 So, in yesterday's class, we discussed what is meant by open balls and the related concept of an interior point which depends on the open ball. Now, let us start with that again. So, suppose x d is any matrix space and we took subset A in x and a point A in x and we said that A is an interior point of A that is small a is an interior point of A. If there exist some open ball with center at A which is completely inside A which is completely inside A. We can to understand the concepts we can give a diagrammatic or representation for this. I mean we can draw some pictures to understand the concepts. So, suppose this is the set A, let us say this is the set A then we will say that the point is interior point if suppose this is the point then we can find some open ball with center at A such that that ball is completely inside this A. So, that is an interior point. Of course, you are you are used to hearing that one picture is superior to thousand words etcetera. But what can happen is that in since in mathematics or in particular the balls may or may not look like that. So, sometimes pictures may mislead also. So, you should be carefully using, but of course pictures many times help in understanding what is happening. So, that is about the interior point and then we have said that the set of all interior points of A that we had denoted by interior of A. In some books they also use this notation A superscript is 0 for the interior of A. Then we have also seen that interior of A is always a subset of A. Now, what is the other possibility suppose you take any point A in x one possibility is that there exist there exist some r such that ball with center at A and radius r is completely inside A. What is the other possibility it may also happen that there exist some r such that ball with center at A and radius r is completely outside A. That is also possible. So, it may it may that ball may be somewhere here suppose the point is somewhere here then there you can find some ball which is completely outside A. So, such a point is called an exterior point. What is the meaning of that there exist some r such that let me repeat here A is called exterior point of A. If there exist r bigger than 0 such that open ball with center at A and radius r is completely outside A that means it is contained in x minus A x minus A and the set of all such points is called exterior of A. So, exterior of A we shall denote that by e x t of A or exterior of A the set of all exterior points of A and is it clear to you that a point is an exterior point of A is same as say that it is an interior point of x minus A because we are saying that there exist r such that that open ball completely lies in the outside A that means it lies completely in the x in x minus in the complement of A. So, exterior of A is nothing but by definition it is same as interior of x minus A. Now, let us see one more thing now given a point A in x it can be as an interior point of A or it can be an exterior point of A what is the third possibility? What does it mean that is neither is there a ball which lies completely inside A nor is there a ball which lies completely outside A what does it mean that is whatever ball you take that is whatever r you take and if you take a ball with center at A and radius r it will have at least one point in A and at least one point outside A that is every ball with center at A will have non-empty intersection with A as well as x minus A. So, if such a thing happens we say that that we call that point as a boundary point for example, a point like here suppose you take any point here then if you take any ball with center at this there should be it exists at least one point inside A and at least one point outside A. So, it will have non-empty intersection with A as well as. So, let us so we can say that A belonging to x is called a boundary point if for every r bigger than 0 for every r bigger than 0 what should happen u a r intersection A of course, boundary point of A that is important this is non-empty and u a r intersection x minus A is also non-empty that means every ball contains at least one point from A and at least one point from the complement of A if such a thing happens it is called a boundary point and you can easily see the examples of interior points, exterior points and boundary points in various examples that we have seen the other day for example, if you take an interval like this for example, suppose you take the interval 0 to 1 we have seen that open interval 0 to 1 that is interior of this what is exterior exterior will be open interval 1 to infinity and then minus infinity to 0 that is exterior open at both sides and what are boundary points it is just these two points 1 and 0 1 and 0 by the way the set of all boundary points is called boundary of A set of all boundary points of A is called boundary of A set of all boundary points of A. Now, is it clear to you that given any point in x it should be either an interior point of A or it should be exterior point of A or it should be if both of these are false it has to be a boundary point of A. So, it has to be one of these it has to lie in one of these three set interior of A exterior of A and boundary of A. So, what I can say is that x is always a union whatever given any non-empty set A or given any set A and a metric space x x can be always written as a union of these three sets all of which depend on A what are these three sets one is interior of A x is interior of A then union exterior of A and then union boundary of A. Of course some of these sets may be empty I am not saying that all for every A all for example, we have seen the examples yesterday where interior is empty similarly there can be sets where exterior is empty now let us go to the next important concept which depends on the concept of an interior point if it so happens that in a set A every point of A see we all order interior of A is always contained in A if the reverse inclusion is also true that means if every point is an interior point then it is called an open set then it is called an open set. So, let us go to the definition A is set to be an open set or A is set to be open if we can say this is the short way of writing A is contained in the interior of A or to put it in the words every point of A is an interior point of