 So, let us continue with what we were doing. So, this is lecture 3. So, we defined so recall that tau is a subset of the power set of R 2 consisting of subsets u which satisfy this property star and recall what property star was if x is in u then there is epsilon positive such that S epsilon ok. So, instead of x I should write a comma b is point in u is completely contained. So, let us check that tau satisfies the conditions defining topology. So, the first condition is phi and R 2 should be in tau. So, as in the example of the standard topology on R. So, phi is in tau is true is vacuously true because there are no points in the empty set and similarly R 2 is in tau is also clear because if we take any point in R 2 we can always take a square of side length 2 for any a comma b in R 2 we can take S 1 a comma b this is obviously contained. So, obviously the first condition is satisfied. So, let us look at the second condition the second condition requires that if we take finitely many elements in tau then the intersection in the intersection u i is in tau. So, to check this. So, once again so let we take a point a comma b which is in this intersection and the same proof as in R since a comma b is in u i and u i satisfies star there is epsilon i positive such that this S epsilon i of a comma b is contained u i right. So, we have our u i may be some set like this and we can take a point a comma b here and there is some epsilon i this S epsilon i this is u i right. So, we take epsilon to be equal to the minimum of epsilon 1 epsilon 2 up to epsilon n and clearly S epsilon a comma b is contained in S epsilon i a comma b right and. So, therefore, S epsilon a comma b is contained S epsilon i a comma b is contained in u i for all i. So, this shows that S epsilon a comma b is contained in the intersection right. So, thus the intersection satisfies this property star and therefore, thus the intersection u i is in tau. So, therefore, tau satisfies condition 2 and let us quickly check that it also satisfies condition 3. So, condition 3 was given a set i for each i in i an element u i in tau and then we need to check that u i is also in tau. So, once again we let a comma b we need to check that the union satisfies this property star. So, let a comma b be an element in the union which implies that a comma b is in u j for some j in i which implies that there is epsilon positive such that the square of silent epsilon around a comma b is contained in u j as u j satisfies is in tau and. So, satisfies this property star and which implies that S epsilon a comma b is contained in u j which in turn is contained in this union. So, thus the union satisfies and so is in tau. So, this shows that tau also satisfies the third condition to define the topology. So, therefore, tau defines a topology and our sixth example which I will leave it as an exercise is the standard topology. So, this is very similar and let me just give some hints how to do how to define this. So, let us take a vector x let us write a vector in R n as x. So, this is a n double and we define once again this subset S epsilon of x to be those y in R n such that the absolute value of x i minus y i is less than epsilon ok. And once again we define this property star in the same way. So, let u contained in R n be a subset we say u satisfies property star if. So, for every this is this property vector x in u there exists epsilon positive which once again epsilon may depend on x such that this subset S epsilon x is contained. And we let tau be the collection of subsets u contained R n which satisfy and I will leave it as an exercise that show that tau defines a topology ok. And so, this is one exercise and I want to give another exercise. So, before I write the exercise let me just. So, in the example of R 2 or even R n. So, in the example of we could we can ok first define. So, let us write like this. So, define for x in R n epsilon positive define the subset B epsilon of x as this is those y in R n such that summation i could 1 to n y i minus x i square is less than epsilon. So, this is open ball of radius epsilon around and we can define. So, define a property star prime let u be a subset of R n we say that u satisfies property star prime if. So, this is this property for every x in u there is epsilon positive which may once again depend on this point x such that this B epsilon of x is continue ok. And we let tau prime be the collection of subsets u such that u satisfies ok. So, then there are two exercises here one show that tau prime defines a topology and two show that tau. So, this is tau in example 6. So, example 6 is this one. So, the standard topology on that we define is equal to. So, what is what do I mean by the second exercise? So, both tau and tau prime are subsets of R n show that these subsets are the same I am sorry are subsets of the power set of R n. So, show that tau is equal to both the subsets are the same. So, this the second exercise let me make a remark exercise 2 above shows that the same topology can be defined in different in different ways. So, we have seen quite a few examples of topological spaces now. And we want to give more examples. So, we want to explain now explain some tools or some methods which we can use to define topologies on topological spaces. So, for that so next we want to explain some ways we can use or some methods we can use to define topologies. So, our aim is to construct a large collection of topological spaces and study the topological properties. So, first we have to construct a large collection of topological spaces. So, we are headed in that direction. So, let us give a definition before we proceed a very basic definition of open sets. So, let x comma tau be a topological space. So, a subset u of x is set to be open if open in tau if u belongs tau ok. So, if we have a set a set x and we have given a topology to x. So, subsets of x which are in tau will often be called open subsets. So, whether u is open or not depends on the topology we have given to x. So, with this so let us define a basis basis for a topology. So, let x be a set let tau be a topology on x. So, a collection B. So, B is a subset of tau is called a basis for tau if it satisfies following two conditions the following condition that is not true. So, given any u in tau and x in u there is an element w in B such that x belongs to w and w is containing u. So, let us see an example of a basis we have already seen it in a form, but let us just make it more explicit. So, example let x be the real line let tau be the standard topology. So, let B be the collection of intervals a comma b. So, first of all we claim that every interval a comma b is open in the standard topology. So, what do we mean by this statement we just we simply mean that that is every interval a comma b is in tau. So, this is an easy check it is yeah we already did it in the example of 0 1. So, you can do this you can try this exercise. So, we have it is the same we showed that the interval 0 1 satisfies property star and the same way. So, hint use the same method or I do not know idea that we used to show that 0 1 satisfies property star to show that the interval a comma b satisfies. So, recall that the standard topology on the real line was defined using the property star and we showed that 0 1 satisfies the open interval 0 1 satisfies this property star. So, the same proof will can be easily modified to show that this interval a comma b also satisfies property star ok. So, the conclusion is that thus B is indeed a subset of tau ok and we want to check that we want to check that B satisfies this condition. So, let us check that. So, suppose so, we claim. So, let me write as a claim yeah. So, claim B is a basis for tau. So, let us prove this claim. So, suppose so, let u be in tau then u satisfies property star which implies that for any x in u there exists epsilon positive there exists epsilon positive such that x minus epsilon comma x plus epsilon is contain u right. So, we take. So, let w be the interval x minus epsilon comma x plus epsilon then x belongs to w w is an element is an interval and therefore, it is an element of the set B and clearly w is contain u right. So, this shows that B is a basis ok. So, similarly so, similarly the collection let me call it B 2 consisting of those sets S epsilon a comma b where a comma b is a point in R 2 and epsilon is positive form forms a basis for the standard topology ok. So, this can be checked in the same way that we checked for R and once again similarly the collection B n this is the collection of those S epsilon x where x is an element of R n and epsilon is positive forms a basis for the standard topology. So, we have seen we have defined the basis for a topology and you can familiarize yourself with the definition of a basis and these three examples how these three examples of standard topology on R and R 2 and R n and the basis for these and in the next lecture we will see two important properties that these basis have that a basis has and using those two properties we shall explain how given any subset B of the power set which has those two properties we can use that set B to define a topology. So, we will end this lecture here.