map in the coordinate projection of theta a theta a projection of alpha. So, pi of theta alpha operating upon alpha theta a operating upon alpha. So, that is true for every alpha inside z power b. So, this just means that this pi a composite theta star is nothing but pi of theta a. So, all coordinate projections are continuous. So, that completes the proof that theta star is continuous. Now, I am making this statement about theorem 5.23 x and y be any two Tickanoff spaces e x bar and e y bar I will denote our standard stone check computation in our notation. But just to be careful I am taking e prime bar here. So, that is there is no confusion that e e etcetera e how can the same maps they are after all not same maps the domain of e is x here the domain of y is e prime here that is why I have domain of y prime is domain of e prime is y here. So, I do not do not use the same notation either. So, take e x bar e prime y bar these are the stone check compactifications of x and y respectively. Now, given any map f from x to y there is a unique map S f from e x bar to e prime y bar such that S f this is the new function the existence of which is stated here as well as uniqueness it has the property that S f composite e is e prime composite f. So, look at the diagram f is here e is going into e x bar here e prime is going to e x prime these are two stone check complications that will be a continuous function here S f which makes this diagram commutative. Same thing as saying you come with f of f here and follow a e prime is same thing as first you go by e and then take S f. There may be many functions no there is only one function like this a unique map f f. So, that is the S f that is the conclusion of the first part here. Second part says in particular if y itself is compact host of space we get a unique map again f f from e x bar to y which is an extension of f again extension means what now f hat composite e is equal to f. The function is from e x bar if you restrict it to just e x which is the copy of x that is the function f itself. So, that is why this f hat is called extension that is all. So, that is the second part here I will explain this one once we complete the proof of the first part. Let us start with the uniqueness part of the existence of this function S f. Suppose you have any topological space any house door space B and two continuous functions from some space into B such that if you look at the set A A belong to A that is h 1 of A equal to h 2 of A that set is always a closed subset of A. So, that is the property of the house doorness of B of the code of A. So, this is one thing which we have several times used at least in the part one. So, here also you are using it. This is one of the properties of house doorness. Therefore, so now I am going to use for the uniqueness part I am going to use the stone check computation e prime of y bar this is house door space. So, if you apply this conclusion here to g 1 and g 2 two functions from e x bar to e prime y bar such that g 1 of g 1 composite e equal to g 2 composite e both of them are worked by the very definition and by the very condition they must be equal to e prime composite f g 1 composite e must also have this property g 2 composite must also which is a property therefore, g 1 of e equal to g 2 of e. That just means that set of all points wherein set of all points of e x bar let us call such that wherein g 1 of z equal to g 2 of that that is a closed subset because what is the meaning of g 1 of z equal to g 2 of z they are functions from they are points from e x bar functions from e x bar to y bar. This is a closed subset for this is a house door space therefore, this is a closed subset. So, that is by this observation, but now what is the meaning of g 1 z equal to the set of points which are inside e x bar automatically functions for e x will have this property because of this one. So, that means all points of e x will satisfy this property right g 1 of e x I put z equal to e x equal to g 2 of e x right z is e x. So, all points of e x have this property which means that e x is contained inside this set and this is a closed set. So, e x is contained inside this closed set which means its closure is also contained inside closed set the closure is the whole space. If a some subset is contained in a closed set the closure will be also contained inside that that is all I am saying. So, it is a whole space. So, what is the meaning of that set of all points e x bar wherein g 1 equal to g 2 is whole space just means g 1 is equal to g 2 that is all. This means that if there are two functions like this they must be equal like this means all commuting continuous functions which have this commuting property namely s f composite e equal to e prime composite x. Now, I have to show the existence because of uniqueness then I can name it it just depends upon f. So, I am calling it as s f s for stone check computation f is any any given continuous function between just these must be taken off space so that I can talk about stone check specifications here. So, let us go to the existence part. Given a function f from x to y first we define f star from f y to f x remember f y denotes all continuous functions from y into 0 1 similarly f x denotes all continuous functions from x into 0 1 starting with a function from y to 0 1 you composite with f you get a function from x to 0 1. So, f star of alpha is just alpha composite f. Next thing is I do one more composition here let f double star from 0 1 raise to f x into 0 1 raise to f y just the way we defined it in the lemma that is z raise to b to z raised to a. So, that is what we are going to do this f star is stated as a theta then theta star will be this f star star all that I have to do is this map is phi going to phi composite f star. So, the f star is theta is replaced by f star here. So, once you have that this is a diagram is shown here starting with alpha you composite with f to get f star of alpha. So, codomain here is always i once you have that f star of alpha. So, the function f star is here starting with any function phi again you composite f star you get f double star. So, you have done the compositions twice here this f double star is now from set of all functions from you know f x to i namely the product space i raised to f x to i raised to f y these are both product spaces this is continuous continuity of this is the statement of this lemma 5.22 I am going to use that lemma now. So, I have got starting with f from x to i I have got a function in the product spaces all that I have to do is that this same function will restrict itself to the stone check comparisons of f x and f x y f y respectively sorry of x and y respectively inside these product spaces. So, that is the next step. So, we claim that in this following picture you start with f here and you have got f star