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

Transcription

1 Topology A Profound Subtitle Dr.

2 Copyright c 2017

3 Contents I General Topology 1 Compactness of Topological Space Introduction Compact Space Compact Space Tube Lemma Product of Compact Spaces Limit Point Compactness Limit Point Compactness vs Compactness Limit Point Compactness vs Compactness for Metric Space Locally Compact Space One-point Compactification Key Notes Important Results 21 Bibliography Books 39 Articles 39 Index

4

5 I General Topology 1 Compactness of Topological Space Introduction 1.2 Compact Space 1.3 Limit Point Compactness 1.4 Locally Compact Space 1.5 One-point Compactification 1.6 Key Notes 1.7 Important Results Bibliography Books Articles Index

6

7 1. Compactness of Topological Space 1.1 Introduction Now we are going study third C" for the topological space as compactness properties. The compactness properties of the space is not as intuitive as continuity or connectedness. Compactness is a property that generalizes the notion of a subset of Euclidean space being closed (that is, containing all its limit points) and bounded (that is, having all its points lie within some fixed distance of each other). Examples include a closed interval, a rectangle, or a finite set of points. The interval A = (, 2] is not compact because it is not bounded. The interval C = (2,4) is not compact because it is not closed. The interval B = [0,1] is compact because it is both closed and bounded. In R n, the A B C R ) [ ] ( ) Figure 1.1: Compactness of subsets of R compact sets are the closed and bounded sets, but in a general topological space the compact sets are not as simple to describe. In fact, a formal definition for compactness in a general topological space took some time for topologists to establish. Several definitions were suggested during the development of topology in the early twentieth century. Ultimately, topologists settled on the definition of compactness proposed in 1923 by Pavel Sergeevich Alexandroff ( ) and Pavel Urysohn; it is presented here. 1.2 Compact Space Compact Space We start with some important definitions associated with compactness.

8 8 Chapter 1. Compactness of Topological Space Definition 1 Cover for Space. A collection C of subsets of a space X is said to cover X, or to be a covering of X, if the union of the elements of C is equal to X. The collection C is said to open covering if its all subsets are open subsets of X. Example 1.1 Let X = {a,b,c} be a space with topology τ = {X,φ,{a},{b},{a,b}} Now we have three collections of subsets of X as C 1 = {{a},{b,c}}, C 2 = {{a},{b},{c}}, C 3 = {X} The collection C 1 and C 2 are not open cover of X since {b,c} and {c} are not open. But The collection C 3 is open cover of X since X is equal to union of elements of C 3. Example 1.2 Let X = {1,2,3,4} be a space with topology τ = {X,φ,{1},{4},{1,4},{2,3},{1,2,3},{2,3,4}} Now we have three collections of subsets of X as C 1 = {{1},{4},{2,3}}, C 2 = {{1,4},{2,3}}, C 3 = {{1},{2,3,4}} The collection C 1 is open cover of X since X = {O: O C 1 } and each member of C 1 are open. Similarly C 2 and C 3 are open cover of X. {( ) } n Example 1.3 The collection O = 0, : n N is a cover of (0,1). n + 1 Let X = (0,1) and {( )} n O n = 0, n N n + 1 Then ( O 1 = 0, 1 ) (,O 2 = 0, 2 ),...O = (0,1) 2 3 ( ) n Since lim n = 1 so, O = (0,1). n + 1 Evidently X = O 1 O 2... O. Therefore, O is cover of X = (0,1). Definition 2 Cover of Subset of Space. Let A be a subset of a topological space X, and let C be a collection of subsets of X, then the collection C is said to 1. Cover A or to be a cover of A if A is contained in the union of the sets in C. 2. Open cover of A, each set in C is open. Definition 3 Subcover of Subset of Space. The subcollection C of C is said to be subcover of a subset A of space X if C cover A. Example 1.4 By the definition of a basis, it follows that every basis for a topological space X is an open cover of X. Example 1.5 Open Covers of R. The two collections of intervals C 1 = {...,( 1,1),(0,2),(1,3),...} and C 2 = {(,1),(0, )} are both open covers of R.

9 1.2 Compact Space 9 From above example (1.5) C 1 and C 2 both are open covers of R, but C 1 consist of infinitely many sets, and C 2 consist of finitely many sets. Now we are interested in spaces for which every open cover containing infinitely many sets can be reduced to a subcover containing finitely many sets and in this way we can defined the compactness of space as follows. Definition 4 Compact Space. A topological space X is compact if every open cover of X has a finite subcover. Example 1.6 The real line R in the standard topology is not compact since C = {...,( 1,1),(0,2),(1,3),...} is an open cover, but no finite subcollection of C covers R. Example 1.7 Let X = {x 1,...x n } be a topological space that contains only finitely many points. Then X is compact since there can be only finitely many open sets in the topology on X, and therefore every open cover of X is already finite. We extend the definition of compactness to subsets of a topological space, as follows: Definition 5 Compactness to Subsets of a Space. A subset A of space X is said to be compact in X if A is compact in the subspace topology inherited from X. The following lemma will give the necessary and sufficient condition for any subset of compact space to be compact. Lemma 1.1 Necessary and Sufficient Condition for Compactness of Subspace. A subspace Y os a space X s compact if and only if every cover of Y by sets that are open in X has a finite subcover. Proof. Necessity As given that Y is compact subspace of X and let A = {A α } α J is cover of Y, where A α are open in X. Then we want to show that A has a finite subcover of Y. Since A covering Y and Y is compact subspace of X, then the collection {A α Y : α J} cover Y by the sets that are open in Y. Hence finite subcollection {A α1 Y,...,A αn Y } cover Y. This show that {A α1,...,a αn } is subcollection of A that cover Y. Sufficiency If every cover of subspace Y by sets that are open in X has a finite subcover of Y, then we want to show that Y is compact. Let A α = {A α} be cover of Y, where A α are open in Y. Since Y is subspace of X, therefore by definition, for each α J, there exist A α in X such that A α = A α Y. Now as given A = {A α } is cover of Y, where A α are open in X, then by given hypothesis there exist finite subcollection {A α1,...a αn } such that it cover Y. Hence {A α 1,...A α n } is subcollection of A cover Y also. Therefore by definition Y is compact. Example 1.8 The subset A = {0} { 1 n : n Z} is compact in R. Let C be a cover of A by open sets in R. There exists at least one open set U 0 in C that contains the point 0. Such an open set contains all but at most finitely many of the points in A. If U 0 contains all of the points in A, then U 0, by itself, is a finite subcover of C. Otherwise, let 1/m be the smallest of the points in A that are not in U 0. For each point 1/i,vthere is an open set U i in C that contains it. It follows that the finite collection {U 0,U 1,...,U m } is a subcover of C. Thus, A is compact in R. Example 1.9 The subspace (0,1] of R is not compact. The collection C = {( 1 n,2): n Z} is a cover of (0,1] by sets that are open in R. There is no finite subcollection of C that covers (0,1], and therefore (0,1] is not compact as a subspace of R. The next two theorems show that being closed and being compact are closely related properties.

10 10 Chapter 1. Compactness of Topological Space Theorem 1 Closed Subset in Compact Set. Let D is compact in space X, and C is closed subset in X and contain in D, then C is compact in X. Proof. Let D be compact in the topological space X. Suppose that C is closed in X and C D. Further, suppose that C is a cover of C by sets that are open in X. The set X C is open. Add X C to the collection C to obtain a new collection C = {C} {X C}. The collection C is an open cover of X and therefore is an open cover of D. Since D is compact in X, there is a finite subcollection of C that covers D. The set C is covered by those sets in the finite subcover of C that were originally in C. Therefore there is a finite subcollection of C that covers C, implying that C is compact in X. R The above Theorem (1) indicates that a closed subset of a compact space is compact. The converse relationship, however, does not generally hold. In other words, a compact set in a topological space is not necessarily a closed set, as demonstrated in the example. Example 1.10 Consider R f c, the real line in the finite complement topology. Every subset of R f c is compact. Aside from the whole set itself, every other infinite set A R f c is not closed, since the complement of such a set A is not open. Therefore there are subsets of R f c that are compact but not closed. Theorem 2 Closed Subspace of a Compact Space. Every closed subspace of a compact space is compact. Proof. Let Y is closed subspace of compact space X, then we want to prove that Y is also compact. Let {A α } be open cover of Y, then by definition of cover Y A α (1.1) α Since Y is closed so X Y is open and also ( ) (X Y ) A α = X (1.2) α Here result (1.2) show that the family consisting of the sets (X Y ) and A α s are open cover of X and X is compact, so there exist finite sub-collection, consisting of X A,A α1,a α2...,a αn also cover the X. Therefore, X = (X Y ) ( ) n A αr (1.3) Now we claim Y r=1 A αr (1.4) r=1 Assume result (1.4) not hold. So, there exist a Y such that a / A αr anda / X Y r=1 This show that a / (X Y ) A αr r=1

