 So, let us recall that in the previous lecture given topological spaces x and y with topologies let us say tau x and tau y we define a topology on x cross y. So, how do we do this? We define we first defined a collection B is a subset of power set of x cross y and it is defined as follows is a set of u cross v says that u is open in x in the topology in x and v is open the topology in y. And we check that B satisfies the two conditions required to generate a topology namely these were a when we take the union overall subsets w in b we should get the entire space in this case that is x cross y and the second is given w 1 and w 2 in b and a point x in the intersection then there exists a w in b such that x is in w and w is contained in w 1 intersection. So, we check that B satisfies these two conditions and using these we define the topology on x cross y with b as basis and let us observe that the same idea can be used to put a topology first on a final product i equal to 1 to n x i's. So, here the x i's are topological spaces with topologies tau x i's or just tau i's. So, what we do we define B contained in the power set of these product as B is equal to a product of u i's where each u i is in tau x i and then I will leave this as an exercise and it can be done in the same way check that. So, we have already done this when n is equal to 2 and the same proof will show that B satisfies the two conditions required to generate this product. So, this topology that is the topology generated by B is called the product topology. So, right now we took a finite product of topological spaces we can ask what happens in the case of a infinite product. So, let us consider. So, next let us consider the case i B set possibly infinite and assume that for each i in i we are given topological spaces for each i in i we are given a topological space x i right. So, how can we put topology on the set product i in i x i's. So, now here there are two possible candidates for topologies. So, let us see. So, we can define B in two ways. So, in this case we can define the collection B in two ways. So, first is we define B to be is the most naive one i in i where each u i is in tau x i this is the first candidate. So, one can check that it is easy to check again. So, let us call this B 1 this collection B 1 satisfies the two conditions required to generate topology and. So, this defines the topology let us say tau 1 on this product. So, this topology is called the box topology the other candidate for B is as follows let us call this B 2 right. So, this is product of i in i u i such that what we want is we put the extra condition that u i is equal to x i for all, but finally many i's. So, let me just write this. So, u i belongs to tau x i and when we look at this is condition this is the first condition let us call this A and the second condition we want is when we look at the collection of i in i such that u i is not equal to x i recall that each u i is contained in x i right. So, we look at the collection of those indices for which u i is a proper subset of x i and what we want is should be finite. So, this collection this collection over here that should be finite which means that except for finitely many i's indices all the u i's are equal to x i. So, this topology. So, once again check that B 2 satisfies the two conditions to define the topology tau 2 and this topology is called the product topology on this product of x i's ok. So, just some remarks. So, the following remarks. So, for us the box topology is not useful and we will see reasons for this very soon and throughout this course when we on this product of topological space is when i is infinite set we shall always be considering the topology tau 2 and tau 2. So, when we say the product topology on an infinite product of topological space x i we shall always mention this we shall always be referring to the second topology defined using the basis B 2 ok. So, let us make some remarks. So, the following two remarks are obvious one is when the index set is finite. So, that is if we take a set i says that cardinality of i is finite right. So, then it is clear that the box topology is equal to the product topology. In fact, B 1 is equal to B 2 if this index set is finite ok. So, that is a easy check when cardinality of i is infinite we have B 2 is contained in B 1 and B 1 we know is contained in tau 1. So, recall that we have to dilemma that if x is a topological space and tau 1 and tau 2 are two topologies on x with basis B 1 and B 2 and if B 1 is contained in tau 2 then we get that tau 1 is contained right. So, using this lemma in our situation we have B 2 is contained in tau 1. So, this will imply that tau 2 is contained in tau. So, this is the product topology and this is the box topology. In fact, tau 1 is much larger ok. So, as we had as I had mentioned before we will almost never use the box topology in this course we will always use the product topology. So, so far. So, since now that we have introduced box topology I am sorry not box topology product topology and subspace topology we have several examples we can construct several examples of topological spaces. So, let us see some of the most important examples which we will encounter in this course. So, the first is we have the standard topology on the real line we have the standard topology on R 2 R n ok. So, on this on R n let me make a remark on R n we have the standard topology let us call this S and the product topology because we can write R we can think of R n as the product of R n times right and each of these R carries the standard topology and where each factor R carries the standard topology right. So, claim. So, let us call the product topology tau claim S is equal to tau. So, the standard topology on R n is equal to the product topology on R n and an easy way to prove this prove this is to show that let B be a basis for S be a basis not let B 1 we had already defined a basis for that B 1 be the basis for S using the sets I do not remember the notation now may be S epsilon x these epsilon squares around point of x and let B 2 be the basis for the product topology for tau which we B 2 is the basis which we had used to define tau right. So, then show that B 1 is equal to 2 right. So, this will automatically imply that S is equal to 2. So, ok. So, in other words this means that on R n we have put 2 topologies the first is the standard topology and the other is the