A or to expand it little further given any point A there exists R such that open wall with center at A and radius R is completely contained in A all that is set in this A is contained in interior of A. Now, let us see what are the open sets that we come across very frequently when we define for example, open ball close ball etcetera that time we had not defined what is meant by open set and so we had no justification why at all it is called an open ball but now we can give that justification but even before going to that let us dispose of some of the obvious things which are very obviously open sets. What about the full metric space x that is that has to be always because whatever ball you take it has to be always inside x. So, x that is always an open set so x is open x is open always. Now, what should happen for set suppose the set is not open then what should happen let us take this example again this example 0 to 1. We have already seen that this point 1 is a point in the set but that is not an interior point. So, this is not an open set this is not an open set whereas open intervals are open sets because there every point is an interior point. So, if a set is not open what it means that there should exist at least 1 point in A which is not an interior point. Now, does it follow from this that empty set is an open set because if the set is not open then there should exist at least one point which is not an interior point. So, such a thing is not possible for empty set. So, empty set is also an open set so empty set is open. Now, let us go to the main proof that we shall now prove that every open ball is an open set. Let us write this as a theorem every open ball in metric space is an open set. So, let us consider some open ball. So, let us say that let u a r be an open ball that this means what a a is in x and r is bigger than 0 and u a r is recall it is a set of all points whose distance from a is strictly less than r. So, let us again recall that u u a r is the set of all x in x such that distance between a and x is strictly less than r. So, suppose this is a and this is r then this is u a r we want to show that this is an open set. So, what is the way of showing you know something is open there is only one way you have to show that every point is an interior point. So, take some point so let x let x belong to u a r that is let us say there is some x. So, what is the property of this x distance between a and x this distance is strictly less than r distance between then distance between a and x is strictly less than r. What do we want to prove that such an x is an interior point of this u a r that means there exists some number let us call it rho there exists that we want to say there exists some rho such that a ball with center at x and radius rho is completely inside this ball u a r that is that is this is what we want to show to show there exists rho bigger than 0 such that ball with center at x and radius rho is completely inside u a r. So, there are two things here first thing is to decide what this rho should be to decide what this rho should be and then prove this inclusion. Now, let us go to the first step what is the choice for rho obviously see what we want is that a ball see this is x what we want is ball with center at this and radius rho that should be completely inside this it should not go somewhere outside. So, how do we choose rho for that because what we know is that see remember what we do is that d a x is strictly less than r d a x is strictly less than r now when there are two when one number is strictly less than r we you can always find some number some positive number such that if you add that positive number to this then the sum is still less than r. For example, you can take r minus d a x divided by 2 that can be one choice r minus d so what we can say that choose any rho bigger than 0 choose rho bigger than 0 such that this d a x plus rho in fact we can say that is less than r is it clear to that such a rho exists that is this number this d a x is strictly less than r. So, I can always find some small number such that if you add it to this the sum is still less than r. So, that number there will be many one choice I already told you so there will be many such choices choose any such rho satisfying this then we should prove this inclusion that is u x v rho is contained in u a r anyway how does one prove that once it is contained in the other this only one take a point here and show that that point lies in this. So, let y belong to u x r then we want to show that we want to show that y belongs to u a r. So, to show that anything is in u a r you want to say the distance between y and a must be strictly less than r. So, look at distance between y and a distance between a and y we can always now here we shall use the triangle equality because we know something about distance between a and x and we know something about distance between x and y. So, distance between a and y should be less than or equal to distance between a and x plus distance between x and y plus distance between x and y and. So, we know that distance between. So, this is less than or equal we know that this is less than rho because y is in u x r. So, distance between y and x is less than rho. So, this is in fact this is strictly less than distance between or less than or equal to distance between a and x plus rho and we have chosen rho in such a way that that is less than that is the reason for choosing the rho in such a way. So, this is less than r. So, that shows this that open ball with center at x and radius rho is completely inside the open ball with center at a and radius r. So, every open ball is an open set every open ball in any metric space is an open set and in yesterday's class we have seen several examples of open balls in various metric spaces. So, you have several examples of open sets now we shall see some more examples of course these two are already there the full metric space x and empty set is always an open set. Then we shall see some more properties of these open sets for example, how can we come across or how can we form new open sets from the known open sets. For example, things like what happens if we take unions or intersections whether they are open sets. It is what is known in this case is that if you take