11 1.2 Compact Space 11 This show that the family consisting of sets X Y,A α1,...,a αn is not an open cover of X. This is contradiction. Hence our initial assumption is wrong, therefore the family {A αr : r = 1,...,n} is finite open cover of Y. Finally, we can conclude that if {A α } is a cover of Y, then there exist finite subcover {A αr : r = 1,...,n} of Y also. Hence by definition, Y is compact. Although compact sets in a topological space are not necessarily closed, the following theorem indicates that there are general circumstances under which compact sets are automatically closed. Theorem 3 Necessary and Sufficient Condition for Compactness of Subspace. Let (Y,τ Y ) be a subspace of (X,τ), then Y is compact with respect to topology τ Y if and only if Y is compact with respect to topology τ on X. Proof. Necessity Let (Y,τ Y ) be a subspace of (X,τ) and Y is compact with respect to topology τ, then we want to show that Y is also compact with respect to τ Y. Let {H i } be a τ Y -open cover of Y, then Y i H i. H i Y G i τ such thath i = G i A G i (By definition of subspace topology) G i τ such thath i G i i H i i G i But Y i H i, so that Y i G i. This show that {G i } is a τ-open cover of Y which is know to be compact with respect to τ and hence the cover {G i } must be reducible to a finite subcover, say {G ir : r = 1,2,...}. Then we can say that Y n r=1 G i r. From which Y Y Y ( n r=1g ir ) = n r=1(y G ir ) = n r=1h ir This show that A n r=1 H i r. So, {H ir : 1 r n} is a τ Y -open cover of Y. Finally, we can say that we are able to show that the cover {H i } of Y is reducible to a finite subcover {H ir : 1 r n} of Y. Therefore Y is τ Y -compact. Sufficiency suppose that (Y,τ Y ) is a subspace of (X,τ) and Y is τ Y -compact, then we want to show that Y is τ-compact. Let {G i } be a τ-open cover of Y, so that Y i G i. Hence Y Y Y ( G i ) or, Y (Y G i ). On taking H i = G i Y, we get Y i H i. G i τ H i = G i Y τ Y Now from above result it is clear that {H i } is τ Y -open cover of Y which is know to be τ Y - compact and hence this cover must be reducible to a finite sub-cover, say {H ir : 1 r n}. This show that Y n r=1h ir = n r=1(g i Y ) Y n r=1(g ir Y ) n r=1g ir Y n r=1g ir This prove that {G ir : 1 r n} is a finite subcover of the cover {G i }. Starring from an arbitrary τ-open cover of Y, we are able to show that this open cover is reducible to a finite subcover thereby Y is τ-compact.

12 12 Chapter 1. Compactness of Topological Space Theorem 4 Compact Subspace of Hausdorff Space. A compact subspace A of Hausdorff space X, is closed in X. Proof. Let A be compact in the Hausdorff space X. To show that A is closed, we prove that X A is open. Thus, let x X A be arbitrary. We show that there is an open set U such that x U X A. Since X is Hausdorff, we know that for each a A, there exist disjoint open sets U a and V a such that x U a and x V a. Then C = {V a } a A is an open cover of A. Because A is compact, there is a finite subcover {V a1,...,v an } of C. Let V = n i=1 V ai and U = n i=1 U ai. Then U and V are open sets such that A V and x U. Furthermore, since U ai and V ai are disjoint for each i, it follows that U and V are disjoint as well. Thus U and A are disjoint, and therefore there exists an open set U such that x U X A, as we wished to show. Hence X A is open, implying that A is closed. The following theorems provides some results on the compactness of unions and intersections of compact sets: Theorem 5 Unions of Compact Sets. If C 1,...,C n are each compact in X, then n j=1 C j is compact in X. Proof. Theorem 6 Intersections of Compact Sets. If {C α } α A is a collection of compact subsets of Housdroff space X, then α AC α is compact in X. Proof. R 1. An arbitrary union of compact sets need not be compact. 2. An intersection of compact sets need not be compact. Now we want to study relation between continuity and compactness of topological spaces. Theorem 7 Continuous Image of Compact Space. Let f : X Y be continuous, and let A be compact in X. Then f (A) is compact in Y. Proof. As given that a function f : X Y is continuous and A is compact in X, then we want to show that f (A) is compact in Y. Let V = {V 1,V 2,...,} is collection of open set in Y which cover f (A). This show that f (A) i V i Now our aim is to show that f (A) V i i=1

13 1.2 Compact Space 13 As given Ais compact Acovered by collection of all open sets of the form f 1 (V ),V V A f 1 (V 1 ) f 1 (V 2 )... f 1 (V n ) (since A is compact) A f 1 (V 1 V 2... V n ) f (A) V 1 V 2... V n f (A) V i Hence f (A)is compact i=1 Now we will show that compact space is homeomorphic to Housdorff space through the following theorem. Theorem 8 Homeomorphism between Compact and Housdroff Space. Let f : X Y be a bijective continuous function. If X is compact and Y is Housdorff, then f is a homeomorphism. Proof. As given that X and Y are compact space and Housdorff sapce respectively and f is continuous bijection map. We want to show that f is homeomorphism. Let g = f 1. We need to show that g: Y X is continuous. For any V X, we have g 1 (V ) = f (V ). We are to show that if V is closed in X, then g 1 (V ) is closed in Y. Suppose V is closed in X. Since X is compact, V is compact by closed subspace of compact space is compact. So, f (V ) is compact from continuous image of compact space is compact. Since Y is Housdorss, f (V ) closed by compact subspace of Housdorff space is closed. But f (V ) = g 1 (V ), so g 1 (V ) is closed. From continuity defined from closed sets, it follows that g is continuous. Thus by definition, f is a homeomorphism. Corollary Let f : X Y be a injection continuous function. If X is compact and Y is Housdorff, then f determine a homeomorphism form X to f (X). That is, f is an embedding of X into Y. We recall that a function f : X Y from a topological space X to a topological space Y is said to be an identification map if it is surjective and satisfies the following condition: a subset U of Y is open in Y if and only if f 1 (U) is open in X. Theorem 9 Identification Map between Compact and Housdroff Spaces. A continuous surjection map f : X Y from a compact topological space X to a Hausdorff space Y is an identification map. Proof Tube Lemma The Tube Lemma is a useful tool in working with Cartesian products of finitely many compact spaces. A general discussion is followed by applications of the lemma. Let X and Y be topological spaces. A slice in the Cartesian product X Y is a subspace of the form {x} Y or X {y} where x X and y Y. A tube is an open subset of the Cartesian product that is of the form G Y or X H where G is open in X and H is open in Y. In the Euclidean plane, a slice would be either

14 14 Chapter 1. Compactness of Topological Space a vertical line or a horizontal line and open strips (vertical or horizontal) are examples of tubes. Tubes are one type of open subsets of the cartesian product X Y. The Tube Lemma is applicable when one of the factors is compact. Let Y be the factor that is compact. A good way of thinking about the lemma is that when you consider the slices {x} Y as points", the tubes G Y, where x G, behave like a base. The following is a statement of the lemma. Lemma 1.2 The Tube Lemma. Let X and Y be topological spaces, and assume that Y is compact. If x X, and U is an open set in X Y containing {x} Y, then there exists a neighborhood W of x in X such that W Y U. Proof. For each y Y pick open sets W y in X and V y in Y such that (x,y) W y V y U. The collection of sets {V y } y Y is an open cover of Y. Since Y is compact, finitely many of these sets cover Y, say V y1,...,v yn. Let W = n i=1 W yi. Then W is open in X, and W contains x because each W y contains x. Note that W Y (W yi V yi ) U Therefore W Y contains {x} Y and is contained in U, as desired Product of Compact Spaces i=1 Theorem 10 Product of Compact Spaces. If X and Y are compact topological spaces, then the product Z = X Y is compact. Proof. Let G be an arbitrary open cover of Z. So every member G of G is a union of the sets of the form U α V α, where U α and V α are open in X and Y respectively. Now let U = {U α V α : α I}. Here every members of U is a subset of every member of G. So, we can say that U is an open cover of Z. Let y Y be arbitrary. Then X {y} is compact (because X {y}cngx). Since X {y} Z and U is open cover of Z U is open cover of X {y}. But X {y} is compact so cover U of X {y} must be reducible to a finite subcover say Note and So, {U i V i : i = 1,...,n} X = U i i=1 y V i,1 i n y n V i = V y say Being a finite intersection of open set, V y is also open and it is nbd of y. Now the family {U i V y : i = 1,...,n} i=1 is also open cover of X {y}. And this show that {V y : y Y } is open cover of Y. But Y is compact so {V y : y Y } must be reducible to a finite subcover say {V y j : j = 1,...,m}

15 1.3 Limit Point Compactness 15 So the family {U i V y j : i = 1,...,n& j = 1,...,m} open cover of Z. Recall the set U i V y j is subset of some G G and let this is G i j and here G i j is finite. So, we can say that G is open cover of Z and it is reducible to finite subscover. Hence by definition Z is compact. Corollary Let X 1,...,X n be topological spaces, and let A i be a compact subset of X i for each i = 1,...,n. Then A 1... A n is a compact subset of the product space X 1... X n. 1.3 Limit Point Compactness As we know there are other useful formulation of compactness as like limit point compactness of topological space. In this section we will discuss the limit point compactness which is weaker form compare to compactness and it coincides with the compactness of metrizable space. Definition 6 Limit Point Compactness. A topological space X is said to be limit point compact if every infinite subset of X has a limit point. Example 1.11 In R, the subspace A = {0} { 1 n : n Z} is limit point compact. Let B be an infinite subset of A. Then B must contain values of the form 1 n where n is arbitrarily large. Therefore 0 is a limit point of B. Thus, every infinite subset of A has a limit point, implying that A is limit point compact. Example 1.12 Let X be an infinite set with the finite complement topology. Then X is limit point compact. Let B be an infinite subset of X. We claim that every point of X is a limit point of B. If x X, and U is a neighborhood of x It intersects B in infinitely many points. (since U contains all but finitely many points of X) In particular, U intersects B in points other than x The point x is a limit point of B. Therefore every infinite subset B of X has a limit point, implying that X is limit point compact Limit Point Compactness vs Compactness As we proved in example that subspace A = {0} { 1 n : n Z} of R is limit point compact but it is compact also. So, we can establish a relation between these two. Theorem 11 Limit Point Compactness vs Compactness. If a topological space X is compact, then it is limit point compact. Proof. We want to show that X is compact X is limit point compact. But we will show that if X is not limit point compact X is not compact. Suppose that X is not limit point compact X has an infinite subset B that does not have a limit point Since each x B is not a limit point of B for every x B there exists a neighborhood U x of x such that U x B = {x}. Also, since B has no limit points B is closed (by standard result). Hence, X B is open. Let O be the collection of open sets {U x : x B} {X B}. Then O is an open cover of X. Furthermore O has no finite subcover, since each of the infinitely many points x in B is contained

