Sanjay Mishra. Topology. Dr. Sanjay Mishra. A Profound Subtitle

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1 Topology A Profound Subtitle Dr.

2 Copyright c 2017

3 Contents I General Topology 1 Topological Spaces Introduction Topological Space Topological Space Examples of Topological Spaces Exercises:Topological Space Base and Subbase for a Topology Basis for a Topology Topology Generated by Basis Subbasis for a Topology Difference between Basis and Subbasis for a Topology Exercise:Basis and Subbasis for a Topology The Topology on R and R The Topology on R The Topology on R Exercise:The Topology on R and R New Topological Spaces from Old The Order Topology The Product Topology Product Topology Generated by Projection Map The Subspace Topology Exercise:New Topological Spaces from Old

4 1.6 Subsets of Topological Space Closed Set Limit Point of a Set Closure, Interior and Boundary of a Set Exercise:Limit Point, Interior, Closure, and Boundary for Topological Space The Metric Topology The Quotient Topology 53 II Algebraic Topology Bibliography Books 57 Articles 57 Index

5 I General Topology 1 Topological Spaces Introduction 1.2 Topological Space 1.3 Base and Subbase for a Topology 1.4 The Topology on R and R New Topological Spaces from Old 1.6 Subsets of Topological Space 1.7 The Metric Topology 1.8 The Quotient Topology

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7 1. Topological Spaces 1.1 Introduction In this chapter we are going to introduce the topological space which is primary object of study in field of topology. 1.2 Topological Space Topological Space There are several equivalent definition of topological space have been given by the mathematician on basis of various utility of a topology in different branch of mathematics. The topological space can be defined in term of open set, closed set, course operator, interior operator and neighborhood system. The most commonly used and most elegant, definition of topological space is in terms of open sets. In this section, we are going to introduce this definition. Definition 1 Let X be a any non-empty set. A topology on X is a collection τ of subset of X such that: (i) The set X and empty set φ belong to τ. (ii) The union of any arbitrary (finite or infinite) number of subsets of τ belongs to τ. In other words if G i belong to τ is arbitrary, then i I G i τ, where I is any set. (iii) The intersection of finite many subsets of τ belong to τ. In other words assume that for each i I if G i belong to τ, then i I G i τ, where I is finite set. In short, a topology on a set X is a collection of subsets of X which includes empty set and X and is closed under unions and finite intersections. The subsets of collection τ are called open. A set together with a topology is called a topological space and denoted by (X,τ). There are two things that make up topological space : a set X, and a collection, τ, of subsets of X that forms a topology on X. To be properly formal, we should refer to a topological space as an ordered pair (X,τ), but to simplify notation we follow the common practice of refereing to the set X as a topological space, leaving it implicitly understood that there is a topology on X.

8 8 Chapter 1. Topological Spaces R 1. The statement intersection of finitely many subsets of τ is equivalent to the statement intersection of two subsets of τ. 2. One should be careful when using the word "open". Open intervals in R are generally not the same thing as open sets in a topological space. Open sets are a generalization of the concept of open intervals. Definition 2 Weaker and Stronger Topologies. Let τ 1 and τ 2 are two topologies on the set X. τ 2 is said to strictly stronger or stronger than τ 1 if τ 2 properly contain τ 1 or τ 2 τ 1 respectively. Similarly, τ 2 is said to strictly weaker or weaker than τ 1 if τ 1 properly contain τ 2 or τ 2 τ 1 respectively. Both topologies are said to be comparable if τ 2 τ 1 and τ 1 τ Examples of Topological Spaces In this section, we give various examples by which can better understand about the topological spaces. Example 1.1 If X be the three-point set X = {a,b,c}, the there are many possibilities of topologies on X. Here we considered some of them schematically in Figure 1.1. (i) The set X with topology τ 1 = {X,φ}. (ii) The set X with topology τ 2 = {X,φ,{a},{a,b}}. (iii) The set X with topology τ 3 = {X,φ,{b},{a,b},{a,c}}. (iv) The set X with topology τ 4 = {X,φ,{b}}. (v) The set X with topology τ 5 = {X,φ,{a},{b,c}}. (vi) The set X with topology τ 6 = {X,φ,{b},{c},{a,b},{b,c}}. (vii) The set X with topology τ 7 = {X,φ,{a,c}}. (viii) The set X with topology τ 8 = {X,φ,{a},{b},{a,b}}. (ix) The set X with topology τ 9 = {X,φ,{a},{b},{c},{a,b},{b,c},{a,c}}. On the set of X there are many topologies can be defined but all the collections of subsets of X are not necessary to be topologies on X. The collections {X,φ,{a},{b}} and {X,φ,{a,b},{b,c}} of subsets of X are not topologies on X. a b c (i) (ii) (ii) (iv) (v) (vi) (vii) (viii) (ix) Figure 1.1: Topologies on a Set Example 1.2 Discrete Topology. If X is non-empty set, then the collection of subsets of X is denoted by τ dis is a topology on X and called it discrete topology. Clearly τ dis is a topology, since arbitrary unions and finite intersections of subsets of X are themselves subsets of X and therefore are in the collection τ dis. This is the largest topology that we can define on any X. Example 1.3 Non-discrete Topology. If X is a non-empty set, then the collection of subsets of X as τ non dis = {φ,x} is a topology and called non-discrete topology on X. The non-discrete

9 1.2 Topological Space 9 topology is smallest topology on any set. R The topology on any non-empty set is always lies between discrete and non-discrete topology. Now here we can raise a question that if (X,τ) is topological space such that τ contains every singleton set, then τ is necessarily the discrete topology or not? The answer is yes", and this is proved as follows Theorem 1 If (X,τ) is a topological space such that for every x X, the singleton set {x} is open, then τ is the discrete topology. Proof. As we know that every set is a union of its singleton subsets so if A be any subset of X, then A = x A {x} Since as given that each subset {x} of X is open set, then by definition of topological space and above result we can say that A τ. Since A is arbitrary subset of X so, τ is discrete topology. Example 1.4 Finite Complement Topology. If X be any non-empty set and τ f c is the collections of all subsets U of X such that X U either is finite or empty or all of X i.e. τ f c = {U X : X U either finite or empty or all ofx} Then the collection τ f c is topology on X and called finite complement topology on X. Here we can check that above collection τ f c is in fact a topology on X as follows. (i) Since X X = φ is finite and X φ = X is all of X, so X and φ belong to τ f c. (ii) Let {U α } be a collections of open sets of X i.e. member of τ f, then we want to show that Uα is in τ f. For this we will show that X U α is finite. Now find X U α = (X U α ) = (X U 1 ) (X U 2 )... Since {U 1,U 2,...} are open so by the definition, {(X U 1 ),(X U 2 ),...} are finite and by the properties of finite set that intersection of finite sets are finite so {(X U 1 ) (X U 2 )...} is finite i.e. X U α is finite. Hence by the hypothesis, U α is in τ f c. (iii) Assume we have a finite collection {U 1,U 2,...,U n } of open sets. If anyone of them is empty, then the intersection is empty and therefore is an open set. Thus assume that each U i is nonempty open set of X and we want to show that n i=1 U i is also open set i.e. n i=1 U i is in τ f c. Now find X n U i = i=1 n (X U i ) i=1 = (X U 1 ) (X U 2 )... (X U n ) Since finite union of finite sets are finite so X n i=1 U i is finite and therefore by the definition ni=1 U i is in τ f c. Example 1.5 If X be a set and let τ c be the collection of all subsets U of X such that X U either is countable or is all of X i.e. τ c = {U X : X U either countable or all ofx} Then τ c is a topology on X. Here we can check that above collection τ c is in fact a topology on X as follows.

10 10 Chapter 1. Topological Spaces (i) As the complement of X in X is the empty set, which is finite and hence countable, so X τ c. The complement of φ is all of X, which is in τ c. (ii) Now we check that the union of {U i } i I τ c is contained in τ c. Let U = i I U i. Then by DeMorgan s laws X U = X i IU i = i I(X U i ) = (X U 1 ) (X U 2 )... If all of the complements X U i equal X, then so does their intersection. If one of the complements X U i is countable, then so is X U X U i, as any subset of a countable set is countable. Thus X U is either countable or all of X, and so U is in τ c. (iii) Finally we check that the finite intersection of elements from τ c is contained in τ c. Let U 1,...U n τ c, and let U = n i=1 U i. Then by DeMorgan s laws n X U = X U i i=1 n = (X U i ) i=1 = (X U 1 ) (X U 2 )... (X U n ) If any of the sets X U i is the total set X, then so is the union X U, as X U i X U. If all of the sets X U i are countable, then so is the finite union of countable sets X U, by result that a countable union of countable sets is countable. Thus X U is either all of X or countable, and so U τ c. Example 1.6 Complement Topology on R. On the real line R, define a topology whose open sets are the empty set and every set in R with a finite complement. See figure. For example, U = R {0,3,7} is an open set. We call this topology complement topology on R and denote it by R f c. Example 1.7 Let (X,d) be a metric space. Define O X to be open if for any x in O, there exist an open ball B d (x,ε) = {y X : d(x,y) < ε,ε > 0} lying inside O. Then, the collection τ d = {O X : Ois open} {φ} is a topology on X. This collection τ d is called the topology induced by the metric d. Example 1.8 If R n is a metric space with the usual metric d((x 1,...,x n ),(y 1,...,y n )) = n i=1 (x i y i ) 2 then R n has a topology τ u induced by usual metric d and this topology on R n is called usual topology induced by usual metric. Example 1.9 Partition Topology. Let X is non-empty set, a partition of X is a set P of mutually disjoint subsets of X whose union is all of X. The partition topology τ P is the set of unions of elements of P together with the empty set. For example, on X = R we can choose P = {[n,n + 1 : n Z}. The open sets of the partition topology associated to this topology are intervals of the form [a,b (with a and b being integers) as well as [a, and,b] (with a and b again being integers).

11 1.2 Topological Space 11 Example 1.10 Particular Point Topology. Let X be a set and τ be collection of subsets of X consisting empty set, X and all subsets of X containing any particular element p of X, then τ is topology on X called particular point topology on X. Example 1.11 Excluded Point Topology. Let X be a set and τ be collection of subsets of X consisting empty set, X and all subsets of X not containing any particular element p of X, then τ is topology on X called excluded point topology on X. Example 1.12 Let N be the set of all natural numbers and let τ consist of N,φ and all finite subsets of N. Then the collection τ is not a topology on N, because the infinite union of members of τ does not belong to τ Exercises:Topological Space {2} {3}... {n}... = {2,3,...,n,...} Exercise 1.1 List all the possible topologies on X = {a,b}. Solution : Write the solution. Exercise 1.2 Let X = {a,b,c,d,e}. Determine whether or not each of the following classes of subsets of the X is a topology on X. (i) τ 1 = {X,φ,{a},{a,b},{a,c}}. (ii) τ 2 = {X,φ,{a,b,c},{a,b,d},{a,b,c,d}}. (iii) τ 3 = {X,φ,{a},{a,b},{a,c,d},{a,b,c,d}}. Solution : Write the solution. Exercise 1.3 Let X be an infinite set and τ be a topology on X in which all infinite subsets of X are open. Show that τ is discrete topology on X. Solution : Write the solution. Exercise 1.4 Let A be a subset of a topological space (X,τ) in such a way that each point x of A belong to an open set G x contained in A, then A is open. Solution : As given each x A and x G x A. Hence {G x : x A} = A and so A is union of open sets. Hence A is open. Exercise 1.5 Let X be a topological space; let A be a subset of X. Suppose that for each x in A there is an open set U containing x such that U A. Show that A is open in X. Solution : As given that for each x in A there exist an open set U x such that x U x A, then A = x A U x. Since union of arbitrary open set U x is open, therefore A is open. Exercise 1.6 Let X be a set; let τ be collection of all subsets U of X such that X U infinite or empty or all of X i.e. τ = {U : X U is infinite or empty or all ofx} Then show that τ is a topology on X. Solution : The collection τ is no topology on X. For each x X, the singleton {x} τ if X is infinite. Yet for any fixed y X the union x y {x} is a union of sets in τ which is not contained in τ, as its complement is the finite set {y}. Exercise 1.7 If {τ α } is a family of topologies on X, show that τ α is a topology on X. Is τ α a topology on X? Exercise 1.8 Let {τ α } is a family of topologies on X. Show that there is s unique smallest topology on X containing all collection τ α, and a unique largest topology contained in all τ α. Solution : Note that every collection of topologies on a set X is itself a set: A topology is a subset of P(X), i.e. an element of P(P(X)), hence a collection of topologies is a subset of P(P(X)), i.e. a set. Let {τ α } be a nonempty set of topologies on the set X.