any family of open sets then the union is always an open set. So, let me just write this as a theorem let x d be a metric space and so let me write this as a. So, first property is that if let us say g alpha alpha belonging to lambda is a family of open sets is a family of open sets in x that union g alpha alpha it lambda is open. In other words you take any arbitrary family of open sets finite infinite countable infinite does not matter their union is always open. But a similar thing is not true about intersection in case of intersection we have to take only a finite family. So, second thing is if g 1 g 2 g n are open open open sets then intersection g g g going from 1 to n is open and proof is more or less straight forward. Let us look at this proof first let g alpha be a family of open sets and let us call this union as g let g be equal to union g alpha. Now, we want to show that g is open we have to say g is open if g is empty there is nothing to be proved if g is empty it is already open. So, let us say that it is non empty and some points belongs to g. So, and let us say a belongs to g then we want to show that there exists r such that open wall with center at a and radius r is contained in g. But again that is trivial since a belongs to g and g is union of this g alpha it is in one of these alphas since g is union. So, we can say that there exists some alpha there exists some alpha in lambda such that a belongs to g alpha a belongs to g alpha and g alpha is an open set that is how we started g alpha is an open set. So, there exist some r such that wall with center at a and radius r is inside g alpha. So, since g alpha is open there exists r bigger than 0 such that wall with center at a and radius r is contained in g alpha and g alpha is contained in g because g is union of all such g alpha. So, g alpha is contained in g and that proves that a is an interior point. So, that shows that g is open again let us look at two. So, we are given that g 1 g 2 g n are open sets and we want to prove that their intersection is open again if the intersection is empty there is nothing to be proved. So, let us assume that there is something in that and then we have to say that point is an interior point. So, let a belongs to intersection g j going from 1 to n, but if a belongs to intersection g j means it belongs to each of the g j this means a belongs to g j for all. And what follows from this that for each for each j there will exist some positive number suppose I call it number r j such that open wall with center at a and radius r j is inside g j. So, for all. So, we can say that for each j there exists r j bigger than 0 such that open wall with center at a and radius r j is contained in g j what is to be done after that is clear you take the minimum of all this. So, let r be equal to minimum of this r 1 r 2 r n there are n such number and since each of this r j is strictly positive that is the most important argument here since each of this r j is strictly positive r also is strictly positive this is strictly bigger than 0. Then you look at u at center at with center at a and radius r take open wall with center at a and radius r. And since r is less not equal to r j u a r is contained in u a r j. So, this is contained in u a r j and u a r j is contained in g j and this is for true for each j for each j this argument will work this is true for all j equal to 1 to n. So, and if something is in g j for each j it has to be in the intersection also. So, this implies u a r is contained in the intersection g j j going from 1 to n. So, we started from a point in this intersection and then we found an r such that u a r is in this intersection g j j going from 1 to n. So, this shows that the finite intersection of open sets is open. So, that theorem is usually expressed by saying this that arbitrary union of open sets is open and finite intersections of open sets is open. Now, looking at this proof you will also understand why we cannot take an arbitrary intersection because suppose you follow the same line here they will suppose the suppose this had infinitely many sets here there will be infinitely many r g's. Then of course minimum would exist we will have to look at infimum and infimum of set of positive numbers can be 0. So, this proof would not work, but again if you want to show that something is false saying that some proof does not work is not a correct argument. Even if this proof does not work in principle you can say that somebody may be able to use some other proof that is not the only way of proof. So, in order to say that this u cannot replace finite by infinite in this part 2 here there is only one way of disposing of that question you have to give an example. Example of what? Example of an infinite family of open sets such that the intersection is not open intersection is not open. What is an example? Let us let me call it g a is minus 1 by n 2 1 by n 5 and for n is equal let us say n belongs to n and what is the intersection g n? Intersection g n is just singleton 0 and it is clear that the singleton 0 is not an open set. So, you cannot replace a finite family here by an arbitrary family all right. Now, that we have learned this various properties of this open sets let us just go to a slightly more general concept. I will just introduce that concept here, but we will not go further about that concept than that, but before that let me also make one more point here or let me give this as an exercise. In a metric space the following thing is 2. Suppose you are given two different points suppose you are given two distinct points then you can always find open sets two disjoint open sets containing those two points. So, let me just write here what is the sense suppose let x y belong to x with x not equal to y then there exists disjoint open sets open sets suppose I call them g 1 and g 2 g 1 and g 2 such that x belongs to g 1 and y belongs to g 2 that is given two distinct points you can always find two disjoint open sets such that one point is in one of the set and second point is in the second set. Is it clear you can prove it all in fact you can find disjoint open balls if the two distinct points are given you can find disjoint open balls such that x is contained in g 1 and y is contained in g 2. Now, suppose x is any arbitrary set