16 16 Chapter 1. Compactness of Topological Space in only the open set U x in O. Therefore X has an open cover with no finite subcover. It follows that X is not compact. The converse of Theorem above does not hold. As the following example demonstrates, limit point compactness does not necessarily imply compactness. Example 1.13 Let Z have the topology generated by the basis B = {( n,n): n Z}. In this topology, Z is not compact but it is limit point compact. Solution : Since the basis B is an open cover of Z and that has no finite subcover so it will compact. Now we claim, that Z is limit point compact in this topology. In fact, we show that every nonempty subset of Z has a limit point. Thus, let A be a nonempty subset of Z. Every subset of Z has an element with minimum absolute value; let s be such an element of A. For every t Z such that t > s every open set containing t must contain s. Therefore every t satisfying t > s is a limit point of A. Hence, A has a limit point, implying that Z is limit point compact in the topology generated by B Limit Point Compactness vs Compactness for Metric Space Now we will establish a relation between compactness and limit point compactness for metric space. But first we will discuss a couple of lemmas which help to establish this result. Lemma 1.3 Let (X,d) be a metric space. If A X and c is a limit point of A, then for every ε > 0 the open ball B d (c,ε) intersects A in infinitely many points. The next lemma that we need is a Lebesgue Number Lemma for limit point compact metric spaces. We are in a somewhat peculiar situation here. As we know that the Lebesgue Number Lemma for compact metric spaces, and compactness is equivalent to limit point compactness for metric spaces. So Does that give us a Lebesgue Number Lemma for limit point compact metric spaces? The answer is no. We have not yet shown that compactness and limit point compactness are equivalent for metric spaces; in fact, that is what we are trying to prove here. Thus, we need to go through the work of proving a Lebesgue Number Lemma for limit point compact spaces, even though it is ultimately equivalent to the Lebesgue Number Lemma that we already proved. Lemma 1.4 Let (X,d) be a limit point compact metric space, and let C be an open cover of X. Then there is a λ > 0 such that for every x X there exists a U C satisfying B d (x,λ) U. Theorem 12 Limit Point Compactness vs Compactness for Metric Space. If (X, d) is a metric space, then X is compact if and only if it is limit point compact. Proof. First part of theorem is already proved, we only need to prove that if X is a limit point compact metric space, then X is compact. Given: The space X is a limit point compact metric space with metric d. And let C be an open cover of X. Aim: We will show that C has a finite subcover. By Lemma there exists a Lebesgue number λ for C. Consider the cover of X given by C = {B d (x,λ): x X}

17 1.4 Locally Compact Space 17 We claim that C has a finite subcover. We use such a subcover of C to help us obtain a finite subcover of C. To prove that C has a finite subcover, we assume that it does not and derive a contradiction. Thus assume that no finite subcollection of C covers X. Pick x 1 X. Then B d (x 1,λ) does not cover X, and therefore we can pick x 1 X B d (x 1,λ). Now, {B d (x 1,λ),B d (x 2,λ)} does not cover X either, so we can pick x 3 B d (x 1,λ) B d (x 2,λ). Continuing this process, we define a set Y = {x n } n Z X that is such that k 1 x k / B d (x j,λ), k Z j=1 The points in Y are distinct. Therefore Y is an infinite set. Let y be a limit point of Y. Then by Lemma, B d (y, λ 2 ) contains infinitely many points from Y. Let x p and x q be two such points and assume that p < q. Since x p and x q both lie in B d (y, λ 2 ), it follows that d(x p,x q ) < λ. But this contradicts the fact that k 1 x k / B d (x j,λ) Therefore C has a finite subcover. 1.4 Locally Compact Space Definition 7 Locally Compact Space. A topological space X is locally compact if every x X has a neighborhood that is contained in a compact subset of X. j=1 Example 1.14 Every compact space is automatically locally compact since each x X has X both as a neighborhood and as a compact set containing the neighborhood. Example 1.15 The real line R is locally compact since for each x R we have and [x 1,x + 1] is compact. x (x 1,x + 1) [x 1,x + 1] On other hand, R is not a compact space, for example, the class A = {...,( 3, 1),( 2,0),( 1,1),(0,2),(1,3),...} is an open cover of R but contains no finite subcover. Thus from this example a locally cpmpact space need not be compact. on the other hand, since a topological space is always a neighborhood of each of its points, the converse is true. That is Theorem 13 Every compact space is locally compact. Example 1.16 The subspace Q of R in the standard topology is not locally compact. There are two important spaces in mathematic are Metrizable spaces" and Compact Housdorff Spaces". These two are very useful for proving some important results and new constriction. If any space is not of these types, the next best thing one can hope for is that is a subspace of one of these spaces As we know that subspace of a metrizable space is itself metrizable, so one does not get any new spaces in this way. But a subspace of a compact Housdorff space need not be compact. Thus arises the question Under what condition is a space homoeomorphic with a subspace of a compact Housdorff space?.here we will find the answer of this question.

18 18 Chapter 1. Compactness of Topological Space 1.5 One-point Compactification Now, we are going to discuss the property of local compactness and a construction, called the one-point compactification, that allows us to add a single point to a locally compact Hausdorff space X in order to obtain a compact Hausdorff space Y containing X as a subspace. But before discussing one-point compactification we will discuss the meaning of compactification. A topological space X is said to be embedded in topological space Y if X is homeomorphic to a subspace of Y. Furthermore, if Y is a compact space, then Y is called a compactification of X. Frequently, the compactification of a space is accomplished by adjoining one or more points to X and then defining an appropriate topology on the enlarged set so that the enlarged space is compact and contains X as a subspace. Example 1.17 Consider the real line R with usual topology R u. We adjoin two new points, denoted by and to R and call the enlarged set R = R {, } the extended real line. The order relation in R can be extended to R by defining < a < for any a R. The class of subset of R of the form (a,b) = {x: a < x < b},(a, ] = {x: a < x}and[,a) = {x: x < a} is base for a topology R u. Furthermore (R,R u) is a compact space and contains (R,R u ) as a subspace and so it is a compactification of (R,R u ). Definition 8 One-point Compactification. Let X be a Hausdorff space. Set Y equal to the union of X and a single additional point, denoted. Define the open sets for a topology on Y = X { } to be subsets of the following two types: 1. Open sets in X, and 2. Sets of the form Y C, where C is a compact subset of X. We call the resulting topological space Y the one-point compactification of X. Of course, we need to verify that the collection of open sets just described is a topology. Theorem 14 Let X be a Hausdorff space. Then show that collection of subsets of Y = X { } in the definition of the one-point compactification of X is a topology on Y. Proof. The empty set is open in Y since it is an open subset of X. The entire set Y itself is open in Y since it is the complement of φ in Y, and φ is a compact subset of X. To prove that finite intersections of open sets in Y are open in Y, it is enough to check intersections of pairs of open sets U and V. The result for arbitrary finite intersections then follows by induction. Thus assume that U and V are open sets in Y. We need to check three separate cases. 1. First, if both U and V are open sets in X, then U V is an open set in X, making it an open set in Y. 2. Second, assume that U = Y C 1 and V = Y C 2, where C 1 and C 2 are compact subsets of X. Then U V = Y (C 1 C 2 ). Since finite unions of compact sets are also compact,c 1 C 2 is a compact subset of X. It follows that U V = Y C for a compact subset C of X, and therefore U V is open in Y in this case as well. 3. Finally, assume that U is an open set in X and V = Y C, where C is a compact subset of X. Then since is not in U, it follows that U V = U (X C). Now C is closed in X by Theorem, since it is a compact set in the Hausdorff space X. Therefore X C is open in X, implying that U (X C) is open in X. Thus, U V is open in X, making it open in Y in this case, too. It follows that if U and V are arbitrary open sets in Y, then U V is also open in Y, as we wished to show.

