# CHAPTER 1 BASIC TOPOLOGY

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1 CHAPTER 1 BASIC TOPOLOGY Topology, sometimes referred to as the mathematics of continuity, or rubber sheet geometry, or the theory of abstract topological spaces, is all of these, but, above all, it is a language, used by mathematicians in practically all branches of our science. In this chapter, we will learn the basic words and expressions of this language as well as its grammar, i.e. the most general notions, methods and basic results of topology. We will also start building the library of examples, both nice and natural such as manifolds or the Cantor set, other more complicated and even pathological. Those examples often possess other structures in addition to topology and this provides the key link between topology and other branches of geometry. They will serve as illustrations and the testing ground for the notions and methods developed in later chapters Topological spaces The notion of topological space is defined by means of rather simple and abstract axioms. It is very useful as an umbrella concept which allows to use the geometric language and the geometric way of thinking in a broad variety of vastly different situations. Because of the simplicity and elasticity of this notion, very little can be said about topological spaces in full generality. And so, as we go along, we will impose additional restrictions on topological spaces, which will enable us to obtain meaningful but still quite general assertions, useful in many different situations in the most varied parts of mathematics Basic definitions and first examples. DEFINITION A topological space is a pair (X, T ) where X is a set and T is a family of subsets of X (called the topology of X) whose elements are called open sets such that (1), X T (the empty set and X itself are open), (2) if {O α } α A T then α A O α T for any set A (the union of any number of open sets is open), (3) if {O i } k i=1 T, then k i=1 O i T (the intersection of a finite number of open sets is open). 5

2 6 1. BASIC TOPOLOGY If x X, then an open set containing x is said to be an (open) neighborhood of x. We will usually omit T in the notation and will simply speak about a topological space X assuming that the topology has been described. The complements to the open sets O T are called closed sets. EXAMPLE Euclidean space R n acquires the structure of a topological space if its open sets are defined as in the calculus or elementary real analysis course (i.e a set A R n is open if for every point x A a certain ball centered in x is contained in A). EXAMPLE If all subsets of the integers Z are declared open, then Z is a topological space in the so called discrete topology. EXAMPLE If in the set of real numbers R we declare open (besides the empty set and R) all the half-lines {x R x a}, a R, then we do not obtain a topological space: the first and third axiom of topological spaces hold, but the second one does not (e.g. for the collection of all half lines with positive endpoints). EXAMPLE Example can be extended to provide the broad class of topological spaces which covers most of the natural situations. Namely, a distance function or a metric is a function of two variables on a set X (i,e, a function of the Cartesian product X X of X with itself) which is nonnegative, symmetric, strictly positive outside the diagonal, and satisfies the triangle inequality (see Definition 3.1.1). Then one defines an (open) ball or radius r > 0 around a point x X as the set of all points at a distance less that r from X, and an open subset of X as a set which together with any of its points contains some ball around that point. It follows easily from the properties of the distance function that this defines a topology which is usually called the metric topology. Naturally, different metrics may define the same topology. We postpone detailed discussion of these notions till Chapter 3 but will occasionally notice how natural metrics appear in various examples considered in the present chapter. The closure Ā of a set A X is the smallest closed set containing A, that is, Ā := {C A C and C closed}. A set A X is called dense (or everywhere dense) if Ā = X. A set A X is called nowhere dense if X \ Ā is everywhere dense. A point x is said to be an accumulation point (or sometimes limit point) of A X if every neighborhood of x contains infinitely many points of A. A point x A is called an interior point of A if A contains an open neighborhood of x. The set of interior points of A is called the interior of A and is denoted by Int A. Thus a set is open if and only if all of its points are interior points or, equivalently A = Int A.

