 Welcome to NPTEL NOC, an introductory course on point set topology part 2. So, today we shall begin chapter 11, a short chapter on compact open topology. Here we will discuss the question of topologizing a family of functions from one set to another set, but with some specific topologies on both sides. For example, we have discussed the Banach space, the Banach algebra of all real and real or complexed functions from one set to another set which are bounded. In its subspace, wherein X is some topological space, you take just continuous functions, continuous and bounded functions, right? Later on, we even studied some specific properties of these spaces such as, Weistra's theorem, Ascoli's theorem and so on, right? So, function spaces have a lot of importance in mathematics of law. Arguably, we may just begin with the set Y power X. This notation is just for set of all functions from X to Y. It also is equivalent to another picture namely product of copies of Y as many as there are elements of X. So, you can think of each copy of Y indexed by alpha and then take the product, index by X and then take the product. So, that is the same thing as all set theoretic functions from X to Y. That will involve only the topology on Y, product topology has nothing to do with the indexing set X. But we want to bring the topology of Y also in the picture. So, let us restrict ourselves to subsets of all continuous functions. So, that is a set theoretically is a subset of set of all functions from X to Y. Now, the central problem here is to approximate a given continuous function with some special properties maybe or totally arbitrary continuous function by some subclass of functions such as polynomial functions. So, that is Stone Weistra's theorem or you may say just Weistra's theorem in the case of functions defined on the closed intervals or something like that. Or it could be smooth function or it could be embeddings and various things are there or you may want to approximate by just what are piecewise linear functions and so on. So, what is the meaning of approximation? Approximations of functions is nothing but the study of convergence properties in the ambient space namely all continuous functions X to Y. Of course, this will demand that we have a topology on CX1. See X is a topology, Y is a topology and you have to take all continuous functions and one way is that you can take a product topology and restrict it. Still it will not involve the topology of X itself only in the choice of subset you have got this one. What about other subset? What are open sets? There also you want to involve the topology of X in some way. That is what we want to concentrate upon. So, the simple answer provided by taking the product topology because CXY is after all a subset of Y power X. The subspace topology has some decent properties but for example what happens if you have a sequence fn from X to Y which converges to a function from X to Y in the product topology that just means that each coordinate function is fn of X that is point-wise convergence fn of X converges to fx. This is point-wise convergence. We have seen that point-wise convergence of continuous functions need not imply the limit is continuous. So, we will be going out of our ambience namely CXY. So, what we need is in this particular point of view product topology is not quite satisfactory. Before we carry on with this discussion let us fix up some notation so that we can discuss it more carefully. Note that there is a canonical function which we will denote by E representing the evaluation map X cross Y power X to Y take a point X here and take a function from X to Y. You evaluate that function at this point E of XF which affects. Next we can define little more complicated function here. By the way this is just the exponential law as far as sets are considered. I am just taking a set of all functions here. All functions from Z to all functions from X to Y so it is Y power X whole power Z to all functions from X cross Z to Y. So, I am going to define a function here namely take a function from Z to Y power X G product with the identity map of X then composite with evaluation. See the map is from Z to Y power X identity function from X to X itself. So, you get a map X to Y power Z to X cross Y power X then you can composite with E. So, that composition we will denote by psi G. So, that will be automatically a function from X cross Z to Y and for every G inside a map from Z to Y power X. This is the exponential correspondence. It is easy to see that it is a bijection. Conversely whenever you have a function from X cross Z to Y you can restrict it to you know sectional maps namely partial maps. Fix a Z and look at it as a function from X to Y. So, you get for each Z you get a function from X to Y that function you can call it G and come back here. So, if you write the cardinalities of each capital X capital Y and so on in little X little Y so on then this is nothing but Y power X power Z is Y power X Z the cardinality of the product is product of a cardinality. So, this is the law of exponentials here Y power X power Z is Y power X Z. So, the law of exponentials for multiplication of natural numbers is a consequence of this one actually that is a very special case when all the sets involved are finite. So, indeed so I have described this one also let me describe it again given any function from X cross Z to Y fix the second slot Z for each Z let F Z denote the function X going to F of X Z. So, this is the partial function right then I have a function phi from Y power X cross Z that means this function is Y power X cross Y Z X cross Z to Y right to Y power X raise to Z given by the other way around phi of F for each function here you associate F Z F Z F underscore Z is this function for each Z take F of Z F of Z here. So, that is the function that will be a function from Z to Y power X. So, you can just check that this phi is the inverse of this side ok. So, do that so that the rest of the material here will be easy for you to understand all right. So, you go through this that this is purely set theory you might have done it elsewhere, but now right now get familiar with these notations. So, I claim that this is a straight forward check that phi is the inverse of psi at least as a just a set function now. Now, when we have take product topology for example, on take Y as a topology of space Y power X cross Z is a product topology Y power X is a itself is a product topology and then take the product topology on Y power X power Z also to take product topology everywhere starting with any topological space Y the point is that this psi and this phi they may not satisfy this condition namely set of continuous functions within them they may not go to continuous functions there one way or the other ok we cannot ensure that. So, even with that respect in that respect product topology is not satisfied all right. So, we have to look for better answer we are also familiar with a partial answer in the case of when you take real or complex word functions on a topological space X ok. Usually X itself was also compact metric space and then