 So, welcome to chapter 2 of the course. I have named it as creating new spaces. We shall discuss some standard procedures of constructing new topological spaces out of the given ones as well as constructing totally new ones also. Of course, we shall not just construct them, we shall also put them in proper prospective, study them later more. That is all this chapter is about. We begin with two fundamental concepts. These concepts apart from helping the topological study elsewhere, give immediate methods of constructing all topological spaces in a general fashion. Some simple examples that we have considered in the previous chapter in an ad hoc fashion will now become motivating examples for some systematic studies. I will point out to them whenever we come across those things. So, welcome to module 22, Basis and Subbasis. You may as well take one of them today. We will see, but I am already laying foundation for the second one also. So, today maybe we will just discuss basis because lack of time. So, the first observation is that start with any set and a family of topologies on X. Take the intersection of all these families. Which means what? All members of Px, may be subsets of X, which are in every member of this family, tau j's. So, there is an intersection. That family will become a topology on X. So, this is what I meant by saying that creating new topologies out of old topologies for example, this comes into that category. Tau j's are a families of topologies. Now, I am taking the intersection of them. Perhaps this one is a new one. It is not always. If you take say tau 1 contains a tau 2, then this intersection will be just tau 1. So, it is not always the new one. That is why it is a general method. So, this size of topology after all is very important concept for us. The proof is very easy. Let me just see what we have to do. Empty set and the whole set is inside psi. Why? Because it is there inside every topology. If you take u 1 and u 2 inside psi, they will be there in each tau j. So, their intersection will be in each tau j. So, intersection is here. Similarly, if you have a family of open sets as u alpha, members of psi, that means what? Each tau j contains all of u alpha. Therefore, the union of u alpha is inside tau j for every j. So, that means that union of y is a member of psi. So, that completes the proof of this topology. This is just like in a vector space. If you have a family of vector subspaces, then the intersection is a vector space. Similarly, if you have a family of subgroups of a group, then the intersection is a subgroup. However, in either of the cases, the union utterly fails to be a subgroup here and some space there, vector some space there. The same thing is happening here. If you take two topologies, the union may not be a topology. Of course, if one is contained in the other, union is the other one. So, that is a very special case. In general, union may not be a topology. But we do not want to give up just like that. So, we will work harder to get topologies out of these things also. So, what we will do? So, here is another general method, which is actually an application of our previous theorem here. So, this theorem is a foundational. It is like it keeps helping us again and again. So, start with any family of subsets. Instead of taking topologies and so on, just any family of subsets of x. In the collection of all topologies on x, which contain this s, there is a unique topology tau s, which is the smallest. First of all, the discrete topology, the entire px contains this one. So, this family of all topologies is non-empty. Logically, it is not necessary, but we can fix it up that it is non-empty. Now, I am taking intersection of all these topologies, which contain s. This theorem says that such intersection is a topology. And since all of them contain s, this will contain s also. So, one part is fine. Namely, we have got a topology which contains s now. But this is the smallest one. Why? Because it is contained in the each of them because this being the intersection of all of them. So, that is all. So, one step application of this one gives you this result. Namely, given any set, there is a smallest topology. That is a unique smallest topology. It is the intersection of all topologies containing that, unique one. Why unique one? Because there are two of them, the intersection, intersection must be equal to both of them. So, we can make a definition. What is that? The topology tau s, namely the smallest topology containing s, is called the topology generated by s. If s itself is a topology, what happens? Tau s will be equal to s because s is already a topology. So, tau s and everything else contains it. So, intersection will be s. So, if s is already a topology, no problem. But if s is not a topology, then this is. So, this is again similar to what we do in linear algebra. You have a vector space. You have a set of vectors there. How do you create a vector subspace which will contain all these points, all these vectors? So, that is the vector space generated by this. So, same word we have taken here. The same terminology. So, this one step consequence of our first theorem is going to be extremely useful method of creating new topologies because I may start with any subset of the power, any subset of x, any collection from the power set that will give you a unique topology, the smallest one that contains that. Of course, what may happen is two different sets, two different subsets of x may give you the subsets of p x I am taking, collection of subsets of x may give you same topology that is possible. Taoist may be Taoist brand. So, here is a simple exercise. Take