3 1.1. TOPOLOGICAL SPACES 7 A point x is called a boundary point of A if it is neither an interior point of A nor an interior point of X \ A. The set of boundary points is called the boundary of A and is denoted by A. Obviously Ā = A A. Thus a set is closed if and only if it contains its boundary. EXERCISE Prove that for any set A in a topological space we have A A and (Int A) A. Give an example when all these three sets are different. A sequence {x i } i N X is said to converge to x X if for every open set O containing x there exists an N N such that {x i } i>n O. Any such point x is called a limit of the sequence. EXAMPLE In the case of Euclidean space R n with the standard topology, the above definitions (of neighborhood, closure, interior, convergence, accumulation point) coincide with the ones familiar from the calculus or elementary real analysis course. EXAMPLE For the real line R with the discrete topology (all sets are open), the above definitions have the following weird consequences: any set has neither accumulation nor boundary points, its closure (as well as its interior) is the set itself, the sequence {1/n} does not converge to 0. Let (X, T ) be a topological space. A set D X is called dense or everywhere dense in X if D = X. A set A X is called nowhere dense if X \ Ā is everywhere dense. The space X is said to be separable if it has a finite or countable dense subset. A point x X is called isolated if the one point set {x} is open. EXAMPLE The real line R in the discrete topology is not separable (its only dense subset is R itself) and each of its points is isolated (i.e. is not an accumulation point), but R is separable in the standard topology (the rationals Q R are dense) Base of a topology. In practice, it may be awkward to list all the open sets constituting a topology; fortunately, one can often define the topology by describing a much smaller collection, which in a sense generates the entire topology. DEFINITION A base for the topology T is a subcollection β T such that for any O T there is a B β for which we have x B O. Most topological spaces considered in analysis and geometry (but not in algebraic geometry) have a countable base. Such topological spaces are often called second countable. A base of neighborhoods of a point x is a collection B of open neighborhoods of x such that any neighborhood of x contains an element of B.

4 8 1. BASIC TOPOLOGY If any point of a topological space has a countable base of neighborhoods, then the space (or the topology) is called first countable. EXAMPLE Euclidean space R n with the standard topology (the usual open and closed sets) has bases consisting of all open balls, open balls of rational radius, open balls of rational center and radius. The latter is a countable base. EXAMPLE The real line (or any uncountable set) in the discrete topology (all sets are open) is an example of a first countable but not second countable topological space. PROPOSITION Every topological space with a countable space is separable. PROOF. Pick a point in each element of a countable base. The resulting set is at most countable. It is dense since otherwise the complement to its closure would contain an element of the base Comparison of topologies. A topology S is said to be stronger (or finer) than T if T S, and weaker (or coarser) if S T. There are two extreme topologies on any set: the weakest trivial topology with only the whole space and the empty set being open, and the strongest or finest discrete topology where all sets are open (and hence closed). EXAMPLE On the two point set D, the topology obtained by declaring open (besides D and ) the set consisting of one of the points (but not the other) is strictly finer than the trivial topology and strictly weaker than the discrete topology. PROPOSITION For any set X and any collection C of subsets of X there exists a unique weakest topology for which all sets from C are open. PROOF. Consider the collection T which consist of unions of finite intersections of sets from C and also includes the whole space and the empty set. By properties (2) and (3) of Definition in any topology in which sets from C are open the sets from T are also open. Collection T satisfies property (1) of Definition by definition, and it follows immediately from the properties of unions and intersections that T satisfies (2) and (3) of Definition Any topology weaker than a separable topology is also separable, since any dense set in a stronger topology is also dense in a weaker one. EXERCISE How many topologies are there on the 2 element set and on the 3 element set?