product topology and both these topologies agree. So, which is a nice thing. So, second we can take S 1 this is the set of those this is the circle the unit circle in R 2 in R 2 such that x square plus y square is equal to 1 with the subspace topology. So, this is the unit circle ok and similarly we can define S n these are the unit spheres this is x naught up to x n in R n plus 1 such that summation i equal to 0 to n x i square is equal to 1 again with the subspace topology. So, here the subspace topology is from R 2 obviously, with the subspace topology from R n plus 1 ok. The fourth example is the set of n cross n matrices over R this is the set of n cross n matrices with real coefficients right. So, this set is in bijection with R n square right and R n square carries the product topology the product topology. So, using the topology on R n square since M n r is in bijection with R n square we can transfer the topology from R n square to M n r. So, in other words we can take right we can take a map phi from M n r to R n square and for every open set over here. So, this is a bijection. So, every open set over here we can take phi inverse u ok. So, in other words tau we define tau to be phi inverse u where u is open in R n square in the standard topology right. So, then tau defines a topology on M n r ok. So, in fact, we can do this. So, if x is a topological space we can do this generally is a topological space and phi from y to x is a bijection then we can define that topology on y using phi by tau let us say y comma phi this topology will depend a priori it will depend on phi this defined as phi inverse u where u is open. So, check that tau y comma phi is a topology ok phi. Let us look at this set G l n r this is the set of those matrices A in M n r. So, is that determinant of A is not equal to 0 and we give this the subspace topology from M n r. So, M n r we have identified with R n square and using that we give a topology to M n r and G l n r is a subset of M n r. So, we can take the subspace topology from M n r and put to G l n r. So, this makes G l n r a topological space and similarly we have we can take various sets of M n r and put this subspace topology on all of these we have the orthogonal groups. So, A transpose A should be identity then we have the special orthogonal groups determinant of A is equal to 1 ok. So, next we can put a topology on complex numbers as follows. So, in the same way that we put the topology on a topology on tau on R sorry. So, we take tau is the collection of sets u in C such that for all which have the property that which satisfy the following condition for all x in u there exists epsilon which depends on x such that let me use a notation z for complex numbers which depends on z such that this open ball of radius epsilon around z which is defined to be those complex numbers y such that absolute value of y minus the modulus of y minus z is less than epsilon. So, we want for every z in u there should be an epsilon positive such that this ball is containing u right. So, in other words this complex numbers and this some u. So, this we say that it is open in this topology if for any point there exists a small ball a small disc around z which is completely contained inside u right. Alternatively we could have done the following there is a bijection phi from C ok let us say R 2 2 right which is x comma y maps to x plus i y and we can take the standard topology on R 2 and since this is a bijection and use it to define a topology on C. So, this is tau is equal to phi of u where u is open in R 2 in the standard topology. So, let me call this tau 1 and let me call this tau 2 right and easy exercise show that both these topologies are going to be the same ok. So, we have put a topology on complex numbers and once again. So, we can put. So, using the topology on C a topology on m and C which is the set of n cross n matrices with complex coefficients. So, how do we do this it is in the same way. So, we identify m and C is in bijection with C n square. So, exactly as in the case of R what we can do is we can take a matrix a 1 1 a 1 n a 2 1 a 2 n and so on and we can send this to first we can write the let us say rho vector a 1 1 up to a 1 n then we can write the next row a 2 1 up to a 2 n as goes on and finally, we have the last row a n 1 to a n n yeah. So, we got done the same thing for real also for m and R also. So, let me say m and R to R n square we have the same bijection and then we can pull back. So, using this bijection. So, C n square has the product topology right where each copy of C has a standard topology. So, by standard topology I mean the topology defined over here as a standard topology and then and using this bijection we can put a topology on m and C. So, just as we considered subsets of m and R we can consider the following subsets of m and C we can take G l and C the set of those a in m and C says that determinant of a is not equal to 0 with the subspace topology from m and C then we can take S l and C the same. So, here determinant of a is equal to 1 then we have the unitary groups a in m and C says that a star a is identity and finally, we have the special unit groups and determinant of a. So, we have constructed various examples of topological spaces and for all the above examples we can study their topological properties. So, the main topological properties we will consider in this course are compactness well first connectedness path connectedness these will be defined later on in this course. So, and now that we have constructed all these examples that sort of brings the first part of this course to an end what we have basically seen is the definition of topology and how to construct examples of topological spaces new topological spaces from ones which we already know. So, maybe we can trace a new topological spaces from old ones and in order to when we study in order to study topological properties of these spaces it is often good if we can relate them is just in the same way that when we study groups in order to study group theoretic properties of groups it is very it can be very convenient if we have group homophisms. So, we can study homomorphisms from one group into other groups and that helps us to say some things about the properties of our original group. So, in the same way we will next we shall see introduce the notion of continuous maps between topological spaces. So, we will end this lecture here.