x is any arbitrary set non-empty set and suppose we consider its power set that is 2 power x is the thing but the set of all subsets of x and I consider some subset of this that means some family of subsets of x suppose I call that set as tau tau is a family of so tau is a family of subsets of x with certain properties I am not taking any arbitrary but some property. So, such a family is called topology on x this is what I want to define it is called topology on x if certain properties are satisfied if the first property is that empty set belongs to tau and the full space also belongs to tau and second property is that if a family of sets is in top then their union is also in top. So, if g alpha belongs to tau for all alpha in some indexing set lambda then union of g alpha alpha belong to lambda this also belongs to tau in other words if this family contains any arbitrary if this tau contains any family of sets then its union over that family is also a member of top and third property is that if it contains a finite if you take any finite number of sets then the intersection is also in top. So, if g 1, g 2, g n belongs to tau then intersection of this that is intersection g j going from point to n this also belongs to tau. So, any family of sets satisfying these three properties or these three or I mean if you write these two things as separate things it will be four properties is called a topology on x is called a topology on x and this pair x tau that is called a topological space. So, what is a topological space topological space is a pair x tau where x is a non empty set and tau is a topology on x topology means family of subset satisfying these properties. Have you already proved that every metric space is a topological space see suppose x is a suppose x has some metric on it suppose x has some metric on it then take this set of all then take tau to be the set of all open sets in x then does tau satisfy all these properties this is what we proved just now that is phi and x are open sets and if you take any arbitrary family of open sets then the union is also an open set and intersection of a finite family of open sets is again open all these things. So, in short what we proved was that if you take the family of open sets in a metric space that forms a topology that forms a topology such that that means every metric induces a topology every metric induces a topology then here one can ask a very obvious question whether converse is also true whether every topology is whether every topology is given by induced by a metric of course the answer is false because if that were true then we would have said there is no difference between a topological space and metric space that is not the case topological space is a much bigger class and the study of this topological spaces and functions between the topological spaces that forms a subject on its own called topology. And as you say as I have said right in the beginning that this real analysis is a very basic subject it has connection to several other subjects for example we have discussed what is normed linear spaces. So, normed linear spaces and the maps between normed linear spaces that forms a topic in what is called functional analysis and here you have topology. So, basic concepts in analysis are used in practically all subjects all right. So, let me just before proceeding further let me just give an example of a topology which is not induced by any metric. Now, the whole question is this how does one show that a topology is not induced by a metric etcetera. We will come to that little later, but let us before that let us dispose of some trivial examples of topologies. For example, I can take tau as this whole thing 2 power x suppose I take all the subsets of x then obviously all these properties are satisfied right that is called discrete topology. And by the way I should mention here that once you consider topological space members of this set tau or curl open sets open sets with respect to this topology open sets with respect to this topology. Because here basically what we are doing here is it we are taking the abstract concept of open sets and then starting from starting from there and then in topological space what happens is that subsequently you define everything in terms of open sets just as in case of metric spaces you will define all concepts in terms of metric. So, similarly in topological spaces you will define all concepts in terms of open sets. So, whichever concepts can be defined using only open sets those are called topological concepts. So, this is one example of a topology which is called discrete topology you can also see why it is called discrete topology. We have discussed what is meant by a discrete metric space. In discrete metric space what are the open sets let me begin it is a singleton set open in a discrete metric space. Because we have seen that a ball with centre at x and let us say radius half is singleton x. So, singleton it is an open ball and hence an open set. But once you say that a single set is open does it follow from there that every subset is open given any subset of x you can write it as a union of singleton sets. And we have already seen that arbitrary union of open sets is open. So, in a discrete metric space every subset is open is it clear in a discrete metric space every subset is open. So, the topology induced by a discrete metric is just this discrete topology what we call discrete topology. Let us now take another extreme suppose I take just these two sets phi and x I will just take two sets empty set and x. Now, that is the minimum for it these two sets have to be there that is how that is the first requirement. And suppose I include nothing else will this also satisfy all these properties because if you take say G alpha G alpha has to be either empty or x. So, it is similarly intersection also. So, is it clear this is also topology this is also topology that is called indiscreet topology. So, in the discrete topology you have all subsets to be open sets and in the indiscreet topology only empty set and a full space these are the only two open sets. Now, let us come to this question look at this exercise here what I have said here is that given any two in a metric space given any two distinct points you can find two disjoint open