Survey

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Transcript

Set Topology MTH 251 • Lecture # 11 to 32 • Lecture Notes • Introduction: • One way to describe the subject of Topology is to say that it is qualitative geometry. • The idea is that if one geometric object can be continuously transformed into another, then the two objects are to be viewed as being topologically the same • For example, a circle and a square are topologically equivalent. • Physically, a rubber band can be stretched into the form of either a circle or a square. • A figure eight curve formed by two circles touching at a point is to be regarded as topologically distinct from a circle or square • A qualitative property that distinguishes the circle from the figure eight is the number of connected pieces that remain when a single point is removed 1 • When a point is removed from a circle what remains is still connected, a single arc, whereas for a figure eight if one removes the point of contact of its two circles, what remains is two separate arcs, two separate pieces. • The term used to describe two geometric objects that are topologically equivalent is homeomorphic. • Thus a circle and a square are homeomorphic. • if we place a circle C inside a square S with the same center point, then projecting the circle radially outward to the square defines a function f :C→S , and this function is continuous: small changes in x produce small changes in f (x). The function f has an inverse : S→C obtained by projecting the square radially inward to the circle, and this is continuous as well. One says that f is a homeomorphism between C and S . • One of the basic problems of Topology is to determine when two given geometric objects are homeomorphic. • Our first goal will be to define exactly what the ‘geometric objects’ are that one studies in Topology. • These are called topological spaces. 2 • Definition #1: (Topological) • Let X be a non-empty set. A collection of subset of X is called a topology on X, if it meets the following requirements: ( ) and X belong to . ( ) the union of any number of sets in belongs to ; ( ) the intersection of any two sets (and hence of any finite number of sets) in belongs to ; • Definition #2: (Topological Space) • A pair (X, ), where X is a non-empty set, and is a topology on X is called a topological space. • The set X is called its ground set and the elements of X are called its points. • Members of a topology subsets of X. on a set X are called open • If ∈ then is open subset of X and closed subset of X. − is known as 3 • Example #1. • Let X={a,b}. Consider the following four collection of subsets of X: ={ , { , } } = { , { }, { , } } = { , { }, { , } } = { , { }, {b }, { , } } Then each • Remark : (for = 1,2,3,4) is a topology on X. • From the above example it is clear: • There may be more than one topologies on a set X • therefore there exist more than one topological spaces of which X is the ground set. • Example #2: • Let X={a,b,c,d}. Consider the following collection of subsets of X. = { }, { , }, { , , , } = = , { }, { , } , { }, { , }, { , , }, { , , , } 4 (i) = , { }, { }, { , , , } is not a topology on X, because to does not belong (ii) is not a topology, because X does not belong to (iii) is not a topology, because { } = { , } ∩ { , , } does not belong to ; and so ( ) is not satisfied. (iv) is not a topology, because { } ∪ { } = { , } does not belong to ; and so ( ) is not satisfied. • Remark: • The above example shows that not every collection of subsets of X is a topology on X. • Example #3: (Co-finite topology) • Let X be a non-empty set and let be the collection of the null set and all subsets of X, whose complements are finite. Then is a topology on X. • This topology is called co-finite topology or (The significance of -topology will appear in later Lectures). 5 Example #4: • Let X be a non-empty set • let = P(X) where P(X) is power set of X, i,e., the set of all subsets of X. • Then is a topology on X. This topology is called discrete topology . • (X, ) is called discrete topological space. • Example # 5: • The Co-finite topology on X is discrete if X is a finite set. • Example # 6: (Indiscrete Space) • For any non-empty set X, • = { , } is a topology on X. This topology is called Indiscrete topology and (X, ) is called indiscrete space. • Note: If 1. 2. and ∪ ∩ are two topologies on X. Then: may not be a topology on X. is always a topology on X. 6 Example #7: (metric topology) Let ( , ) be a metric space. Let subsets of X such that, for each consist of all those ∈ , there is an open sphere is a topology on X. This ( ) ⊆ . Then topology is called the metric topology on X. • Example # 8: • Let . = {1,2,3, … } and . Then = { , • Example # 9: , ={ , , …, + 1, + 2, … } for , … } is a topology on • Let = and a collection of subsets of which are unions of open intervals in . Then is a topology on . This topology is called usual topology on . • Example # 10: • Let = and a collection of subsets of which are unions of open discs in . Then is a topology on . This topology is called usual topology on . • Example # 11: • Let = { , } and = { , { }, }. Then is a topology on X. This topology is call Sierpinski topology and (X, ) is called Sierpinski space. 7 Lecture # 12 • In this Lecture we will define • Neighbourhood of a point in • Coarser and Finer topologies • And discuss results on open sets and • Give characterizations of neighbourhood of a point • Remember: • If is a topology on X, then: • Union of any collection of open sets is again open set in X • Intersection of finite number of open sets is again open set in X • If is open set in X then − is closed set in X. • Union of finite number of closed sets is a closed set in X. • Intersection of any number of closed sets is a closed set in X. • Problem # 1: Let : → ( , ) be a function from a non-empty set X into a topological space ( , ) . Further let be the class of inverses of open subsets of Y: 8 ={ ( ): Show that is a topology on X. • Proof. ∈ } Open sets and Neighbourhoods • Definition # 1. Let (X, ) be a topological space. A subset of X is said to be open if and only if it belongs to . The conditions ( ) − ( ) imply that (A ) the union of any number of open sets is open; (A ) the intersection of any two open sets (and hence of any finite number of open sets) is open; (A ) ∅ and X are open. Definition # 2 (Neighbourhood of a point) • Let (X, ) be a topological space and let be a point of X. • A subset N of X is called a neighbourhood of there exists an open set for short, x ∈ ⊆ N). such that ∈ if and only if and U⊆ N (or, • In other words, N is a Neighbourhood of , if and only if it contains some open set to which belongs. • Example # 1: Let X= { , , } and = {∅, { }, { , }, }. Find neighbourhoods of the points of X. 9 • Example # 2: In an indiscrete topological space, each point has a single neighbourhood which is the ground itself. • The following example illustrate that a neighbourhood of a point may not be an open set. • Example # 3: Let X = { , , , } and = { , { }, }. Clearly the set { , } is a neighbourhood of , but it is not open because it is not a member of . • Example#4: Let [ − , ∈ . Then each closed interval + ] with centre at is a neighbourhood of . • Theorem # 1. If (X, ) is a topological space, then a subset A of X is open, if and only if A is a neighbourhood of each of its points. Proof. Suppose that A is open. • We shall show that A is a neighbourhood of each of its points. • Let • ∈ ⊆ , where A is open. • It follows that A is neighbourhood of each of its points. • Conversely, 10 • Let A be a neighbourhood of every point belonging to it, • than for each x ∈ A there exists an open set • • • that x ∈ ⊆ A A = U {x: x∈ A} ⊆ U{ A ⊆ U{ A = U{ , such :x ∈ A} ⊆ A. :x ∈ A} and U { :x ∈ A} : x∈ A} ⊆ A Since the union of open sets is open it follows that A is open. This completes the proof. The most important properties of neighbourhoods in a topological space are established in the following Theorem. Theorem # 2: Let (X, ) be a topological space. Then ( ) each x ∈ X has a neighbourhood, ( ) If A is a neighbourhood of x and A⊆B, then B is also a neighbourhood of x, ( ) If A and B are neighbourhoods of x; then A ∩ B is also a neighbourhood of ( ) If A is a neighbourhood of x ∈ X , then there exists a set B such that B is also a neighbourhood of x and A is a neighbourhood of each point of B. 11 • . ( ): Since X is open, it follows from theorem 1, that X is a neighbourhood of each of its points. • In other words, each point of X has at least one neighbourhood, i.e., X itself. ( • ): since A is neighbourhood of x, there exists an open set • Since B is a neighbourhood of x, • such that x ∈ U ⊆ A. there exists an open set V such that x ∈ • x∈ ∩ ⊆ B). Hence x ∈ and U ∩ V ⊆ A ∩ B (because ∩ ⊆A∩B. ⊆ . ⊆ A and V But U ∩ V , being the intersection of two open sets is open. Therefore A ∩ B is neighbourhood of x. • ( • ): Since A is a neighbourhood of x, there exists an open set U, such that x∈ U ⊆A. • But A⊆B so x∈ U ⊆ B. • Hence B is a neighbourhood of x. ( • ) Since A is neighbourhood of x, there exists an open set B such that x∈ B⊆ A. 12 • Since x is a point of B and B is