5 1.2. CONTINUOUS MAPS AND HOMEOMORPHISMS 9 EXERCISE On the integers Z, consider the profinite topology for which open sets are defined as unions (not necessarily finite) of arithmetic progressions (non-constant and infinite in both directions). Prove that this defines a topology which is neither discrete nor trivial. EXERCISE Define Zariski topology in the set of real numbers by declaring complements of finite sets to be open. Prove that this defines a topology which is coarser than the standard one. Give an example of a sequence such that all points are its limits. EXERCISE On the set R { }, define a topology by declaring open all sets of the form { } G, where G R is open in the standard topology of R. (a) Show that this is indeed a topology, coarser than the discrete topology on this set. (b) Give an example of a convergent sequence which has two limits Continuous maps and homeomorphisms In this section, we study, in the language of topology, the fundamental notion of continuity and define the main equivalence relation between topological spaces homeomorphism. We can say (in the category theory language) that now, since the objects (topological spaces) have been defined, we are ready to define the corresponding morphisms (continuous maps) and isomorphisms (topological equivalence or homeomorphism) Continuous maps. The topological definition of continuity is simpler and more natural than the ε, δ definition familiar from the elementary real analysis course. DEFINITION Let (X, T ) and (Y, S) be topological spaces. A map f : X Y is said to be continuous if O S implies f 1 (O) T (preimages of open sets are open): f is an open map if it is continuous and O T implies f(o) S (images of open sets are open); f is continuous at the point x if for any neigborhood A of f(x) in Y the preimage f 1 (A) contains a neighborhood of x. A function f from a topological space to R is said to be upper semicontinuous if f 1 (, c) T for all c R: lower semicontinuous if f 1 (c, ) T for c R. EXERCISE Prove that a map is continuous if and only if it is continuous at every point. Let Y be a topological space. For any collection F of maps from a set X (without a topology) to Y there exists a unique weakest topology on Categorical language: preface, appendix reference?

6 10 1. BASIC TOPOLOGY R ] 1, 1[ FIGURE The open interval is homeomorphic to the real line there is a problem with positioning this figure in the page X which makes all maps from F continuous; this is exactly the weakest topology with respect to which preimages of all open sets in Y under the maps from F are open. If F consists of a single map f, this topology is sometimes called the pullback topology on X under the map f. EXERCISE Let p be the orthogonal projection of the square K on one of its sides. Describe the pullback topology on K. Will an open (in the usual sense) disk inside K be an open set in this topology? Topological equivalence. Just as algebraists study groups up to isomorphism or matrices up to a linear conjugacy, topologists study (topological) spaces up to homeomorphism. DEFINITION A map f : X Y between topological spaces is a homeomorphism if it is continuous and bijective with continuous inverse. If there is a homeomorphism X Y, then X and Y are said to be homeomorphic or sometimes topologically equivalent. A property of a topological space that is the same for any two homeomorphic spaces is said to be a topological invariant. The relation of being homeomorphic is obviously an equivalence relation (in the technical sense: it is reflexive, symmetric, and transitive). Thus topological spaces split into equivalence classes, sometimes called homeomorphy classes. In this connection, the topologist is sometimes described as a person who cannot distinguish a coffee cup from a doughnut (since these two objects are homeomorphic). In other words, two homeomorphic topological spaces are identical or indistinguishable from the intrinsic point of view in the same sense as isomorphic groups are indistinguishable from the point of view of abstract group theory or two conjugate n n matrices are indistinguishable as linear transformations of an n-dimensional vector space without a fixed basis. EXAMPLE The figure shows how to construct homeomorphisms between the open interval and the open half-circle and between the open half-circle and the real line R, thus establishing that the open interval is