12 12 Chapter 1. Topological Spaces (i) Since every {τ α } is a topology on X it is clear that the intersection τ α is a topology on X. The union τ α is in general not a topology on X: Let X = {a,b,c}. It is straightforward to check that τ 1 = {X,φ,{a},{a,b}} and τ 2 = {X,φ,{c},{b,c}} are topologies on X. But τ 1 τ 2 is not a topology on X, since {a,b} {b,c} = {b} / τ 1 τ 2. (ii) The intersection of all topologies that are finer than all τ α is clearly the smallest topology containing all τ α. The intersection of all τ α is clearly the largest topology that is contained in all τ α. (iii) The topology τ 3 = τ 1 τ 2 = {X,φ{a}} is the largest topology on X contained in τ 1 and τ 2. The topology τ 4 = {X,φ,{a},{b},{a,b},{b,c}} is the smallest topology that contains τ 1 and τ 2. Exercise 1.9 If X = {a,b,c}, let two topologies on X are as follows τ 1 = {φ,x,{a},{a,b}}andτ 2 = {φ,x,{a},{b,c}} Find the smallest topology containing τ 1 and τ 2, and the largest topology contained in τ 1 and τ 2. Exercise 1.10 Let τ be a family of subsets of N consisting of empty set and all subsets of N of the form A n = {n,n + 1,n + 2,...}, where n N, then show that τ is topology on N and also find the open sets which conain positive integer 6. Solution : The empty set φ and A 1 = {1,2,3,...} = N lies in τ. Since τ is totally ordered by set inclusion so finite intersection of members of τ also belong in τ. Now let A be subfamily of τ {N,φ}, that is, A = {A n : n I}, where I is some set of positive integer. Note that I contain smallest positive integer n 0 and i {A n : n I} = {n 0,n 0 + 1,n 0 + 2,...} = A n0 which belong to τ. Hence τ is topology on N. Exercise 1.11 Prove that (R 2,τ) is a topological space where the elements of τ are φ and the complements of finite sets of lines and points. Exercise 1.12 Give an example of a topological space different from the discrete space in which open sets are exactly the same as closed set. Solution : Let X = R and τ be family of subsets U of R such that x U x U (1.1) Now we shall show that (R,τ) is required topological space. (i) Since φ does not contain any element and so that condition (1.1) is automatically satisfied. Since x R this show that x R, theretofore R T. (ii) Let U 1,U 2 τ anyx U 1 U 2 x U 1,x U 2 Therefore U 1 U 2 τ. (iii) Let U i τ for all i J. x U 1, x U 2 x U 1 U 2 anyx i U i x U i for somei J x i U i Therefore i U i τ. From all above conditions we can say that τ is topology on X. Let U be open set. We shall show that U is closed set or R U is open set. anyx R U x / U x / U asu / τ x R U

13 1.3 Base and Subbase for a Topology 13 Therefore R U is open. Exercise 1.13 Let τ be a family consisting of empty set, R and all infinite open intervals A x = (x, ), where x R, the real. Verify τ is topology or not on R. Solution : The sets R and φ is in τ since as already given. The family τ is totally ordered by set inclusion, therefore finite intersection of members of τ definitely lies in τ. Now let A is a subclass of τ {R,φ}, which is defined as A = {A i : i I}, where I is some set of real numbers. We want to show that i A i is in τ. If I is bounded from below say in f (I) = i 0, then i A i = (i 0, ) = A i0. If I not bounded from below, so in f (I) =, then i A i = R. In both cases, i A i τ. Finally, τ is topology on R. Exercise 1.14 Let τ be a family consisting of empty set, R and all infinite open intervals A x = (x, ), where x Q, the rational. Verify τ is topology or not on R. Solution : As given A x = (x, ) is members of family τ. Observe that A = {A x : x Q,x > 2} = ( 2, ) is the union of members of τ, but A / τ since 2 is irrational. Therefore family τ is not a topology. Exercise 1.15 Let τ be a collection of empty set, R and all intervals (, p) for p R, then show that τ is a topology on R. Exercise 1.16 Let τ = {R 2,φ} {G k : k R} be the class of the subsets of the plane R 2 where G k = {(x,y): x,y R,x > y + k} (i) Prove that τ is topology on R 2. (ii) If τ is topology on R 2 if k R is replaced by k N, by k Q. Solution : τ = { f 1 (G): G τ} Here we want to show that τ is topology on X. By definition of τ, Y and φ are in τ, then by definition of τ we can say that X = f 1 (Y ) τ and φ = f 1 (φ) τ. Let G i are the open sets Y i.e. elements of τ, then by the definition of τ there exists open sets A i of X in τ such that A i = f 1 (G i ). SinceG i is open in Y i G i is open in Y f 1 ( i G i )is open in X i f 1 (G i )is open in X i (A i )is open in X Finally, we can say that if A i are open in X, then i (A i ) also open in X. Exercise 1.17 Show that circle is topological space. Solution : The set of points S 1 = { (x,y) R 2 : x 2 + y 2 = 1 } is called the circle, or the 1-sphere, and is a topological space with the subspace topology of R Base and Subbase for a Topology We are familiar with the notion of a basis for a finite dimensional vector space. It is a minimal collection of vectors that spans the vector space. If we know a basis, we can always recover the vector space. A basis for a topology is in a quite similar fashion a family of open subsets that span the topology. If we know a basis, then we can find the topology for given set. There is no good notion of minimality for a basis here, so there is no such requirement, but it is often convenient to have a basis with as few elements as possible.

14 14 Chapter 1. Topological Spaces In lots of examples of topological space, we are able to specify the entire collection of open sets. But in general this is difficult to do. Instead this we can specify a smaller collection of open sets, called a basis and then generate the rest of the open sets from this collection. The advantage of such collection is that it allow us to easily define a topology on given set. There are two ways by which we can relate basis and topology as first to find basis for a topology and second generation of topology by a basis. In first case we have a topology on any set X then we would like to make a smaller collection of subsets of X which will be basis for given topology. In other hand if we have a smaller collection of subsets of X as basis, then we can generate the topology on X with the help of this collection. Topology Smaller collection of subsets of X as basis Smaller collection of subsets of X as basis Topology This section is divided into two parts where in first part we discuss how a smaller collection called basis, of subsets of non empty set X is able to define a topology on X. We focus on defining this smaller collection and finding require condition under which this collection will give the topology on X. In second section, we learn basic procedure of finding the topology with the help of smaller collection of subset of X Basis for a Topology Consider a set X = {a,b,c} and any topology on X can obtain by selecting randomly some subsets of X in this way that all of will hold there conditions for topology. Let us this selection is τ = {X,φ,{a},{b},{a,b}}. Now think what procedure we are adopting for each subsets of X while considering them as a members of collection τ. At each time, actually we are verifying that subset which we considered for τ is satisfying the require condition for topology or not. This procedure is not convenient if number of elements in X is large. In this situation, now question raise that how to find out the topology while the X is very large as like R,Z and Q etc. We can solve this problem by constructing a smaller collection of subsets of X in this way that it help to generate the topology on given set. So in this section we learn to construct that collection by which we can defined the topology. Now we are going to consider two types of definitions of base for a topology τ on X. The first definition tell us global concept of base where subsets of X is playing important role to define the base while second definition focus local concept where element of X is responsible for making a base. Definition 3 Base in term of Open Set. Let (X,τ) be any topological space, then subcollection B of τ is said to be base for topology if every element of τ can be obtained by union of some elements of B. Example 1.13 Let X = {a,b,c} be topological space with topology τ = {X,φ,{a},{b},{a,b}}, then B = {{a},{b},{c}} is base for the topology τ because every elements of τ can be write as the union of members of B. In other word we can define the base for topology in term of elements of set X as follows. Definition 4 Base in term of Element of Open Set. A sub-collection B of τ is said to be base for a topology τ if for every element x U τ, there exist B in B such that x B U. Example 1.14 Let X = {a,b,c} be topological space with topology τ = {X,φ,{a},{b},{a,b}}, then B = {{a},{b},{c}} is base for the topology τ because for every points a,b,c which contain in member of τ, there exists a members of B which contain these point and also contained in members of τ.

15 1.3 Base and Subbase for a Topology 15 R The base for topology is not unique. Many bases, even of different sizes, may generate the same topology. For example, the open intervals with rational endpoints are also a base for the standard real topology, as are the open intervals with irrational endpoints, but these two sets are completely disjoint and both properly contained in the base of all open intervals. In next result we will see how both above definitions of base for a topology are related to each other. Theorem 2 If B is a base for (the open sets of) a topology τ on a set X, then every open set in X is a union of sets from B i.e. union of some basis element. Proof. Let U be open subset of X (that is U τ), then for each x U there exists B B which contain x and B U. Let us pick one such B for each x can call it B x. Now we have a family {B x : x U} of basic open sets with x U. This show that x B x U. Consider the union V = x U B x. Since B x U for every x U, then V U. But we have U V since x U implies x B x V or x V. Hence we can say that U = V or open set is equal to the union of basic open sets. Theorem 3 Global and local definitions of base for a topology is equivalent. Proof. See in exercise Topology Generated by Basis Now we are going to learn how to find the topology on any set X if the smaller collection of subsets of X is given. But before this first we will again define basis which help to generate the topology on set. Definition 5 Let X be a set and B be a collection of subsets of X, we say B is basis (for a topology on X) if the following conditions hold: (i) For each x in X there exists B in B such that x B. (ii) If B 1,B 2 B and x B 1 B 2, then there exists B 3 in B such that x B 3 B 1 B 2. Here B 1,B 2 and B 3 are called basis elements of basis B. This definition gives necessary and sufficient condition for family of subset to be basis for topology on X. If B is the basis i.e. the x B 1 B 2 B 3 Figure 1.2: Basis for a Topology smaller collection of subsets of X which satisfied the above two condition of Definition 5, then we define the topology generated by B as follows: Definition 6 Topology Generated by a Basis. A subset U of X is said to be open in X (that is, to be element of τ) if for each x U, there is a basis element B B such that x B U.

16 16 Chapter 1. Topological Spaces R Each basis element is itself open set i.e. an element of τ. Example 1.15 If X is any set, the collection of all one-point subsets of X is a basis for the discrete topology on X. Theorem 4 Let X be a topological space and C is a collection of open sets of X such that for each open set U of X and each x U, there is an element C of C such that x C U, then C is a basis for the topology of X. Proof. In order to show that C is a basis, need to show that C satisfies the two properties of basis. To show the first property, let x be an element of the open set X. Now, since X is open, then, by hypothesis there exists an element C of C such that x C X. Thus C satisfies the first property of basis. To show the second property of basis, let x X and C 1 and C 2 be open sets in C such that x C 1 and x C 2. This implies that C 1 C 2 is also an open set in C and x C 1 C 2. Then, by hypothesis, there exists an open set C 3 C such that x C 3 C 1 C 2. Thus, C satisfied the second property of basis too and hence, is indeed a basis for the topology on X. The above result tell us that how to go from a topology to its basis and also gives the tool to obtain a basis for a given topology. Example 1.16 The collection B of all circular regions in the plane generates the same topology as the collection B of all the rectangular regions. Figure 1.3 illustrates the proof. For each element x of an arbitrary circular region in B, there always exists a rectangular region in B containing x which fits within the said circular region and vice versa. x B B B x Figure 1.3: Bases for a Topology The next result provide necessary and sufficient condition for any collection of subsets of non-empty set to be a basis for any topology. Theorem 5 Let X be an arbitrary set and B is a non-empty subset of the power set P(X) of X. Then B is a base for some topology on X if and only if (i) {B: B B} = X. (ii) If x B 1 B and x B 2 B, then there exists B 3 in B such that x B 3 B 1 B 2. Proof. Let B be basis for some topology τ on arbitrary set X, then we prove given two conditions satisfied well. The set X is open as well closed in X and by the definition of base, X τ can be expressed as X = {B: B B}. Hence first condition hold. Let x be arbitrary element of X and B 1 and B 2 are arbitrary basis elements of B such that x B 1 B 2. Since B 1 and B 2 are in B, then B 1 B 2 τ. By definition of base, if x B 1 B 2 τ, then there exist B 3 B such that x B 3 B 1 B 2. Hence second condition hold good. Conversely, suppose the collection B of subsets of X hold given two conditions, then we want to prove that there exists a topology τ on X having B is as basis for topology. Let B 1 is sub-family of B and take τ as collection of all subsets of X which hold both given conditions. Now our aim is to prove that τ is topology on X which of base is B. B

17 1.3 Base and Subbase for a Topology 17 (i) Suppose B = φ, then {B: B B} = {B: B φ} = φ τ. Taking B 1 = B, then by given first condition, {B 1 : B 1 B 1 } = {B 2 : B 2 B} = X τ. (ii) By the construction of τ, each U α is the union of any sub-family of B and therefore α J U α is also the union of members of a sub family of B and hence α J U α τ. (iii) Let U 1 and U 2 are two arbitrary elements in τ such that U 1 U 2 = φ, then obviously U 1 U 2 τ. Now suppose U 1 U 2 φ, then there exists B 1 and B 2 contained in B such that U 1 = {B: B B 1 } and U 2 = {B: B B 2 }. Now U 1 U 2 = [ {B: B B 1 }] [ {B: B B 2 }] = {B 1 B 2 : B 1 B 1,B 2 B 2 } By the second given condition show that B 1 B 2 can be expressed as a union of the members of B. From the above result, we can say that U 1 U 2 is expressible as a union of the members of B and hence U 1 U 2 τ. From above results we conclude that τ is topology on X having B as a base. Theorem 6 Let (X,τ) be a discrete space and B = {{x}: x X}, then family B of subsets of X is a base for τ on X if and only if B contain B. Proof. Let B contain B, then we want to show that B is base for a topology on X but before this first we show that given family B is basis. Since X is discrete space, then every subset A of X is open set. Now consider the family of singleton subset of X as A = {{x}: x A}. This construction show that every non-empty open set is expressible as a union of some members of B. So, by definition B is base for topology on X. Now B is as a base and let U is any non-empty open set, then there exist family B 1 B such that U = {B: B B 1 }. But as given B B, then every non-empty open set U can be expressed as the union of member of B 1. Therefore, B is basis for a topology on X. Conversely, B is base for a topology on discrete space X, then we prove that B contain in B. Let {x} arbitrary element of B and as given B is basis so {x} must be expressed as a union of some of members of B. Since any singleton set is a union of itself or with empty set showing that {x} B. Finally, B contain B. Example 1.17 Find out a sub-base S for the discrete topology τ on X = {a,b,c} such that S does not contain any singleton set. Solution : If we write B = {{x}: x X}, then by above result B is a base for a topology τ on X. Any family B of subsets of X. S does not contain any singleton set. Hence S is the required subbase. On many occasions it is much easier to show results about a topological space by arguing in terms of its basis. For example, to determine whether one topology is finer than the other, it is easier to compare the two topologies in terms of their bases. Here we are going to find the method by which we can compare two topologies on same set when their bases are given. Theorem 7 Comparison of Topologies. Let B and B are bases for topologies τ and τ respectively on X, then following are equivalent. (i) The topology τ is finer than τ. (ii) For each x X and each basis element B B containing x, there is a basis element B B such that x B B. Proof. Given: Two topologies τ and τ defined on X. Two bases B and B which generate the topologies τ and τ respectively. (i) (ii): If τ is finer than τ i.e. τ τ then we wish to show that for each x X and each basis