open, by theorem 1, B is a neighbourhood of x. • If y ∈ B, then because B is open. (Theorem 1). B is a neighbourhood of y . • Since B⊂A therefore A is also a neighbourhood of y by ( ). • A is therefore a neighbourhood of each point of B. This completes the proof. • Definition # 3. • If and • then are two topologies on X, and ⊆ is said to be coarser (or weaker) than • alternatively • is said to be finer or stronger than , : . • Definition# 4. Two topologies on X are incomparable if neither is coarser than the other. called • Note: • The indiscrete topology is contained in any topology on X and is, therefore, the coarsest topology on X. 13 • At the other extreme is the discrete topology which contains any topology on X and is , therefore, the finest topology on X. • Example # 5. The topologies and in example below are not comparable: on X = { , } given ={ , { , } } = { , { }, { , } } = { , { }, { , } } = { , { }, {b }, { , } } • Example # 6. Let X= { , , } and let = {∅, { , }, } and = {∅, { , }, { , }, { }, } then τ and τ are topologies on X. Since every member of is also a member of . τ is coarser than τ ⊆ τ is finer than τ P1=page 44. Bsc book • Theorem#3.Let and ∩ be any two topologies on a non empty set X. Then is also a topology on X. 14 Proof. (O ) : Let {X ∶ ∈ I} be any subcollection of ∩ . Then X ∈ and X ∈ , for each ∈ ∪ ∪ I and so X ∈ and X ∈ because ∈ I ∈ I ∪ and are topologies on X. Hence X ∈ ∩ . ∈ I Thus ∩ satisfies (O ). (O ) : Let X and X be any two sets of and X both belong to and . and ∩ Then X both being topologies on X, satisfy (O ). Hence X ∩ X ∈ X ∩ X ∈ so X ∩ X ∈ ∩ . Therefore satisfies (O ). (O ) : Condition (O ) for belong to both ∩ Hence and ∩ implies that X and ∅ . Hence X and ∅ belong to and (O ) is satisfied. ∩ and and is a topology on X. • Example # 7. Let X= {a,b,c,d} . Then each of the collections: • on X. • ∪ = {∅, { , }, } and = {∅, { }, } is a topology is not topology on X. 15 Lecture#13. Lecture Plan • Neighbourhood systems • Properties of nbd system • Closed sets in • Characterizations of closed sets • Exercises • Definition: Neighbourhood of a point • Let be a point in a topological space (X, ). Then the set of all neighbourhoods of is called the neighbourhood system of the point and it is denoted by ( ) . • Example #1. • Let X= { , , } and of , , • Solution. = {∅, { }, { , }, } find nbd systems • Let X= { , , } and = {∅, { }, { , }, } • • • Then ( )= ( )= ( ) ={ { }, { , }, { , }, ,{ , } } 16 • The main properties of the neighbourhood systems are summed up in the following. Theorem #3. Let ( , ) be a topological space and let be the neighbourhood system of the point ∈ X. Then: (1). is not empty (2) the intersection of any two members of to . (3) If a subset of X contains a member of subset also belongs to . • . ( ) From ( ) it follows that neighbourhood, namely X, Therefore • (2) From ( belongs then this has at least one is non-empty. ) it follows that the intersection of any two neighbourhoods of x is a neighbourhood of x . Hence the intersection of any two members of ( ) is a member of ( ). • (3) From ( ) , it follows that if a subset A of X contains a neighbourhood of x . Then A must also be a nbd of x . In other words if a subset of X contain a member of ( ) then it must also a member of ( ). 17 • Definition: Closed sets • A subset of X is said to be closed in a topological space (X, ) if its complement w.r.t. X is open or equivalently. If it is the complement of an open set. • Example #3 • Let X={ , , , } and let = closed sets in (X, ) are X, { , the complements of ∅,{ },{ , {∅, { }, { , }, }. Then the , }, { , } and ∅, which are } and X respectively. • The following examples indicate that unlike doors, sets in a given topological space may be both open and closed, or neither open nor closed. Example #4. In the topological space whose ground set is X={ , , } whose topology is {∅, { }, { , }, }, • the set { } is open as well as closed , being the complement of { , }. • But { } and { } are neither open nor closed. • Example #5. In any topological space, the empty set and the ground set are both open and closed, because both belong to the topology and are complements of each other. 18 • Example #6. In a discrete topological space (X, P(X)) every set is both open and closed. Because, if A is a subset of X, then A ∈ P(X). Also A is the complement of the set X-A, which being a subset of X, is also open. • Hence A is both open and closed. • Example #7. In an indiscrete topological space, any non empty set different from the ground set is neither open nor closed, • It can be easily verified. • The collection G of all closed sets in any topological space has three basic properties, • These are summed up in the following theorem. • Theorem #5. Let (X, ) a topological space. Then the collection G of closed sets has the following properties: • • • • Let { ( ) The intersection of any number of closed sets is closed. ( ) The union of any two closed sets (and hence of any finite number of closed sets is closed. ( ) X and ∅ are closed sets. .( : ): ∈ 1} be any number of closed sets in G 19 • We shall show that ∩ is closed. • By proving that for each • By De Morgan's Law (∩ • Since each , and • Now ∪ • By condition ( • ( ∈ is open is closed • Then • Hence , is open. • Hence ∩ ): Let ) =∪ is closed, therefore each • Hence ∪ • ( Complement is open. ∈ ∪ be any two closed sets are open. = ) ∩ . ∩ is open. is closed , as required. ): The complement of X is the open set ∅ • the complement of ∅ is the open set X. • Hence X and ∅ are closed sets. i. ii. • Q#2. Explain why each of the following collection is not a topology on X = { , , , , , } = {∅, { , }, { , }, X} = {∅, { , , }, { , , }, X} 20 iii. iv. = { }, { , }, { , } , { , , } , X = ∅, { } Q#3. Explain why each of the following collections is not a topology on the set Z of all integers: i. The collection of all finite subsets of Z, ii. The collection of all infinite subsets of Z, iii. The collection of all subsets of Z whose complements are finite, iv. The collection of all subsets of Z of which 0 is an element, v. The collection of all subsets of Z of which 0 is not an element. Q#4. Show that following collections of subsets of X are topologies on X. X = { , , , } are topology on X, = {∅, { }, { , }, X} = {∅, { }, { }, { , }, X} = {∅, { }, { , }, { , , }, X} Q#5. Write down the discrete topological space of which X= { , , } is the ground set. Q#6. Are the topologies comparable? 21 X={ , , , } = {∅, { }, { , }, } = {∅, { }, { }, { , }, } = {∅, { }, { , }, { , , }, } Q#7. Find the neighbourhoods and the neighbourhood systems of , , , for each of the topologies . X = { , , , } are topology on X, = {∅, { }, { , }, X} = {∅, { }, { }, { , }, X} = {∅, { }, { , }, { , , }, X} Q#8. List the closed sets in each of the three topological spaces: ={ , , , } = {∅, { }, { , }, X} = {∅, { }, { }, { , }, X} = {∅, { }, { , }, { , , }, X} Q#9. Prove that there are twenty-nine different topologies on the set { , , }. 22 Lecture# 14. Lecture Plan • Interior Points • Theorems • Worked Examples Definition # 1. Let A be a subset of a given topological space X. Then a point • ∈ is called an interior point of if and only if there exists a neighbourhood N of that ∈ N ⊆ A. • 1. Every neighbourhood N of containing , such , contains an open set • 2. Since every open set V in X is a neighbourhood of each of its points, and x ∈ V⊆ V, ∀ x ∈ V, It follows that every point of an open set V is an interior point of V. 23 • . (Interior of a set) • The set of all interior points of a subset A of a topological space X is called the interior of Int(A) and shall be denoted by • Example 1. • Let R be the topological space with the usual metric topology. Let A =[0,1[ be a subset of R. • Then every point of A except 0 is an interior point of A. Hence Int (A)= ] 0,1 [ which is an open interval of R. • . • Let be the topological space with its usual metric topology. Let • A= {( , ): ( , ) ∈ , • Let ( , ) ∈ A such that • Then ( , ) ∈ + + ≤ 1} be a subset of . = ( , ) ⊆ A, Where r=1− , • This implies that ( , ) ∈ Int(A). • Example # 3: • Let be the topological space with its usual metric topology. Let 24 • A= {( , ): ( , ) ∈ , + • Let ( , ) ∈ A be such that + ≤ 1} be a subset of . =1. • Then it is easy to see that there is no sphere centered at ( , ), and contained in A. • Hence, it follows that • Int(A) ={( , ): ( , ) ∈ • # . , + • Let X={ , , } and the topology = { , }. Find Int((A) < 1} = • Example # 5. Let X = { , , , } and For • = { , }. Find X. Then ( ) , ∅, { }, { , } , Let = , ,{ , } . # . Let A be any subset of a topological space (1) Int (A) ⊆ A; (2) Int (A) is the largest open set of X which is contained in A • . ( ) Let x ∈ Int (A). • It follows from definition, that A contains a neighbourhood of x and hence, ∈ . Therefor we conclude that 25 Int(A) ⊆ A. (2). To prove that Int(A) is the largest open set contained in A, we first establish that every open set contained in A, is contained in Int(A). (This means: Int(A) is largest open set containing A) Let x∈ , where U ⊆ is an open set. Now being an open set, is a nbd of x and x∈ It follows that x is an interior point of A. Thus every point of ⊆ A. is an interior point of A and hence U⊆ Int(A). • Now we show that Int(A) is open, • we shall prove that Int(A) is union of open sets of X • Let ∈ Int (A). • It follows from definition that there exists a nbd N( ), (say) of such that ∈N( ) ⊆ A ……………..