 Welcome to module 30 of concept topology part 1. Last time we introduced the study of product sets basically and started just the meaning of putting some topology on the product set. I will recall this theorem which we did last time and then we will go ahead. So this was the theorem on xj which is the product of the family x little j's. There is a unique topology satisfying these two conditions. The projection maps pj's are all containers and given any topological space y a function is continuous from y to xj if and only if pj composite f which are so-called coordinate functions of pf they are all continuous for every shape. So this was proved last time and we observed that this proof gives you two descriptions of this topology. Namely you can just say that the topology that we have got here is the smallest topology which is the which all the pj's are continuous or you can just describe it by the sub-base take any opens of set in tau j take pj inverse of that put all such elements in one single collection that will be a sub-base for this topology so it is tau s means what generated by s. So this is another description. So now let us continue the study of the product spaces to some extent whatever we can do in half an hour that is all. Of course the entire course we will keep coming to study of product spaces again and again. So start with a family of topological spaces and take the cartesian coordinate space which we have defined as set of all functions from the indexing set j into the union of xj's with certain property. Let tau be the smallest topology such that all the projection maps are continuous. So this is the theorem that we had. So I can call this as a definition now. Then tau is called the product topology and xj, tau is called the product space of what of the collection tau j or xis or xj's. Whenever we are dealing with a family of topological spaces by the word product space we shall always mean this topology space unless mentioned otherwise. Why I am telling this one because though xis are given there may be many different ways of putting topology on x. When you say product space you should take this that is a convention now just like when we are taking r and then we say usual topology it is a topology induced by the distance function there. So that is a convention. Another important aspect of the product topology is the following. First recall this definition of convergence of sequences in a topological space. What is that? Here any topological space a sequence xn is converges to x that is what it if and only for every neighborhood u of x there exists a case such that n is bigger than equal to k implies xn is inside u. I am just recalling this definition then I am going to use it here now. So a sequence xn inside product space okay. This will convert to x belonging to xj now I am taking the product topology here and the previous definition which we just saw. If and only if each coordinate sequence see evaluate each element at the jth coordinate then you get a sequence x and j right in xj so that must converge to xj. What is xj? xj is the jth coordinate of x okay. So this is one way of more or less characterizing the topology but that is not what we are going to do we are going to only one way characterization is a little more stringent they do not work in complete generality. So if you take the product topology then it has this property is what we are going to see. The necessity of the condition follows from the continuity of projection maps. Once this sequence is convergent pj of that sequence must be convergent because pjs are continuous pj of this sequence is precisely this sequence and where they should convert to pj of x. So that is continuity of pjs will ensure this if this is a convergent sequence then each x and j is a convergent in xj. Converse is what we have to take care of it that is also one line proof. First of all you must observe that for every u containing x something happens that is what we have to verify but it is enough to do it for sub basic open sets. Once you have done it for sub basic open sets finite intersections of that will be verified by taking that corresponding n0 to be maximum of this n1 and n2 in case we have taken. So that will give you the verification for all basic open sets. Once for basic open sets you have to give you have done to each open set and each point belonging to that open set there will be always a basic open set containing of that. So it will come for all the elements. So this is the elaborate way of telling you but we have all seen all these things the role of basis and sub-basis. So here I am going to use it for the first time. So this is the way economically we can do a lot of work that is the whole idea of base and sub-base after all. So let u be a sub-basic open set in xj such that x belongs to u. Then I must find an n, n0 so set for n bigger than n0 all the accents are inside inside what? Inside this open set. So that is what I have to do. Since it is a basic open set that a basic open set u will look like pj inverse of vj for some open set uj in xj for sum j. For sum j that is all right. Now x belongs to pj inverse of uj is the hypothesis because x is inside u, u is this one. So implies that pj of x is inside uj, pj of x is not xj but now convergence of the sequence x and j will tell you that there is an n0 such that all the accents are in pj inverse of uj because all the x and j's are inside uj for n bigger than n0. But that is same thing as again if you take xj is what pj of xn. So xn belongs to pj inverse of uj and that is u. So u is arbitrary open subset, base sub-basic open set we have verified for every sub-basic open set containing the point x. So this is done okay. So let us go ahead with the number of remarks here. In the language of function spaces the Egoterm tells us that product topology is the same as topology of point-wise