18 18 Chapter 1. Topological Spaces element B B containing x, there exist a basis element B B such that x B B. Since as given B is basis for the topology τ on X and x X such that x B B, then we can say that B τ. But as given τ finer than τ so B τ. Now B is open in τ and also as given that B is basis which generate τ, so by the definition of topology generated by basis, there exists B B such x B B. (ii) (i): If for each x X and each basis element B B containing x, there exist a basis element B B such that x B B, then we want to show that τ τ. Let U τ be arbitrary open set and x U is arbitrary element. By as given that B generate the topology τ so there exist B B such that x B U. Now by given there is B B such that x B B. This show that x B B U. Therefore x B U. Hence by definition of generation of topology through basis, we can say that U τ. Finally we consider that U τ and we proved that U τ, so τ is finer than T. Theorem 8 Metric Topology. If (X,d) is metric space and B be the collection of all open spheres B d (x,ε),r > 0, then B is base for some topology τ on X. Proof. gg In 1976, A. Rosenfeld and A. C. Kak [rose76] discussed about image processing which is growing discipline with broad applications in document reading in business, automated assembly and inspection in industry, radiology, hematology in medicine and meteorology, geology, land-use management in the environmental sciences, among many other fields [rose1976] also. Digital topology is the study of topological relationship of a digital image display for example a computer serene. This techniques is apply to study the digital image processing. On the set of integers Z, the standard topology turns out to be the discrete topology but non-discrete topologies on Z play a important role in the application of topology. Here we are going to introduce the topology on Z which is useful for the study of digital image processing. Definition 7 Digital Line Topology. For each n Z, define { {n}, if n is odd; B(n) = {n 1,n,n + 1}, if n is even. The collection B = {B(n): n Z} is called basis for a topology on Z and the topology generated by this basis is called digital line topology on Z. We refer to Z with this topology as the digital line. Theorem 9 The collection B = {B(n): n Z} is a basis for a topology on Z, where B(n) for each n Z is defined by { {n}, if n is odd; B(n) = {n 1,n,n + 1}, if n is even. Proof. For proving that B is a basis for a topology on Z, we just verify following two conditions. (i) It is clear by the definition of B(n) if that for integer n there exist an element B(n) B which contain n. (ii) Now suppose B(n) B(m) is a non-empty intersection. Clearly n and m cannot both be odd since B(n) B(m) is nonempty. If one of n and m, say m, is odd and the other is even, then m = n + 1 or m = n 1, and B(n) B(m) = {m} = B(m). If both n and m are even, then m and n are consecutive even integers, and B(n) B(m) = {n + 1} = B(n + 1) if n < m and

19 1.3 Base and Subbase for a Topology 19 R B(n) B(m) = {m + 1} = B(m + 1) if n > m. We can visualize the digital line by encircling its basis elements: Figure 1.4: Basis element for the digital line topology All sets {n} where n is even are closed sets in the digital line. Definition 8 Local Base. Let x be any arbitrary point in a topological space (X,τ), then a family B x of open sets containing x is called a local base at x if and only if for each open set U containing x, there exists U x B x such that x U x U. Example 1.18 If X = {a,b,c,d,e} be space with topology τ = {φ,x,{a}, {a,b},{a,c,d}, {a,b,c,d},{a,b,e}} then B a = {{a}}, B b = {{a,b}},b c = {{a,c,d}} are local bases at a,b,c are respectively for topology τ on X. Example 1.19 The family of all open intervals (a δ,a + δ) in the line R with center a R is a local base at the point a. Example 1.20 Let a plane R 2 with usual topology R s and any point p R 2. Then the family B p of all open disc centered at p is a local base at p. For, as we know that any open set U containing p also contains an open disc D p whose center is p. Make figure also The next result show the relationship between a base for a topology and a local base at a point in topological space. Theorem 10 Let B be a base for a topology τ on X and let x X be arbitrary, then the members of the base B which contain x form a local base at the point x. 1 Theorem 11 A point x in a topological space (X,τ) is a limit point of subset A of X if and only if every members of any local base B x at x contains a point of A different from x. Proof. Let x be a limit point of A and B x be local base at x, then we want to show that (B {x}) A φ for all B B x. Since x is limit point of A, then there exist open set U which contain x such that (U {x}) A φ. Now by the definition of local base if U B x, then this show that U τ with x U. Then the forgoing statement takes the form (U {x}) A φ U B x (B {x}) A φ B B x Conversely, let B x is a local base at x for a topology τ on X and (B {x}) A φ B B x

20 20 Chapter 1. Topological Spaces Now want to show that x is a limit point of A. Let U be any arbitral open set which contain x, then by definition of local base, there exist B B x such that x B U. Consequently (U {x}) A (B {x}) A φ. Thus, x is limit point of A. Theorem 12 Let (X,τ) and (Y,τ 1 ) be two topological spaces and B and B 1 are bases for τ and τ 1 respectively. If each open set U τ is expressible as the union of members of B 1, then τ is weaker than τ 1. Proof Subbasis for a Topology As we know that topology generated by a basis B may be described as the collection of arbitrary unions of elements of B, but what happens if we start with a given collection of sets and take finite intersection of them as well as arbitrary unions? This question leads to the notion of a subbasis for a topology. Definition 9 Subbasis for a topology. A subbasis S for a topology on X is a collection of subsets of X whose union equals to X. Symbolically, a collection S of subsets of X is said to subbasis for a topology on X if S S S = X. Definition 10 Topology generated by subbasis. The topology generated by the subbasis S is defined to be the collection τ of all unions of finite intersections of elements of S. Alternatively, the topology generated by S is defined by the rule: a subset U of X is open if for each x in U there exists S 1,...,S n in S such that x S 1..., S n U. Example 1.21 Every basis for a topology is sub-basis. Example 1.22 Consider the set X = {a,b,c,d,e, f } with the topology τ = {φ,{a},{c,d}, {a,c,d},{b,c,d,e, f },X}. Show that the subset is a subbasis of τ. S = {{a},{a,c,d},{b,c,d,e, f }} τ Solution : The collection of all finite intersections of elements from S is: B S = {φ,{a},{c,d},{a,c,d},{b,c,d,e, f }} Every set in τ apart from X is a trivial union of elements in B S and X = {a} {b,c,d,e, f }, so B S is a basis of τ so S is a subbasis of τ. Example 1.23 Consider the set X = {a,b,c,d,e} with the topology τ = {φ,{a},{b},{a,b},{b,d}, {a,b,d},{a,b,c,d},x}. Show that is not a subbasis of τ. Solution : Consider the following set: S = {{a},{b}{a,b},{a,b,d},{a,b,c,d},x} τ The set of all finite intersection of sets from is: S = {{a},{b}{a,b},{a,b,d},{a,b,c,d},x} B S = {φ,{a},{b},{a,b},{a,b,d},{a,b,c,d},x} All sets except {b, d} can be expressed as trivial intersections. However, {b, d} cannot be expressed as a union of elements from B S, so B S is not a basis of τ and hence S is not a subbasis of τ.

21 1.3 Base and Subbase for a Topology 21 Example 1.24 Let X = {a,b,c,d,e} and let A = {{a,b,c},{c,d},{d,e}}, then the topology on X generated by A can be obtained as follows: First compute the class B of all finite intersection of sets in A : B = {X,{a,b,c},{c,d},{d,e},{c},{d},φ} (Note that X B, since by definition X is the empty intersection of members of A.) Taking unions of members of B gives the class τ = {X,{a,b,c},{c,d},{d,e},{c},{d},φ,{a,b,c,d},{c,d,e}} which is the topology on X generated by A. Example 1.25 Let S = {(a, ): a R} {(,b): b R} is the set of all infinite rays in R. Then S is clearly not a basis for the standard topology on R. For example: (0,1) is open in R, but we can t find an infinite ray containing 1/2 which is contained in (0,1). Other hand, S is a sub-basis, for (a, b) = (a, ) (, b), so if U is an open set containing the point x, and x (a,b) U, we have x (a, ) (,b) U. Example 1.26 Determine the topology τ on the real line R generated by the class A of all closed interval [x,x + 1] with length one. Solution : Let p be any point in R. As we know that closed intervals [p 1, p] and [p, p + 1] belong to A as they have length one. Hence [p 1, p] [p, p + 1] = {p} belongs to the topology τ or all singleton sets {p} are τ-open and so τ is the discrete topology on X. Example 1.27 The intersection of a vertical and a horizontal infinite open strip in the plane is an open rectangle as shown in Figure 1.5. The open rectangles form a base for the usual topology on R 2 and the collection of all infinite open strips (horizontal and vertical) is a subbase for the usual topology on R 2. O Y Figure 1.5: Subbasis for a topology on R 2 X Now we are going to check in next result that topology generated by a subbasis is actually a topology of given set. Theorem 13 Let S be a non-empty collection of subsets of a non-empty set X, then S is sub-base for a unique topology τ for X, that is finite intersection of members of S from a base for τ.

22 22 Chapter 1. Topological Spaces Proof. Suppose B is the collection of all finite intersections of members of S, then we would like to show that B is a base for a unique topology on X. For the we will show the necessary and sufficient conditions for any family of subsets to be base. (i) Since X is intersection of empty collection of members of S, it show that X B and so X = {B: B B}. (ii) Let B 1,B 2 B and x B 1 B 2, then B 1 and B 2 are finite intersections of members of S. Hence B 1 B 2 is also a finite intersections of members of S and so B 1 B 2 B. Theorem 14 If S is subbase for two topologies τ and τ 1, then τ = τ 1. Proof. Let U τ be arbitrary open set in X and as given S is a subbase for topology τ on X, then U = i (S i1 S ini )wheres ik S But S is also a subbase for topology τ 1 so S τ 1. Hence each S ik τ 1. Since τ 1 is a topology, S i1 S in1 τ 1 or U τ 1. Thus τ τ 1. Similarly τ 1 τ and so τ = τ Difference between Basis and Subbasis for a Topology The difference between basis and subbasis in a topology are as follows: (i) Bases and subbases generate" a topology in different ways. (ii) Every open set is a union of basis elements. (iii) Every open set is a union of finite intersections of subbasis elements. (iv) Every basis for a topology is sub-basis also but its converse is not true. For this reason, we can take a smaller set as our sub-basis, and that sometimes makes proving things about the topology easier. We get to use a smaller set for our proof, but we pay for it; with a sub-basis we need to worry about finite intersections, whereas we did not have to worry about that in the case of a basis. Let X = {0,1,2} be any set and consider the family of subsets of X as S = {{0,1},{0,2}}, then we can verify by the definition that S is sub-basis which generate the topology τ S = {φ,{0},{0,1},{0,2},{0,1,2}}. But same time S is not basis which generate the topology τ S because we cannot write {0} as a union of any eleme Exercise:Basis and Subbasis for a Topology Exercise 1.18 Let (X,d) is a given metric space, then show that the collection B defined as B = {B(x,r): x X,r > 0} is a basis for τ d. Exercise 1.19 Show that if B is a basis for a topology on X, then the topology generated by B equals the intersection of all topologies on X that contain B. Solution : Let (X,τ) be a topological space, B basis for τ and let {τ α } be the set of topologies on X that contains B. Now we will show that τ = {τ α }. Let U τ, then U is an union of elements of B. Since τ α is a topology for all α, it follows that U τ α for all α, i.e. U τ α. Since B τ α τ. Exercise 1.20 Show that the countable collection B = {(a,b): a < b,aandbrational} is a basis that generates the standard topology on R. Solution : Let B = {(a, b): a, b Q, a < b}. It is straightforward to check that B is a basis. Let τ be the standard topology on R generated by the basis: {(r,s): r,s R}. Let U τ and let x U. Then (by definition of an open set in a topology generated by a basis) there exists