7 1.2. CONTINUOUS MAPS AND HOMEOMORPHISMS 11 homeomorphic to the real line. EXERCISE Prove that the sphere S 2 with one point removed is homeomorphic to the plane R 2. EXERCISE Prove that any open ball is homeomorphic to R 3. EXERCISE Describe a topology on the set R 2 { } which will make it homeomorphic to the sphere S 2. To show that certain spaces are homeomorphic one needs to exhibit a homeomorphism; the exercises above give basic but important examples of homeomorphic spaces; we will see many more examples already in the course of this chapter. On the other hand, in order to show that topological spaces are not homeomorphic one need to find an invariant which distinguishes them. Let us consider a very basic example which can be treated with tools from elementary real analysis. EXAMPLE In order to show that closed interval is not homeomorphic to an open interval (and hence by Example to the real line) notice the following. Both closed and open interval as topological spaces have the property that the only sets which are open and closed at the same time are the space itself and the empty set. This follows from characterization of open subsets on the line as finite or countable unions of disjoint open intervals and the corresponding characterization of open subsets of a closed interval as unions of open intervals and semi-open intervals containing endpoints. Now if one takes any point away from an open interval the resulting space with induced topology (see below) will have two proper subsets which are open and closed simultaneously while in the closed (or semi-open) interval removing an endpoint leaves the space which still has no non-trivial subsets which are closed and open. In Section 1.6 we will develop some of the ideas which appeared in this simple argument systematically. The same argument can be used to show that the real line R is not homeomorphic to Euclidean space R n for n 2 (see Exercise ). It is not sufficient however for proving that R 2 is not homeomorphic R 3. Nevertheless, we feel that we intuitively understand the basic structure of the space R n and that topological spaces which locally look like R n (they are called (n-dimensional) topological manifolds) are natural objects of study in topology. Various examples of topological manifolds will appear in the course of this chapter and in Section 1.8 we will introduce precise definitions and deduce some basic properties of topological manifolds.

8 12 1. BASIC TOPOLOGY 1.3. Basic constructions Induced topology. If Y X, then Y can be made into a topological space in a natural way by taking the induced topology T Y := {O Y O T }. FIGURE Induced topology EXAMPLE The topology induced from R n+1 on the subset {(x 1,..., x n, x n+1 ) : n+1 x 2 i = 1} produces the (standard, or unit) n sphere S n. For n = 1 it is called the (unit) circle and is sometimes also denoted by T. EXERCISE Prove that the boundary of the square is homeomorphic to the circle. EXERCISE Prove that the sphere S 2 with any two points removed is homeomorphic to the infinite cylinder C := {(x, y, z) R 3 x 2 +y 2 = 1}. EXERCISE Let S := {(x, y, z) R 3 z = 0, x 2 + y 2 = 1}. Show that R 3 \ S can be mapped continuously onto the circle Product topology. If (X α, T α ), α A are topological spaces and A is any set, then the product topology on α A X is the topology determined by the base { } O α T α, O α X α for only finitely many α. α O α EXAMPLE The standard topology in R n coincides with the product topology on the product of n copies of the real line R. i=1

9 1.3. BASIC CONSTRUCTIONS 13 Y FIGURE Basis element of the product topology EXAMPLE The product of n copies of the circle is called the n torus and is usually denoted by T n. The n torus can be naturally identified with the following subset of R 2n : with the induced topology. {(x 1,... x 2n ) : x 2 2i 1 + x 2 2i = 1, i = 1,..., n.} EXAMPLE The product of countably many copies of the two point space, each with the discrete topology, is one of the representations of the Cantor set (see Section 1.7 for a detailed discussion). EXAMPLE The product of countably many copies of the closed unit interval is called the Hilbert cube. It is the first interesting example of a Hausdorff space (Section 1.4) too big to lie inside (that is, to be homeomorphic to a subset of) any Euclidean space R n. Notice however, that not only we lack means of proving the fact right now but the elementary invariants described later in this chapter are not sufficient for this task either Quotient topology. Consider a topological space (X, T ) and suppose there is an equivalence relation defined on X. Let π be the natural projection of X on the set ˆX of equivalence classes. The identification space or quotient space X/ := ( ˆX, S) is the topological space obtained by calling a set O ˆX open if π 1 (O) is open, that is, taking on ˆX the finest topology for which π is continuous. For the moment we restrict ourselves to good examples, i.e. to the situations where quotient topology is natural in some sense. However the reader should be aware that even very natural equivalence relations often lead to factors with bad properties ranging from the trivial topology to nontrivial ones but lacking basic separation properties (see Section 1.4). We postpone description of such examples till Section X

