 Okay, so we shall continue from where we left off in the previous lecture. So in the previous lecture we saw how we saw how to give different topologies on the same set x and the set the set x could be any set. So but today in this lecture we will see some examples which are going to be more important in this course. So so the next example we are going to see is the standard topology on the set of real numbers. So the real line so recall that so recall that the by the interval a comma b in r. So this is a subset of r we mean the set we mean the subset definition is those real numbers x such that x is strictly less than a and strictly less than b. So here the assumption is that a is strictly less than b and we allow a and b to be minus infinity and plus infinity. So if a is equal to minus infinity then this is those x in r such that x is strictly less than b. And similarly if b is equal to plus infinity then this is those x in r such that x is greater than a. So in particular r can be written as minus infinity comma. So we want to define a topology on the real line. So first let us define a property which we will denote by star. So let u in r be a subset. So we shall say u satisfies this property star. So let us see what star is if for every point x in u there is an epsilon positive. So I just want to emphasize that the epsilon depends on x such that when we take the interval x minus epsilon comma x plus epsilon this is continue. So let us see some 2 easy examples for example the interval 0 comma 1 satisfies. So why is this? So if we take so let us say this is 0 and this is 1 and if we take any x which lies in this interval so x lies between. So we can always find a small neighborhood ok. So let me so let this distance so this distance is x and this distance is 1 minus x. So if we take epsilon to be minimum of x by 2 and 1 minus x by 2 then so for us it will be something like this. Then it can be checked easily the interval x minus epsilon x plus epsilon is contained in 0 1. Let us see one more example. So however the interval if we take this interval 0 1 so this is those x in r so that x is greater than equal to 0 and strictly less than 1 right. If we take this interval so does not satisfy. So why is that? So if we take so for the point 0 for the point 0 in this set there is no epsilon positive such that 0 minus epsilon comma 0 plus epsilon. So this interval is simply equal to minus epsilon comma plus epsilon this contain in 0. So there is no epsilon positive for which minus epsilon this interval minus epsilon comma epsilon is contained in this half open interval 0 1. So therefore this interval does not satisfy this property star. So that sort of gives us an idea of ok. So what kind of set satisfies this property? So we define a topology so tau in power set of x so tau is equal to those u those subsets of r such that u satisfies. So we need to check so to check that tau defines a topology we need to check it satisfies the three defining conditions for a topology. So let us check these one by one. So recall that the first condition was that phi the empty set and the entire set should be in tau. So clearly the empty set is in tau because there is nothing to check there is no x in the empty set and therefore there is no condition to check and so it is this is vacuously true and it is also clear that clearly r is also in tau as for any x in r we can simply take epsilon to be 1 and clearly x minus 1 comma x plus 1 it is a contain. So therefore this first condition first defining condition for being a topology is satisfied. So for the second condition we need that if u 1 u 2 up to u n in tau are finitely many subsets of r then the intersection i equal to 1 to n u i should also be. So let us check that this condition is satisfied. So we need to check that this intersection u i satisfies. So let us choose some x in the intersection. So in particular so this implies that x belongs to u i for all i yeah i equal to 1 2 and since each u i satisfies property star there exists some epsilon i positive such that the interval x minus epsilon i and x plus epsilon i is contained in u i ok. So then so now let epsilon be equal to the minimum of epsilon 1 epsilon 2 up to epsilon n right. So clearly epsilon is positive because we are finitely many positive real numbers and we take the smallest one among them so that is also going to be positive. So and it is also clear that x minus epsilon comma x plus epsilon this interval is contained in x minus epsilon i comma x plus epsilon right because this is x and let us say this is x minus epsilon i and this is x plus epsilon i. So epsilon is the smallest one among all these epsilon i's. So therefore x this will be x plus epsilon and this will be x minus epsilon. So this implies that this x minus epsilon comma x plus epsilon this is a subset of x minus epsilon i x plus epsilon i which is contained in u i and this happens for all i. So in particular this implies that this interval x minus epsilon this is a subset of intersection of the u i's. So this shows that this intersection u i satisfies. So therefore the second condition for being a topology this is also satisfied and finally we have to check one more condition. So let i be a set and suppose for each i we are given u i in. So then we need to show that the union i belongs to i is in tau that is it satisfies. So once again this is easy. So we apply the same method that we used in to show proper the second case. So we just take any x in in this union then there is some j j in i such that x belongs to u j. Now since u j satisfies star there is an epsilon positive such that x minus epsilon x plus epsilon is a subset of u j. So in particular so this implies that x plus epsilon sorry x minus epsilon comma x plus epsilon is contained in u j which in turn is contained in the union of all these u i's right. So thus this we approve that this u i's it satisfies that is thus this union is also in tau. So therefore tau also satisfies the third condition. So all these implies that tau defines a topology on r and which is which we call the standard topology. So in the same way that we define this topology on r we can define a topology on r 2 and more generally r n. So let us do the example for r 2 also and the example of r n will be left as an exercise. So this is our fifth example the standard topology. So this example is similar to the previous one. So we define so but before that first just the way we have intervals we are going to define some we define an open square side length to epsilon around a point a comma b in r 2 as follows. So we define so we will call this s epsilon a comma b this is defined to be those points in r 2 such that the absolute value of x minus a is less than epsilon and the absolute value of y minus b is less than epsilon. So in terms of a diagram so this looks like maybe I can make it on the next page. So suppose we take a point a comma b here. So this distance this length is epsilon this length center is point a comma b this length is epsilon this length is epsilon this length is epsilon. So this is the picture of this open square and so now that we have defined this. So let us define the analog of property star we say that a subset u in r 2 satisfies property star if so this is a property star. So maybe I can if you want you can denote it by star 2 so that you do not confuse it with the same star earlier but I will be lazy and I will call this property star for every point a comma b in u there is an epsilon positive such that s epsilon a comma b is completely contained inside u and once again this epsilon epsilon depends on the point or it may depend on the point a comma b it does not have to be the same for all points in u. So as before let us take an example of let see two examples of one set which satisfies this property and another which does not. So for example if we take the set u to be those a comma b in r in r 2 this is that a square plus b square is strictly less than 1 then we claim that u satisfies. So here we have r 2 this is r circle so this has radius 1 so this is the point let us say 1 comma 0. So the set u is the region inside the circle and the point to note is that if you take any point over here which is in this green region the interior of this. So then we can always find an s epsilon so such that so for any point a comma b in u we can find an epsilon positive such that this set is contained in u. So this is left as an exercise and so maybe I should have this is the open interval so and similarly similar to the non example we had seen. However if we take the set v this is those a comma b in r in r 2 such that a square plus b square is less than equal to 1 right. So then we claim that v does not satisfy so why is that because so our set v is this is 1 0. Now if we take the point 1 0 then no matter which no matter how small we take epsilon this s epsilon will always go outside this region v. So if we take the point 1 0 in v for any epsilon positive s epsilon a 1 0 is not going to be contained in v. So this is also left as so we will stop here and in the next lecture we will continue with the proof that tau defines a topology on r 2.