23 1.3 Base and Subbase for a Topology 23 a basis element (r,s),r,s R, such that x (r,s). Furthermore there exists a,b Q such that r a < x < b s, hence x (a,b) (r,s). It follows, that B is a basis for τ. Exercise 1.21 Show that the collection C = {[a,b) : a < b,aandbrational} is a basis that generates a topology different from the lower limit topology on R. Solution : Let C = {[a, b) : a < b, a and b rational}. It is straightforward to check that C is a basis. Let τ C be the topology on R generated by C. Consider [ 2,2) R l. There are clearly no basis element [a,b) C such that 2 [a,b) [ 2,2), hence τ C is not finer than R l. Since R l is clearly finer than τ C, it follows that R l is strictly finer than τ C. Exercise 1.22 Consider the following topologies on R: (i) τ 1 = the standard topology, (ii) τ 2 = the topology R K, (iii) τ 3 = the finite complement topology, (iv) τ 4 = the upper limit topology, having all sets (a,b] as basis, (v) τ 5 = the topology having all sets (,a) = {x: x < a} as basis. Determine, for each of these topologies, which of the others it contains. Solution : We know that τ 1 and τ 2 are bases for topologies on R. Furthermore τ 3 is a topology on R. It is straightforward to check that the last two sets are bases for topologies on R as well. The following table show the relationship between the given topologies on R. τ 1 τ 2 τ 3 τ 4 τ 5 τ 1 = (1) (2) (3) (4) τ 2 (5) = (6) (7) (8) τ 3 (9) (10) = (11) (12) τ 4 (13) (14) (15) = (16) τ 5 (17) (18) (19) (20) = (1) Using direct theorem??. (2) Since R (0, 1) not finite. (3) Given basis element (a,b) τ 1 and x (a,b) then the basis element (a,x] τ 4 satisfy x (a,x] (a,b), hence τ 4 is finer than τ 1, by theorem??. (4) Given a basis element (a,b) τ 1 and x (a,b) then there are clearly no basis element ( 1,c) τ 5 such that x ( 1,c) (a,b), hence τ 5 is not finer than τ 1, by theorem??. (5) Using direct theorem??. (6) Since R (0, 1) not finite. (7) Given basis element (a,b) k τ 2 and x (a,b) k. If x (0,1) then there exists m Z such that 1 m < x < 1 m 1, hence x [ 1 m,x) (a,b) k. If x / (0,1) then x (a,x] (a,b) k. It follows from (4) and theorem?? that τ 4 is finer than τ 2. (8) Since τ 1 τ 5 and τ 1 τ 2. (9) Let U τ 3, U is non-empty i.e. R U = {r 1,...,r n },r 1 < < r n Since ( ) ( ) ( ) n 1 U = (r 1 i,r 1 ) (r j,r j+1 ) (r n,r n+k ) r=1 j=1 k=1 it follows that U τ 1. (10) Since τ 3 τ 1 τ 2. (11) Let U τ 3, U is non-empty i.e. R U = {r 1,...,r n },r 1 < < r n and let x U. If d = min{ x r i : i {1,...,n}} > 0 then x (x d 2,x + d 2 ] U. It follows from theorem?? that τ 4 is finer than τ 3.

24 24 Chapter 1. Topological Spaces (12) Consider U R {0} τ 3. There are no basis element (,a) τ 5 such that 1 (,a) U, hence τ 5 is not finer than τ 4 by the theorem??. (13) Given basis element (c,x] τ 4 there is clearly no basis element (a,b) τ 1 such that x (a,b) (c,x], hence τ 1 is not finer than τ 4, by the theorem??. (14) Given basis element (c,x] τ 4 there is clearly no basis element B k τ 2 such that x B k (c,x], hence τ 2 is not finer than τ 4 by the theorem??. (15) Since R (0, 1) not finite. (16) Given basis element (c,x] τ 4 there is clearly no basis element (,a) τ 5 such that x (,a) (c,x], hence τ 5 is not finer than τ 4 by the theorem??. (17) Since (,a) = i=1 (a i,a) τ 1 for all a R. (18) Since τ 5 τ 1 τ 2. (19) Since R (, 0) not finite. (20) Given basis element (,a) τ 5 and x (,a) then clearly ( ] x a x x x a,x + (,a) 2 hence τ 4 is finer than τ 5, by the theorem??. Exercise 1.23 Consider the upper limit topology R u on the real line R which has a base the class of open-closed interval (a,b], then (i) Show that open infinite interval (4, ) and the closed infinite infinite interval (,2] are open sets. (ii) Show that any open infinite interval (a, ) and any closed interval (,b] are open sets. (iii) Show that any open-closed interval (a,b] is both open-closed set. Solution : (i) Observe that (4, ) = (4,5] (4,6] (4,7] (,2] = (0,2] ( 1,2] ( 2,2] Since each is the union of members of the base for τ, therefore each sets is open. (ii) Similarly, (a, ) = (a,a + 1] (a,a + 2] (a,a + 3] (,b] = (b 1,b] (b 2,b] (b 3,b] hence each is open sets. (iii) (a,b] c = (,a] [b, ) and the two intervals on the right are open, so their union is open and therefore (a,b] is closed. But (a,b] belongs to the base for R u and so is also open. Exercise 1.24 Determine wether or not the following classes of subsets of the plane R 2 is a base for the usual topology R s on R 2 (i) The class of open equilateral triangles. (ii) The class of open squares with horizontal and vertical sides. Solution : Both of the classes are base for the usual topology on R 2. Let U be an open subset of R 2 and let x U, then there exist an open disc D x centered at x such that x D x U. observe that either an equilateral triangle or a square can be inscribed in D x as given in the diagram. Hence each class satisfied the definition of base. Exercise 1.25 Consider the discrete topology D on X = {a, b, c, d, e}. Find a subbase S for D which does not contain any singleton sets. Solution : As we know that any class B of subsets X is a base for the discrete topology D on X if and only if it contains all singleton subsets of X. Hence S is subsets for D if and only if finite

25 1.4 The Topology on R and R 2 25 O Y x D x U Figure 1.6: Bases for a Topology intersection of members of S gives {a}, {b}, {c}, {d}, and {e}. So, S ={{a,b}, {b,c}, {c,d}, {d,e}, {e,a}} is subbase for D. Exercise 1.26 Prove that if S is a basis of all topologies on X, then the topology generated by S equals the intersection of all topologies on X that contain S. Prove the same if S is a sub-basis. Solution : Let (X,τ) be a topological space, S basis for τ and let {τ i } be the set of topologies on X that contains S. Here we are going to show that τ = τ i. First we assume that S is base for topology τ on X. Let U τ be arbitrary open set, then by standard result U is an union of elements of S. Since τ i is a topology for all i, it follows that U τ i for all i, i.e. U τ i. Hence τ τ i. Since S τ i τ. So, finally τ = τ i. Now we assume that S is sub-base for topology τ on X. Let U τ be arbitrary open set, then by the definition of a subbasis and the standard result, U is an union of finite intersections of elements of S. Since τ i is a topology for all i, it follows that U T i for all i, i.e. U τ i. Hence τ τ i. Since S τ i τ. So, finally τ = τ i. 1.4 The Topology on R and R The Topology on R In mathematics, the real line, or real number line is the line whose points are the real numbers. That is, the real line is the set R of all real numbers, viewed as a geometric space, namely the Euclidean space of dimension one. It can be thought of as a vector space, a metric space, a topological space, a measure space. We now define three topologies on real line R, all of which are of interest. Theorem 15 Usual Topology on R. If τ u the collection of subsets of R defined as X O Y x D x U X τ u = {S R: (x ε,x + ε) S x Rwhereε > 0} φ then this collection τ u is topology on R and call usual topology or standard topology or Euclidean topology on R. Proof. We will show that collection τ u hold all three conditions for topology.

26 26 Chapter 1. Topological Spaces (i) By the definition of τ u, clearly τ u P(R) so we can say that φ τ u is vacuously true and also R τ u since each x R, (x 1,x + 1) R. In fact, for any ε > 0, (x ε,x + ε) R. (ii) Let {A α J }, be a family of sets over indexed set J such that for each α J, A α τ u. Now we want to show that α J A α τ u. Let W = α J A α. For each x in W, there exists α in J such that x in A α. So, by hypothesis, there exists ε > 0 such that (x ε,x + ε) A α W. Thus, α J A α in τ u. (iii) Let A,B in τ u, then we must show that A B in τ u. By hypothesis, if A and B in τ u, then there exists non-zero ε a and ε b such that (x ε a,x + ε a ) A and (x ε b,x + ε b ) B. If we choose ε = min{ε a,ε b }, then (x ε,x + ε) A and (x ε,x + ε) B, that is (x ε,x + ε) A B. Thus, A B in τ u. Theorem 16 Standard Topology. The collection of open intervals as B s = {(a,b) R: a < b} is basis for a topology on R. The topology generated by the basis B s is called standard topology on R which is denoted by R s and corresponding space is denoted by (R,R s ). Proof. To show given collection B s is basis for a topology on R we will verify following two conditions: For all x R, there exist an certainly open interval B = (a,b) such that x (a,b) because every point of R is contained in open interval. Furthermore, consider two open intervals (a,b) and (c,d) such that c < b and b < d. Let arbitrary element x (a,b) (c,d), then there exist open interval x (a 1,b 1 ), where c < a 1 and b 1 < b and (a 1,b 1 ) (a,b) (c,d). This can be verify by particular element as x = 2.5 (1,3) (2,4), then certainly exist an open interval B = (2.1, 2.7) which contain x = 2.5 and also (2.1, 2.7) (1,3) (2,4). (See figure). Thus given collection is B s is basis for a topology on R. R Open sets in the standard topology R s on R which is generated by the basis B s can be obtain by taking unions of open intervals. We can also check that the collection R s generated by the basis B s is, in fact, a topology on R. Theorem 17 Upper Limit Topology. The collection of open-closed intervals as B u = {B = (a,b] R: a < b} is basis for a topology on R. The topology generated by the basis B u is called upper limit topology on R which is denoted by R u and corresponding space is denoted by (R,R u ). Proof. To show given collection B u is basis for a topology on R we will verify following two conditions: Since every real numbers belong to some open-closed interval so R = B B B. So, for every x R, there exist an open-closed interval B = (a,b] which contain x. Furthermore, the intersection of two open-closed interval is either empty set or open-closed set. For example if (a,b] and (c,d] are two open-closed intervals such that a < c < b < d, then (a,b] (c,d] = (c,b]. If x (a,b] (c,d] = (c,b], then certainly there exist open-closed interval (c 1,b 1 ] which contain x and also (c 1,b 1 ] (c,b], where c < c 1 and b 1 < b. Thus B u is basis. Theorem 18 Lower Limit Topology. The collection of closed-open intervals as B l = {B = [a,b) R: a < b} is basis for a topology on R. The topology generated by the basis B u is

27 1.4 The Topology on R and R 2 27 called lower limit topology on R which is denoted by R l and corresponding space is denoted by (R,R l ). Proof. We can proof as like last theorem. Theorem 19 k-topology. If k = {1/n : n N}, then collection B k of all open interval (a, b), along with all sets of the form (a, b) k, form basis for topology on R. This topology is said to be k-topology on R and corresponding space is denoted by (R,R k ). Proof. As given the collection B k can be write as B k = {U : U = (a,b)oru = (a,b) k for somea,b Rwitha < b} Let x R, then x (x 1,x + 1) B. Therefore, for all x R there exist a Now the next result establish the relation between above these topologies as follows : Theorem 20 The topologies of R l and R k are strictly finer than the standard topology R s on R, but are not comparable with one another. Proof. 1. To prove that R l is strictly finer than the standard topology R s on R. Consider an interval (0,1) in R. Here we can see that (0,1) is open in R with respect to topology R l or we can say that (0,1) is as union of sets of the form [a,b). For example [ ) [ ) [ ) (0,1) = 2,1 3,1 n,1 So, any open in R with respect to standard topology R s is also open in R with respect to lower limit topology R l. Now consider an interval [0,1) in R with respect to standard R s. Can[ we write [0,1) as a union of sets of the form (a,b)?. No, if we attempt to include zero, say 1 ) 100,1 will include zero, but as well as erroneous values less than zero. Every set open in R is open in R l, but not the converse. Thus, R l is finer than R s. 2. To prove that R k is strictly finer than the standard topology R s on R. All sets open in R are open in R k. Are sets open in R k open in R as well? Or ( ) ( 1 1 ( 1,1) k = ( 1,0) 2,1 3, 1 ) 2 This is very close because 0 belong left-hand side, but 0 not belong right-hand side. So R k is finer than R s. 3. To prove that R l and R k are not comparable. (a) First we show that R l does not have a finer topology that R k. The set U = ( 1,1) { 1 n : n N} is open in R k, as it is a basis element. However, this set is not open in R l, as any half-open interval [a,b) which contains 0 also contains a small half-open interval of the form [0,ε), where ε > 0. Such a set must contain elements of k = { 1 n : n N} is not open in the R l, so that the R l is not finer than the R k.

28 28 Chapter 1. Topological Spaces The Topology on R 2 (b) Now we show that the R k is not finer than the R l. Let U = [0,1). Then U is open in R l as it is a basis element. But [0,1) is not open in the R k, as a basis element of the R k looks like (a,b) or (a,b) k, and if either of these contains 0 it also contains ( ε,0] for some ε > 0. Thus we cannot find an open set in R k which contains 0 and is contained in [0,1), so that the latter set is not open in R k. After learning some important topologies on R now we are going to discuss topology on Euclidean plan R 2. As we know that Euclidean space (R 2,d) is also a metric space with metric d which is defined as d(p,q) = (p 1 q 1 ) 2 + (p 2 q 2 ) 2, where p = (p 1,q 1 ),q = (p 2,q 2 ) R 2 The subset B(x,ε) of metric space (R 2,d) as called open ball of radius ε with center x is defined as B(x,ε) = {p R 2 : d(x, p) < ε} Before discussion for topology on R 2 we prove following lemma which tell us that that if a point x is in some open ball B in R 2, then there is an open ball centered at x contained in B as well. Lemma 1.1 If y be in R 2 and suppose r > 0, then for every x B(y,r) there exists an ε > 0 such that B(x,ε) B(y,r). x d(x, y) ε y B(x, ε) r B(y, r) Figure 1.7: Every point x in B(y,r) is the center of some open ball as B(x,ε) contained in B(y,r). Proof. If x B(x,ε), then d(x,y) < r. Choose ε such that 0 < ε < r d(x,y). Now we show that B(x,ε) B(y,r). Let z B(x,ε), then we have d(y,z) < d(y,x) + d(x,z) d(y,z) < r < d(y,x) + ε < d(y,x) + r d(x,y) R 2 Therefore z B(y,r), and it follows that B(x,ε) B(y,r). Theorem 21 Standard Topology on R 2. The collection B = {B(x,ε): x R 2,ε > 0} is a basis for a topology on R 2.