convergence. If you do not know this terminology from function spaces you are excused from understanding this okay. When you come to that and your analysis teacher says this point-wise convergence then he will say oh this I know that is the other way around it is actually what is happening. Since we have not studied any function space topology yet you may not be able to make much sense out of this remark at this stage okay. So therefore let this remark hang for a while okay. Don't throw it away it will be useful when you study function analysis or any function space topology and so on. In order to facilitate the discussion of relationships between you know x i and x j where i is a subset of j let us introduce okay an important topological concept here. This is just a stop-gap definition but the concept itself will be useful elsewhere also. So I am taking this opportunity to introduce that one here. Start with any two topological spaces okay by an embedding of x and y we mean a function f from x to y such that f from x tau to f is always from x to f x. First part I am taking this subspace fx the subspace topology is tau prime restricted to fx. So this is the notation for subspace topology remember that. So now instead of y comma tau prime I am taking just fx comma tau prime restricted to fx. So this must be a homeomorphism okay. So let me elaborate on this one namely if f is a continuous injection okay first of all why because the same f is there it is a homeomorphism it must be injection and it must be continuous from x to fx from fx to y there is an inclusion map composite with that that is also f because there are two different definitions this is f restriction right. So I am using the same notation f for both of them. So the original f this one is f to fx here and then fx to y inclusion map. So if this is continuous that will be also continuous. So f must be continuous first of all and injective map right. So this must be necessary it may not be an open map may not be a closed map may not be surjective all these things are true when you come down here because it's a homeomorphism if you are here only continuity and injectivity is still there right. So this is what you have chosen. So it may not be an open map or a closed map or may not be a surjective however if you take the codomain of f to be equal to the image of fx with the subspace topology then it satisfies all these conditions that is the definition of an embedding. The simplest example of an embedding is the inclusion map itself from a subspace into the original space. If we identify x with fx via the map f namely x goes to fx then we can think of x as a subspace of y because fx is a subspace of y now fx is replaced by x and so on. Thus an embedding is a direct generalization of the concept of a subspace. You can almost confuse it with a the subspace but don't confuse it because you want to have some separate identity for x. So whenever it is convenient you can identify this x with fx that is the whole idea here because they are after all homeomorphisms homeomorphic to each other. Let me continue with some more remarks given a subset i of j this was my aim of the indexing set. So I am taking a subset of the index inside we can also define xi what is xi product of all xj is j is j is inside i. This is some independent product topology product set and so on. So what is the relation between the xi and xj then there is the obvious projection maps from the largest this one xj to xi what does it do just ignore all the other coordinates. Suppose this is just a only two elements that x1 and x2. So in x1 cross x2 to x1 or x1 cross x2 to x2 you had the projection maps. Similarly what do you do you just drop out one of the coordinates or several of the coordinates retain the rest of them as they are. So this is what is pi is doing here pi of xj retains all the i coordinates inside capital i ignores all of them inside j minus i pi of x equal to xi this is all i because there are no other coordinates here all those coordinates which are here namely i all those j's inside i they are there okay. Once again from the definition of the product topology this topology this product is nothing but the induced topology on xj with respect to the family pj of coordinate projections okay you take the whole space xj but now you take the subspace of all so that will give you this xi it follows from the above lemma that all the pi's are continuous okay no matter which what is capital i okay this i is fixed now this pi is continuous why because the i th coordinate all the j th coordinate of this one is nothing but the old pj so all the j th coordinate j is j inside capital i they are continuous so pi is continuous we shall refer to these maps also as coordinate projections it is like x1 x2 x3 going to x1 x2 okay so we have been using all these things in the case of rn and cn and so on so only those practices we are putting in a general setup that's all okay so we shall refer to these maps also as coordinate projections if k is j minus i that means the left out indices here and let us assume that both i and k are non-empty this is a standing assumption of course it is easily seen that the map pi comma pk from x capital j to xi cross xk this is a homeomorphism this is a one-one map is clear on two is also clear so bijection is obvious here set theoretically why this is continuous look at any coordinate projections here okay they are all the old coordinate one of the pis and pk pk k belong to k so they are continuous if you go from here to this way xj the same coordinate functions will give you that that's also continuous okay so this is a homeomorphism that preserves the projection maps the i th projection here the i th projection here the same thing it is more or less like the identity map okay so this is the way you identify r2 cross r2 with r4 right the beauty of this notation is now it is