19 1.5 One-point Compactification 19 Last, we prove that arbitrary unions of open sets are open. We can express such an arbitrary union in the form ( U α ) ( (Y C β )) α A β B where each U α is open in X and each C β is a compact subset of X. The set α AU α is open in X; we denote this by V. Furthermore, (Y C β ) = Y β B and since each set C β is a compact subset of the Hausdorff space X, by Theorem implies that β BC β is a compact subset of X. Letting C = β BC β, we see that β B(Y C β ) = Y C and C is a compact subset of X. Therefore we only need to verify that U (Y C) is open in Y where U is open in X and C is a compact subset of X. Let C = X U, the complement of U in X; then C is closed in X and therefore β B C β U (Y C) = (X C ) (Y C) = (Y C ) (Y C) = Y (C C ) Now C is a compact subset of the Hausdorff space X, so C is closed in X. Therefore C C is closed in X, and since C C is a subset of the compact set C, it follows that C C is a compact subset of X. Thus Y (C C) is an open set in Y, implying that U (Y C) is an open set in Y. It follows that an arbitrary union of open sets in Y is an open set in Y. Thus the collection of subsets of Y described in the definition of the one-point compactification is a topology on Y. Now, as we know that X is a subset of the one-point compactification Y = X { }. Therefore X inherits a subspace topology from Y. The next theorem indicates that this subspace topology is the same as the original topology, and therefore we can view X as a subspace of its one-point compactification. Theorem 15 Let X be a Hausdorff space, and let Y = X { } be its one-point compactification. Then the subspace topology that X inherits from Y is equal to the original topology on X. Now we will justify the how one-point compactification is compact. Theorem 16 Let X be a Hausdorff space. Its one-point compactification Y = X { } is compact. Proof. Let C be an open cover of Y. Define C X to be the collection of subsets of X given by {V X : V C}. The sets in C X are open in the subspace topology that X inherits from Y ; therefore, by Theorem they are open sets in X. It follows that C X is an open cover of X. Now, C is an open cover of Y, and therefore there exists a U C such that U. It must be that U = Y C where C is a compact subset of X. The collection C X covers C. Since C is a compact subset of X it follows that there is a finite collection {V 1 X,...,V n X} C X that covers C. Therefore {O,V 1,...,V n } is a finite subcover of C, implying that Y is compact. Although we can construct the one-point compactification of any Hausdorff space, the result is not necessarily Hausdorff. For example, the one-point compactification of the set of rational numbers, Q, as a subspace of, is not a Hausdorff space. On the other hand, we have the following theorem:

20 20 Chapter 1. Compactness of Topological Space Theorem 17 Let X be a locally compact Hausdorff space. Then Y = X { }, the one-point compactificatial of X, is Hausdorff. Proof. To see that Y is Hausdorff, let x and y be points in Y. In the first case, assume both x and y are in X. Since X is Hausdorff, we can find disjoint open sets U and V in X that contain x and y, respectively. The sets U and V are also open sets in Y, and therefore there exist disjoint neighborhoods of x and y in Y. In the second case, suppose that x = and y X. By the local compactness of X, there is a compact set C in X that contains a neighborhood U of y. Now, Y C and U are open sets in Y, are disjoint, and contain x and y, respectively. Thus, in this case too, there exist disjoint neighborhoods of x and y in Y. It follows that Y is Hausdorff. 1.6 Key Notes 1. A collection C of subsets of a space X is said to cover X, or to be a covering of X, if the union of the elements of C is equal to X. The collection C is said to open covering if its all subsets are open subsets of X. 2. Let A be a subset of a topological space X, and let C be a collection of subsets of X, then the collection C is said to (a) Cover A or to be a cover of A if A is contained in the union of the sets in C. (b) Open cover of A, each set in C is open. 3. A subset A of space X is said to be compact in X if A is compact in the subspace topology inherited from X. 4. A topological space X is compact if every open cover of X has a finite subcover. 5. A topological space X is said to be limit point compact if every infinite subset of X has a limit point. 6. A topological space X is locally compact if every x X has a neighborhood that is contained in a compact subset of X.

21 1.7 Important Results Cover and Subcover 2. Compact Space 3. Compact Subspaces of the Real Line 4. Limit Point Compact Space 5. Locally Compact Space 6. Lindelöf Space 7. Sequentially Compact Space 8. Compactification 1.7 Important Results 1. Show with example that locally compactness of topological space need not be compactness. Proof. Let us consider a discrete topological space (X,D) with X as infinite set. Since the collection of all singleton sets is an infinite open cover of X which is not reducible to a finite subcover so X is not compact. In other hand, let x be an arbitrary element in X. Since every subset of X is open and hence {x} is an open neighbourhood of x. Evidently {x} is a compact subset of X, on account of the fact that {x} is a finite open subset of X. Thus x has a compact neighborhood {x}. Consequently X is locally compact. 2. Show that infinite subset of a discrete topological space is not compact or no finite discrete space is compact. Proof. Let us consider a discrete topological space (X,D) and A is infinite subset of X. Now we want to show that A is not compact. By the definition of discrete topology, every finite subset of X is open and hence {x} is open foe every x in X. Thus we can write as A = {{x}: x A}. Clearly A is an open cover of A. Evidently A is an infinite set. Hence any finite subfamily of A is not cover of A, showing thereby A is not compact. 3. Show that (R,D) is not compact, where D is discrete topology on R. Proof. As we showed that discrete topological space (X, D) is not compact, then it easily show that (R,D) is also not compact just by replacing X by R. 4. Every indiscrete space is compact. Proof. Let (X,τ) is indiscrete topological space, then τ = {X,φ}. Consequently the only open cover of X is {X} which is finite cover. Since it consists of only one member X. Hence X is compact. 5. Show that (R,τ) is compact, where I is indiscrete topology. Proof. As we already proved that indiscrete topological space (X,τ) is compact so similarly we can show that (R,τ) also compact just by replacing X by R. 6. Show with example compact space which is not Hausdroff. Proof. The topological space X = {a, b, c} with topology τ = {X, φ,{a},{a, b}} is compact but not Hausdroff space. 7. Show that (R,U) is not compact, where U is usual topology on R. Alternatively, any open interval is not compact with respect to its relativised-topology.

22 22 Chapter 1. Compactness of Topological Space Proof. Let G = {G n = ( n,n): n N} is a family of open subset G n of R. Now if any x in R then there exists r in N such that r > x. This show that x ( r,r) = G r G. This proves that G is open cover of R. Let any finite subfamily of G is G 1 = {G ni : 1 i n}. Suppose n 0 = max{n 1,n 2,...,n k }, then n 0 / {G ni : 1 i n}. Consequently, G 1 is not open cover of R. Hence an open cover of G of R is not reducible to finite subcover. Therefore (R,U) is not compact. Since compactness is a topological property and R is homeomorphic to any open interval so this show that an open interval is not compact for R is not compact. 8. A closed subset of a compact space is compact. Proof. Done 9. Let (A,U) be subspace of (X,τ), then A is compact with respect to U if and only if A is compact with respect to the topology τ on X. Proof. Let (A,U) be a subspace of (X,τ) and A be a compact with respect to the topology τ, then we want to show that A is compact with respect to U. Let {H i } be U-open cover of A, then A i H i. By the definition of subspace if H i U, then there exist G i τ such that H i = G i A G i. This show that H i G i. Therefore, i H i i G i. But A i H i so that A i G i. This proves that {G i } is a τ-open cover of A which is known to be compact with respect to topology τ and hence the cover {G i } must be reducible a finite subcover, say {G ir : r = 1,2,...,n}. Then A n r=1 G ir. From which ( ) n A A A G ir = (A G ir ) = r=1 r=1 H ir r=1 This show that {H ir : r = 1,2,...,n} is a U-open cover of A. Starting from an arbitrary U-open cover {H i } of A, we are able to show that the cover is reducible to a finite subcover {H ir : r = 1,2,...,n} of A, so A is U-compact. Conversely, let (A,U) is a subspace of (X,τ) and A is U-compact, then we want to show that A is τ-compact. Let {G i } is τ-open cover of A, so that A i G i. From which A A A ( G i ) of A (A G i ). On taking H i = G i A, we get A H i. Since G i τ show that H i = G i A U. Now it is clear that {H i } is U-open cover of A which is know to be U-compact and hence this cover must be reducible to a finite sub-cover, say {H ir : r = 1,2,...,n}. This show that A H ir = or r=1 A or r=1 (G ir A) r=1 (G ir A) A G ir r=1 G ir r=1 This proves that {G ir : r = 1,2,...,n} is a finite subcover of the cover {G i }. Starting from an arbitrary τ-open cover of A,we are able to show that this open cover is reducible to a finite sub-cover, showing thereby A is τ-compact. 10. A topological space is compact if every basic open cover has a finite subcover. Proof. Let B be an open base for space (X,τ) and let G = {G α } be basic open cover for X and also let every basic open cover for X have a finite subcover, then we want to show that Xis

23 1.7 Important Results 23 compact. Now by the definition of base,each G α is expressible as a union of some members of B and the collection of all such members of B is evidentely a basic open cover for X. By assumption, this collection of members of B has a finite subcover say, {B αi : i1,2,...,n}. So that X = B αi (1.5) For each B αi in this finite subcover, we can select G αi from G such that i=1 B αi G λi (1.6) This follows from the definition of base. From the results (1.5) and (1.6) X G λi i=1 But Xis the universe set so that n i=1 G λi X. Now combining las two results,we get X = n i=1 G λi. This show that {G λi : i = 1,2,...,n} = G 1 is a cover of X. Thus the open cover G is reducible to a finite subcover G 1, showing thereby X is compact. 11. The intersection of the members of an arbitrary family of closed and compact subsets is also closed and compact. Proof. Let C = {C α : α = 1,2,...} be an arbitrary family closed and compact subsets of space X. Let C = {C α : α = 1,2,...} An arbitrary intersection of closed sets is closed and hence C is closed and also {C α : α = 1,2,...} C α for all α. Therefore C C α for all α. Thus C is a closed subset of a compact set C α and hence C is compact. Thus C is compact and closed. 12. If (X,τ) is compact, then space (X,τ 1 ) is compact if τ 1 is coarser than τ. Proof. If (X,τ) ic compact space and τ 1 is coarser than τ so that τ 1 τ. Now we want to show that (X,τ 1 ) is compact. Let {G α : α } be τ 1 -open cover of X, then {G α : α } is τ-open cover of X. For τ 1 τ. Also X is τ-compact. Hence {G α : α } is reducible to finite subcover which is also τ 1 -open. So (X,τ 1 ) is compact. 13. Every cofinite topological space is compact. Proof. Let X be cofinite topological space and G be an open cover for X and G G be arbitrary, then G contains all points of X except finitely many, a 1,a 2,...,a n. To cover these members of X, we need at most n members G 1,G 2,...,G n of G. Thus G,G 1,G 2,...,G n is a finite sub-cover of G. Ny the definition of cofinite space this proves that (X,τ) is compact. 14. Show with example compact subset of topological space need not be closed. Proof. Let (X,I) is an indiscrete space such that X contains more than one element. Let A be a proper subset of X and let (A,I 1 ) be a subspace of (X,I). Here we have I 1 = {φ,a} for I = {φ,x}. Hence the only I 1 -open cover of A is {A} which is finite. Hence A is compact. But A is not I-closed. For the only I-closed sets are φ,x. Thus A is compact but not closed. 15. Countable compact topological space has a BWP.