(1) It follows from the definition of neighbourhood that there exists an open set U( ). (say), such that ∈ ( ) ⊆ N(y) ……………..(2) • From (1) and (2) we get 26 ∈ U( ) ⊆ N ( ) ⊆ A Since U( ) is an open set contained in A, it follows that U( ) ⊆ Int(A) Let =U ( ) ⊆A. It follows that ∈ Int(A) Since U is open. Therefore U ⊆Int(A). Also, Int(A)=U{ } ⊆ U ( )= . Hence we conclude that ∈ Int A IntA= and consequently Int(A) is open. • Corollary 1. A subset A of a topological space X is open if and only if Int(A) =A . Suppose A is open. Then A is a neighbourhood of each of its points and as a consequence every point of A is an interior point of A. Hence Int(A)=A Conversely, If Int(A)=A, then A is open • Corollary 2. Int(A) is the union of all the open sets contained in A. • • Then . Let { } be family of all open sets contained in A. ⊆ Int(A) and so ∪ ⊆ Int(A) 27 But Int(A) is open and Int A ⊆ A . Hence Int (A) ⊆ U , hence it follows that U =Int(A)- • : . • Let X= {1,2,3,4} and = , ∅, {1}, {3}, {1,3} Let A ={1,2,3} and B={2,4}. Then Int{A}={1,3} and Int (B)=∅ • topology, . Let R be the topological space with the usual Let A=[1,2]. Then Int (A)=]1,2[ Which is clearly open and is the largest open set contained in A. • . Let A and B be any two subsets of a topological space. Then 1. Int (Int (A)=Int (A), 2. Int (AUB) ⊇ Int (A)UInt (B), 3. Int (A∩B)= Int (A)∩Int (B), . . Since Int (A) is open, therefore Int (Int (A))=Int (A) 28 Proof 2. • Let x ∈ Int (A) U Int (B). • Then either x ∈Int (A) or x ∈ Int (B). • For the sake of definiteness, we assume that x ∈ Int (A). • It follows from the definition of an interior point that there exists a neighbourhood N of x such that • x ∈ N ⊆ A. • Since A ⊆ AUB, therefore, we have • x ∈ N ⊆ AUB • ⇒ x ∈ Int (AUB) • Hence Int (AUB) ⊇ Int(A) U Int (B) • Proof 3. • Let x ∈ Int (A∩B). • Then there exists a neighbourhood N of x such that • • • • x ∈ N ⊆ A∩B ⇒ x ∈ N ⊆ A and x ∈ N ⊆ B ⇒ x ∈ Int (A) and x ∈ Int (B) ⇒ x ∈ Int (A) ∩ Int (B) 29 • Therefore Int (A ∩B) ⊆ Int (A) ∩ Int (B) (1) • Now let x ∈ Int (A) ∩ Int (B). Hence • x ∈ Int (A) and x ∈ Int (B) • ⇒ x ∈ N ⊆ A and x ∈ N ⊆ B, where • N and N are neighbourhoods of x • ⇒ x ∈ N ∩ N ⊆ A∩B • ⇒ x ∈ Int (A ∩B) • ⇒ Int (A) ∩ Int (B) ⊆ Int (A∩B) (2) From (1) and (2) the required result follows . To illustrate the fact that Int (AUB) is not, in general, equal to Int(A) U Int(B), we cite the following example. Let X= { , , , } and = X, ∅, { , }, { , } Let A = { , } and B = { , }, Then it is evident that Int (A) = ∅, Int (B) = ∅, Int (AUB)= Int (X)=X. And Int (A) U Int (B)= ∅. Hence Int (AUB)≠Int (A) U Int (B) 30 Lecture#15. Lecture Plan • Exterior Points • Closure of a set • Theorems • Worked Examples • . (Exterior Point) • Let A be a subset of a topological space X. A point x is called an exterior point of if there exists neighbourhood N of x such that x ∈ N ⊆ X−A . A point x ∈ X is an exterior point of A if i. x ∈ X−A and ii. There exists at least One neighbourhood of x which does not intersect A • . Let X= { , , } and = X, ∅, { , } If A= { , }, then A has no exterior point because is the only point which belongs to X−A but there is no neighbourhood of contained in X−A. 31 • . The set of all exterior points of A is called the Exterior of and we shall denote it by Ext (A) • . Let X be any non-empty set and = {X, ∅} be the trivial topology on X. Then for every non-empty subset A of X, Ext(A)= ∅ • because there is no point of X for which there exists a neighbourhood contained in X−A • . Let X={ , , } be the discrete topological space Let A = { , } Then Ext (A)= { }=X−A. This result is true for all subsets of X (why)? • Let X= = X, ∅, { }, { , }, { , , } . Let • and = { , }, Then verify that Ext (A)= { } . Let R be the topological space with its usual metric topology. Let A= ] , ] Then Ext (A)={x: x∈ R. x< • { , , , } or x>b} . Let R be the topological space with its usual metric topology. Let = ( , ): ( , ) ∈ R , + ≤ 4 Then 32 Ext • = ( , ): ( , ) ∈ R , . + >4 • Let A be a subset of a topological space X. Then the intersection of all closed subsets of X containing A, i,e., the smallest closed set of X containing A, is called the closure of A, and is denoted by CI (A) • . Let X={ , , } and = X, ∅, { , } Let A={ , }, The A =X because the smallest closed subset containing A is X itself. • topology. • Let • . Let R be the topological space with the usual A= ( , ): ( , ) ∈ R , + ≤ 1 . Then A = Cl(A) . Let X= {1,2,3,4} and let = X, ∅, {1,2} and two topologies on X. = X, ∅, {1}, {1,2}, {1,2,3} Let A={4}. Find Cl(A) w.r.t and be 33 • . Let X be a topological space. Then 1. A subset A of X is closed if and only if A=A; 2. A=A; 3. A ∪ B = A ∪ B for any two subsets A and B of X; 4. x =X, ∅ = ∅ Proof 1. Suppose A is closed. Then A itself is a member of the family of closed sets containing A. Hence A being the intersection of this family, is contained in A. But A ⊆ A. Hence A = A . Conversely,, if A = A then A is closed because A is closed. 2. This follows from (1) since A is closed. 3. By definition of the closure, A is the smallest closed set containing A. Since A ⊆ A∪B ⊆ A ∪ B, It follows that A ⊆ A ∪ B because A ∪ B is a closed set which contains A. Similarly B ⊆ A ∪ B. Hence A∪B ⊆ A ∪ B (i) Since A ⊆ A , B ⊆ B it follows that A ∪ B ⊆ A ∪ B. which is a closed set. Hence A ∪ B ⊆A∪B (ii) From (i) and (ii) we get A ∪ B = A ∪ B. 34 • . Let A and B be any two subsets of a topological space X such that A⊆B. Prove that A ⊆ B . Since A ⊆B ⊆ B and B is closed, it follows that • A ⊆ B . Cl(B) is any closed set containing A and Cl(A) is smallest closed set containing A. Thus Cl(A) ⊆ Cl(B) . We give an example to show that, in general, A∩B ≠A∩B • Let • Let = { , , } and = { , , { }, { }, { , }}. = { } and = { , }. Then A ∩ B ≠ A ∩ B . A point x ∈ A if and only if every open set containing x contains a point of A. Proof. Let x ∈ A and be an open set containing x. Since A = A ∪ . It follows that either x ∈ A or x ∈ . If x ∈ A, then ∩ A is non-empty because it contains x. If x ∈ , then , being an open set containing x is a neighbourhood of x, and hence by the definition of , U ∩ A ≠ ∅. Therefore, U contains a point of A. Conversely, we suppose that every open set U containing x satisfies A ∩ U ≠ ∅. If x ∈ A, then clearly x ∈ A ⊇ A. if x ∉ A, then every open set containing x 35 • contains a point of X−A, viz., x itself. That is to say, every open set containing x contains a point of A and a point of X−A. Hence x ∈ A ⊆ A. Therefore x ∈ A. • A point x ∈ X is called a limit point of if every open set containing x contains a point of A other than x . . Let A be a subset of a topological space X. . The set of all limit points of A is called the derived set of • and is usually denoted by A . . Let X={ , , } and = {X, ∅} Let A={ , } Then it is clear that • • A =X . Let = X , ∅, { }, { } be a topology on X= { , }. Let A={ }. Then A =∅ because A has no limit points. . Let R be the topological space with its usual metric topology • let A={x: x∈ R, x = 1 + Then clearly A ={1} 1 , ∈ N} 36 Lecture#16 Lecture Plan • Limit point of a set • Derived set • Examples • Theorem on • Limit point • Exercise problems • • . Let A be a subset of a topological space X. A point x ∈ X is called a limit point of if every open set containing x contains a point of A other than x . . The set of all limit points of A is called the derived set of • and is usually denoted by A . . Let X={ , , } and = {X, ∅} Let A={ , } Then it is clear that A =X 37 . Let • = X , ∅, { }, { } be a topology on X= { , }. Let A={ }. Then A =∅ because A has no limit points. . Let R be the topological space with its usual metric topology • let A={x: x∈ R, x = 1 + • Then clearly A ={1} • For any • • , ∈ N} . A point x ∈ A if and only if every open set containing x contains a point of A. • Proof. Since • 1 ∈ ∈ , ( ) ( )= ∪ ∈ ∪ ∈ ∈ for each open set • This completes the proof. containing ; ∩ ≠ 38 • Q#1. Let X={1,2,3,4,5,6} and = X, ∅, {1}, {1,2,3,4}, {2,3,4}, {1,5,6}, {5,6}, {2,3,4,5,6} Let A = {1,2,3}, B={1,3,4},C={5,6,2} and D={1,2,4,5}. Find the interior, exterior, closure and the derived sets for each of A,B,C, D. • Q#2. Let A be a subset of X. Prove that i. ii. Int (X−A)= X −A; X − A = X − Int (A). Q#3. If A and B are any subsets of a topological space X then prove that A∩B ⊆A∩B • Q#4. Let R be the topological space with its usual topology Let A={( , ): ( ) ∈ R , x ≥ 0, Find Int (A), Ext (A) and A . ≥ 0, + ≤ 1} 39 Lecture#17 Lecture Plan • Bases • subbases • • • base for . . A subcollection of a topology is called a if every member of is a union of members of . Let X={ , , , , } and let = {∅, { , }, { , }, { , , , }, X} be a topology on X, = { , }, { , }, ∅, X is a subcollection of , which meets the requirement for a base . . Let = ∅, { }, { , }, { , , }, { , , , }, { , }, { , , }, X be a topology on X={ , , , , }. Consider the following two subcollections of : = ∅, { }, { , }, { , , , }, { , } = ∅, { }, { , }, { , } Each • form base