10 14 1. BASIC TOPOLOGY EXAMPLE Consider the closed unit interval and the equivalence relation which identifies the endpoints. Other equivalence classes are single points in the interior. The corresponding quotient space is another representation of the circle. The product of n copies of this quotient space gives another definition of the n torus. EXERCISE Describe the representation of the n torus from the above example explicitly as the identification space of the unit n cube I n : {(x 1,..., x n ) R n : 0 x i 1, i = 1,... n. EXAMPLE Consider the following equivalence relation in punctured Euclidean space R n+1 \ {0}: (x 1,..., x n+1 ) (y 1,..., y n+1 ) iff y i = λx i for all i = 1,..., n + 1 with the same real number λ. The corresponding identification space is called the real projective n space and is denoted by RP (n). A similar procedure in which λ has to be positive gives another definition of the n sphere S n. EXAMPLE Consider the equivalence relation in C n+1 \ {0}: (x 1,..., x n+1 ) (y 1,..., y n+1 ) iff y i = λx i for all i = 1,..., n + 1 with the same complex number λ. The corresponding identification space is called the complex projective n space and is detoted by CP (n). EXAMPLE The map E : [0, 1] S 1, E(x) = exp 2πix establishes a homeomorphism between the interval with identified endpoints (Example 1.3.6) and the unit circle defined in Example EXAMPLE The identification of the equator of the 2-sphere to a point yields two spheres with one common point. FIGURE The sphere with equator identified to a point

11 1.3. BASIC CONSTRUCTIONS 15 EXAMPLE Identifying the short sides of a long rectangle in the natural way yields the lateral surface of the cylinder (which of course is homeomorphic to the annulus), while the identification of the same two sides in the wrong way (i.e., after a half twist of the strip) produces the famous Möbius strip. We assume the reader is familiar with the failed experiments of painting the two sides of the Möbius strip in different colors or cutting it into two pieces along its midline. Another less familiar but amusing endeavor is to predict what will happen to the physical object obtained by cutting a paper Möbius strip along its midline if that object is, in its turn, cut along its own midline. FIGURE The Möbius strip EXERCISE Describe a homeomorphism between the torus T n (Example 1.3.3) and the quotient space described in Example and the subsequent exercise. EXERCISE Describe a homeomorphism between the sphere S n (Example 1.3.1) and the second quotient space of Example EXERCISE Prove that the real projective space RP (n) is homeomorphic to the quotient space of the sphere S n with respect to the equivalence relation which identifies pairs of opposite points: x and x. EXERCISE Consider the equivalence relation on the closed unit ball D n in R n : n {(x 1,..., x n ) : x 2 i 1} which identifies all points of D n = S n 1 and does nothing to interior points. Prove that the quotient space is homeomorphic to S n. EXERCISE Show that CP (1) is homeomorphic to S 2. i=1

12 16 1. BASIC TOPOLOGY DEFINITION The cone Cone(X) over a topological space X is the quotient space obtained by identifying all points of the form (x, 1) in the product (X [0, 1] (supplied with the product topology). The suspension Σ(X) of a topological space X is the quotient space of the product X [ 1, 1] obtained by identifying all points of the form x 1 and identifying all points of the form x 1. By convention, the suspension of the empty set will be the two-point set S 0. The join X Y of two topological spaces X and Y, roughly speaking, is obtained by joining all pairs of points (x, y), x X, y Y, by line segments and supplying the result with the natural topology; more precisele, X Y is the quotient space of the product X [ 1, 1] Y under the following identifications: (x, 1, y) (x, 1, y ) for any x X and all y, y Y, (x, 1, y) (x, 1, y) for any y Y and all x, x X. EXAMPLE (a) Cone( ) = D 1 and Cone(D n 1 ) = D n for n > 1. (b) The suspension Σ(S n ) of the n-sphere is the (n + 1)-sphere S n+1. (c) The join of two closed intervals is the 3-simplex (see the figure). FIGURE The 3-simplex as the join of two segments EXERCISE Show that the cone over the sphere S n is (homeomorphic to) the disk D n+1. EXERCISE Show that the join of two spheres S k and S l is (homeomorphic to) the sphere S k+l+1. EXERCISE Is the join operation on topological spaces associative? 1.4. Separation properties Separation properties provide one of the approaches to measuring how fine is a given topology.