29 1.5 New Topological Spaces from Old 29 Proof. For proving that given collection B is basis for topology on R 2 we will verify following two conditions. 1. Since each x R 2 is contained in the basis element B(x,1), the first condition for a basis is satisfied. 2. If x is in the intersection of two basis elements, there is then a basis element containing x and contained in the intersection. Let B(p,r 1 ) and B(q,r 2 ) in B such that x B(p,r 1 ) B(q,r 2 ), then by the above Lemma 1.1 there exist ε 1,ε 2 > 0 such that B(x,ε 1 ) B(p,r 1 ) and B(x,ε 2 ) B(q,r 2 ). Let us choose ε = min{ε 1,ε 2 }, then B(x,ε) B(x,ε 1 ) B(x,ε 2 ) B(p,r 1 ) B(q,r 2 ) and it follows that B satisfies the second condition for a basis Exercise:The Topology on R and R New Topological Spaces from Old The Order Topology Topologies often arise naturally from other types of structures, such as in analysis where almost all topologies are derived from metrics, norms and inner products. Order structures also induce topologies in various ways, but here we consider only one: the order topology on a simply ordered set. If X is a simply ordered set 1, there is a standard topology for X, defined using the order relation. It is called the order topology. Here we will consider it and study some of its properties. Suppose that X is set having a simple order relation <. Given elements a and b of X such that a < b, there are four subsets of X that are called the intervals determined by a and b. They are following ; (a,b) = {x: a < x < b},(a,b] = {x: a < x b}, [a, b) = {x : a x < b},[a,b] = {x: a x b} Definition 11 Order Topology. Let X be a set with a simple order relation and assume that it has more than one element. Let B be the collection of all subsets of the following types : (i) All open intervals (a,b) in X. (ii) All intervals of the form [a 0,b),where a 0 is the smallest element (if any) of X. (iii) All intervals of the form (a,b 0 ],where b 0 is the largest element (if any) of X. The collection B is a basis for a topology on X, which is called the order topology and denoted by τ O. The set X with topology τ O is called order topological space and denoted by (X,τ O ). R If X has not smallest element, there are not sets of type (2), and if X has not largest element, there are not sets of type (3). Furthermore, we can define order topologies on X which are generated by the different basis are as follows. Definition 12 Order Topologies. Let X is simply order set with order relation, then (i) If X has no least element and no greatest element, then order topology τ O1 on X is generated by the basis B o1 = {(a,b): a,b X,a < b} 1 A relation C on a set A is called an simple order set if it has hold three properties as like Comparability, Non-reflexivity and Transitivity.

30 30 Chapter 1. Topological Spaces (ii) If X has a least element a 0 but no greatest element, then order topology τ O2 on X is generated by the basis B o2 = {(a,b): a,b X,a < b} {[a 0,b): b X} (iii) If X has a greatest element b 0 but no least element, then order topology τ O3 on X is generated by the basis B o3 = {(a,b): a,b X,a < b} {(a,b 0 ]: a X} (iv) If X has a least element a 0 and a greatest element b 0, then order topology τ O4 on X is generated by the basis B o4 = {(a,b): a,b X,a < b} {[a 0,b): b X} {(a,b 0 ]: a X} In every case, it is clear that B oi covers X, and any non-empty intersection of two sets in B oi covers X again, so B oi covers X is a basis for a topology on X. Example 1.28 Order Topology on R. The standard topology on R is order topology derived by usual order relation on R. Example 1.29 Order Topology on R 2. Let us consider the set R 2 with dictionary order, then it has neither largest element not smallest element. The order topology on R 2 has as basis the collection of all open intervals of the form ((a,b),(c,d)) for a < c and for a = c and b < d. The subcollection consisting of only second type of intervals is also a basis for a opology on R 2. R (a,b) (c,d) Figure 1.8: Basis of Order Topology of R 2 R (a,d) R (a,b) Figure 1.9: Basis of Order Topology of R 2 R Example 1.30 The positive integer Z form an ordered set with a smallest element and it is

31 1.5 New Topological Spaces from Old 31 discrete topology. Example 1.31 If X = [0,1], with the usual simple order relation, then order topology on [0,1] is generated by the basic τ O - nbd system at x is B τox = {(x ε,x + ε): 0 < ε < min{x,1 x}}. Here B τo0 = {[0,ε): 0 < ε < 1} and B τo1 = {(0,ε]: 0 < ε < 1}. Note that [0,1) is open in the order topology on [0,1], but not in the order topology on R. Example 1.32 Let X = { 5} (0,1) [4,8). A basis for the order topology on X: (i) All open intervals (a,b) in X. (0,b), ifa = 5,b (0,1); (0,1), ifa = 5,b = 4; (0,1) [4,b), ifa = 5,b (4,8); (a,b) = (a,b), ifa (0,1),b (0,1); (a,1), ifa (0,1),b = 4; (a,1) [4,b), ifa (0,1),b (4,8); (a,b), ifa [4,8),b (4,8). (ii) All intervals of the form [a 0,b), where a 0 is the smallest element (if any) of X. { 5} (0,b), ifb (0,1); ( 5,b) = { 5} (0,1), ifa = 5,b = 4; { 5} (0,1) [4,b), ifa = 5,b (4,8); (iii) All intervals of the form (a,b 0 ], where b 0 is the largest element (if any) of X. The X does not have a largest element hence there are no basis elements of this type. Hence a basis for the order topology on X is The Product Topology If X and Y are two topological space, then what happen if we want to define the topology on cartesian product of X Y. This is another way to define the new topology on set X Y. Actually there are two way to define the product topology on X Y. First we can define the product topology on (x,y) with the help of open set of given topologies on X and Y respectively. Other hand we can define the product topology on (x,y) with the help of the bases for the topologies on X and Y respectively. Definition 13 Product Topology in Term of Open Set. Let (X,τ 1 ) and (Y,τ 2 ) be topological spaces. The product topology (let τ) on X Y is topology having as basis the collection B defined as follows : B = {U 1 U 2 : U 1 τ 1,U 2 τ 2 } Now we will show that the collection B of products of open sets as defined in the above definition 13 is actually as basis which generated the product topology on X Y. Theorem 22 Product Topology in Term of Open Set. If (X 1,τ 1 ) and (X 2,τ 2 ) are two topological spaces, then the collection B = {U 1 U 2 : U 1 τ 1,U 2 τ 2 }

32 32 Chapter 1. Topological Spaces V 2 V 1 is a bases for some topology on X 1 X 2. Y (U 1 V 1 ) (U 2 V 2 ) U 1 V 1 U 1 U 2 U 2 V 2 Figure 1.10: Union of open sets Proof. We will show that B is a base for some topology on X 1 X 2. (i) We prove that {B: B B} = X 1 X 2. Since X 1 τ 1 and X 2 τ 2, so by definition of given collection B, X 1 X 2 B. Thus, X 1 X 2 = {B: B B}. (ii) Let U 1 U 2,V 1 V 2 B and let (x 1,x 2 ) U 1 U 2 V 1 V 2, then we will show that there exist W 1 W 2 B such that (x 1,x 2 ) W 1 W 2 (U 1 U 2 ) (V 1 V 2 ) If(x 1,x 2 ) U 1 U 2 V 1 V 2 (x 1,x 2 ) U 1 U 2,(x 1,x 2 ) V 1 V 2 x 1 U 1,x 2 U 2 ;x 1 V 1,x 2 V 2 x 1 U 1 V 1,x 2 U 2 V 2 x 1 W 1,x 2 W 2 (x 1,x 2 ) W 1 W 2 (LetW 1 = U 1 V 1,W 2 = U 2 V 2 ) So, ifu 1 U 2 B,V 1 V 2 B U 1 τ 1,U 2 τ 2 ;V 1 τ 1,V 2 τ 2 U 1 V 1 τ 1,U 2 V 2 τ 2 W 1 τ 1,W 2 τ 2 W 1 W 2 B Thus, we have shown that there exist W 1 W 2 B such that (x 1,x 2 ) W 1 W 2. Let (y 1,y 2 ) be arbitrary element of W 1 W 2, then (y 1,y 2 ) W 1 W 2 y 1 W 1,y 2 W 2 X y 1 U 1 V 1,y 2 U 2 V 2 y 1 U 1,y 1 V 1 ;y 2 U 2,y 2 V 2 (y 1,y 2 ) U 1 U 2,(y 1,y 2 ) V 1 V 2 (y 1,y 2 ) (U 1 U 2 ) (V 1 V 2 ) This show that W 1 W 2 (U 1 U 2 ) (V 1 V 2 ). Finally from result (i) and (ii) we can say that B is base for some topology τ on X 1 X 2.

33 1.5 New Topological Spaces from Old 33 In definition (13), the basis B that we use to define the product topology is relatively large since we obtain it by paring up every open set U 1 in X 1 with every open set V 1 in Y 1. Fortunately, as the next result indicated, we can find a smaller basis for the product topology by using bases for the topologies on X 1 and X 2, rather than using whole topologies themselves. Theorem 23 Product Topology in Terms of Basis Elements. If B 1 and B 2 are bases for the topological spaces (X 1,τ 1 ) and (X 2,τ 2 ) respectively, then collection is a base for the topology τ on X = X 1 X 2. B = {B 1 B 2 : B 1 B,B 2 B 2 } Proof. Let C = {G 1 G 2 : G 1 τ 1,G 2 τ 2 } is a base for the topology τ on X, then by the theorem (22). We have to show that B is a base for τ on X. By the definition of base, if any (x 1,x 2 ) G τ, then this show that there exist G 1 G 2 C such that (x 1,x 2 ) G 1 G 2 G (1.2) Applying again definition of base, we get and If (x 1,x 2 ) G 1 G 2 C x 1 G 1,x 2 G 2 x 1 G 1 τ 1 B 1 B 1 such thatx 1 B 1 G 1 (1.3) x 2 G 2 τ 2 B 2 B 2 such thatx 2 B 2 G 2 (1.4) B 1 B 1,B 2 B 2 B 1 B 2 B The results (1.3) and (1.4) show that there exist B 1 B 2 B such that (x 1,x 2 ) B 1 B 2 G 1 G 2 G (x 1,x 2 ) B 1 B 2 G Thus we have shown that if (x 1,x 2 ) G τ, then there exists B 1 B 2 B such that (x 1,x 2 ) B 1 B 2 G Hence by definition of bases this show that B is base for topology τ on X. Example 1.33 Let X 1 = {1,2,3} and X 2 = {a,b,c,d} are spaces with topologies τ 1 = {X 1,φ,{1}} and τ 2 = {X 2,φ{a},{b},{a,b},{c,d},{a,c,d},{b,c,d}} respectively, then bases for the product topology τ of X 1 X 2 will be as: Suppose B 1 = {X 1,{1}} and B 2 = {X 1,{a},{b},{c,d}} are bases for τ 1 and τ 2 respectively, then basis for the product topology τ on (x,y) is D = {B 1 B 2 : B 1 B 1,B 2 B 2 } i.e. D = {{(1,a)},{(1,b)},{(1,c),(1,d)},{(1,a),(2,a),(3,a)}, {(1,b),(2,b),(3,c)},{(1,c),(2,c),(3,c)},{(1,d),(2,d),(3,d)}} Now we are going to find out the sub-base of product topology when the sub-basis of topological spaces are given.