for . . If X is a non-empty set, then the collection of all singletons of X is a base for discrete topology on X. 40 The following theorem gives a condition under which a collection of sets is a base for given topology. • . A subcollection of a topology is a base for • if and only if • for any points exists B ∈ Proof. belonging to an open set U∈ , there such that x ∈ B ⊆ . Proof. Let be a base for and let x ∈ . Where Then = ∪ B , where each B ∈ Since x ∈ =∪B x ∈ B , for some Hence x ∈ B ∈ . = ⊆∪B = Conversely, if for each x ∈ such that x ∈ B ⊆ , then , there exists a set B ∈ = ∪ {{x} : x ∈ } ⊆ ∪ {B : x ∈ } But ∪ { B : x ∈ } ⊆ (1) (2) Hence from (1) and (2), if follows that = ∪ {B ∶ x ∈ } 41 Hence every open set ∈ is expressile as a union of members of . Thus is a base for . • Note. Bases for the usual topology on the line and the plane. I. Consider the usual topology on the real line R. Then a base for this topology is the set of all open intervals. II. Consider the usual topology on the plane R . Each of the following collections is a base for the usual topology of the plane: i. The collection of all open discs, ii. The collection of all open rectangles, bounded by the sides, which are parallel to the co-ordinate axes. Note: An alternate definition for Base of a topology . A collection of subsets of a non-empty set X is a base for some topology on X if meets the following requirements: (B ) if , ∈ ∈ such that ∈ ⊆ and ∩ ∈ ∩ , then there exists a set ; (B ) ∪ = , where ∪ members of denotes the union of all 42 • • Let . (Base at a point) be any point of a topological space X • A collection the point of open sets containing is called a base at • (or a local base at ), i.e., • if for every open set V to which set B ∈ such that ∈ B ⊆ V • . A collection space X is a base, belongs, there exists a of open sets in a topological • if and only if • contains a base at each point of X. Proof. Let be a base for a topology on X. We shall show contains a base at each point of x. Let x be any arbitrary point of X and let V be any open set containing x. ∪ B ,B ∈ ∈I ∪ Hence x ∈ V ⇒ x ∈ B ∈I Since is a basis, V = ⇒ x ∈ B for some for some index set I. ∈I 43 But B ⊆ V. Thus there exists an open set B ∈ B ⊆ V. Hence contains a base at each point x ∈ X. Conversely suppose that such that x ∈ contains a base at each point of X. We shall prove that is a base for the topology on X. Let be any open set and let x ∈ . Since contains a base at the point x, there exists some B ∈ such that x ∈ B ⊆ . Hence =∪{{x} : x ∈ } ⊆ But each B ⊆ ∪ ∈ B Hence ∪ B ⊆ (1) (2) From (1) and (2), it follows that Therefore Thus ∪ ∈ B = is expressible as a union of members of . is a base and the theorem is proved. 44 • . Let X= { , , , , } and let = {∅, { }, { }, { }, { , }, { , }, { , }, { , , }, { , , }, { , , , }, { , , , }, X} be a topology on X,then β , a base at point a, is { }, • because the open sets containing are { },{ , }, { , }, { , , }, { , , , } and X, { } is a subset of each. . The set of all open discs with Centre is a base at w.r.t. the usual topology on the plane, because if is any open set containing , then we can find an open disc D with Centre at such that x ∈ D ⊆ . . In a discrete topological space X, the base at each point is {{ }}. For if ∈ X, then the singleton set { }, which is open serves as a base at , because { } is a subset of every open set which contains . Hence ={{ }}. • Subbases: • . If is a topology on X, a subcollection called a subbase for , of is • if and only if • finite intersections of members of • form a base for . X={ , , , , } and let = ∅, { }, { , }, { , , }, { , , , }, { , }, { , , }, X . Let 45 • Be a topology on X. A subbase collection: • for is given by the sub- = { }, { , , }, { , , , } , { , , } • Q#1. Let X={ , , , }, show that the collection = { , }, { , } cannot be a base for any topology. • Q#2. Find a subbase (with as few members as possible) for each of the following topologies on X ={ , , }. • {∅, { }, X}, • {∅, { }, { , }, X}, • {∅, { }, { , }, { , }, X} • Q#3. Let X= { , , , , }. Find the topologies on X generated by the following collections of subsets of X. • • = { , }, { , }, X , = { , }, { , , }, { , } , X 46 Lecture#18 Lecture Plan • Continuity • Homeomorphism • . (continuous function) • Let f : X→Y be a function from topological space X to a topological space Y, • Let x ∈ X. Then f is said to be continuous at x if and only if for every open set V containing f(x ) in Y, there exists an open set • Y={1,2} , . in X such that x ∈ Let X={ , , }, ⊆f (V). = X, ∅, { }, { , } and = Y, ∅, {1}, {2} . Let f : X →Y be defined by f(a)=1, f(b)=1, f(c)=2. Then f is continuous at x = a, x = b but not continuous at x =c. • . Let X be an indiscrete topological space and let Y be a discrete topological space. Then every function • f : X →Y which is not a constant function, is discontinuous at all points of X. • . Let X be a arbitrary topological space and let Y be an indiscrete topological space. The every function 47 • f : X →Y is continuous at all points of X (prove)!. • . A function f: X → Y from a topological space X to topological space Y is called a (or ) if it is continuous at all points of X. • . Let f : (X, ) → (Y, Then f is continuous if and only if f ∈ . Proof. ) be a given function. (V) ∈ Let us assume that f is continuous. Let V ∈ for every V and let A = f (V). Let x be any point of A. Since f is continuous and f(x) ∈ V ∈ , it follows that there exists an open set (say), such that Now f f x∈ ⊆ f (V) f (V) =∪ {{ } ∶ ∈ ∪{ ∶ ∈ ( )} ⊆ ( )} ⊆ f (V), and therefore, f (V)={ ∶ x ∈ (V) which is open because each is open. Hence (V) ∈ . Conversely, we assume that f (V) ∈ for every V ∈ . Let x be any point of X and let V be an open set in Y which contains f(x). Since x ∈ f (V) and f (V), being an open set by the hypothesis, is a neighbourhood of x, it follows that there exists an open set in X such that 48 x∈ (V) ⊆ f This implies that f is continuous at x. Since x is an arbitrary point of X, it follows that f is continuous from X to Y. • Let . Y={ , , , }, X={1,2,3}, • and let = Y, ∅, { }, { }, { , } let f : X → Y be defined by continuous. • = X, ∅, {1,2} f(1) = f(2) = f(3) = a. Then f is . Let X={1,2}, = X, ∅, {1} and Y={ , }, = Y, ∅, { }, { } . Let f be given by f(1) = Then f is not continuous. , f(2) = . . Let X be any topological space and let Y be an indiscrete topological space. Then every function from X to Y is continuous because there are only two members of namely, Y, ∅ whose inverse images under any f are • respectively X, ∅ which belongs to • x ∈ X. Then i is continuous. . Let i : (X, )→ (X, . Let f : (X, ) →(Y, be a base of ∈ for every . ) be given by i (x) = x, ∀ ) be a function and let . Then f is continuous if and only if f B∈ . (B) 49 Proof. Suppose f is continuous. Then f (V) ∈ for every V ∈ Since ⊆ , it follows, in particular that f every B ∈ . Conversely we assume that f Let V ∈ . Since (B) ∈ is a base of (B) ∈ for every B ∈ . for . ∪ B , ∈ for all ∈ I. , we have V = for some index set I, where B ∈ ∪ Therefore, it follows that f (V)= f (B ) which is ∈ open by (O ). Hence f (Y) ∈ for every V ∈ and consequently f is continuous. . Let f : X →Y and g : Y →Z be two continuous functions. Then g of : X →Z is a continuous function. Proof. Let h= gof. To prove that h is continuous we have to establish that for every open set W in Z, h (W) is open in X. This fact may be established as follows: Let W be an open set in Z. Since g: Y →Z is continuous (given), it follows that g (W) is open in Y. Now g (W) is open in Y and f: X →Y is continuous (given), therefore f (g (W)) is open in X. Since h (A)= (gof) (A)= f (g (A)) for all A ⊆ Z, 50 f (g (W))= (gof) (W) = h (W) Therefore h (W) is open in X for every open set W in Z and hence h=gof is continuous. • . (Homeomorphism) • Let f : X→Y be a function from a topological space X to a topological space Y which meets the following requirements. (H ) f is continuous, (H ) f is bijective (i.e f is both one-one and onto), (H ) f is continuous , where f function of f. denotes the inverse Then f is called a homeomorphism • • • . Let X= {1,2}, = X, ∅, {1} and Y={ , }, = Y, ∅, { } . Let f : X →Y be given by f(1) = , f(2) = . Then f is a homeomorphism between X and Y. . Let i : X →X be the identity function on a topological space X. Then i is homeomorphism. . Y={ , , }, Let X={ , , }, i = X, ∅, { }, { , } and = Y, ∅, { } . Let f : X →Y be given by f( ) = 51 , f( ) = , f( )= . Then it can be seen that f is continuous and bijective. • . Let X and Y be two discrete topological spaces and let f : X → Y be a bijective function. Prove that f is a 52 Lecture#19 Lecture Plan • Subspaces • Relative Topologies • Note: Let X be a topological space with topology . Let A be a non-empty subset of X. Although it is possible to assign many different topologies to A without making any reference to the topology , We would like to assign to A a definite topology, which it inherits from . • In this Lecture we shall construct such topologies and study some of the properties of the resulting topological space. • Relative Topology: • Definition. Let be a subset of a space ( , ). Then a topology = { ∩ : ∈ } is called a relative topology on and ( , • Example. ) is called a subspace of ( , ). • Let = { , , , , } with = { , { }, { , }, { , , }, { , , , }, }. • Let = { , , }. Then 53 • . Let X be a topological space with a given topology and let A be a given non-empty subset of X. Then prove that the collection of subsets of A, defined as follows: = A∩ : Proof. In order to prove that the following ( ∶ ) : Let {A ∩ ∪ . Now (A ∩ ∈I being a topology Hence Thus ( ∪ (A∩ ∈I ∈ I } be any subcollection of ∪ )=A ∩ by distributive law ∈I ∪ = ∈ ∈I ∈ )A∩ ∈ and C= A ∩ ( ) : let B = A ∩ where and belong to . Since =(A ∩ ( and ( ) : Since X, ∅ ∈ A∩X=A∈ ) ∩ (A ∩ ∩ ∈ be any two members of . ) ) belong to , Hence B ∩ C = A ∩ is a topology on A. is a topology on A, we verify ) is satisfied. Then B ∩ C = (A ∩ , ∈ = ∩ ∈ and the condition ( and A ∩ ∅ = ∅ ∈ ) is satisfied. 