13 1.4. SEPARATION PROPERTIES 17 x y T 1 Hausdorff T 4 FIGURE Separation properties T1, Hausdorff, and normal spaces. Here we list, in decreasing order of generality, the most common separation axioms of topological spaces. DEFINITION A topological space (X, T ) is said to be a (T1) space if any point is a closed set. Equivalently, for any pair of points x 1, x 2 X there exists a neighborhood of x 1 not containing x 2 ; (T2) or Hausdorff space if any two distinct points possess nonintersecting neighborhoods; (T4) or normal space if it is Hausdorff and any two closed disjoint subsets possess nonintersecting neighborhoods. 1 It follows immediately from the definition of induced topology that any of the above separation properties is inherited by the induced topology on any subset. EXERCISE Prove that in a (T2) space any sequence has no more than one limit. Show that without the (T2) condition this is no longer true. EXERCISE Prove that the product of two (T1) (respectively Hausdorff) spaces is a (T1) (resp. Hausdorff) space. REMARK We will see later (Section 1.9) that even very naturally defined equivalence relations in nice spaces may produce quotient spaces with widely varying separation properties. The word normal may be understood in its everyday sense like commonplace as in a normal person. Indeed, normal topological possess many properties which one would expect form commonplaces notions of continuity. Here is an examples of such property dealing with extension of maps: THEOREM [Tietze] If X is a normal topological space, Y X is closed, and f : Y [ 1, 1] is continuous, then there is a continuous On the picture the interior does not looks closed but the exterior does 1 Hausdorff (or (T1)) assumption is needed to ensure that there are enough closed sets; specifically that points are closed sets. Otherwise trivial topology would satisfy this property.

14 18 1. BASIC TOPOLOGY extension of f to X, i.e., a continuous map F : X [ 1, 1] such that F Y = f. The proof is based on the following fundamental result, traditionally called Urysohn Lemma, which asserts existence of many continuous maps from a normal space to the real line and thus provided a basis for introducing measurements in normal topological spaces (see Theorem 3.5.1) and hence by Corollary also in compact Hausdorff spaces. THEOREM [Urysohn Lemma] If X is a normal topological space and A, B are closed subsets of X, then there exists a continuous map u : X [0, 1] such that u(a) = {0} and u(b) = {1}. PROOF. Let V be en open subset of X and U any subset of X such that U V. Then there exists an open set W for which U W W V. Indeed, for W we can take any open set containing U and not intersecting an open neighborhood of X \ V (such a W exists because X is normal). Applying this to the sets U := A and V := X \ B, we obtain an intermediate open set A 1 such that (1.4.1) A A 1 X \ B, where A 1 X \ B. Then we can introduce the next intermediate open sets A 1 and A 2 so as to have (1.4.2) A A 1 A 1 A 2 X \ B, where each set is contained, together with its closure, in the next one. For the sequence (1.4.1), we define a function u 1 : X [0, 1] by setting 0 for x A, u 1 (x) = 1/2 for x A 1 \ A, 1 for X \ A 1. For the sequence (1.4.2), we define a function u 2 : X [0, 1] by setting 0 for x A, 1/4 for x A 1 \ A, u 2 (x) = 1/2 for x A 1 \ A 1, 3/4 for x A 2 \ A 1, 1 for x X \ A 2.