24 24 Chapter 1. Compactness of Topological Space 16. A closed subset of a countably compact space is countably compact. Proof. Let Y be a closed subset of a countable compact space (X,τ) and {G n : n N} be a countable τ-open cover of Y, then Y n G n. But X = Y Y, therefore X = Y {G n : n N}. This show that the family consisting of open sets Y G 1,G 2,... forms an open countable cover of X which is know to be countably compact. Hence this cover must be reducible to a finite subcover, say Y G 1,G 2,...,G n so that ( ) n X = Y G i This show that Y n i=1 G i i.e. {G i : i = 1,2, ldots,n} is finite subcover of the countable cover{g n : n N}. Hence Y is countably compact. 17. A continuous image of a sequentially compact set is sequentially compact space. Proof. Pending 18. Finite Intersection Property: A topological space (X,τ) is compact if and only if each family of closed set with finite intersection property has a non-empty intersection. Proof. We are going to prove this result in contrapositive method. So we will show that a topological space is not compact if and only if for every family of closed sets with finite intersection property has an empty intersection. Let (X, τ) be a topological space which is not compact. To prove that there exist a family F of closed subbsets of τ with finite inntersection property such that {F : F F } = φ. Let {G α : α } be an open cover of X. Then X = {G α : α }. Taking complements of both sides, we get in virture of De-Morgan s law, φ = {(X G α ): α } (1.7) Let F α = X G α which is closed since G α is open. Now any finite subfamily of {G α : α } has the form G α : α 1, 1 is finite Since X is not compact and hence any open cover of X, in particular {G α : α } must both be reducible to a finite subcover. Thus we can say that {G α : α 1 } X and {G α : α 1 } = X. From this it follows that i=1 X {G α : α 1 } = φ {X G α : α 1 } = φ (1.8) Let F be the family generated by the closed sets F α s. Then F = {F α : α }. Let F 1 = {F α : α 1 }. Now the results (1.7) and (1.8) respectively take the form {F : F F } = φ (1.9) {F : F F 1 } = φ (1.10) The results (1.9) and (1.10) taken together imply that F is a family of closed sets with finite intersection property such that {F : F F } = φ. Conversely, let (X,τ) be a topological space and there exist a family F of closed subsets of X with the finite intersection property such that {F : F F } = φ (1.11)

25 1.7 Important Results 25 Now we want to show that X is not compact. The result (1.11) show that {X F : F F } = X (1.12) This show that the family {X F : F F } is open cover of X. Any finite subfamily of this cover has the form X F : F F 1 F,F 1 is finite If this were to be open cover of X, then we shall have {X F : F F 1 } = X (1.13) Taking complements of both sides, we get, in virtue of De Morgan s Law {F : F F 1 } = φ (1.14) The result (1.14) show that the family F does not have the finite intersection property and this a contradiction. Therefore (1.13) does not hold. Showing thereby {X F : F F 1 } is not an open cover of X. From what has been done it follows that the open cover {X F : F F } of X is not reducible to a finite subcover. Consequently X is not compact. 19. A topological space is compact if and only if every class of closed sets, which has empty intersection has a finite subclass with empty intersection. Proof. Let (X,τ) be a topological space and there exist a family such that {F i X : F i is closed for alli } F i = φ (1.15) i We want to show that there exist a finite subfamily {F ir : 1 r n} of the given family {F i : i } such that n F ir = φ (1.16) r=1 Taking complements of both sides of result (1.15) and using De Morgan s Law, we get {X F i : i } = X (1.17) Since F i are closed so X F i are open so the result (1.17) shows that the family {X F i : i } is an open cover X which is known to be compact. Hece this cover must be reducible to a finite subcover, say {X F ir : 1 r n}. Then {F ir : 1 r n} = X. Taking complements of both side, we get require result n F ir = φ r=1 Conversely, let (X,τ) is a topological space and given family {F i : i } of closed subsets of X such that {F i : i } = φ. A finite subfamily of this is {F ir : 1 r n} such that nr=1 F ir = φ. Now we are going to prove that X is compact space. As given {F i : i } = φ {X F i : i } = X {G i : i } = X, sayg i = X F i n And F ir = φ {X F ir : 1 r n} = X r=1 n G ir = X r=1

26 26 Chapter 1. Compactness of Topological Space Since F i is closed so X F i is open. Thus we have shown that given a family of open subsets {G i : i } of X such that i G i = X, there exist a finite subfamily {G ir : 1 r n} such that n r=1 G ir = X. This proves that {G i : i } is an open cover of X and is reducible to finite subcover {G ir : 1 r n}, showing thereby X is compact. 20. If A is a compact set in a regular space (X,τ) and G X is a neighbourhood of A, then there exist a neighbourhood V X of A such that V G. Proof. Let A be compact space in regular space (X,τ) and G is neighborhood of A, then we want to show that there exist neighborhood V of A such that V G. Let a be arbitrary element of A and G is neighborhood of A, then it will neighborhood of a also. Now by regularity of X, there exist another V a of a A such that a V a V ) a G for all a A. Form which we can say that A {V a : V a is a neighborhood ofa A} V ai G (1.18) Evidently {V a : V a is a neighborhood ofa A} is a cover of A which is known to be compact. Hence this is reducube to a finite subcover, say {V ai : a i A,1 i n}. Then result (1.18) is reduced to A {V ai : a i A} V ai G i=1 Let n i=1 V ai = V, we get A V V G or V G. Also V is neighborhood of A. 21. If A is a compact subset if a regular space (X,τ) and G is an open subset containing A, then there exist a closed set H containing A such that A H G. Proof. Prove this as proved above result that there exist a neighborhood V of A such that A V V G. Take V = H, then H is closed and A H G. 22. Every compact regular space is normal. Proof. Let (X,τ) be a compact regular space, then we are going to show that if is normal also. Let A and B are disjoint closed subsets of X and X is compact so both will be compact in X also. Since A and B both are closed so X A and X B are open sets. SinceA B = φ A (X B) i=1 a i A (X B) is an open set containing A (X B)is an neighborhood of the compact seta there exist a neighborhood G of A such that A G G (X B) Let U = G,V = X G, then U and V are open sets. Now U V = G (X G) = (G X) (G G) = G G = φ As we shown that A G G (X B) implies that B X G = V i.e. B V. As G is neighborhood of A show that G is an open set containing A, then it implies that A G = U i.e. A U. Finally, given pair of disjoint closed subsets A and B of X, there exist disjoint open subsets U and V such that A U and B V. So, by the definition X is normal space.

27 1.7 Important Results A compact Hausdorff soace is normal. Proof. Let (X,τ) be compacct Hausdroff space, then we are going to show that it will normal also. Let C 1 and C 2 are two disjoint closed subsets of compact space X, then by weakly hereditary property both will compact in X also. Let a C 1 be arbitrary and since C 1 and C 2 are disjoint so a / C 2. By the Hausdorff space property there exist disjoint open sets G a and H a such that a G a and C 2 H a. Consider the family {G a : a C 1 } which is open cover of C 1 and since C 1 is compact so this cover must be reducible to a finite subcover, say {G ai : i = 1,2,,...,n}, then C 1 n i=1 G ai = G (say), then C 1 G. Furthermore C 2 H ai for 1 i n, therefore C 2 n i=1 H ai = H, then C 2 H. Since a finite union of open sets and hence G is open and H is also open because a finite intersection of open sets are open. Since H ai G ai = φ for all i = 1,2,,...,n. So ( ) n G ar H = G ar H ai i=1 = (G ar H a1 ) (G ar H a2 )... (G ar H ar )... (G ar H an ) = φ φ... φ... φ G ar H = φ for all 1 r n ( ) n G H = H G ai G H = φ i=1 = (H G a1 ) (H G a2 )... (H G ar )... (H G an ) = φ φ... φ... φ Given a pair of disjoint closed sets C 1 and C 2 in X, we are able to get a pair of disjoint open sets G and H in X such that C 1 G and C 2 H. Therefore by the definition of normality, X is normal space. 24. A compact Hausdorff space is regular. Proof. Let (X,τ) be a compact Hausdorff space and we want to show that it is regular. As we know that compact Hausdorff space is normal so we will only show that normal space is regular. If X is Hausdorff then it will T 1 also which show that {x} is closed for all x X. Let x X and F a closed subset of X such that x / F. Since X is normal, therefore given a pair of disjoint closed set {x} and F, there exist disjoint open set G and H in X such that {x} G,F H i.e. given a closed set F X and x X such that x / F, then there exist open sets G and H in X such that x G and G H. So by the definition X is regular. 25. Every compact Hausdorff space is T 3 -space. Proof. As we know that if (X,τ) is compact Hausdorff space then it will regular and also T 1 -space. Any regular space and T 1 -space is always T 3 -space. 26. A compact subset of a Hausdorff space is closed. Proof. Let (X,τ) be Hausdorff space and A be compact subset of X, then we will prove that A is closed or X A is open. Let a A and p (X A) be two arbitrary elements where p is fixed element. Since a / (X A) so a p. By the Housdorffness, given a, p X such that a p, there exist disjoint open sets G a and H a such that a G a, p H a. This is true for each a A.