54 Hence A, ∅ ∈ Thus • and ( ) is satisfied. is a topology on A. . Let (X, ) be a topological space and let A be a non-empty subset of X. Then the topology defined by = A∩ : ∈ is called the relative topology or the induced topology on A. • . A non-empty subset A of a topological space (X, ) is called a subspace of X if and only if it has been assigned the relative topology . • . Let X be a non-empty set and let be the indiscrete topology on X. Let A be a non-empty subset of X. Since ={∅, X}, ={A ∩ ∅, A ∩ X}={∅, A} • Thus the relative topology on A is the indiscrete topology. • Note. • A subspace of a discrete space is discrete • A subspace of indiscrete space is indiscrete • . • Let X={ , , , , } and let • = {∅, { , }, { , }, { }, { , , }, X} 55 • Find the relative topology on A={ , , } and the resulting subspace. • . • Let A be a straight line • where , + + = 0, are real constants • one of the constants is not zero. • Let be the usual topology on the plane R of which A is a subset. Describe the relative topology . • . Let be the usual topology on a real line R. Determine whether or not the following subsets of the subspace [1,2] are open w.r.t the relative topology on [1,2]: • [1, 4/3[ , • ]11/6, 2], iii. ]1, 3/2]. • . If (A, ) is a subspace of a topological space (X, ), then a subset B of A is closed in the subspace, if and only if B= A∩F, where F is some closed set in (X, ). 56 Proof. Suppose that B ⊆ A is closed in the subspace. Hence B is the complement relative to A of some open set in A. i.e., A-B=A ∩ , where ∈ we have B =A −(A ∩ Where F=X− II. • )=A− =A ∩ (X− )=A ∩ F is a closed set of X Conversely, let B= A ∩ F, where F is a closed set of (X, ). Now F=X− where U is some open set in X. Hence = ∩ ( − ) = − = −( ∩ ) Since ∈ , ∩ ∈ Hence B, being the component of an open set of the subspace is closed in the subspace. . Let X={ , , , , } • let ={∅, { , }, { , }, { }, { , , , }, { , , }, X} topology on X. be a • Find the closed sets of the subspace whose ground set is A={ , , }. • . • Let a< < < and let A=[ , ] ∪ ] , [ be considered as a subspace of the real line R with the usual topology . 57 • Show that [ , ] and ] , [ are both open and closed in the subspace. 58 Lecture#20 Lecture Plan • Topological property • Subspaces Continuity w.r.t. spaces and subspaces • . Let be a base for a topological space ( , ) • If A is a subspace of X, then prove that the collection = {A ∩ : B ∈ } is a base for A. Proof. To show that = {A ∩ : B ∈ } is a base for A, we need to prove that every open set of A can be expressed as union of members of . Let = be any open set of the subspace A. Then ∩ for some open set U in X. Since B is a base for X, therefore ∈ for each . Hence Since = ∈ , ∩ (∪ ∩ : ∈ Hence every open set as union of members of =∪ ) = ∪ {( ∩ for each . ): : , where } of the subspace A is expressible . 59 Hence is a base for relative topology on A. This completes the proof. Example. Let ={ , , , , } and let τ = {ϕ, {a, b, c}, {c, d}, {d, e}, {c}, {d}, {a, b, c, d}, {c, d, e}, X} and let = { , { }, { }, { , }, { , }, { , , }} be a base for . Find the base for the subspace whose ground set is ={ , , } • Q#1. Let X be a non-empty set and let be the discrete topology on X. Let A be a non-empty subset of X. show that the relative topology on A is also discrete. • Q#2. Let X={ , , , , } and let = X, ∅, { }, { , }, { , , }, { , , , }, { , , } be a topology on X. Find the relative topology on each of the following subsets of X: i. ii. iii. iv. A={ , , } B={ , , , } C={ , , , } D={ , } Indicate the sets which are open (closed) in the subspace so obtained, but are not open (closed) in the given topological space. 60 • Q#3. if A is a subspace of X and B is a subspace of A, prove that B is subspace of X. • Q#4. Let be the usual topology on the real line R. Determine whether or not the following subsets of the subspace [1,3] are open in the relative topology: ii. 3 [1, 2 [, 4 iii. ]1, 5/2]. i. ]1 5/7, 3], • Q#5. Let X={ , , , , , } and let = {∅, { , }, { , , }, { }, { , }, { }, { , , }, { , , , }, { , , , }, { , , , , }, { , , , , }X} be a topology on X. Let = { }, { }, { , }, { , , }, { , , , , } be a base for . Find a base for each of the subspaces whose ground sets are : i. ii. iii. A= { , , } B= { , , , } C= { , , , } Hence find the relative topologies , . 61 TOPOLOGICAL Property • Recall that if X, Y are homeomorphic, then open sets of Y are the images of the open sets of X and open sets of X are the inverse images of the open sets of Y. Therefore any property of X expressed entirely in terms of set operations and open sets is also possessed by each space homeomorphic to X. • length angle, boundedness, Cauchy sequences, straightness and being triangular or circular are not topological properties. • Whereas limit point, interior, nbd, boundary and first/second countability are topological properties. Definition (Topological property) • A topological property is a property which, if possessed by a topological space, is also possessed by all topological spaces homeomorphic to that space. Example • Let X =(-1,1) and f: X→R defined by f(x)=tan x/2. • Then f is a homeomorphism and therefore (-1,1)≃R • Note that (-1,1), R have different lengths, • therefore ‘length’ is not a topological property. 62 • Also X is bounded and R is not bounded, • therefore ‘boundedness’ is not a topological property. Example. • Let f : (0, ∞) → (0, ∞) defined by f(x) =1/x, then f is homeomorphism. Consider the sequences • (x ) = (1, 1/2,1/3, … ). Then (f(x )) = (1,2,3, … ) in (0, ∞). (x ) is a Cauchy sequence, where (f(x )) is not. Therefore ‘being a Cauchy sequence is not a topological property. • Example. Consider f : D → D defined by f(r, )=(2r, ) • Then f is a homeomorphism. Note that the area of D is half of the area of D . Therefore ‘area’ is not a topological property. • Example. Being ‘triangular’ is not topological property: • since a triangle can be continuously deformed into a circle and conversely. • Whereas limit point, interior point, boundary, nbd and first/second countability are topological properties Theorem#1. If A is a subspace of a space X, then G is open in A if and only if G= A ∩ O, O ∈ . 63 Theorem#2. Let A be a subspace of a space X, then H is closed in A if and only if H= A ∩ C, C is closed in X. Proof. Let H be closed in A. Then H = A−W, where W is open in A. Since W = A ∩ O, O ∈ , we have H= A −(A ∩ O) = A−O = A ∩ (X –O) = A ∩ C, where C is closed in X. Conversely, H= A ∩ C, where C is closed in X. Now C = X−O, O ∈ . Then H= A ∩ (X−O) = A−O = A−(A ∩O). Since O ∈ , A ∩ O ∈ and hence H is closed in A. Theorem#3. Let A be a subspace of a space X, then N is a nbd of x in A if and only if N =A ∩ U , U is a nbd of x in X. Proof. • Suppose is a nbd of • We show that • Suppose ∩ = is a nbd of • Then ∃ an open set • Also in X. is a nbd of ∈ ∩ in in such that ⊆ in X s.t. ∩ in . ∈ in X, • There exists an open set • This gives is a nbd of ∈ ⊆ ⊆ 64 • Since • Implies ∩ is open in ∩ is a nbd (say) of in and conversely. Theorem#4. Let A be a subspace of a space X, if x ∈ A and is a nbd base at x in X, then {A∩B: B ∈ } is a nbd base at x in A. Proof. • By theorem 3, • If • ∩ is any nbd of = ∩ , • Then, • So • Or = ∩ is a nbd of in , then is a nbd of ⊆V, for some ∩ ⊆U ⊇ • This gives { ∩ : ∩ ∈ ∈ in X ∈ } forms a nbd at in Theorem#5. Let A be a subspace of a space X, and is a base for the topology on X, then {A∩B: B ∈ } is a base for the relative topology for A. Proof. • We show that each open set in can be expressed as a union of members of { ∩ : ∈ } • Let ∈ 65 • Then = • Since where ∩ ∈ is a base for • Therefore • Thus =∪ , for some = ∩ = ∈ ∩ (∪ ) =∪ ( ∩ ), • This completes the proof ∈ Theorem#6. Let A be a subspace of a space X, and E ⊆ A, then CI (E) = A ∩ CI (E), where CI (E) denotes the closure of E in A. Proof. • Let ∈ ∩ • Then for all • Since ⊆ , • Therefore • Or ( , Theorem#7. Let ∩ ≠ ∩( ∩ )≠ ∩ )∩ • This implies ( ) ∈ ≠ ( ) and conversely. f : X → Y be continuous and A ⊆ X. Then f : A → Y is continuous, where f denote the restriction of f to A. 66 Proof. Since f (G) = A ∩ f (G). G ⊆ Y, we have if G ∈ , then f (G) ∈ . Therefore A ∩ f (G) ∈ , gives f is continuous. This completes the proof. 