independent of or independent of the order on the indexing sets therefore i can write xi cross xj or xj cross xi it will be the same xj up to homeomorphism is what we want to say okay in particular for any point y in xk we get fix one point we get an embedding of xi in xj defined by x going to x comma y this x is varying over xi y is fixed which is inside x xk so the whole thing will be in xj so this is just like the example wherein you take x going to x comma 0 or x going to x comma 1 or x going to x comma 1500 or the other way around you can take y going to 0 comma y right so all those are embeddings so that is why the word embedding was defined just before okay these are all embedding so the partial products can always be thought of as subspace is in various ways you may say horizontally and vertically and so on so these these coordinates can be thought of as horizontal and these are vertical and if there are only two of them all right as a further special case of this assume that xj has at least two elements then given any i belonging to j take the singleton i as capital i okay then you have to fix an element y in the rest of them xk then what it will give you it will give you a parallel various embedding parallel embeddings of xi x smaller i right this i inside xj instead of capital i to do a smaller i inside so this is special case which I have already explained with examples alright there is some more interesting things we have to tell you by taking finite intersection of members of a subbase we of course get a base for the topology how does a typical member of this base look like in the case of finite products we know that these are nothing but proto types of rectangular boxes interval cross interval cross interval and so on right right especially when each xj is r however observe that in the case of infinite products at most only finitely many of the coordinates will be restricted what is pj inverse of let us say what is p1 inverse of u1 where u1 is an open subset of x1 it will be just u1 first coordinate cross the entire of xj minus 1 in our notation here okay take k equal to j minus 1 right one particular case here j minus i you have taken i is single point to fix one point there then you have got embeddings now I am looking at inverse mesh so the second coordinate onwards it will be completely free so that it will look like u1 cross the entire of the space built upon j except 1 suppose similarly take u2 u1 now x1 will be free again only u2 will be restricted u2 coordinate will be restricted p2 inverse of u2 will have only u2 restricted all other coordinates are correct therefore the intersection will have only u1 cross u2 and then the rest of the coordinates are all free no conditions there so this is not a box this could be especially when r cross r cross r you are taking all of them are r the other coordinates are freely varying infinitely you know and unbounded right so so this is what you have to be careful here except for finitely many it is fine just like the box but in the case of infinite product at most only finitely many of the coordinates will be restricted in other words if a is a non-empty member of b then pj of a just assume that a is non-empty that is all pj of a okay equal to xj for all but finitely many j belong to j what I am assuming a a is a sub base basic open set okay I am making the comment on basic open sets here in the view of the remark 3 this can be put as follows this I already told you every member of b looks like that means what homomorphic tau u cross xk where k contain inside j is such that i which is j minus k is finite so this is co-finite and u is an open subset of xi because i is finite u itself will look like u1 cross u2 cross uk or again countable union basic open sets only u1 cross u2 cross uk but every member of b looks like that u1 cross u2 cross uk cross the whole of x so coming back to the finite case suppose j is finite the sub base s gives rise to a base b which coincides with the base that we have taken for box topology namely take all u1 cross u2 cross uk okay for instance when j is just two element set you have to note that p1 inverse of u1 intersection p2 inverse of u2 is just u1 cross u2 just intersection which this will says the first coordinate is restricted inside u1 second coordinate is restricted to u2 therefore it is u1 cross u2 however as soon as the indexing set j is infinite the two topologies may be different in any case you may check that the box topology is finer than the product topology this is a handicap in getting continuous functions into the product of xj's infinite products with box topology the if part of the statement two of theorem 2.70 does not hold as seen by the following example see if this where you are given given this one the product topology then we have the criterion number two there a function is continuous if and only if all the coordinate functions pj composite f those are continuous okay so I will show you now an example very easy example which will violate this one in the case of box topology okay so put xn equal to r with the usual topology for all n so I am taking a countable family here okay indexed by n take fn from r to xn to be the function fn of t is n times t okay I think I wanted 1 by n times t here let us see which one will work n times t 1 by n times t is what I wanted then each fn is continuous okay however the function f from r to xn the product space given by ft of n equal to fn of t is not continuous is the claim for instance the set u which is a product of infinite copies of the interval minus 1 plus 1 okay look at the interval minus 1 plus 1 is a neighborhood of f0 in the box topology what I am taking infinite product of minus 1 minus 1 plus 1 is a neighborhood of 0 in every every one so it will be product of it will be neighborhood of the product box topology 000 all coordinates 0 and that is f0 because