28 28 Chapter 1. Compactness of Topological Space Consider the family {G a : a A} which is open cover of A and A is compact so this cover must be reducible to a finite subcover, say {G ai : i = 1,2,...,n} so that A n i=1 G ai = G (say). So we get A G. Moreover p H ai for 1 i n, therefore p n i=1 H ai = H (say). Being a finite intersection of open sets, H is open. As G, is finite union of open sets so it will open also. Then H neighborhood of p. H ai, where i = 1,2,...,n is disjoint from each G ai and hence n i=1 H ai = H is disjoint from each G ai and therefore n i=1 G ai = G. Therefore G H = φ. Thus A F,G H = φ, p H. Since G H = φ G X H A G X H A X H H X A Given any p X A show that there exist open set H with p H such that H X A. Hence X A is an open subset of X. 27. If X is Hausdorff space and A is compact subset of X not containing x, then there exist disjoint open sets U and V of X such that x U and A V. Alternatively, In a Hausdorff, any point and disjoint compact subspace can be separated by open sets in the sense that have disjoint neighbourhoods. Proof. Not done 28. Every compact subset of a Hausdorff space is closed but compact space is not Hausdorff space. Proof. Not done 29. If A and B be disjoint compact subsets of a Hausdorff space (X,τ), then there exist disjoint open sets G and H such that A G and B H. Proof. Let a be arbitrary but fixed element of A and x be an arbitrary in B. Since A and B are disjoint so a x. By Hausdorffness there exist disjoint open sets G x and H x such that a G x and x H x. The collection {H x : x B} is a cover of B which is know to be compact. Hence there exists finitely many points x 1,x 2,...,x n B such that {H xi : i = 1,2,...,n} is a finite cover of B so that B n i=1 H xi = H a (say). Let G a = {G xi : i = 1,2,...,n}, then G a and H a are disjoint open sets such that a G a and B H a as H xi G xi = φ. Now let a A be arbitrary, then {G a : a A} is an open cover of A which is know to be compact. Hence there exist finitely many points a 1,a 2,...,a m A such that A {G ai : i = 1,2,...,m} = G (say). Let H = {H ai : i = 1,2,...,m}. Then G and H are disjoint open sets such that A G and B H. 30. A topological space (X,τ) is Hausdorff space if and only if disjoint compact subsets of X can be separated by disjoint open sets. Proof. If (X,τ) is Hausdorff space and A and B are disjoint compact subsets of X, then by the above result we can prove that A and B can be separated by disjoint open sets i.e. there exist disjoint open sets G and H such that A G and B H. Other hand, if every pair A and B of disjoint of compact sets and if there exist disjoint open sets G and H such A G and B H, then by assumption we can easily prove that X is Hausdorff space. 31. A continuous image of a compact space is compact.

29 1.7 Important Results 29 Proof. Let (X,τ 1 ) be a compact topological space and f : (X,τ 1 ) (Y,τ 2 ) be a continuous map. Let f (X) = Z and (Z,τ 3 ) be a subspace of (Y,τ 2 ). Now we are going to show that Z is compact. Let {G α : α } be an open cover of Z, then Z = α G α. From which ( ) f 1 (Z) = f 1 G α = f 1 (G α ) Let f 1 (G α ) = O α, we get α f 1 (Z) = α α O α (1.19) Since G α is open in Z and f is continuous, then it show that f 1 (G α ) is open in X i.e. O α is open in X. Now f (O α ) = f ( f 1 (G α )) = G α. But f 1 (Z) = X, therefore X = α O α, where each O α is open in X. This proves that the family {O α : α } is an open cover of X which is already compact hence this cover must be reduce to a finite subcover, say {O αi : 1 i n}. Then X = n i=1 O αi. From which, we get ( ) n f (X) = f O αi = f (X) = Z = i=1 f (O αi ) i=1 G αi i=1 This proves that {G αi : 1 i n} is a finite open cover of Z. Thus the open cover {G α : α } of Z is reducible to a finite subcover {G αi : 1 i n}. So, by definition Z is compact. 32. If (X,τ 1 ) and (Y,T 2 ) are topological spaces and f : (X,τ 1 ) (Y,τ 2 ) be a continuous map and A is a compact subset of X, then f (A) is compact. Alternatively, compactness is a topological invariant. Proof. As given that f : (X,τ 1 ) (Y,τ 2 ) be a continuous map and A is a compact subset of X, then we are going to show that f (A) is compact. Let {G α : α } be an open cover of f (A), then f (A) G α (1.20) From which Let f 1 (G α ) = O α, then α A f 1 ( f (A)) f 1 ( or A α α f 1 (G α ) A α G α ) = α f 1 (G α ) O α (1.21) Therefore, since f is continuous and f 1 (G α ) = O α, then by (1.21) shows that {O α : α } is an open cover of A which is already compact hence this cover must be reduce to a finite subcover, say {O αi : 1 i n}. So clearly A n i=1 O αi. From which f (A) f (O αi ) (1.22) i=1

30 30 Chapter 1. Compactness of Topological Space Now f (O α ) = f ( f 1 (G α )) G α, so f (O α ) G α. Since By the result of (1.22), we get f (O αi ) i=1 f (A) G αi i=1 G αi i=1 This proves that {G αi : 1 i n} is an open cover of f (A). Thus an arbitrary open cover {G α : α } of f (A) is reducible to a finite subcover {G αi : 1 i n}, therefore by the definition f (A) is compact. 33. Compactness is a topological property. Proof. The topological property is one which remains invariant under a homeomorphism. First we show that if f : (X,τ 1 ) (Y,τ 2 ) is a continuous map and A is any subset of X, then f (A) is compact. Second, we show that f is homeomorphism and for this we show that f and f 1 are continuous maps. 34. A one-one continuous map of a compact space onto Hausdorff space is homeomorphism. Proof. Let (X,τ 1 ) and (Y,τ 2 ) compact topological space and Hausdorff space respectively. Let f : X Y be one-one onto continuous map, then we are going to show that f is homoeomorphism. According to definition of homeomorphism it is enough to show that f 1 is continuous. Now for this we show that f (F) is closed in Y for any closed set F subset of X. Now if F X be arbitrary closed set. Being a closed subset of a compact set X, F is compact by standard result. The f (F) is compact, being a continuous image of a compact set. Furthemore f (F) is a compact subset of a Hausdorff space Y and hence f (F) is closed in Y. Thus we have shown that any subset F of X is closed impels that f (F) Y is closed. Finally, this show that f 1 is continuous. 35. A continuous image of a BWP set is BWP. 36. A continuous image of a sequentially compact set is sequentially compact. 37. Bolzano Weirstrass Theorem: Every infinite bounded set of real numbers contains at least one accumulation point. 38. A closed and bounded subset(subspace) of R is compact. Alternatively, for a closed and bounded subset A of R, every open covering of A admits a finite sub-covering. Proof. Let I 1 = [a 1,b 1 ]be a closed and bounded subset of R and G = {(c i,d i ): i } be an open covering of I 1. We will show that there exist finite covering of the original cover G. We will show this with the help of contradiction method so suppose it is not true. Then there exist no finite subcover of cover G. Divide I 1 innto two equal closed intervals as [ a 1, a ] 1 + b 1 2 and [ ] a1 + b 1,b 1 2 Then by assumption, at least one of these two intervals will not be covered by ant finite subclass of the cover G. Let that interval is I 2 = [a 2,b 2 ]. Then [ [a 2,b 2 ] = a 1, a ] 1 + b 1 2 [ ] a1 + b 1 or [a 2,b 2 ] =,b 1 2