67 Lecture#21 Lecture Plan • First Countable Space • Second Countable Space • Definition. First Countable Space • A space X is said to be first countable (or to satisfy the first axiom of countability) if, there exists a countable base at every point ∈ . • Example. • Let X be a discrete space. Each { } is open in X and is contained in every open set containing every discrete space is first countable. . This implies that (second Countable Space) • A space X is said to be second countable If it has countable base for . Remark: • Every second countable space is first countable. • For if • if is a countable base for a space X, and consists of members of which contain x ∈ X, 68 • then is a countable base at x. • on the other hand, • any uncountable discrete space is first countable without being second countable. Example • The usual line R is second countable, • since it has countable base consisting of open intervals (a,b), a,b ∈ Q. Example • Let R have the discrete topology. • Then any base for this topology contains singletons • and hence cannot be a second countable space. Example. with the usual topology is second countable. Theorem. The continuous open surjective image of a second countable space is second countable. Proof. Let f: X →Y be an onto continuous open map. We shall prove that f( ) = {f(B): B ∈ } is a base for , where is a base for . Let V ∈ and b ∈ V. then f (V) is open in X and if we pick a ∈ f (b) ⊆ f (V), then for some basic open set B ∈ , a ∈ B ⊆ f (V). it follows that b = f(a) ∈ f(B) ⊆ ff (V) ⊆ V or b ∈ 69 f(B) ⊆ V and thus sets f(B), B∈ completes the proof. , form a base for . This • Corollary. Second countability is a topological property Theorem. Prove that first countability is a hereditary property. Proof Let A ⊆ X be a subspace of a space X. Then x ∈ A gives x ∈ X. By hypothesis, X is first countable, therefore there exists a countable base ={B : n ∈ N} at x in X. Then , = {A ∩ B : n ∈ N} is a base at x in A. which is countable. This completes the proof. 70 Lecture#22 Lecture Plan • First Countable Space • Hereditary Property • Separation Axioms Theorem. First countability is preserved under open continuous surjection OR The continuous open surjective image of a first countable space is first countable. Proof. • Definition. A property of a space X is said to be hereditary if every subspace of X also possesses the same property. • Definition ( −Space). If for x, y ∈ X, x ≠ y, there exists open set U of X such that x ∈ U, y∉ U. Then X is said to satisfy T – axiom. • X with − axiom is called a − Space. Example. Let X = { , , } and ={∅, { }, { }, { , }, X} Then , : , : ∈ { }, ∈ { }, ∉{ } ∉{ } 71 , : ∈ { }, ∉ { } This shows that X is a T − Space. • Definition ( −Space). If for x, y ∈ X, x≠y, • there exist open sets U,V of X such that x ∈ U, y ∉ U and x ∉ V, y ∈ V , then X is said to satisfy T − axiom. • X with − axiom is called − space. • Remark. − space is a − space but not conversely as the following example shows: • If X= {1,2} and ={∅, {1}, X}, then X is a T − space • But not T − space. Example. The cofinite space is T − space. (verify) • Definition ( − space). if for x, y ∈ X, x ≠ y, there exist open sets U, V of X such that x ∈ U, y ∈ V and U∩V=∅, • then X is said to satisfy T − axiom or Hausdorff property. • X with T − axiom is called a T − space or Hausdorff space. • Remark. Every conversely − space is a • as the following example shows: − space but not 72 Example. Let R be with the cofinite topology. Then R is a T − space but not T . • Example. Let X= { , , } ={∅, X, { }, { }, { }, { , }, { , }, { , }, X}. and Then X is a T − space as well as T − space. • Remark. Note that: • → → • reverse implications do not hold in general Theorem (Characterization of − Space). A space X is a T −space if and only if each singleton in X is closed. Proof. Necessity. Let X be a T −space. We show that each singleton {x} is closed, that is, X−{x}, simply we write X−x, is open. Let y ∈ X−x, x ≠ y. Then there exists an open set U such that y ∈ U , and x ≠ Y. Then y ∈ U ⊆ X−x and hence ∪ {U ∶ y ∈ X−x} = X−x. This gives X−x is open. Sufficiency. Suppose each {x} is closed. We show that X is a T −space. Let x, y ∈ X, x ≠y. then X−x, X−y are open in X. Thus x ∈ X−y, y ∉ X−y and x ∉ X−x, y ∈ X−x imply X is a T −space. This completes the proof. 73 Theorem. The property if being a T − space, topological property. = 0,1, is a Proof. Let f: X → Y be a homeomorphism and X, a T −space. We show that Y is a T −space. For this, we show that each singleton in Y is closed. Let y ∈ Y. Then there exists x ∈ X such that f(x) = y or f({x}) = {y}. Since each singleton {x} in X is closed, therefore f is closed implies {y} is closed in Y. This gives that Y isT . This completes the proof. 74 Lecture#23 Lecture Plan • We will continue our study on Separation Axioms • Hereditary Property Theorem. The property if being a T − space, = 0,1, is a 1.hereditary property. 2.Productive property. Proof. (1)Let Y ⊆ X be a subspace of a T −space X. We show that each singleton {y} is closed in Y or X −{y} is open in Y. Since y ∈ X, and X is T , therefore X−{y} is open in X, Thus y ∈ Y ⊆ X ⇒ Y ∩ (X−{y}) = Y – {y}. Hence Y−{y} is open in Y or {y} is closed in Y. This proves that Y is a T −space. (2) Let X , X be T −spaces, we show that X × X is T . It suffices to show that each {(x , x )} is closed in X × X . since each X is T ,therefore by theorem 32.1, each {x } is closed in X . Therefore {(x , x )} = {x } × {x } gives CI({(x , x )})= CI({x })× CI({x }) = {x } × {x } ={(x , x )} or CI({(x , x )}) = {(x , x )} implies {(x , x )} is closed in X × X . Hence by theorem 32.1, X × X is T . This completes the proof. 75 Theorem. (1) A bijective closed or open image of a Hausdorff space is Hausdorff. Hence the property of being a Hausdorff is topological. (2) Hausdorff property is hereditary. Proof. (1). Since each bijective closed map is open, therefore the image of disjoint open nbds are disjoint open nbds and this proves (1). (2). Let ⊆ be a subspace of X and , , ≠ . Since X is Hausdorff, there exist disjoint open nbds U, V respectively. Then ∩ , ∩ are also disjoint open nbds of x and y respectively. This proves that A is Hausdorff. Theorem. If f : X → Y is a continuous injection and Y is T , Then X is T . Proof. Since f: X→ Y is a continuous injection, therefore f ∶ f(X) →X is bijective and open. Since f(X) is T , by the result “bijective open image of Hausdorff space is Hausdorff” we get, X is T . This completes the proof. 76 Lecture#24 Lecture Plan • Regular spaces • spaces Definition. (Regular Space) A space X is said to be a regular space if it satisfies the axiom: (R) for any closed set A and x ∉ A, there exist disjoint open sets U, V such that x ∈U, A⊆ V. • Example. Let X= { , , } and ={∅, { }, { , }, X} • then X is regular space. • Remark. Note that X is not a not closed. − space. Since each Example. An indiscrete space is regular, it is neither T nor T is . • Definition ( − Space). A space is said to be a T − Space, if it is regular and T . • thus we have the following : 77 Theorem. Every −space X is . Proof. Let a, b ∈ X, a ≠ b. Since each {x} is closed in X, therefore {a} is closed and b ∉ {a}. X is regular implies there exist disjoint open sets U, V such that b ∈ U, {a} ⊆ V or b ∈ U, a ∈ V. This gives that X isT . This completed the proof. • Theorem. The following are equivalent for a space X: 1. X is regular. 2. If U is open in X and ∈ U, then there exists an open set V such that ∈ V ⊆ CI(V) ⊆ U. Each ∈ has a nbd base consisting of closed sets. 3. Each ∈ X has a nbd base consisting of closed sets. • Proof. (1) • (2) (2) (3). • (2) gives that every open set containing contains (V) which is closed nbd of . Therefore, closed nbds of forms the nbd base. • (3) (1). 78 • Suppose each sets. has a nbd base consisting of closed ∈ • Let be a closed set s.t. of . ∉ . Then • By (3), there exists a closed nbd or • ⊆ − • This implies • Thus • ∈ ( )⊆ ( ) and ( ) and − ⊆ of ( − )= − s.t. is an open nbd ∈ ⊆ − − are disjoint open sets s.t. − • This implies that X is a regular space. Theorem. Every subspace of a regular space (T −space) is regular (T −space) . Proof. Suppose X is regular and A is a subspace of X. We show that A is regular. Suppose Y is closed in A. Then = ∩ , Z is closed in X. Now if ∈ and ∉ , then ∉ . Then X is regular implies there exist disjoint open sets U, V such that , ⊆ . This gives ∩ , ∩ are disjoint open sets in A such that ∈ ∩ , ⊆ ∩ . This proves that A is regular. 79 Lecture#25 Lecture Plan • Characterization of Normal Spaces • Urysohn Lemma Definition (Normal Space). A space X is said to be normal if it satisfies the axiom: • (N) for any disjoint closed sets U, V of X, there exist disjoint open sets G, H such that U ⊆ G and V ⊆ H. • Example. Let X= { , , } and = {∅, { }, { , }, X}. • Then X is normal. • Note that X is neither nor regular. • All discrete spaces are normal. Theorem. T −space (normal withT ) is a regular space. Proof. Let X be a T −space. Suppose F is a closed set of X such that x ∉ F. By T −axiom, each {x} is closed. Therefore {x}, F are disjoint closed sets. Since X is normal, therefore, there exist disjoint open sets U, V such that {x} ⊆ U, F ⊆ V or x ∈ U, F ⊆ V. This proves that X is regular. This completes the proof. 