x equal to z or any countable infinite set and take s equal to x minus one single point, n belong to z. One point is missing from each subset here. Look at that is the collection. What is the Tao s corresponding to this s? What is the topology generated by this family on the set of integers? It turns out to be something familiar to you. Can you see what is happening? Okay, I do not mind. This is precisely my point, namely we need to work harder to identify the topology generated by s s s. Like in the case of vector spaces, we give a collection of sets, collection of elements. What are all the elements in the vector space generated by that? They will be finite linear combinations of elements from this set. So, that is a description. So, we would like to have similar description here. So, let us try that. So, in order to be able to identify Taoist, identify members of Taoist, it is necessary to establish or invent a method to describe the elements of the topology generated by s in terms of the sets inside s. Okay, so here again, you see the motivation comes from matrix spaces. So, that is why certain things there we have done just to add up fashion perhaps, whatever motivated them there, but now they will motivate to enter new constructions here. Recall that a metric space x d, okay, open walls where basic objects in define the topology for Tao. It is Tao d, right? Open walls, right? That is a good example that will guide you, okay. So, how was the Tao d defined then arbitrary union of open walls, all right? So, based on that, we will do something now, namely we will define another term here, base for a topology. So, that was the title of the today's topic after all. A subset of px that means the collection of subsets of x is called a base for a topology on x if the following conditions are satisfied. First condition is that if you take union of members of b, that should cover the whole of x. Second condition is that given two members of b and a point inside the intersection, you must have a third member in b, okay, such that x is inside b3 contained in b1 intersection b2. When I say third member, it is I am not saying that they are all distinct, no. But the point is neither b1 intersection b2 itself may not be there, that is the point, okay. If b1 is contained in b2, intersection will be just b2, it is just b1 and so on. So, those are easy cases. So, you may have two different subsets, they may not intersect. So, then also I do not have any moderation, but if they intersect for every point inside b intersection b2, I must have a member of b such that that member is contained in the intersection and contains the point, okay. So, this is pay attention to this one, this may take little more time. But look at the set of open balls in a matrix place. This is precisely what we had proved there. If you take intersection of two open balls, it need not be an open ball. But every point inside it contains an open ball, every point is around that point that is a open ball contained in the intersection, okay. So, that is why we have proved this condition. So, that is motivation is from open balls inside a matrix place. Once b1 and b2 are satisfied, look at the topology generated by b now, tau b, that is what the least topology which contains b. Then we say that this b is a base for tau, okay. Now, here is a theorem. First of all, to characterize what kind of collections will be a base for some topology, okay. So, start with a collection b and let tau be any topology. Will b be a base for tau? This is the question we want to address now. So, this theorem says that the following three conditions are equivalent. The first condition is b is a base for tau, that is our motivation. The second condition is that first of all b is contained inside tau which is very obvious if b is going to be a base for tau and every member of tau can be expressed as a union of members of b. So, this is precisely the result that we proved for open balls. In fact, sorry, the definition that we had for topology from the open balls, okay. For tau b was precisely defined this way, right. So, that condition comes here. Third condition is of course, b contained inside tau is common given any x inside u where u is inside tau. tau is what? tau is the given topology. So, u is an open subset in this topology. I must find a b inside b such that x belongs to b contained inside u. These three conditions are equivalent in the statement of this theorem. Let us go through this proof. One implies two, okay. If b is a topology, b is a base for tau, okay, by the definition tau has to contain inside, contain b. So, that is for the second part is, second part is what? That every member of tau can be expressed as union of members of b. Okay. So, take tau prime with the family of all subsets of x which can be expressed as a union of members of b, okay. Let this be the family. Then I want to show that this tau prime itself is a topology. The proof is exactly similar to what we did for the open balls. There we had to work harder there. Here we have made whatever that work into a definition here, namely the condition, okay. So, let us see tau prime be the family of all subsets of x which can be expressed as union of members of b, okay. So, why tau prime is a topology? Again, the axiom Au, Au is what arbitrary union is easy because union of union, union over the families of union is again another union. To see the finite intersection, suppose you verify it for two of them, then again the intersection of unions will be unions of intersections. Therefore, it is enough to verify it for only two of them anyway. But I have written complete proof here, U equal to