31 1.7 Important Results 31 Now divide I 2 into two equal closed intervals [ a 2, a ] 2 + b 2 and 2 [ ] a2 + b 2,b 2 2 Again, by assumption, at least one of these two intervals will not be covered by any finite sub-family of the cover G. Let that interval is I 3 = [a 3,b 3 ]. Repeating this precess an infinite numbers of times, we get a sequence of intervals I 1,I 2,I 3,... with the following properties. (a) For all n N, I n I n+1. (b) For all n N, I n is closed set. (c) The interval I n is covered by any finite sub-family of G. (d) The lim n I n = 0, where I n is length os interval I n. From above properties it is clear that sequence of interval I n satisfies all the conditions of nested closed interval property. So this show that i=1 I n φ so that there exist a number p 0 i=1 I n. Hence p 0 I n for all n N. Now the properties (d) show that there exist n 0 N such that n n 0 i.e. I n < ε. In particular I n0 < ε, p 0 I n0. Choose ε = min{ p 0 a 1, p 0 + b 1, p 0 a 2, p 0 + b 2,..., p 0 a n, p 0 + b n } Then I n0 < ε implies that I n0 (p 0 ε, p 0 + ε). This violates the property (c) of the intervals I n. It is a contradiction, hence the required result hold. 39. The closed interval [0,1] is compact. 40. Every closed interval [a,b] is compact with respect to therelative topology for [a,b]. 41. The space (R,U) is not compact and no open interval is compact in its relative usual topology U. Proof. Consider the family of open set G = {G n : n N}, where G n = ( n,n). Let x R, then we may choose a positive integer n x > x and so x ( n x,n x ) G. Thus each point of R is contained in some member of G and therefore G is open cover of R. Consider a finite subset G 1 of G defined by G 1 = {( n k,n k ): k = 1,2,...,m}. Let p = max{n 1,n 2,...,n m }, then p / m k=1 n k,n k ). It follows that no finite sub-family of G cover R. Hence R is not compact. Consequently a open interval (a,b) is not also compact. R The Cantor set is a set of points lying on a single line segment that has a number of remarkable and deep properties. It was discovered in 1874 by Henry John Stephen Smith and introduced by German mathematician George Cantor in The Cantor set is an specific subset of R which has lots of weird and surprising properties. Let E 0 = [0,1], then remove the open middle third from the interval E 0 to get the set E 1. In other words, [ E 1 = 0, 1 ] [ ] 2 3 3,1 Now to get E 2, we remove the open middle third from each of the two segments [ 0, 1 ] 3 and [ 2 3,1 ]. In other words, [ E 2 = 0, 1 ] [ 2 9 9, 1 ] [ 2 3 3, 7 ] [ ] 8 9 9,1 We keep going like this. Inductively, we construct a sequence of sets E 0 E 1 E 2...

32 32 Chapter 1. Compactness of Topological Space such that E n is the union of 2 n disjoint closed intervals, each of length 3 n. The Cantor set is the set E = E n n N and we regard it as a metric space by restricting the euclidean metric on R. 42. Cantor s set is compact. Proof. Let the Contor s set is denoted by Γ, then Γ = i=1 E n. where E 0 = [0,1] [ E 1 = 0, 1 ] [ ] 2 3 3,1 [ E 2 = 0, 1 ] [ 2 9 9, 1 ] [ 2 3 3, 7 ] [ ] 8 9 9,1 Here E n is the union of 2 n disjoint closed intervals each of length 1/3 n. Since finite union of closed sets, E n is closed so consequently Γ is closed. Also Γ is bounded as Γ [0,1]. Therefore Γ is compact. 43. Heine-Borel Theorem for R: A subset of Euclidean n-space is compact if and only if it is closed and bounded. In other word, a subsapce (Y,U 1 ) of (R,U) is compact if and only if Y is bounded and U-closed. Proof. (a) If (Y,U 1 ) is subspace of (R,U) such that Y is bounded and U-closed, then we will show that Y is compact. As given Y is bounded then there exist closed interval [a,b] of R such that Y [a,b]. Since Y is U-closed and Y = Y [a,b] follows that Y is U-closed in [a,b] which is compact. Consequently Y is compact because closed subset of compact set is compact. (b) Let (Y,U 1 ) is compact subset of (R,U), then we will show that Y bounded and U-closed. As we know that (R,U) is T 2 -space and Y is a compact subset of T 2 -space so it follows that Y is U-closed. Now we will show that Y is bounded. Let for each y Y, we write G y =Y (y 1,y+1), then G = {G y : y Y } is U 1 -open cover of Y, which is itself compact. Hence this cover is reducible to a finite subcover {G yi : i = 1,2,...,n}. Therefore Y = n i=1 G yi. Let M 0 = max{y 1,y 2,...,y n } and m 0 = min{y 1,y 2,...,y n } Then Y = G yi [m 0 1,M 0 + 1] i=1 This show that Y is bounded. 44. A continuous real valued function, whose domain is closed and bounded interval of R, attains its supremum and infimum. Alternatively, a continuous real map of a compact space into any metric space is bounded. In other words, a real continuous function defined on a compact space attains its supremum and infimum. Proof. Let f : [a,b] R be a continuous real valued function where by Heine-Borel theorem [a,b] is compact. Now we will show that f attains its supremum and infimum. Let f ([a,b]) =

33 1.7 Important Results 33 A and since continuous image of compact space is compact so A is compact. Then f : [a,b] A R is continuous onto map. Now by standard result A is bounded and closed being a compact subset of R. First we shall show that f attains its supremum. Every real valued function has its supremum which may be either a real number or, therefore supremum of f exists. Let s be the supremum of f, then given ε > 0, there exist p [a,b] such that f (p) > s ε. Since f (x) A for all x [a,b], particular f (p) A. Now as f (p) A and f (p) > s ε show that there exist an element of A which lies in (s ε,s + ε). Therefore s is a contact pint of A. But A is closed and hence s A. Now s A and map f : [a,b] A is onto then this implies that there exist x 0 [a,b] such that f (x 0 ) = s. This prove that f attains its supremum. Similarly we can show that f attains its infimum also. 45. Let (X 1,τ 1 ) and (Y,τ 2 ) are topological spaces and f : (X 1,τ 1 ) (Y,τ 2 ) be a open continuous onto map, then if X is locally compact, then Y is also. Alternatively, every open continuous image of a locally compact space is locally compact. Proof. If f : (X 1,τ 1 ) (Y,τ 2 ) be a open continuous onto map where X is locally compact space then we will show that Y is also locally compact space. Let y be arbitrary element of Y and U Y is neighborhood of y, then there exist x in X such that f (x) = y. Since f is continuous, therefore given an neighborhood Uof y, there exist a neighborhood V X of x such that f (V ) U. Now As X is locally compact Xis locally compact at x and V is a neighborhood of x a compact set A such thatx A A V f (x) f (A ) f (A) f (V ) U f is open anda X is open f (A ) Y is open y f (A ) f (A) U (1.23) f (A ) = ( f (A )) (1.24) Now from result (1.23) f (A ) f (A) so ( f (A )) ( f (A)). Form the result (1.24) we get f (A ) ( f (A)). Therefore f (A ) f (A)) f (A). Using the result (1.23) y f (A ) f (A)) f (A) U i.e. y f (A)) f (A) U. Let B = f (A) is continuous image of a compact set A so B will be also. Therefore we obtain y B B U and B is compact. Finally we have shown that given any y Y and a neighborhood Uof y, there exist a compact set B Y such that y B B U. Hence by standard result Y is locally compact at y so that Y is locally compact. 46. Every locally compact T 2 -space is a regular space. Proof. Let (X,τ) is locally compact T 2 space, then we want to show that it is regular. By the definition of locally compact space there exist compact subset A of X such that x A A G. If A is compact and X is T 2 space, then it show that A is closed i.e. (A ) A = A. Since x A A G, therefore x A (A ) A G. Let A, then x U U G. Thus we have shown that given any neighborhood G of x, there exist a neighborhood U of x such that x A (A ) A G. Consequently X is regular by standard result. 47. Any open subspace of a locally compact is a locally compact. Proof. Let (Y,τ 1 ) be an open subspace of a locally compact space (X,τ) so that Y is open in X. We shall show that Y is locally compact. Let x Y X be arbitrary and G is T 2 - neighborhood of xin Y, then x X and G Y. As X is locally compact show that X is locally

34 34 Chapter 1. Compactness of Topological Space compact at x. If G is a τ 1 of x in Y, then there exist G 1 T 1 such that x G 1 G. Since Y is open in X, then G 1 exist in τ such that x G 1 G i.e. G is a τ-neighborhood of x in X. Also X is locally compact, then there exist a compact set A X such that x A A G. Thus A Y, A is τ 1 for A is τ-compact. As G is a neighborhood x in Y such that x A A G, This shown that Y is locally compact at any any x in X and hence y Y. Hence the result follow. 48. Every closed subspace of a locally compact is locally compact. Proof. Let (Y,τ 1 ) be a closed subspace of a locally compact space (X,τ), then Y is τ-closed. Let y be an arbitrary element in Y X, then we are going to show that Y is locally compact i.e. it is locally compact at y. Since X is locally compact then it is locally compact at y. This show that there exist τ-open neighborhood U of x such that U is τ-compact. So U Y is τ 1 -open neighborhood of y. Since N Y N, then by closure property N Y Y. Thus N Y is a closed subset of a compact set N. Hence N Y is compact by standard result. Now Y is τ-closed implies that τ-closure of N Y = τ 1 -closure of N Y. Thus N Y is τ 1 -open neighborhood of y such that N Y is compact, showing thereby Y is locally compact at y. 49. Locally compact space need not be compact. Proof. Let (X,D) be a discrete space where X is infinite. Evidently {{x}: x X} is an infinite open cover of X and which has no finite subcover. Consequently X is not compact. Let x be an arbitrary element of X, then {x} is an open set such that {x} = {x} is compact, being finite set. Hence every point x has open neighborhood whose closure is compact. Hence X is locally compact. 50. A compact metric space is separable. Proof. A topological space is called separable if it contains a countable, dense subset i.e., there exists a sequence {x n } n=1 of elements of the space such that every nonempty open subset of the space contains at least one element of the sequence. Let (X,d) be a compact metric space, then we are going to show that it is separable. Let n is fixed positive integer. As we know that in metric space each open sphere form an open set. Consider family of open sphere S = {S (x,1/n) : x X}. Clearly is an open cover of X which is known ro be compact. Hence this cover must be reducible to a finite subcover, let S = {S (xr,1/n) : r = 1,2,...,k n }. Let A = {x nr : r = 1,2,...,k n } be a set constructed for each n N such that (a) The set A n is finite. (b) For given x in X there exist x nr A n such that d(x,x nr ) < 1 n. Let A = n N A n. Here A is countable union of countable sets so A us enumerable. Since A X and as we know that A X = X. Now we are going to show that A = X and for this we will show that X A. Let x X be arbitrary and let G be open subset of X such that x G. By the property (ii) of A n given x in X, there exist A nr A such that d(x nr,x) < ε, where 1 n < ε. By the definition of open set in metric space if x G and G is open, then there exist a positive real number r such that S x,r G. In particular S (x,ε) G.