80 Theorem. The following are equivalent in a space X: (1) X is a normal space. (2) For each closed set F and open set H containing F, there exists an open set G such that F ⊆ G ⊆ CI(G) ⊆ H. Proof. (1) ⇒(2) Suppose X is normal. Also F ⊆ H, F is closed and H is open. Then X−H is closed and F ∩ (X−H) = ∅. Since X is normal, there exist disjoint open sets G and G such that F ⊆ G and X−H ⊆ G . But G ∩ G = ∅ implied G ⊆ X−G . X−H ⊆ G implies X−G ⊆ H. Where X−G is closed. Thus F ⊆ G ⊆ CI(G) ⊆ CI(X−G ) = X−G ⊆ H or F ⊆ G ⊆ CI(G) ⊆ H. This proves (2) (2) ⇒ (1) Let F , F be disjoint closed sets. Then F ⊆ X−F and X−F is open by (2), there exists an open set G such that F ⊆ G ⊆ CI (G) ⊆ X−F But CI(G) ⊆ X−F implies F ⊆ X−CI(G).G ⊆ CI(G) implies G ∩ (X−CI(G)) = ∅. Thus F ⊆ G, F ⊆ X−CI(G) and G ∩ (X−CI(G)) = ∅. This proves (1). Hence the proof. 81 Theorem. The continuous closed surjective image of a normal space is normal. In other words, normality is invariant under continuous closed surjections and hence is a topological property. Proof. Let f: X→ Y be a continuous closed surjection and X a normal space. Let F , F be disjoint closed sets in Y. Then f ( F ), f ( F ), are disjoint closed sets in X. Since X is normal, there exist disjoint open sets U, V such that f ( F ) ⊆ U, f ( F ), ⊆ V. Since f is closed, therefore G= Y−f(X−V), H= Y−f(X−V) are open in Y. Then it is easy to see that G, H are disjoint. This completes the proof. 82 Lecture#26. Lecture Plan • We will continue our study on normal spaces • Hereditary of Normal Spaces • Urysohn Lemma Theorem. Normality is a closed hereditary property. Proof. If C is closed in X and A, B are disjoint closed sets in C, then, A, B are closed in X. Since X is normal, therefore there exist disjoint open sets U, V in X such that A ⊆ U, B ⊆ V. Then U ∩ C, V ∩ C are disjoint open sets in C such that A ⊆ U ∩ C, B ⊆ V ∩ C. This proves that C is normal. Theorem (Urysohn Lemma). Let X be a normal space. If F , F are any disjoint closed sets in X, then there exists a continuous map f : X→ [0,1] with f( )=0, f(F )=1. Proof. Step I. F ∩ F = ∅ implies F ⊆ X−F and X−F is open. Therefore there exists an open set U F ⊆U / ⊆ CI(U / ) ⊆ X−F / such that 83 Step II. Now U / and X−F are open subsets of closed set F and CI(U / ) respectively. Therefore, in the same way, there exist open sets U F ⊆U / ⊆CI(U / and U / )⊆U / such that ⊆ CI(U / / )⊆U / ⊆ CI(U / ) ⊆ X−F Step III. Let d=m/2 , n=1,2,… , m=1,2,… , 2 −1. Numbers of the form d are called dyadic rational numbers. If we continue this process, we obtain an open set U such that < ⇒F ⊆U ⊆CI(U ) ⊆ U ⊆CI(U ) ⊆ X− F . Step IV. We define f : X →[0, 1] as : f(x)=0, x ∈ U and f(x) = sup{d: x ∉ U }. It is clear that f(F ) = 0 and f(F )=1 Step V. We show that f is continuous. Consider the family {[0, ), ( , 1],0 < < 1}, which is a subbase for [0,1]. We simply show that f We note that f(x)< ⇔ x ∈ U , for some This gives that f Similarly, f(x) > f [0, ) and f ( , 1] are open . < . [0, ) ={x : f(x) < }= ∪ U , which is open. ⇔ x∉CI(U ), for some ( , 1] = {x : f(x) > }= > . Then ∪ (X−CI(U )), which is open. > 84 This proves that f is continuous . Hence the proof . 85 Lecture#27 Lecture Plan • Compactness • Definition • Examples • Theorems Definition ( Open Cover ) • Let X be a topological space. A class F= {F : i ∈ I} of open subsets of X is called an open convering or cover of X, If ∪ F =X, i ∈ I. • A class G= {F : i ∈ I }, I ⊆ I, of open subsets of X is called a subcover of the cover F, If ∪ F =X, i∈ I . • It is called a finite (respt. countable ) subcover, if I is finite (respt. countable). • Example. Consider the class of = : ∈ × , is the open disc in the plane with radius 1 and centre = ( , ), , ∈ . Then is a cover of , i.e., every point in belongs to at least one member of . 86 • On the other hand , the class of open discs × } Where cover of . ( , )∈ ={ has centre at p and radius ∗ : ∈ is not a but not belong to any member of H. • Definition (Compactness). A space X is said to be compact, • if every open cover of X has a finite subcover. • Example. The interval = (0,1) on the real line the usual topology is not compact. with • Definition (Compact subset). A space X is said to be compact, if every open cover of X has a finite subcover. • A subset A of a space X is compact, if it is compact as a subspace of X. • That is , if A is compact and ⊆ Example. ∈ A⊆ ∈ , Then , where , I is finite and each F is open in X. (1) All finite spaces are compact and are called trivial compact spaces. (2) Every cofinite space is compact. (verify) (3) R with the usual topology is not compact. (4) No infinite discrete space is compact (verify) 87 Lecture#28 Lecture Plan. • We will continue the study of Compactness • Union of compact subsets • Image of compact subsets • Hereditary property • Finite Intersection Property Theorem. Any finite union of compact subsets of a space X is compact. Proof. Theorem. A continuous image of a compact space is compact. Proof. Let f: X→ Y be a continuous map and X, a compact space. We show that f(X) is compact as a subspace of Y. Let {G : i ∈ I} be an open cover of f(X). Then each G = f(X) ∩ H , H open in Y. Clearly F= {f (H ): i ∈ I} is an open cover of X. Since X is compact, therefore F has a finite subcover {f (H ): i ∈ I }, I is finite, that is to say, f (H ) ∪ f (H ) ∪ …∪ f (H ) = X. 88 Then the corresponding finite family (G , G , … , G ) is an open cover of f(X) which is elaborated as: H ∪ H ∪… ∪ H ⊇ ff G H ff (H ) = f(X). ∪ ff (H ) ∪… ∪ Or ∪ H ⊇ f(X). Now G = f(X) ∩ H gives =1 ∪ G ∪ … ∪ G = f(X) ∩ ( ∪ H ) ⊇ f(X) ∩ f(X) = f(X) =1 ∪ G ⊇ f(X). This gives that {G , j= 1, 2, … , n} is a finite =1 open cover of f(X). This completes the proof. Or Theorem (Closed Hereditary Propety) A closed subspace of a compact space is compact. Proof. Let A be a closed subspace of a compact space X and {G = i ∈ I} an open cover of A. Then G = A ∩ H , i ∈ I, where H is open in X. Clearly {X−A, H : i ∈ I} Is an open cover of X. Since X is compact, therefore X has a finite open cover {X−A, H , i ∈ I }, I is finite. It follows that the corresponding finite class {G = i ∈ I } is a finite open cover of A. This proves that A is compact. Hence the proof. 89 Example. (1) All finite spaces are compact and are called trivial compact spaces. (2) Every cofinite space is compact. (verify) Example. (1) All finite spaces are compact and are called trivial compact spaces. (2) Every cofinite space is compact. (verify) (3) R with the usual topology is not compact. (4) No infinite discrete space is compact (verify) Theorem. Any finite union of compact subsets of a space X is compact. Proof. 90 Lecture#29 Theorem. Let be a subspace of . Then is compact if and only if every covering of by sets open in contains a finite subcollection covering . • Note. Remember that: • Every class of closed sets with f.i.p has a non empty intersection • is equivalent to • Every class of closed sets with empty intersection has a finite subclass with empty intersection Definition (Finite Intersection Property). • Let F= F : i ∈ I be a class of subsets of space X. • Then F is said to have a finite intersection property, if every finite subclass F , F , … , F of {F } has a nonempty intersection, that is, F ∩ F ∩ … ∩ F ≠ ∅. • Example. Let X= R and • F={…,(−∞, −2), (−∞, −1), (−∞, 0), (−∞, 1), (−∞, 2),…} is a class of open intervals. • Then F has a finite intersection property. 91 • Example. Let X = R and F = {(0,1), (0, ½), (0,1/4), …}. • Then F has a finite intersection property. Theorem (Characterization of Compactness). following are equivalent in a space : The (1) X is compact (2) Every class of closed sets with empty intersection has a finite subclass with empty intersection. Proof: (1) ⇒ (2) Suppose {F } is a class of closed sets with ∩ F = ∅. Then by De-Morgan’s law X = X−∅ =X− ∩ F = ∪ (X−F ) This implies {X−F } is an open cover of X. Since X is compact therefore X has a finite open cover {X−F , X−F , … X − F }, that is, ∪ (X− F ) = X. Again by De-Morgan’s Law i=1 This gives (2). ∩ F = X−X = ∅. j=1 (2) ⇒ (1) is similar and hence is left for the reader. Theorem (Characterization). The following are equivalent in a space X : 92 (1) X is compact. (2) Every class of closed sets having finite intersection property has a nonempty intersection. Proof. Proof is similar as above. Theorem A space X is compact, if every basic open cover of X has a finite subcover. Proof. Let {G : i ∈ I} be an open cover of X and = : j ∈ J an open base for . Then each G is the union of some B ′s. Then class of all such B ′s gives a basic open cover of X. By hypothesis, this class has a finite subcover . then for each ∈ , B ⊆ G , for some i ∈ I. The class of all such G ′s gives us a finite open cover of X. This proves that X is compact. Hence the proof. • Example . Let X= { , , , } Then each of the collections = {∅, { , }, X} and = {∅, { }, X} is a topology of X. • But ∪ = {∅, { , }, { }, } is not a topology on X, because it does not meet the requirement ( ), for { , }, • { }∈ ∪ to ∪ . but { , } ∪ { } = { , , } does not belong 93 Lecture#30 Theorem. If X is a Hausdorff space and C, a compact subspace of X such that ∉ then and can be separated by open sets. Proof Let y ∈ C. Then x ≠y and since X is Hausdorff, there exist disjoint open sets G , H such that x ∈ G , y ∈ H . Thus for each y ∈ C, we obtain an open set H , such that C ⊆ ∪ U , y ∈ C. since C is compact, therefore there exists a finite open cover for C, that is H ,H ,… H such that C ⊆ Correspondingly, we get G , G , … G contain x. put ∪ H . =1 ∩ G =1 = G and ∪ H = H. Then x and C can be separated by open =1 sets G and H. This completes the proof. Theorem. Every compact subspace of a Hausdorff space is closed. Proof Let C be a compact subspace of a Hausdorff space X. We prove that C is closed, that is, X−C is open. Let x ∈ X−C. By theorem 37.1, x and C can be separated by open sets G and H. Thus x ∈ G ⊆ X−C implies X−C is the union of open sets and hence is open. This completes the proof. 