here f double star here e x bar is sitting here e prime y bar is sitting here these are the stone check competition. First I want to show that all these solid arrows they are they form a commutative diagram the outermost one this is a dotted figure dotted arrow this also dotted arrow. So, what I do is once you have this you take this subset e x here f double star of e x goes inside e prime y I will show once I show that automatically its closure and the same map f double star will go into this one. So, these dotted lines will come once I show this the entire figure. So, automatically these are nothing but just the restrictions of f double star corresponding. So, automatically what will happen if I take sf here e goes all the way from x to here and take sf that is same thing as first take f then come by a prime. So, let us all that I have to show this is that this map into if followed by f double star is same thing as f followed by this e prime here the evaluation maps t and e prime remember they are actually maps from x into iris to sf y into iris to f y. So, that is a tautological thing actually fixing each x belonging to x I have to show that f double star of e x is equal to e prime of f x. This is the same thing as proving remember what is the definition of f double star f double star of e x is a function operating upon any index alpha is by definition e x operating upon alpha of f right. So, that is same thing as e prime operating upon f x of alpha. So, this is what I have to prove but this is the same thing as proving that now alpha composite f operating upon x this part alpha composite f operating upon x e x of that is equal to alpha of f x this is another evaluation map. So, if I prove this then I will have to prove this one this is by definition I have to prove this it is the same thing as proving this but this is obvious because there is a double composition here alpha composite f is alpha of f x that is all right. So, I repeat now having proved having proved that this solid diagram is commutative all that I have to do is now f double star of e x is contained inside e prime y. But that is obvious because of this commutativity take a point here which comes from here okay. So, wherever it goes this part has to come from here now here e prime y because f of e prime is factoring here e prime y that is all. So, this is purely just a synthetic observation okay. So, any point e x is nothing but coming from e of this one f double star of that by definition is equal to first f of something and then e prime of whatever e prime which is e prime of y and then that is the inclusion map. So, that is the purely observation right. Once you have that when you take the closure and then f double star by continuity of f double star it is contained inside f double star e x then take the closure this is already inside e prime of y so it is contained inside e prime closure all right. So, this dotted arrow is also justified and you just name this for sf it is just a convenient notation extra you can f double star can always make sense but sf makes sense only when you have stone check computation all right. The last part remember what was the last part here in particular when y is a compact half star space what is e prime y e prime y will be a compact subset of a half star space therefore it is closed already. So, e prime y bar will be equal to e prime y which is a copy of y. So, the stone check computation of a compact space is itself itself means what e y is homomorphic to y right. So, the map sf is actually from e x bar to y itself this e y bar is just y itself then when you may say when you say it is y itself you are identifying this e prime y bit y via e prime here okay. So, you are thinking this as an identity map that is why you have simple notation here if you write sf equal to f hat here treat it as a function into y then what you have got f hat composite e which is nothing but sf composite e is just f because e prime is treated as an identity map okay. So, second part just follows because a compact half star space has its stone check computation is the same as itself that is all okay. So, this is obtained very easily all that you have to say f hat you know it is e prime inverse of sf or if you do not want to do all that you think of e prime as identity map that the inclusion that is all. So, I will restate more or less this one what what is the meaning of whatever we have done this theorem has many many way of interpretations. So, the stone check compactification is the maximal compactification among all Hausdorff compactifications of a Tickanoff space only when we have Tickanoff space we can talk about stone check compactification okay and that is a Hausdorff competition. Look at all other Hausdorff's competition this will be larger than that remember we have a partial ordering defined on all compactifications of a given space if you restrict only to Hausdorff competitions among us all of them stone check is largest okay. If eta comma y is a Hausdorff compactification of a Tickanoff space X X is fixed look at any Hausdorff competition then we just take eta hat remember in the previous theorem for any if this is a compact space okay this also compact also space is also Tickanoff space. So, whenever you have a map like this you have eta hat but eta hat is now ex bar to y itself such that e followed by eta hat is eta. So, that is precisely the meaning that this compactification is bigger than this compactification e ex bar is bigger than eta comma y okay. So, that is just a restatement of the previous second part of the previous theorem you may say but that has more content than this one this is just a consequence of that theorem. So, I make a remark here in theorem 5.23 it is easy to check that if y is x itself f from x to y is identity map then this sf is nothing but the identity map of e x bar to e x bar second comment here is if z is another Tickanoff space g from y to z is another continuous map continuous function then you can look at sf g composite f from g composite f will be from where to where from x to z. So, that is nothing but sf composite sf. So, sf g composite f is from e x bar to e z bar. So, in between you have this e y bar. So, e x bar to e y bar to e z bar you have s g composite f. So, I will leave this one this verification this just purely satiric verification to you but the important property importance of this one is that these two properties actually have a very good name they are called canonical properties or factorial property. The adjective canonical that we have used in the map sf is precisely for this reason a and b that is the meaning of this canonical. Finally, we shall end this topic today by giving a characterization of stone check computation you can also call this universal property. For a Tickanoff space s stone check computation e comma e x bar is