34 34 Chapter 1. Topological Spaces Theorem 24 Let (X 1,τ 1 ) and (X 2,τ 2 ) are topological spaces and L and M are sub-base of topology τ 1 and τ 2 respectively, then the collection S of all subsets of the form L X 2 and X 1 M is a sub-base for the product topology τ on X 1 X 2, where L L and M M Proof. For proving S is a sub-base for τ on X = X 1 X 2, we have to show that the collection G of finite intersections of members of S is X = X 1 X 2 and so X 1 X 2 G. Let us suppose that {L 1 X 2,L 2 X 2,...,L p X 2 } {X 1 M 1,X 1 M 2,...,X 1 M q } be a non-empty finite-subcollection of S. This intersection of these elements belong to G, by constructing of G. This element of G is [(L 1 X 2 ) (L 2 X 2 )... (L p X 2 )] [(X 1 M 1 ) (X 1 M 2 )... (X 1 M q )] = [(L 1 L 2... L p ) X 2 ] [X 1 (M 1 M 2... M p )] ( A (B C) = (A B) (A C)) = [(L 1 L 2... L p ) X 1 ] [(M 1 M 2... M p ) X 2 ] = (L 1 L 2... L p ) (M 1 M 2... M p ) [ p ] [ q ] = L r M r r=1 r=1 ( (A B) (C D) = (A C) (B D)) ( L n X 1,M n X 2 ) We suppose that B is base for τ 1 on X 1 generated by the elements of L and C is base for τ 2 on X 2 generated by the elements of M. As we know that finite intersection of sub-base from the base for that topology. In point of view, [ p ] [ q ] L r B and M r C r=1 From the result (1.5), it follows that G is expressible r=1 G = {B C : B B,C C} Then G is a base for the topology τ on X by the Theorem (??). But G is obtained from the finite intersections of members of S. It follows that S is a sub-base for τ on X. R Using an inductive construction we can extend the definition of product topology to a cartesian product of finitely many topological spaces Product Topology Generated by Projection Map It is sometimes useful to express the product topology in terms of a subbasis. To do this, we first define certain functions called projections. Definition 14 Projection Map. The map π 1 : X T X is defined by π 1 (x,y) = x is called projection map of X Y onto X. Similarly the map π 2 : X T Y is defined by π 2 (x,y) = y is called projection map of X Y onto Y. (1.5)

35 1.5 New Topological Spaces from Old 35 R 1. Both maps π 1 and π 2 are surjective map. 2. If U is open inn X, then π1 1 (U) = U Y, which is open in X Y. Similarly if V is open in Y, then π2 1 (V ) = X V is also open in X Y. Thus by the definition of topology, the intersection of above two open sets as π1 1 (U) π 1 2 (V ) = (U Y ) (X V ) = U V indicated in figure 1.11 is also open in X Y The Subspace Topology V Y O π1 1 (U) = U Y U π1 1 (U) π 1(V ) = U V π2 1 X 2 (V ) = X V Figure 1.11: Intersection of Horizontal and Vertical Strips in plan In this section we will discuss how a subset of topological spaces inherits a topology from the space itself. There are various way to define the subspace topology in term of open sets, basis, closed set and induced map. But here we will only first two. Given a subset Y of a topological space X, there is a natural way to define a topology on Y, based on the topology on X. Definition 15 Let X be a topological space and let Y be a non-empty subset of X then the collection τ Y = {Y U : U τ} is a topology on Y, called the subspace topology and with this topology, Y is called a subspace of X. We can say that open sets of subspace topology on Y is consists of all intersection of open sets of X with Y. The above definition 15 can be write in the form of theorem as follows : Theorem 25 Let (X,τ) be space and Y is nonempty subset of X, then the collection τ Y = {Y U : U τ} is topology on Y. Proof. (i) Since φ = φ Y and Y = X Y, therefore φ and Y are both open in Y. (ii) Suppose that V 1,V 2,...,V n are open in in Y, then for each i there exists U i, open in X, such that V i = U i Y. Now, V 1 V 2... V n = (U 1 Y ) (U 2 Y )... (U n Y ) = (U 1 U 2... U n ) Y

36 36 Chapter 1. Topological Spaces Since U 1 U 2... U n is open in X, it follows that V 1 V 2... V n is an open set in X intersected with Y, and therefore is open in Y. (iii) Suppose that {V α } is a collection of sets that are open in Y. Then for each α there exists an open set U α in X such that V α = U α Y. Therefore, V α = (U α Y ) = ( U α ) Y Now, U α is open in X, show that V α is an open set in X intersected with Y. Thus, V α is open in Y. Example 1.34 Let X = {a,b,c,d,e} be space with topology τ = {X,φ,{a},{c,d}, {a,c,d}, {b,c,d,e}}. And let Y = {a,d,e} be subset of X. Now Y X = Y, Y φ = φ, Y {a} = {a},y {c,d} = {d},y {a,c,d} = {a,d},y {b,c,d,e} = {d,e}}. Hence by?? the subspace topology τ Y on Y will be τ Y = {Y,φ,{a},{d},{a,d},{d,e}}. Example 1.35 Let (R,τ) be space, where τ is the standard topology on R and N be the set of natural number subset of R, then subspace topology on N will be ( as follows : As we know that every open interval on R is τ-open set. Let G = n 1 2,n + 1 ), n N. ( 2 As G τ and G N = n 1 2,n + 1 ) N = {n}. As we know that subspace topology on N 2 denoted by τ N is defined as τ N = {G N : G τ} = {{n} : n N}. Finally we can say that every singleton set of N is τ N. As any arbitrary subset of N is an arbitrary union of singleton sets and so every subset of N is τ N. Consequently τ N is a discrete topology on N. Example 1.36 Let I = [0,1] is subset of R with the standard topology. In the subspace topology on I, open sets in I are open sets in R intersected with I. For example, sets of the form (a,b), where 0 < a < b < 1, are open in the subspace topology on I (and open in R as well). Also, for 0 < a < 1, the sets [0,a) and (a,1] are open in the subspace topology on I, even though they are not open in R. 0 1 [0,1] Figure 1.12: The subspace of unit interval Example 1.37 The subspace topology on the X-axis as a subset of R 2 (with its usual topology) is the usual topology on R. Now we discuss the relation between basis and subspace topology. R Theorem 26 Basis and subspace topology. Let B be a basis for topology on X and if Y X, then the collection B Y = {B Y : B B} is a basis for the subspace topology on Y.

37 1.6 Subsets of Topological Space 37 X Y y B G 1 Y y G 1 Y Figure 1.13: Inheriting basis elements for the subspace topology B Y Proof. As given Y is subset of X and B is base for topology τ on X. Now we want to prove that B Y is base for any topology τ 1 on Y. For this we show that if any y G 1 τ 1, then this implies that there exist B 1 B Y such that y B 1 G 1. Let G 1 is open in Y and y G 1 is arbitrary, then Now, So by the definition of base, Thus we have show that G 1 τ 1 G τ such that,g 1 = G Y y G 1 y G Y y G and y Y y G τ B B such that y B G B B such that y B Y G Y B 1 B Y such that y B 1 G 1, where B 1 = B Y y G 1 τ 1 B 1 B 1 such that y B 1 G 1 Theorem 27 Closed set and subspace. Let (X,τ) be a topological space and (Y,τ Y ) be subspace, then a subset A of Y is τ Y -closed if and only if A = U D, where D is τ-closed. Proof. We will show that the closed sets in Y are the intersections of Y and the closed sets in Z. Supposed A is subset of Y closed in Y, then A = Y E, where E is open in Y. Therefore, E = Y G, where G τ. Hence A = Y E = Y Y G = Y (X G) Since G τ, X G is τ-closed. Now, suppose A = Y D, where D is τ-closed. Then Y A = Y Y D = Y (X D) where X D is τ-open and Y A is τ Y. Thus A is τ Y -closed Exercise:New Topological Spaces from Old 1.6 Subsets of Topological Space Closed Set

38 38 Chapter 1. Topological Spaces E A Figure 1.14: Closed set and subspace Figure 1.15: 1 Definition 16 Closed Set. A subset A of a topological space X is said to be closed if the set X A is open. Example 1.38 Let X = {1,2,3} be topological space with topology τ = {X,φ,{1}, {2},{1,2}}, then the closed are X,φ,{2,3},{1,3} and {3}. Example 1.39 The empty set φ and the entire real line R are closed sets since their complements R and φ, respectively, are open sets. Example 1.40 The subset [a,b] of R is closed because its complement R [a,b] = (,a) (b,+ ) is open. Similarly [a,+ ) is closed, because its complement (,a) is open and the subset [a,b) of R is neither open nor closed set. Example 1.41 In the plane R 2, the set {x y : x and y 0} is closed, because its complement is the union of the two sets (,0) R and R (,0) each of which is a product of open sets in R and is, therefore, open in R 2. Example 1.42 In the finite complement topology on a set X, the closed sets consists of X itself and all finite subsets of X.

39 1.6 Subsets of Topological Space 39 Example 1.43 In the discrete topology on the set X, every set is open; it follows that every set is closed as well as open. Example 1.44 Consider the following subset of the real line : Y = [0,1] (2,3) in the subspace topology. Now we can see that : In the given space the set [0,1] is open, since it is the intersection of the open set ( 1 2, 3 2 ) of R with Y. The set (2,3) is open as a subset of Y ; it is even open as a subset of R. Since [0,1] and (2,3) are complements in Y of each other, so we can say that both [0,1] and (2,3) are closed as subset of Y. R From above two examples we raise the question that "How is a set different from a door? ". A door must be either open or closed, and cannot be both, while a set can be open, or closed, or both, or neither!. The collection of closed subsets of space X has properties similar to those satisfied by the collection of open subset of X. Theorem 28 Characterization of Topological Space in Term of Closed Set. Let (X, τ) be a topological space, then the following conditions hold: 1. φ and X are closed. 2. Arbitrary intersection of closed sets are closed. 3. Finite union of closed sets are closed. Proof. 1. φ and X are closed because they are the complements of the open sets X and φ, respectively. 2. Given a collection of closed sets {A α } α J, we apply DeMorgan s law, X A α = A α ) α J α J(X Since by definition, the sets X A α are open, so right side of above equation represents as arbitrary union of open sets, and is thus open. Therefore, A α is closed. 3. Similarly, if A i is closed for i = 1,,n, consider the equation X n A i = i=1 n (X A i ) The set on right side of above equation is a finite intersection of open sets and is therefore open. Hence A i is closed set. i=1 R Instead of using open sets, one could just as well specify a topology on a space by giving a collection of sets (to be called "closed sets") satisfying the three properties of this Theorem 28. Now next theorem provide the necessary and sufficient condition for any subset to be closed in the subspace of of given topological space.

40 40 Chapter 1. Topological Spaces Theorem 29 Necessary and Sufficient Condition for a Closed Set. Let Y be a subspace of topological space X, then a set A is closed in Y if and only if A equal the intersection of a closed set of X with Y. Proof. Assume that A = C Y, where C is closed in X. Then X C is open in X, so that by definition of subspace (X C) Y is open in Y. But (X C) Y = Y A. Hence Y A is open in Y, so that A is closed in Y. Conversely assume that A is closed in Y. Then Y A is open in Y, so that by definition it equals the intersection of an open set U of X with Y. The set X U is closed in X, and A = Y (X U), so that A equals the intersection of a closed set of X with Y, as desired. As set A that is closed in the subspace Y may or may not be closed in the larger space X. As was the case with open sets, there are some condition for A to be closed in X as follows. Theorem 30 Condition for a Set to Closed in Subspace as well as larger Space. If A is closed subset of subspace Y of a topological space X and Y is closed in X, then A is closed in X. Proof Limit Point of a Set Consider a sequence 1,.01,.001,.0001,.00001,... in R. This sequence tell us that each term of it close to 0 compare to previous term i.e. each term near to 0. So we can say that the limit point of this sequence is zero. Now consider another set (0,1], the elements of this set are near to zero so its limit point is zero even it not lies in the set. In both example the distance between two points is very important for both sets but in topological space there is no any role of distance between points to describe the closeness property of set. The limit point of any set help to understand closeness property of any set. In this section we are going to discuss about the limit point of any set in topological point of view. Definition 17 Limit Point and Derived Set. Let A is a subset of the topological space X, then point x in X is called a limit point (or cluster point or point of accumulation) of A if every open set containing a intersect A in some other than a itself. Symbolically, the point x in X is called a limit point of A if for every open set U which contain a such that (U {x}) A φ The set of all limit points of A is called the derived set of A and is denoted by D(A). R 1. If x D(A) for all U τ with x U, (U {x}) A φ 2. If x / D(A) for all U τ with x U, U A = φ or {x}. Example 1.45 Let X = {a,b,c,d,e} with topology τ = {X, φ, {a}, {c,d}, {a,c,d}, {b,c,d,e}}, where the closed subsets of X are φ,x,{b,c,d,e},{a,b,e},{b,e},{a} Consider the set A = {a,b,c} X. (i) The element b in X is a limit point of A since the open sets containing b are {b,c,d,e} and X, and each contains a point of A different from b as c.

41 1.6 Subsets of Topological Space 41 In other way we can say that b is the limit point of A since there exist two open sets {b,c,d,e} and X which contain b such that ({b,c,d,e} {b}) A φ and X {b} A φ (ii) Similarly, the points d and e are the limit points of A. (iii) The point a in X is not a limit point of A since the open set {a}, which contains a, does not contain a point of A different from a. Similarly, c is not a limit point of A. Hence the derived set of A is D(A) = {b,d,e}. There are very interesting results about the limit point of subset of discrete space or indiscrete space as follows. Example 1.46 Every subset of discrete space has no limit point. Let (X,τ) be a discrete space and A a subset of X. Then A has no limit points, since for each x X, {x} is an open set containing no point of A different from x. Example 1.47 Let X is indiscrete space, then X is only open set containing any point a X. Hence a is an limit point of every subset of X except the empty set and the set consisting of a alone i.e singleton set {a}. Example 1.48 Consider the set A = { 1 n R: n Z +} as a subset of R with the standard topology. Is 0 is limit point of A. Let U is open set which contain 0, then there is possibility that there exist an open set (a,b) such that 0 (a,b) U and also (a,b) A φ for all such open set U. So, U A φ. Therefore every neighborhood of 0 intersects A, and since 0 is not in A, the intersection contains points other than 0. It follows that 0 is a limit point of A. Example 1.49 Let us consider subset A = (0,1] of R with standard topology. Then there are two possibilities of limit point of set A as follows. (i) If x [0,1], then every x is limit point of (0,1] since of (0, 1] every open U which contain x [0,1] contains an interval (a,b) containing x. Such an interval intersects (0,1] in a point other than x, and therefore x is a limit point of (0,1]. (ii) If x / [0,1], then there are open intervals containing x that are disjoint from (0,1]. Therefore if x / [0,1], then x is not a limit point of (0,1]. Example 1.50 Consider the subset Q of R with the standard topology, then every x in R is a limit point of Q. Given a real number x, there exist an open set U containing x contains an open interval (a,b) that also contains x. But every open interval intersects Q in infinitely many points, and therefore (a,b) intersects Q in a point other than x. Hence, x is a limit point of Q. Example 1.51 Let N be set of natural numbers and τ is topology on N is defined as it consist of empty set and all subsets of N of the form E n N = {n,n + 1,n + 2,...}, then limit point of set A = {4, 13, 28, 37} is as follows. The open sets containing any point p N are sets E i, where i p. (i) If n 0 36, then every open set containing n 0 also contains 37 A which different from n 0. Hence n 0 36 is a limit point of A. (ii) If n 0 > 36, then the open set E n0 = {n 0,n 0 + 1,n 0 + 2,...} contains no point of A different from n 0. So, n 0 > 36 is not a limit point of A. The derived set of A is D(A) = {1,2,3,...,34,35,36}.