union of Vi, V equal to union of VJs, okay. Let them be members of tau prime, that is by definition. What are UIVJs? They are inside b. Now, we look at intersection, it is Uij, okay. Intersection taken over all i and j, Ui intersection Vj. Now, suppose x is inside the intersection, U intersection V left hand side, then it must be inside one of these, okay. This means Ui intersection Uj contains x for some ij because the union is a union. So, there exists a b belong to b, this is the part of the definition for b is a base. So, that x belongs to b contained inside U intersection V, Uj, Ui intersection Vj. It is a part of the definition for b is a base. So, that is what I am using here. So, this implies that x is inside b intersection U intersection V, okay, because this is the union, U intersection V is a union of this. Therefore, U intersection V is a union of members of b, every point is inside b. So, it will be union. So, therefore, it is an element of tau prime. Once this is a family, this family is a topology, the s tau b, which is the smallest one will be contained inside tau prime. So, that is every member of tau can be expressed as union of members of b. Once it is contained inside tau prime, what is tau prime? It is expressed, every member is expressed as union of members of b, okay. So, this proves one implies two. Now, two implies three. That is very straightforward because one, every member expressed as union of members of b, even any x belong to U, given b is contained inside tau is already there. Even x inside U contains a tau, there is a b such that because U itself is union of the members of b i, okay. So, this part is very easy. Once again, three implies one will be the hardest at all, if I tau, even that is not difficult. So, let us go through that. In addition to proving conditions b1 and b2, we must also prove that tau is the smallest topology containing b. So, this is the meaning of b is a base for tau. First of all, I should condition b1 and b2 should be this way. Then the smallest topology containing b must be equal to tau. So, these two things I have to show, actually three things I have to show, okay. Since every member of x belongs to some member of b, okay, b1 holds. So, this part was starting with what we had, three implies, three implies one is what I have to show, right. Given any x belong to U inside tau, there exists a b belong to b, tau, members of tau cover all of x, right. So, x belong to U, then there is a b belong to U which contains that one. So, union of all members of b must be cover the whole of x. That part comes from here, okay. So, that is first part here, okay. That also I have to observe. So, b1 holds. Secondly, b is contained inside tau, that is given, right. In three, that is given, okay. b is contained inside tau. Given b1 and b2 belong to b, intersection will be member of tau because each of them is a member of tau. You take U equal to b1 intersection b2 and apply this condition 3 for U, b1 intersection b2. You get a b, such that x belong to be contained in the b1 intersection, b2 now in the intersection, okay. So, b2 also holds. Finally, I have to show that tau b is equal to the given tau here, okay. b is contained inside tau. Therefore, tau b is contained inside tau because tau b is the smallest topology. I have to show that tau is contained inside tau b, okay. Only that one remains now here. Okay. So, we want to show that tau prime, any topology on x such that b is contained inside tau prime, okay. Then we want to show that this tau is contained inside tau prime. That will show that tau prime, tau is the smallest topology containing b, that is tau b, tau, right. So, given any element in tau, for each x inside U, we have some bx belonging to b, such that bx is contained inside U. This is given to us. Therefore, U is the union of bx, x belong to U, right. Note that each bx belongs to b but they are all contained, b is contained inside tau prime. Therefore, and tau prime is a topology. Therefore, the union of all bx must be in tau prime. So, starting with U which in element of tau, I have shown that it is in tau prime. That is all, okay. So, tau is the smallest topology containing b, that means tau b is equal to tau. So, equivalence of 1, 2, 3 is established by that. So, that is what I am telling here which I have told already. If x d is a metric space and b is the collection of all open balls in it, clearly this b satisfies b1 and we have proved that it satisfies b2 also in that theorem, chapter 1. Therefore, it is a base for a topology and that topology is nothing but tau d as much we have seen anyway. In particular, the collection of all open intervals in R is a base for the usual topology. I just want to recall all these things. They were the basic things, they were the motivating examples for us, okay. For the future, okay, the purpose is future. Right now, we do not need it because there is this b1 occurring. So, we can make that as a separate definition, namely any collection U of subsets of x satisfying the condition b1 is called a cover for x, okay. Union of members of U, if it is equal to whole of x, then we say U is a cover for x. This we will use later on, okay. So, what we have seen above, okay, we cannot expect any arbitrary cover of b to generate a topology. We need condition b2 also, right. This b2 may not be satisfied by the given b. However, there is one type of families which satisfy this property, okay. It is stronger than b2. That is why I want to pay, I want to call your attention to this one, all right. We will take up that one next time and that will lead to the concept of sub-basis. So, thank you. So, we will meet next time.