all the coordinates f of 0 is is n times 0 so it is 0 but for no open interval i around 0 we have f of y is contained inside u can you see that so for that what I needed here was 1 by nt okay t divided by n so what I want to say suppose you suppose you choose a interval to be minus 50 to plus 50 okay then if I take n you have to choose some some n so if you choose say n equal to 500 then that point will come inside here but here I only have an interval minus we have minus 100 to plus 100 or minus 50 okay so f of i is a bounded interval we take some interval okay so I should I should say that given an open interval here okay f of something must be inside u this u is that is a continuity so that is violated this is not true okay okay so an interesting property of the projection maps is that each f pj from xj to x little j is an open mapping it is enough to show that pj of u is open for each basic open set u inside xj okay so here you have to be careful showing only for sub basic is not enough here because you are taking intersections and image of intersections may not be intersection of you have sub basic unions will be okay so sub basic open set if you show that you are done okay you be careful to pj of u where u is a basic open set I should show that it is open inside xj for every j okay so let u equal to intersection of pki inverse of uk i i range from 1 to n this is a typical way as sub basic open set will look like finite intersections of of basic sub basic open sets so this will be basic one set each uk i is non-empty open subset of k xki now what is pj of u pj of u depends upon what j is if j is one of the pki one of the k i is here then pj of u is precisely uk i you remember in the standard notation this will be nothing but uk 1 cross uk 2 cross uk whatever n cross the rest of all other entire spaces xj therefore its jth projection would be just uk i where j is equal to k i i equal to 1 2 3 up to n if j is not equal to any of these first k i coordinates okay n coordinates if one of the other j then it will be just the whole of xj okay in any case in any case they are all open subsets on the right hand side therefore pj is an open map all right all coordinate projections are subjective maps because i am assuming that all the xj's are non-empty that is important here otherwise the product will be empty even though some of the xj's may not be empty even if one of them is empty the product will be empty so you are assuming that then each pj is a subjective map also we have seen that a subjective open continuous map is a quotient map in other words each coordinate space here xj which you begin they are all quotients of the product space okay so this any surjective continuous function then you can give the product you can give the base the quotient topology if you do that first take this xj's take the product and give quotient space you will get back xj the original topology space that is the meaning of this how did you do this just by showing that each projection map is open map okay the same thing applies instead of one single j you can do it for a bunch of j's xj to xi which we have discussed earlier because that is also surjective open map so I want to discuss a few more things here just as in the infinite in the finite case you have x cross x let us consider the case for example all the xj's are x now one single space okay then we have introduced this notation remember that x power i or x power j whatever now I am taking the indexing set as i so its x power i is what is the product over i but all the x i's are equal to x the same topological space same set same topological space okay just like in the infinite case we can define the diagonal set to be all those x i's such that all of them are equal to each other equal to one single x belong to x okay that is the definition of diagonal the map eta from x to this x power i given by eta i of x see when you want to define a map into x power i what you have to do you have to just mention all the coordinates all the all the coordinate functions here eta i of x which you can write eta x i okay that must be x for all i that is the definition of this map eta so this will give you an identification of x with a diagonal by which I mean a homeomorphism or here in the case it's embedding of x inside x power i okay with respect to this embedding you can identify x as if it is the diagonal of okay so I am just giving you more elaborate explanation here eta is injective and it's on to delta x it is continuous because eta i is identity for all i the coordinate projection eta composite p i is eta i this is what I have defined they are all eta i of x is x so they are all continuous so eta is continuous also for any open set u in x okay pi 1 inverse of u is open in x power i okay that is by definition this is a sub basic open set and we check that eta u okay you start with an open set in x but now you can think of x power i to x there is a projection map also now you take the inverse image of that that is an open set but that open set intersection with delta is precisely equal to eta u eta u is delta x intersection minus hence eta u is an open set inside delta x which means eta is an open mapping into delta x so so x to delta x eta is a homeomorphism okay so let us stop here let me just give you one exercise here namely take subsets aj inside xj especially this this exercise is for infinite products okay look at the product set aj if this is like a box the jth coordinate is aj a 1 cross a 2 cross and so but this infinite product i am taking its closure is the same thing as closure of aj's then take the product you know whenever infinite processes are there you have to be careful with the closures right so you better don't believe this one but try to disprove it or try to prove it only after proving it you believe it all right so this is the exercise any doubts okay let us stop here