94 Theorem. Let : → be a bijective continuous function . If is compact and is Hausdorff, then is homeomorphism. Proof Let f: X → y be a bijective continuous map and X, a compact space and Y a Hausdorff space. It suffices to show that f is closed. Suppose F ⊆ X is closed, then by theorem 36.5, F is compact and by theorem 36.3, f(F) is compact. Since a compact subspace of a Hausdorff space is closed, therefore f(F) is closed. This proves that f is closed. Hence the proof. Definition. (Locally Compact) • A space X is said to be locally compact at x ∈ X • If there is some compact subspace Y of X that contains a nbd of x. • If X is locally compact at each of its points, then X is called a locally compact space. Renark. Every compact space is locally compact but converse is not true in general. • Example. • The real line • The point is locally compact. lies in some interval ( − , + ), which in tern is contained in the compact subspace { − , + } 95 Theorem. Prove that every compact space is locally compact. Proof. Theorem. Let : → be an open continuous surjection. If is locally compact then prove that is locally compact. Proof. Let y ∈ Y. Then there is x ∈ X such that f(x) = y. Let U be a compact nbd of x in X, since X is - compact. Also f(x) ∈ f(U ) ⊆ f(U ) gives f(U ) is a nbd of f(x), since f is open implies f(U ) is open in Y. since f is continuous, therefore by theorem 36.4, f(U ) is compact. So each y ∈ Y has a compact nbd. This proves that Y is - compact. Hence the proof. 96 Lecture#31 Lecture Plan • Connected Spaces • Connected subsets • Some spaces are in a sense ‘disconnected’, being the union of two or more completely separate subspaces. • For example the space X ⊂ R consisting of the two intervals • A = [0, 1] and B = [2, 3] should certainly be disconnected Definition. • A space X is connected if it is not the union of two nonempty disjoint open sets. • In other words, • if X is disconnected, then there exist disjoint nonempty open sets A, B such that A ∪ B= X. • The pair (A, B) is called a disconnection of X. 97 Examples (1) (0,1) – {1/2} is disconnected. (2) {x} ⊆ R is connected. (3) Every indiscrete space is connected. (4) Every discrete space with more than one point is disconnected. Example • Let X= {0,1} and = ∅, {0}, X . • Then X is connected, called Sierpinski space. Example • Let X ={ , , } and • and = {∅, { }, { , }, X} = {∅, { }, { }, { , }, X}. • Then (X, ) is disconnected and (X, ) is connected. Theorem (characterization of Connected Space) In a space X, the following are equivalent : (1) X is connected. (2) The only open and closed subsets of X are ∅, X. (3) There does not exist a continuous map 98 f : X → {0,1} from a space X onto the discrete space {0,1}. Proof ~(2) ⇒ ~(1).Suppose A ⊆ X is both open and closed, and A≠ ∅, A ≠X. Then X = A ∪ (X−A) gives a disconnection of X. ~(3) ⇒ ~(2). Suppose there is a continuous map f: X → {0,1} from a space X onto the discrete space {0, 1}.Then f (0} and f {1} are open and X = f {0} ∪ f {1}. Thus f {0}, f {1} are the nonempty open and closed subsets of X. ~(1) ⇒ ~ (3) If X = A ∪ B, A, B are nonempty disjoin open sets. Then define g : X → {0,1} as : g(x) = 0, x ∉ A 1, x ∈ A. Then g is continuous surjective. This completes the proof. Theorem. The continuous image of a connected space is connected. Proof supposes f: X → Y is continuous and X is connected. We show that f(X) is connected. Suppose f(X) is disconnected. Then there is continuous surjection g: f(X) → {0, 1} (discrete). Then gof: X → {0, 1} is also continuous surjection. 99 This shows that X is disconnected, a contradiction. This completes the proof. Definition • A subset A of a space X is disconnected if there exist open subsets G, H of X such that ∩ ≠ and • (1) ( ∩ ) ∪ ( ∩ ) = • (2) ( ∩ ) ∩ ( ∩ ) = Theorem. If f : X →Y is continuous and A, a connected subset of X, then f(A) is a connected subset of Y. Proof Suppose f(A) is not connected. Then there exist nonempty open sets U, V in Y such that f(A) ⊆ U ∪ V, U ∩ V ⊆ Y−f(A) and U ∩ f(A) ≠ ∅, V ∩ f(A) ≠ ∅. Since f is continuous, therefore f (U), f (V) are open in X. But A ⊆ f f(A) ⊆ f (U ∪ V) = f (U) ∪ f (V) or A ⊆ f (U) ∪ f (V) … (I) Also f (U) ∩ f (V) = f (U ∩ V) ⊆ f f (Y)− f f(A) = X−f f(A) ⊆ X−A Or f (U) ∩ f (V) ⊆ X−A It is easy to verify that f (Y-f(A)) = (U) ∩ A ≠ ∅, f … (II) (V) ∩ A ≠ ∅. From (I) and (II), it follows that A is not connected, a contradiction. Thus f(A) is connected. This completed the proof. 100 101 Lecture#32 Lecture Plan • Components of Connected Spaces • Totally Disconnected subsets • Locally Connected Spaces • Path Connected Spaces • Definition (component) • A component E of a topological space X is a maximal connected subset of X; • that is, E is connected and E is not a proper subset of any connected subset of X. • Example • If X is connected, then X has only one component: X itself • Consider the following topology on • ={ , , , , } = { , , { }, { , }, { , , }, { , , , }} • The components of X are { } and { , , , } • Any other connected subset of X, such as { , , } is a subset of one of the components. 102 • Example • If X is connected, then X is itself a component. • Let X be a discrete space. Then each singleton is a component • Let X={ , , , , }, = {∅, { }, { , }, { , , }, { , , , }, X} • Then connected subsets of X are { }, { }, { }, { , , } , { , , , } • Thus the component of X are : • { }, { , , , } . • Definition (Totally Disconnected) • A space X is totally disconnected, • if for each x, y ∈ X, • x ≠y, there exist disjoint open sets G, H • such that x ∈ G, y ∈ H and X= G ∪ H. • Example • One point space is totally disconnected • Discrete space is totally disconnected 103 • R with the upper limit topology generated by open- closed intervals (a,b] is totally disconnected. For, if x, y ∈ R, x < y, then (−∞, x]= G, (x, ∞)= H are such that R= (−∞,x] ∪ (x, ∞) and x ∈ G, y ∈ H, • R with the usual topology is not totally disconnected. • Theorem. The component of a totally disconnected space are singletons. • Proof. suppose a component C of X is not singleton. • Let x, y ∈ C, x ≠ y. • Since X is totally disconnected, • therefore there exist a disconnection of X, • that is, X = G ∪ H, x ∈ G, y ∈ H. • Consequently, C ∩ G and C ∩ H are open and disjoint subsets of C such that C= (C ∩ G) ∪ (C ∩ H) and x ∈ C ∩ G, y ∈ C ∩ H. • This shows that C is disconnected, • a contradiction. This proves that C is singleton. • Hence the proof. • We know that 104 • every totally disconnected space is Hausdorff, but the converse is not true in general . • Definition (Locally Connected Space) • A space X is said to be locally connected at x ∈ X , • if for every open set U containing x, • there exists a connected open set V such that x ∈ V ⊆ U. • A space is said to be locally connected, • if it is locally connected at each of its points. • A subset A of a space X is called locally connected, • if it is locally connected as subspace. • Example. Every discrete space X is locally connected. • Definition. Paths • Let = [0,1], the closed unit interval. • A path from a point to a point in a topological space X is a continuous function : → X with • (0) = and (1) = . • A is called initial point of the path • B is called terminal point of the path • 105 • Example • For any • ∈ , the function : → defined by is a constant function and hence a path. ( )= • It is called constant path. • Let : → • then g: → be a path from to . defined by ( ) = (1 − ) • g is a path from to . • Definition. Path Connected Space • A space X is path connected, • if each pair of points can be joined by a path in X. • Theorem. Let X be a space and ∈ . Then X is path connected if and only if each ∈ can be joined to by a path in X. • Proof. • Let X be path connected. • By definition, each X. • Converselly • Let , ∈ ∈ can be joined to by a path in 106 • Let : → • g: → be a path from a path from • Then h: → • to and to defined by (2 ), 0 ≤ ≤ 1 2 (2 − 1), 1 2 ≤ ≤ 1 ( )= • Is a path from to in X. • This completes the proof. • Theorem. Every path connected space is connected. • Proof. • Suppose a path connected space X. • We shall prove that X is connected. • Assume contrary • Suppose X is disconnected. • Let = ∪ , where G and H are non empty open sets form disconnection for X. • Let : → • ∈ , • Then ∈ be a path in X from a to b in X such that ( ), ( ) form a disconnection for I, 107 • a contradiction to the fact that [0,1] is connected • This proves that X is connected. • This completes the proof. End. 108