it this were closed and these functions were bounded also. So, this way formed a Banach space Banach algebra on which we have studied many other results also ok. Now, we do not want to restrict to X to be a metric space ok that is all we are going to achieve namely take any topological space there is one which will imitate the supreme topology on this one the topology of uniform convergence. So, it will imitate to a large extent very good imitation and that is called the compact open topology. So, that much motivation let us start today module 51 the earnest study of compact open topology. There are many other topologies on function spaces the study of various topologies itself is a very you know interesting and thriving business in function analysis software. So, some more notation here for any subset K of X and U of Y let us denote this bracket K U to be all functions from X to Y such that F of K is contained inside U. So, this is where we are imposing functions that we take you know some restrictions has been imposed the function should take our Q 1 K inside U ok. Another notation namely this langle-langle bracket like inner product ok. So, I will call this as a bracket after all physicists do call this as just a bracket also bracket K U is all those you know square bracket K U intersect with C X Y condition is the same, but only take continuous function that is all. Now, given X and Y consider family S of all subsets now both X and Y are topological spaces ok. S equal to all K U square bracket K is compact and U is open. So, that is suggestive that is why I have put K here in a subsets of X and subsets of Y have taken U K is compact U is open ok. The symbol represent the same things only now qualification is that K must be compact and U must be open. Take this collection, this collection declare it as a sub base for a topology call that let us denote it by CO call that compact open topology. I can do this one ok on the entire Y power X remember this square bracket you know all functions ok. So, that will be perhaps different from the product topology let us see. So, this is compact open topology and Y power X finally we will come to this subsets C X Y ok. From now onwards on C X Y we should take the induced topology from the compact open topology unless specified otherwise ok there are so many different topologies on C X Y, but right now we are going to take compact open topology. Clear the family if you take round this bracket ok K U that means what only continuous functions are taken here K compact U open topology that will form a sub base for C X Y. If we have base for some larger space intersect it with a sub base for larger space intersect it with a subsets of that that will give you a subsets for the smaller space ok. Note that the collection of all X U now I am specifying little X instead of K ok clearly singleton points are compact. So, what happens to this one we would like to know these are we will give you a sub base for the product topology what is X comma U all functions whose X coordinate is inside U which is if you take pi X as projection map to the X coordinate ok this is nothing but pi X inverse of U. So, those are sub base that we already know. So, therefore you see that at least the definition this compact open topology is some kind of a generalization of the product topology ok thus compact open topology is finer not only generalization in the idea but it is finer because all these sub bases which are the bases for the product topology they are there. Therefore every open subset in a product topology is there that is the meaning of that this topology is finer than the product topology ok. Therefore when you have introduced more open sets convergence becomes stringent for example immediately you can see that convergence with respect to this compact open topology immediately implies convergence with respect to namely the point wise convergence with respect to product topology ok point wise convergence immediate ok. So, what is the meaning of this one X comma U controlled only point whereas, now we are controlling compact subsets or the functions have been controlled over the compact subsets if control over the functions not over the compact subsets ok. So, here is a theorem the CO ok this topology is finer than the product topology on CXY it is house door respectively regular if Y is house door respectively regular ok. So, these two are more or less consequence of the first observation which we have already seen the topology CO is finer than the product topology because it contains the sub base itself contains all the sub base for the product topology as soon as it is finer than a house door topology it will be also house door if Y is house door the product topology house door is old game for us therefore, house door regularity is not that quick. So, let us see why regularity is also true ok. So, assume Y is regular it is a first to show that given F belonging to K comma U these are basic open sets right sub basic open sets there is a closer neighborhood W of F such that W is contained inside K U instead of doing it for all open sets you can just do it for the basic open sets and then you can take intersection ok then that will prove for all open sets of. Now F belongs to K U implies F of K is contained inside F since K is compact F K is compact. So, if you have a compact subset of an open set what happens inside a regular space. So, these are all old game we know that know by regularity there is an open subset V of Y such that this whole compact set is contained inside V contained inside V bar contained inside U ok. Clearly F belongs to this K comma V now because F of K is a constant V but K V bracket is contained K V bar V bar being larger that is going to K U ok. So, now K V bar is nothing but intersection of all X V bar where X runs over K remember X V bar means all functions which take a point X inside V bar ok. If all the point given a function if all the points of K are inside V bar F K is inside V bar and conversely. So, this is the intersection of all these X V bar. Now what are X V bar this is not pi X inverse of V bar in the product topology. So, there is a closed in the product topology therefore it is closed in the finer topology also CO topology also and this is the intersection of closed sets right. So, V X bar is actually closed in the product topology and hence it is closed in CO also. So, this is a closed subset. So, what we have found out we have found out F is inside this open set contained inside this opens inside this closed closure closed subset contained inside the given K. So, that is regularity for the CO topology over. From now on word shall use the notation K U this bracket or the ordinary bracket K V only when K is compact and U is open ok. This is just a lazy way of saying, but sometimes I may forget it. You do not write this kind of notation unless K is compact and U is open alright. So, let us take a break here next time we shall study carefully the exponential correspondence which you have introduced today for only sets. Now we will study it on continuous functions ok on CO with CO topology. Thank you.