characterized by the property this property I am calling it as p given any continuous function f from x to the closed interval 0 1 there is a unique continuous function f f from e x bar to 0 1 which extends f again extends f in the sense f hat composite is f. In the theorem we have proved that this e x bar as a property instead of taking 0 1 here I could have taken any compounds that is the second part of previous theorem. In the characterization we are restricting it to only 0 1 only the code domains are always just 0 1. So, this is a lot you know cutting down if something has every compact space of course it will have this property also you do not have to verify it for all compact spaces this you have to verify it for only 0 1. So, that is the beauty of this characterization the characterization has its own use you do not have to every time use the property that e x bar is sitting in a product space remember this was taken as a subspace of certain product space right you do not have to use that product structure or anything you can just use this property. So, that will be automatically give you the stone chip computation take any computation of a of a ticker of space which has this property it has to be stone chip computation. So, that is the whole idea. So, proof is not all that difficult first of all the stone chip computation itself has this property as a special case of 5.23 which I have just told by taking y equal to 0. Now, I want to prove the corner. Suppose eta z is some half star computation of x which has the above mentioned property property b applying 5.24 we get a continuous map tau from e x bar to z such that tau composite e equal to eta. So, this just means that e e x bar is bigger than equal to eta z. This we have already proved I am just repeating this part every stone chip computation it will adjust this was some computation which will add what we have to prove is this is all larger than that one then because all of them are house door spaces by our 5 point earlier remark two lecture before we have done it will follow that they are equal. So, it remains to prove that eta z is bigger than equal to e comma e x bar which is the same thing as you must find a tau prime from z to e x bar such that tau prime composite eta equal to e. So, you have to reverse the arrow here similar to this one you have to do the other way around z to e x bar you have to find. So, this also not very difficult but it is something new. So, that is why I have to give you a complete proof. So, what is the assumption eta z satisfy the set property p therefore, for each f inside f x what is f x continuous functions from x to the closed interval 0 1. Let us take f hat from z to i the unique map which has f hat composite take out f. So, this is the property p every function there is an f hat. Once you have this defined tau prime from z to the product space 0 1 raised to f x by the formula tau prime of z is that element which has its f coordinate equal to f hat of z. So, tau prime z is defined by this equation for each f this is tau prime of f is f hat of z. So, that defines a point here for each z I want to say that first of all this is continuous why because what is its fth coordinate its f hat. So, if fth coordinate is f hat I mean each coordinate is continuous therefore, the function is continuous. So, tau prime is continuous. So, continuity of this function is fine moreover if you come from eta remember eta is a function from x to z. So, take eta x and look at tau prime of eta x what is it operating upon f its f hat of eta x f hat of eta is nothing but f. So, it is fx but what is fx it is ex of f by definition of evaluation map. Therefore, if this is true for every f tau prime of eta x is nothing but ex. So, that is true for every f it just means that tau prime of eta e. In particular this implies that tau prime of eta x is contained inside e capital X. Therefore, tau prime of z z is what eta x closure. So, I have to take tau prime of eta x closure. So, that is contained inside the closure can be pulled out. So, but that is nothing but ex bar. So, tau prime though it is a starting with a map from z to 0 and raised to fx it is as image inside the stone check compactification. So, take that map tau prime as a function from z into x bar. So, we have already shown that tau prime from eta e could be before that completes the proof of the characterization also. So, I will end this talk today with a general remark now. There are many, many interesting compactifications depending on the kind of spaces and the kind of problems that we are studying, the problems that whatever you are interested. So, each kind of problems there may be some compactifications to go to simplify the problem and try to go to answers and then come back and so on that is the gap. Here we mention a few of them other than the Alexander of one point compactification and stone check compactification the smallest one and the largest one we have discussed. So, Walman compactification is another important one which is much more general than these two compactifications maybe it works for all T1 spaces and stone check compactification is one of the most popular compactifications but is Walman compactification also equally popular that is what I wanted to say. This will be taken in a later chapter. Walman contribution we are not going to lay off. In the study of manifolds you may come across with problems of putting boundary to the manifold. Since you do not know much about manifolds I cannot explain this part much this is only for information which if you remember then you may say this is what was told to us and so on that solved. So, that putting a boundary is actually some kind of a compactification. A further special case is the so called space time compactification which is interesting in the relativity theory. Another point of view taken takes you to the study of ends. For example, you will see that the real number system the real line has two ends whereas the compressed plane has only one end. I do not want to elaborate anything more than that but these are all you know part and parcel of various types of compactification. In algebraic geometry you come across with algebraic compactification. The projective spaces are standard examples of what of affine spaces C n the compactifications of C n they are many of them. While studying topological groups you may come across what is called as Bohr compactifications. Maybe when you compactify you would like to retain the topological group structure itself there see the group structure should also extend and so on. So, this is related to the study of what is called almost periodic functions. So, when you are studying that you will come across Bohr compactification. With this many little bit of remarks. So, let us end today's talk. So, next time we will take a different topic. Thank you.