42 42 Chapter 1. Topological Spaces Theorem 31 Properties of Derived Set. Let A and B are two non-empty subsets of topological space X, then (i) D(φ) = φ (ii) If x D(A), then x D(A {x}) (iii) If A B, then D(A) D(B) (iv) D(A B) = D(A) D(B) (v) D(A B) D(A) D(B) Proof. (i) Let x arbitrary element of X and U be an open set which contain x, then (U {x}) φ = φ This show that x is not a limit point of φ i.e. x / D(φ). Therefore for any x X impels that x D(φ). Hence D(φ) = φ. (ii) Let x be arbitrate element in D(A), then by definition This is true for all open set U which contain x. Now (U {x}) A φ (1.6) (U {x}) (A {x}) = (U {x} c ) (A {x} c ) = U A {x} c So by definition this show that x D(A {x}). (iii) Let x is arbitrary element of D(A). Now = (U {x} c ) A = (U {x}) A φ = (U {x}) (A {x}) φ A B A (U {x}) B (U {x}) B (U {x}) φ (M N,M φ N φ) x D(B) Hence for any x D(A) show that x D(B). Consequently, D(A) D(B). (iv) Since A A B and B A B, then by the result (3) D(A) D(A B) and D(B) D(A B). Therefore, D(A) D(B) D(A B) (1.7) Now we prove D(A B) D(A) D(B) And for this we prove that if x contain in D(A B), then contain in D(A) D(B) also. Here we will prove its contrapositive result as Any Therefore x / D(A) D(B) x / D(A B) x / D(A) D(B) x / D(A)andD(B) (U {x}) A = φ and(u {x}) B = φ, for allu τ,andx U (U {x}) (A B) = φ x / D(A B) D(A B) D(A) D(B) (1.8) Finally, by the result (1.7) and (1.8), we get the desired result.

43 1.6 Subsets of Topological Space 43 (v) Since A B A and A B B, then by the result (3) D(A B) D(A) and D(A B) D(B). Therefor D(A B) D(A) D(B). Theorem 32 Necessary and Sufficient Condition for Closed set in Term of Limiting Points. A subset A of topological space X is closed if and only if A contains each of its limiting points. Proof. As given a subset A of topological space X is closed, then we want to show that D(A) A. Let x be a arbitrary elements of X A and X A is open set, then (X A) A = φ. By the definition of derived set D(A) not contain x, therefore x D(A). Now if any x X A then it show that x D(A) so X A X D(A). This proves that D(A) A. Conversely, suppose that A is subset of topological space X such that D(A) A, then we prove that A is closed. Let x X A be arbitrary, then x / A. Now x / A,D(A) A x / D(A) there existu τ withx U such that(u {x}) A = φ U A = φ ( x / A) U X A Therefore, any x X A show that there exist a open set U which contain x such that G X A, so by the definition of interior point x is interior point of X A. Since x is arbitrary hence each point of X A is an interior point which show that X A is open or A is closed. Theorem 33 Relation Between Closed Set and Derived Set. Every derived set in a topological space is a closed. Proof. Let A be subset of topological space (X,τ), then we want to prove that D(A) is closed. By the theorem (??) D(A) will be closed only if D(D(A)) D(A). Let x D(D(A)) be arbitrary, then x is a limit point of D(A). Therefore by definition of derived set there exist a open set U which contain x such that (U {x}) D(A) φ. This show that x D(A). Theorem 34 Union of any Subset and their Derived Set. The union of any subset of topological space and their derived set is closed. Proof. Let A be any subset of topological space (X,τ) and D(A) is the derived set of A, then we want to prove that A D(A) is closed. For this, we will prove that X (A D(A)) is open. Let x X (A D(A)) be arbitrary, then x / (A D(A)) so that x / A and x / D(A). If x / D(A), then there exist a open set U which contain x such that (U {x}) A = φ. This show that U A = φ since x / A. Now we will show that U D(A) = φ. Let y be a arbitrary element of U. Since U is open set which contain y such that U A = φ, show that y / D(A). Therefore U D(A) = φ. Now U (A D(A)) = (U A) (U D(A)) = φ φ = φ Thus U (A D(A)) = φ show that U X (A D(A)). Therefore any x X (A D(A)) show that there exist a open set U which contain x such that U X (A D(A)), hence by definition of interior point x is interior point of X (A D(A)). Therefore, X (A D(A)) is open since x is arbitrary point of it.

44 44 Chapter 1. Topological Spaces R If A is subset of topological space (X,τ), then A D(A) is smallest closed subset of X containing A Closure, Interior and Boundary of a Set After learning about limit point and derived set we are able to utilize this to find the smallest closed subset as closure of any subset of topological space. In this section we are going to discuss three important definitions of closure of any set like closure in term of closed set, closure with respect to particular element of set and closure in term of derived set. Definition 18 Closure of Set. Let A is subset of a topological space X, then closure of A is defined as the intersection of all closed sets containing A. It is denoted by cl(a) or A. Symbolically, A = {F X : F is closed and A F} We can also say that A = F i, where F i are closed super sets of A. Alternatively, the local definition of closure is as follows. Definition 19 Point of Closure. A point x of a space X is called a point of closure of subset A of X if every open set containing x as point of A. The set of all closure points of A is called closure of A. A point x A is called a point of closure. We can also defined closure of any subset of topological space in term of derived set as follows. Definition 20 Closure Set in term of Derived Set. Let A be a subset of a topological space X, then the set A D(A) consisting of A and all its limit points is called the closure of A. R 1. Since an arbitrary intersection of closed set is closed so the closure of any set is always closed. 2. The closure of any set A contain A i.e. A A. 3. The closure of any set A is smallest closed set. Example 1.52 Let X = {a,b,c,d,e} with topology τ = {X,φ,{a},{c,d},{a,c,d},{b,c,d,e}}, where the closed subsets of X are φ,x,{b,c,d,e},{a,b,e},{b,e},{a} (i) Let A = {b} be subset of X, then collection of all closed which contain {b} is {X,{b,c,d,e},{a,b,e},{b,e}} and intersection of all these closed sets is {b,e}, which is require closure of A. (ii) Similarly, {a,c} = X,{b,d} = {b,c,d,e}. Example 1.53 (i) If (X,τ) is discrete topological space, then any subset A of X is closed. Therefore, A = A. (ii) If (X, τ) is discrete topological space. If A = φ, then A = φ. If A φ, then X is only closed superset of A so A = X. Simply we can say that if A is subset of discrete topological space X, then { φ, if A = φ; A = X, if A φ.

45 1.6 Subsets of Topological Space 45 Example 1.54 Let N be set of natural numbers and τ is topology on N is defined as it consist of empty set and all subsets of N of the form E n N = {n,n + 1,n + 2,...}, then closure of subsets A = {7,24,47,85} and B = {3,6,9,12,...} of N are as follows. Since by definition closure of set is smallest closed superset of A, therefore A = {1,2,...,84,85} and B = N Example 1.55 (i) If A = [0,1) as a subset of R with the standard topology, then A = [0,1]. (ii) If A = [0,1) as a subset of R with the discrete topology, then A = [0,1). (iii) If A = [0,1) as a subset of R with the finite complement topology. Since A is infinite, and the only infinite closed set in this topology is R it follows that A = R. (iv) If A = [0, 1) as a subset of R with lower limit topology. Here R [0, 1) = (, 0) [1, ) is an open set in the lower limit topology. Therefore [0,1) is closed in the lower limit topology and hence A = A. Example 1.56 Let a topological space (R,τ), where topology τ on R is consisting R,φ and all open infinite intervals E a = (a, ), where a R, then we can find closure of A = [3,7),B = {7,24,47,85},C = {3,6,9,12,...} Since closure of set is smallest closed superset, therefore A = (,7], B = (,85], C = R Example 1.57 Consider the subset Q of R with the standard topology, then Q = R. This can be show with contradiction. Let Q R, then there exists an x R Q. As R Q is open in R, there exist a,b with a < b such that x (a,b) R Q. But in every interval (a,b) there is a rational number q; that is, q (a,b). So q R Q which show that q R Q. This is a contradiction, as q Q. Hence Q = R. Definition 21 Interior of a Set. Let A is subset of a topological space (X, τ), then interior of A is defined as the union of all open sets contained in A. It is denoted by Int(A) or A. Symbolically, A = {U : U is open and U A} We can also say that A = U i, where U i are open subsets of A. Definition 22 Interior Point. Let A is subset of a topological space X, then point a of A is called interior point of A if a belong to open set U contained in A i.e. a U A. The set of interior points of A is called interior set of A. R 1. Since an arbitrary union of open sets is open so the interior of any set is always open. 2. The interior of any set A contained in A i.e. A A. 3. The interior of any set A is the largest open set. 4. By the definition of closure and interior, A A A.

46 46 Chapter 1. Topological Spaces Definition 23 Exterior Set. Let A be subset of a topological space X, then a point a X, is said to be an exterior point of A, if there exist an open set U, such that x U A c. In other words, a point x A c is said to be an exterior point of A if there exists an open set U containing x such that U A c. The set of all exterior points of A is called exterior set of A and it is denoted by ext(a). R The exterior set of A is interior of complement of A i.e. int(a c ). Definition 24 Boundary of Set. The boundary of set A is the set of points which do not belong to the interior or the exterior of set A and denoted by b(a). Example 1.58 Let X = {a,b,c,d,e} with topology τ = {X, φ, {a}, {c,d}, {a,c,d}, {b,c,d,e}}. Let A = {b,c,d} is subset of X, then (i) The elements c and d are interior points of A since c, d {c,d} A, where {c,d} is an open set. The point b A is not interior point of A; so A = {c,d}. (ii) The point a in X is exterior to A i.e. interior to the complement A c, hence int(a c ) = ext(a) = {a}. (iii) The boundary of A is consist of the points b and e i.e. b(a) = {b,e}. Example 1.59 If A = [0,1) subset of R with the standard topology, then (i) With the standard topology, A = (0,1). (ii) With discrete topology, A = [0,1). (iii) With finite complement topology, A = φ since there are no nonempty open sets contained in [0,1). Example 1.60 Let A be proper subset of indiscrete space X, then there are only two open sets as X and φ. Since X A so only open subset of A is φ, therefore A = φ and ext(a) = (A c ) = φ. Thus b(a) = X. Example 1.61 Let τ be topology on R consisting of R,φ and all open finte interval E a = (a, ), wehre a R. The interior and exterior and boundary of A = [7, ) are as followes. Interior of set is largest open subset of A, so A = (7, ). The complement of A is A c = (,7) contain no open set except φ. Therefore (A c ) = ext(a) = φ. The bounday consists of those points which do not belong to A or ext(a). hence (,7] Example 1.62 Let (Q,τ) is space with standard topology. Suppose Q = φ. Assume it is not, and suppose that U is a nonempty open set contained in Q. Let x be an element in U, then there is an open interval (a,b) such that x (a,b) U Q. But between every pair of real numbers there is an irrational number. Thus every interval contains elements of R Q, and therefore so does U. This is a contradiction; hence Q = φ. Now we can find int(q c ) = ext(q) = φ, therefore b(q) = R. After defining some important concepts as like closed set, closure set, interior set, exterior set and boundary set for topological space we are going to establish relation among them. Theorem 35 Closure of Closed Set. A subset A of a topological space X is closed if and only if A = A. In other words, closure of any closed subset of topological space is the set itself. Proof. This can be prove using two equivalent definitions of closure of a set. (i) By definition, A = F S S, where S is the set of all closed subsets of X that contain A, so A A since A is an intersection of sets F each of which contains A. And A A since one of

47 1.6 Subsets of Topological Space 47 the sets A whose intersection is A is A itself, since by hypothesis A is a closed subset of X which contains A. (ii) If x A then every open neighborhood U of x clearly meets A (since x is in both U and A), and hence by definition, x A. Thus A A. And if x A, then every open neighborhood of x meets A by definition. If x / A, then x A c and A c is open (since A is closed), so A c is an open neighborhood of x that does not meet A, which is a contradiction. Thus we must have x A, and hence A A. Theorem 36 Interior of Closed Set. A subset A of a topological space X is closed if and only if A = A. In other words, interior of any closed subset of topological space is the set itself. Proof. By definition of interior, Int(A) is open set since arbitrary open set is open. As given Int(A) = A, therefore A is open. Conversely, let A is open subset of X, then by definition interior it is clear that any open set U A show that U A. Since A is open, therefore A Int(A). But for all open set AInt(A) A. Consequently, A = Int(A). Theorem 37 Let (X,τ) be a topological space, A and B are any tow arbitrary subsets of X, then (i) The colure of empty set is empty i.e. φ = φ. (ii) If A B, then A B. (iii) A B = A B (iv) A B A B (v) The closure of and closure set is set itself i.e. A = A Proof. (i) Since φ is closed so φ = φ. (ii) Let A B, then A B B. Since closure of any set is closed and so B is a closed set containing A. Also A is a smallest closed set containing A. (iii) As A A B and B A B, then and A A B, B A B A B A B A A, B B A B A B Since A and B are closed, then A B is also closed. Now A B is closed set continuing A B. Also A B is the smallest closed set containing A B. Consequently, A B A B Hence we can say that A B = A B. (iv) As A B A and A B B, then this show that A B A, A B B Hence A B A B. (v) As we know that A is closed for all A and also B is closed if and only if B = B, then easily we can say that A = A. Now we are going to discuss about the closure of finite union of any subset of topological space with the help of nest result.