sets such that x is contained in G 1 and y is contained in G 2. Whenever topological space has this property it is called a Hausdorff topology whenever topology has this property it is called Hausdorff topology. That means two distinct points can be separated by two disjoint open balls that is called Hausdorff topology. So, every metric space has this property that is topology induced by a metric has this property. Does this space have this property suppose I take this indiscreet topology suppose you take indiscreet topology then can you show that with respect to that topology also you can find given any two point x and y you can find G 1 and G 2 etcetera see remember there are only two open sets empty set and the whole set x. So, G 1 and G 2 can be either empty or x. So, if G 1 has to contain the point x G 1 has to be whole of x similarly G 2 has to be whole of x. So, there cannot exist any disjoint sets there cannot exist any disjoint sets separating two distinct points. So, this indiscreet topology will not satisfy this property. So, this topology is cannot be obtained by any metric this topology cannot be obtained by any metric. We shall not pursue this topic of topology and topological space any further this is just to let you know that the concept of open sets is a very basic concept and using that concept we can define several other things and in metric space also we shall show that initially many of the things we shall define using the distance d, but many of those concepts can be defined at discuss using only the concept of metric spaces and those are the ones which are then taken in the topological spaces and those become topological concepts and topological properties etcetera. Let us now come back to the description of open sets in various metric spaces. We already seen that open balls are open sets and just now we have seen that in the discrete metric space every subset is an open set. Let us now go back to our most familiar space namely R with the usual topology or usual metric. Usual topology means topology given by the usual metric. We already seen that the intervals are open sets open intervals are open balls and every open ball is an open set. So, any arbitrary union of intervals open intervals is an open set any arbitrary union of open intervals is an open set, but in real life we can say something more that every open set can be expressed as a union of open intervals. Not only that we can say something more we can say that it can be expressed as a union of a countable family of mutually disjoint open intervals. So, let us let me write by the way in any metric space given every open set can be expressed as a union of open balls that should be fairly easy to show. If you take any open set and if you take a point in that open set you can always you can always find a ball with the center at that point and that ball is completely contained in that open set. So, suppose you take collect all such balls the union has to be the same as that whole set. So, in every metric space every open set can be expressed as some union of some union of open balls. Only thing is that those open balls which form that union they may not be disjoint they may not be disjoint their number may not be countable, but in case of real line we can say this you can always express every open set as a disjoint union of a countable family of intervals. So, every open set in R is union of a countable family union of a countable family of mutually disjoint or pair wise disjoint intervals pair wise disjoint open intervals. Let us now look at the proof. So, let G be an open set in R I will just give some steps in the proof and then you will complete the details. Again if G is empty there is nothing to be proved we can say that it is union of empty family of open intervals. So, let us assume that G is non-empty if G is non-empty. So, let G be non-empty then to begin with what I will do is that I will define a relation between the points in G. So, suppose we take two points in G let x and y belong to G. So, let us say I will define a relation define x is related to y this means there exists an open interval which contains both these points x and y and it is also contained in G. So, we say that x is related to y if there exists an open interval there is an open interval suppose I call it open interval I such that both of these points x and y belongs to I and I is contained in G. Now, let us quickly see some properties of this relation is this relation reflexive given x can be always find an open interval such that x is contained in I and I is contained in G. In fact, that is that follows because G is open even any point x we can find always find some R such that ball with center at x and radius R is completely inside G and that ball is nothing but an open interval x minus R to x plus R it is reflexive clear is it symmetric that is clear is it also transitive see suppose x is related to y that means there exists some interval suppose you call it I 1 I 1 contains x and y and I 1 is contained in G. Suppose y is related to z then there exists some other interval j such that j contains y and z and j is contained in G I and j both are contained in G. So, I union j is in G but is I union j again interval that is where the properties that we study yesterday will come into picture because I I and j are not disjoint because y is common to both y is common to both. So, I and j are not disjoint so this is an equivalence relation this is an equivalence relation. Now, what does an equivalence relation do? It will partition all the points into what are called equivalence classes. So, this equivalence relation will partition the given set G into what are called equivalence classes and the union of those equivalence classes will be same as G. So, let us say that let this x so denote equivalence class equivalence class containing x then we know that union of these equivalence classes x belonging to G it is same as G. Now, suppose we show that each of this equivalence class is an open interval suppose we said each of this equivalence class is an open interval and suppose we show that their total number is countable their number is countable then we would have completed the proof I think we will complete the proof tomorrow since the time is over I will stop with this today.