35 1.7 Important Results 35 d(x,x nr ) < ε x nr S (x,ε) G Gcontain some points of A other thanx (G {x}) A φ x D(A) A x A Thus we have shown that any x X implies x X which prove that X A. Finally we have shown that there exist a subset A of X such that A is enumerable and A = X. Hence X is separable. 51. Every totally bounded metric space is separable. 52. A finite subset of a topological space is necessarily sequentially compact. Proof. Let A is finite subset of topological space (X,τ) and x n be a finite sequence in A so that x n A for all n N. It follows that at least one element of A, say x 0 must appear infinite number of times in x n. Thus x 0,x 1,x 2,... is a subsequence of x n and this subsequence converges to x 0 A, showing thereby A is sequentially compact. 53. A metric space is sequentially compact if and only if it has the BWP. 54. Every compact metric space has the BWP. 55. A countably compact topological space has BWP. 56. Lebesgue Covering Lemma: Every open cover of sequentially compact metric space has a Lebesgue number. 57. Every sequentially compact metric space is totally bounded. 58. Every compact subset od metric space is closed and bounded. 59. Every sequentially compact metric space is compact. 60. A metric space X is compact if and only if it is sequentially compact. 61. Let A be a compact subset of a maestri space (X,d), then for every B X, there exist p in A such that d(p, B) = d(a, B). 62. A compact set in a metric space is totally bounded. 63. A continuous map from a compact space to another metric space is uniformly continuous. 64. If A be a compact subset of a metric space (X,d) and B X is closed, then d(a,b) > 0 if A B = φ. 65. Lindelöff Theorem: If (X,τ) is second countable space and G be an open cover of a set A X, then G is reducible to countable cover. Alternatively, a second countable space is Lindelöff space. Proof. Let (X,τ) be second countable space and G be an open cover od a subset A of X so that A {G: G G } (1.25) Now we want to show that G is reducible to a countable cover. Since X is second countable implies that there exist a countable base B for the topology τ on X. Since B is countable and hence it members may be enumerated as B 1,B 2,... Let x be an arbitrary element of A, then by the result 1.25 show that x G x for at least one G x G. Since G x is subset of X is open so by the definition of base if x G x τ show that there exist B x B such that x B x G x. We can write A {B x : x A} because {B x : x A} B and hence it can be expressed as A {B n : n N}. Choose G n G such that B n G n for all n N. This show that {B n : n N} {G n : n N}. Hence A {G n : n N}. This show that the open cover G of A is reducible to a countable cover {G n : n N}.

36 36 Chapter 1. Compactness of Topological Space 66. A closed subspace of a Lindelöf space is a Lindelöf space. Proof. Let (Y,τ 1 ) be a closed subspace of a Lindelöf space (X,τ). We are going to show that Y is also. Let {G i } be an open τ 1 -open cover of Y, so that Y i G i. Since G i τ 1 show that there exists H i in τ such that G i Y H i. So Y i G i = i (Y H i ) i H i (1.26) i.e. Y i H i. Evidently (X Y ) i ( H i ) = X. Since Y is closed so X Y is open. Now from the result (1.26) implies that the family consisting of the sets (X Y ) and H i s is τ-open cover of X which is know to be Linelöf space. Hence this cover is reducible to a countable subcover, say (X Y ) and {H ir : r = 1,2,...}. Then {H ir : r = 1,2,...} is countable τ-open cover of Y. Consequently, {A H ir : r = 1,2,...} = {G ir : r = 1,2,...} is countable τ 1 -open subcover of {G i }. Thus we have shown that an open cover of Y is reducible to a countable space. Consequently Y is Lindelöf space. 67. Every Lindelöf metric space is second countable space. Proof. Let (X,d) be a Lindelöf metric space and n N, then collection of open spheres {S 1/n (x): x X} is an open cover of X. Since X is Lindelöf metric space and therefore there exists a countable subcover {S 1/n (x ni ): i = 1,2,...}. We can find such subcover for each n N. Cosider the collection B = {S 1/n (x ni ): n,i N} (1.27) is countable. Now we will show that B is base. Let τ bemetric topology and G τ and x is arbitrary element of X. By the definition of open set, x G show that S ε (x) G, where ε > 0. Let n N such that 1 n < ε 2. Since S 1/n(x ni ) covers some points of X and hence there exist an index k such that x S 1/n (x nk ), then d(x,x nk ) < 1 n. Let y S 1/n(x nk ), then Therefore d(y,x) d(y,x nk ) + d(x nk,x) < 1 n + 1 n = 2 n < ε d(y,x) < ε ory S e (x) If any y S 1/n (x nk ) then it show that y S ε (x). Hence S 1/n (x nk ) S ε (x) and also S ε (x) G. Therefore S 1/n (x nk ) G. Thus if x G τ, then there exist S 1/n (x nk ) B such that S 1/n (x nk ) G. This proves that B is base and also it is countable. Hence B is countable base for τ on X. Consequently X is second countable. 68. A metric space is Lindelöf space if and only if it is second countable. 69. The property of being a Lindelöf space is a topological property. Proof. Let (X,τ) is Lindelöf space and a map f : (X,τ) (Y,τ 1 ) is homeomorphism, then we will show that (Y,τ 1 ) is also Lindelöf space. Since f is homeomorphism, so this are one-one onto, f and f 1 both are continuous. Let C = {G α : α } be an open cover for Y so that {G α : α } = Y. From which f 1 ( {G α : α }) f 1 (Y ) or { f 1 (G α ): α } = X (1.28)

37 1.7 Important Results 37 Since f is continuous and G α is open so f 1 (G α ) is open subset of X. Now by the result (1.28), { f 1 (G α ): α } is an open cover for X which is known to be Lindelöf space. Hence this cover is reducible to countable subcover, say { f 1 (G αi ): i N}. Hence { f 1 (G αi ): i N} = X. This gives f ( { f 1 (G αi ): i N}) = f (X) = Y. Since f ( f 1 (G αi )) = G αi so {G αi : i N} = Y. This show that {G αi : i N} is open cover of Y. Hence the given cover C is reducible to countable subcover. Hence Y is a Lindelöf space. 70. Lindelöfness is not a hereditary property. Proof. This can be prove with an example. Let X be an uncountable set and let x 0 be arbitrary but fixed element in X. Define a topology τ on X by requiring that (a) φ,x τ (b) If A is subset of X and x 0 / A, then A τ Now we claim that (X,τ) is Lindelöf space. Let G be an open cover of X. The only open set which contains x 0 is X. If follows that X G. Obviously {X} os countable subcover of G. Hence (X,τ) is Lindelöf space. Let Y = X {x 0 }. Let τ 1 be the topology on Y relative to the topology τ on X. Suppose y Y be arbitrary, then y x 0 and y X.Evidentely {y} is τ-open. For {y} X and x 0 {y}. If follows that {y} = Y {y} and so {y} is also τ 1 -open. Thus every singleton subset Y is τ 1 -open. Write C = {{y}: y Y }. Then C has no countable sub-cover. Thus Y is not Lindelöf space. Thus X is Lindelöf but its subspace is not Lindelöf, therefore Lindeoöfness is not heredity property. 71. A space is separable but not Lindfelof space. 72. A space is Lindelöf but not separable. 73. Every regular Lindelöf space is normal. 74. Every compact regular space is normal. 75. Every second countable regular space is normal. Main Results from Munker s Topology 1. Let Y be a subspace of X, then Y is compact if and only if every covering of Y by sets open in X contains a finite subcollections covering Y. 2. Every closed subspace of a compact space is compact. 3. Every compact subspace of a Hausdorff space is closed. 4. If Y is a compact subspace of the Hausdorff space X and x 0 is not in Y, then there exist disjoint open sets U and V of X containing x 0 and Y, respectively. 5. The image of a compact space under a continuous map is compact. 6. This result tell us for the verification of a map to be a homeomorphism or not. Let f : X Y be a bijective continuous function and if X is compact and Y is Hausdorff, then f is homeomorphism. 7. The product of finitely many compact spaces if compact.

38

39 Bibliography Books Articles

40

41 Index Compact space, 9 Compactness to subsets of space, 9 Cover for space, 8 Cover of subset of space, 8 Homeomorphism between compact and Housdroff space, 13 Identification map between compact and Housdroff spaces, 13 Limit Point Compactness, 15 Product of compact spaces, 14 Subcover of subset of space, 8 The Tube Lemma], 14