48 48 Chapter 1. Topological Spaces Theorem 38 Let (X,τ) be any topological space and A i are subset of X, then [ ] n A i = i=1 n A i i=1 Proof. Using mathematical induction method we can prove it. As we know that So we can say that Let C = A 1 A 2, so that C = A 1 A 2. Now Hence, by induction it follows that A B = A B, A,B X A 1 A 2 = A 1 A 2 C A 3 = C A 3 = A 1 A 2 A 3 A 1 A 2 A 3 = A 1 A 2 A 3 [ ] n A i = i=1 n A i i=1 Example 1.63 With the help of counter example we can show that A B = A B is not true always. Consider two subsets A = (0,1) and B = (1,2) of R with usual topology on R, then A B = φ and so A B = φ = φ. And Now Hence A B A B. A B = [0,1] [1,2] = {1} A B = φ {1} = A B or A B A B Theorem 39 Let sets A and B are two subsets of a topological space X, then (i) (X A) = X A (ii) (X A) = X A Proof. (i) As we know that that A is closed and contains A, and therefore X A is an open set contained in X A. Hence X A (X A). Let x (X A) be arbitrary. Note that (X A) is disjoint from A, and therefore x is in an open set that is disjoint from A. Therefore it follows that x / A; hence, x X A. Thus, (X A) X A. Hence we can say (X A) = X A. (ii) Let x X A, then x / A.

49 1.6 Subsets of Topological Space 49 Theorem 40 If A is subset of topological space, then A = A D(A). Proof. For the proving of A = A D(A), we will show that A A D(A) and A D(A) A. Since A D(A) is closed, therefore A D(A) = A D(A). Now A A D(A) this show that A A D(A) = A D(A). Hence A A D(A). As we know that A A. Now we will show that D(A) Aor D(A) i {F i X : F i is closed setf i A} Let x D(A) be arbitrary element. x D(A) x is a limit point of A x is a limit point of all those sets which containa x is a limit point of allf i x D(F i ) F i x F i for all i x i {F i X : F i is closed setf i A} x A Hence D(A) A. Now A A and D(A) A, therefore A D(A) A. Theorem 41 Let A be subset of topological space (X,τ), then a point x of A is a interior point of A if and only if x is not a limit point of X A. Proof. Let x is an interior point of A, then we will show that x is not a limit of X A or x / D(X A). x A there exist open setu with x U such thatu A U (X A) = φ (U {x}) (X A) = φ x / (X A) Therefore, we have open set U which contain x such that (U {x}) (X A) = φ, hence by the definition of derives set x / D(X A). Conversely suppose x is not a limit of X A, then we want to show that x A. By given x A, x / D(X A) there exist a open setu withx U such that(u {x}) (X A) = φ U (X A) = φ U A Therefore, if x A, then there exist open set U which contain x such that U A, hence by definition of interior point x A. Theorem 42 If A and B are two subsets of topological space (X,τ), then (i) φ = φ and X = X. (ii) If A B, then A B (iii) (A ) = A (iv) (A B) = A B (v) (A B) A B Proof.

50 50 Chapter 1. Topological Spaces (i) As we know that if A is open then A = A and since φ and X both are open so easily we can say that φ = φ and X = X. (ii) Let x is interior point of A, then there exist an open set U with x U such that x U A. Since A B, therefore x U A B. So, by defintion of interior point x B. Hence A B. (iii) As we know that interior set is always open and also A = A. Therefore (A ) = A. (iv) We will show that (A B) A B and B (A B) A. Let x be arbitrary element of (A B), then by definition there exist open set U which contain x such that U A B. Now x U A B x U Aandx U B ( A B A,A B B) x A, x B x A B Hence (A B) A B. Let y A B, then y A,y B U A,U B τ such thaty U A Aandy U B B Therefore A B (A B). (v) Let x A B, then R y U A U B A B y (A B) x A, x B U A such thatx U A Aor U B such thatx U B B U A U B τ such thatx U A U B A B x (A B) Therefore, (A B) A B 1. (A B) is not subset of A B. 2. In general (A B) A B. Let A = [0,1] and B = [1,2] are subset of topological space (R,R U ) with usual topology R U. Then A = (0,1),B = (1,2) and A B = (0,1) (1,2) = (0,2) {1}. And A B = [0,1] [1,2] = [0,2], therefore (A B) = (0,2). If a set X is given, then to promote it to the topological space (X,τ), we have to define a family τ of its subsets which satisfies certain conditions. There are many ways of defining (or finding) a topology on X. Now we are going to learn such type of the families of subsets which tern into topological space under the certain conditions. Theorem 43 Interior Operator. Let I : P(X) P(X) be an interior operator defined on set X, then there exists a unique topology τ on X such that for each subset A of X, I (A) = A. Proof. By the definition of interior operator, a map I : P(X) P(X) is interior operator on X defined as (i) I (X) = X (ii) I (A) A (iii) I (A B) = I (A) I (B) (iv) I (I (A)) = I (A)

51 1.6 Subsets of Topological Space 51 Where A and B are subsets of X and P(X) is power set of X. Now our aim is to show that there exist a unique topology τ on X such that I (A) = A. The collection τ can be written as τ = {A X : I (A) = A} (1.9) Now we will prove that the collection τ satisfied conditions for topology. (i) By the given definition of interior operator as I (X) = X, therefore X τ. (ii) By the given definition of interior operator as I (A) A, therefore I (φ) φ but as we know that φ I (φ). Hence I (φ) = φ so φ τ. (iii) Now by the definition of collection τ U 1,U 2 τ I (U 1 ) = U 1,I (U 2 ) = U 2 I (U 1 U 2 ) = I (U 1 ) I (U 2 ) I (U 1 U 2 ) = U 1 U 2 U 1,U 2 τ U 1 U 2 τ (iv) Let A B, then A B = A and therefore I (A) = I (A B) = I (A) I (B). So, we can say that if A B, then I (A) I (B). Let U i τ for all i J so that I (U i ) = U i. Suppose {U i : i J} = U. Then U i U I (U i ) I (U) U i I (U) {U i : i J} I (U) U I (U) But I (U) U, therefore I (U) = U. Consequently U τ. Finally the collection τ satisfied all the conditions for topology on X. Now we will show that I (A) = A. As given that I (I (A)) = I (A) and by the definition of collection τ, I (A) τ. Thus I (A) in open set such that I (A) A. Let B be an open set such that it contain A, then I (B) = B and I (B) I (A). Thus B I (A). Finally, I (A) contains any open set B such that B A, it follows that I (A) is the largest open subset of A. Consequently I (A) = A. The Polish mathematician Kazimierz (Casimir) Kuratowski ( ) developed a radically different approach to specifying a topology for a set. Kuratowski considered particular functions from the set of subsets of X to the set of subsets of X. Definition 25 Kuratowski Closure Operator. If X is any non-empty set and A,B be subset of X, then the function C : P(X) P(X) is called closure operator on the subsets of X which is defined as follows: (i) φ = φ (ii) A A (iii) (A) = A (iv) A B = A B The set X with closure operator C is called Kuratowsi Space and it is denoted by (X,C ). R The above conditions are trivially satisfied by any topological space provided the closure of a set is defined in the usual way. With the help of next result we can understand that how any topological space with usual definition of closure is a Kuratowsi Space.

52 52 Chapter 1. Topological Spaces Theorem 44 Kuratowsi Theorem. Let C be closure operator defined on X and F be the collections of all subsets A of X such that C (A) = A. Let τ be the family of all complements of members of F, then τ is a topology on X and C (A) = A Proof. Let C : P(X) P(X) be a closure operator defined on X such that (i) C (φ) = φ (ii) A C (A) The given two family of subsets of X can be write as (iii) C (C (A))C (A) (iv) C (A B) = C (A) C (B) F = {A X : C (A) = A} (1.10) τ = {X A: A F} (1.11) Now we will prove that the collection τ satisfied conditions for topology. (i) Since C (φ) = φ, then this show that φ F. Therefore X φ τ or X τ. (ii) By definition of closure operator X C (X) and since X is universe set then C (X) X, therefore C (X) = X. Hence by definition of collection F, X F. Finally by the definition of collection τ, X X = φ τ. (iii) We shall prove that if U 1 and U 2 are arbitrary members of τ, then U 1 U 2 also in τ. But before proving this first we shall show that A B, then C (A) C (B). A B A B = B C (B) = C (A B) C (B) = C (A) C (B) C (A) C (A) C (B) = C (B) C (A) C (B) (1.12) If U 1 and U 2 are arbitrary members of τ, then X U 1 and X U 2 in F. Therefore, C (X U 1 ) = X U 1, C (X U 2 ) = X U 2 (1.13) C (X (U 1 U 2 )) = C [(X U 1 ) (X U 2 )] = C (X U 1 ) C (X U 2 ) = (X U 1 ) (X U 2 ) (By De-Morgan law) C (X (U 1 U 2 )) = X (U 1 U 2 ) Therefore C (X (U 1 U 2 )) F and finally U 1 U 2 τ. (iv) Now we shall prove that if U i τ for all i J, then {U i : i J} τ. If U i inτ, then X U i F. By definition of closure operator X U i = C (X U i ). Let us suppose {U i : i J} = U. X U = X i U i = i (X U i ) X U i X U X U i for alli J C (X U) C (X U i ) = X U i C (X U) X U i C (X U) i (X U i ) = X i U i = X U C (X U) X U

53 1.7 The Metric Topology 53 But X U C (X U). Consequently X U = C (X U) so that X U F and therefore U T. Finally, the collection τ satisfied all the conditions for topology. Now we shall prove that C (A) = A. By the definition of closure operator that C (C (A))C (A), so C (A) F and therefore X C (A) τ. Hence C (A) is closed set, then A C (A). Thus C (A) is a closed set containing A, therefore C (A) = A. Theorem 45 Exterior Operator. Let E : P(X) P(X) be an exterior operator defined on set X, then there exists a unique topology τ on X such that for each subset A of X, E (A) = ext(a). Proof. As given E : P(X) P(X) be an exterior operator defined on set X such that (i) E (X) = φ and E (φ) = X (ii) E (A) A c (iii) E (A B) = E (A) E (B) (iv) E (E (A)) c = E (A) Where A and B are subsets of X and P(X) is power set of X. Now we shall prove that the collection τ = {U X : E (U c ) = U} is unique topology on X. (i) Since E (φ c ) = E (X) = φ and E (X c ) = E (φ) = X, hence φ and X contain in τ. (ii) Let U 1 and U 2 in τ, then by the definition of operator E (U1 c) = U 1 and E (U2 c) = U 2. As we know that (U 1 U 2 ) c = U1 c Uc 2, then E [(U 1 U 2 ) c ] = E (U c 1 U c 2) = E (U c 1) E (U c 1) = U 1 U 2 This show that U 1 U 2 τ. (iii) Let U i in τ for all i J, then we shall prove that U = {U i : i J} contain in τ. It is easy to show that if A B, then E (A) E (B). By the De Morgan s law we can say that U c = {Ui c : i J}. Now by the definition of operator E (U c ) (U c ) c or E (U c ) U. If U i U U c (U i ) c E (U c ) E ((U i ) c ) U i E (U c ) U i E (U c ) U E (U c ) Finally, U = E (U c ) so that U τ. Now we prove that E (A) = ext(a). By the definition of operator E (E (A)) c = E (A) show that E (A) τ and as E (A) A c given then we can say that E (A) is an largest open set contained in A c. Therefore, E (A) = ext(a) Exercise:Limit Point, Interior, Closure, and Boundary for Topological Space 1.7 The Metric Topology 1.8 The Quotient Topology

54

55 II Algebraic Topology Bibliography Books Articles Index

56

57 Bibliography Books Articles

58

59 Index k-topology, 27 [Base in term of Element of Open Set, 14 [Comparison of Topologies, 17 Base in term of Open Set, 14 Basis for a Topology, 15 Boundary of Set, 46 Closed Set, 38 Closure of Set, 44 Closure Set in term of Derived Set, 44 Complement Topology, 10 Complement Topology on R, 10 Derived Set, 40 Digital Line Topology, 18 Discrete Topology, 8 Excluded Point Topology, 11 Exterior Operator, 53 Exterior Set, 46 Finite Complement Topology, 9 Interior of a Set, 45 Interior Operator, 50 Interior Point, 45 Kuratowsi Theorem, 52 Kuratowski Closure Operator, 51 Limit Point, 40 Local Base, 19 Lower Limit Topology, 26 Metric Topology, 18 Non-discrete Topology, 8 Order Topology, 29 Order Topology on R, 30 Order Topology on R 2, 30 Particular Point Topology, 11 Partition Topology, 10 Point of Closure, 44 Product Topology in Term of Open Set, 31 Projection Map, 34 Standard Topology, 26 Standard Topology on R 2, 28 Subbasis for a topology, 20 Subspace Topology, 35 Topological Space, 7 Topology Generated by a Basis, 15 Topology Generated by Basis, 15 Topology generated by subbasis, 20 Upper Limit Topology, 26 Usual Topology on R, 25 Weaker and Stronger Topologies, 8