15 1.4. SEPARATION PROPERTIES 19 Then we construct a third sequence by inserting intermediate open sets in the sequence (1.4.2) and define a similar function u 3 for this sequence, and so on. Obviously, u 2 (x) u 1 (x) for all x X. Similarly, for any n > 1 we have u n+1 (x) u n (x) for all x X, and therefore the limit function u(x) := lim n infty u n (x) exists. It only remains to prove that u is continuous. Suppose that at the nth step we have constructed the nested sequence of sets corresponding to the function u n A A 1... A r X \ B, where A i A i+1. Let A 0 := int A be the interior of A, let A 1 :=, and A r+1 := X. Consider the open sets A i+1 \ A i 1, i = 0, 1,..., r. Clearly, r r X = (Āi \ A i 1 ) (A i+1 \ A i 1 ), i=0 so that the open sets A i+1 \ A i 1 cover the entire space X. On each set A i+1 \ A i 1 the function takes two values that differ by 1/2 n. Obviously, i=0 maybe insert a picture u(x) u n (x) k=n+1 1/2 k = 1/2 n. For each point x X let us choose an open neighborhood of the form A i+1 \ A i 1. The image of the open set A i+1 \ A i 1 is contained in the interval (u(x) ε, u(x) + ε), where ε < 1/2 n. Taking ε, we see that u is continuous. Now let us deduce Theorem from the Urysohn lemma. To this end, we put r k := 1 ( 2 ) k, k = 1, 2, Let us construct a sequence of functions f 1, f 2,... on X and a sequence of functions g 1, g 2,... on Y by induction. First, we put f 1 := f. Suppose that the functions f 1,..., f k have been constructed. Consider the two closed disjoint sets A k := {x X f k (x) r k } and B k := {x X f k (x) r k }. Applying the Urysohn lemma to these sets, we obtain a continuous map g k : Y [ r k, r k ] for which g k (A k ) = { r k } and g k (B k ) = {r k }. On the set A k, the functions f k and g k take values in the interval ] 3r k, r k [; on

16 20 1. BASIC TOPOLOGY the set A k, they take values in the interval ]r k, 3r k [; at all other points of the set X, these functions take values in the interval ] r k, r k [. Now let us put f k+1 := f k g k X. The function f k+1 is obviously continuous on X and f k+1 (x) 2r k = 3r k+1 for all x X. Consider the sequence of functions g 1, g 2,... on Y. By construction, g k (y) r k for all y Y. The series r k = 1 2 k=1 ( 2 k 3) k=1 converges, and so the series Σ k=1 g k(x) converges uniformly on Y to some continuous function F (x) := g k (x). Further, we have k=1 (g 1 + +g k ) = (f 1 f 2 )+(f 2 f 3 )+ +(f k f k+1 ) = f 1 f k+1 = f f k+1. But lim k f k+1 (y) = 0 for any y Y, hence F (x) = f(x) for any x X, so that F is a continuous extension of f. It remains to show that F (x) 1. We have F (x) = g k (x) k=1 ( 2 3 k=1 ) k = 1 3 r k = k=1 ( 2 k 3) k=1 ( 1 2 3) 1 = 1. COROLLARY Let X Y be a closed subset of a normal space Y and let f : X R be continuous. Then f has a continuous extension F : Y R. PROOF. The statement follows from the Tietze theorem and the Urysohn lemma by appropriately using the rescaling homeomorphism g : R ( π/2, π/2) given by g(x) := arctan(x). Most natural topological spaces which appear in analysis and geometry (but not in some branches of algebra) are normal. The most important instance of non-normal topology is discussed in the next subsection.

17 1.4. SEPARATION PROPERTIES Zariski topology. The topology that we will now introduce and seems pathological in several aspects (it is non-hausdorff and does not possess a countable base), but very useful in applications, in particular in algebraic geometry. We begin with the simplest case which was already mentioned in Example DEFINITION The Zariski topology on the real line R is defined as the family Z of all complements to finite sets. PROPOSITION The Zariski topology given above endows R with the structure of a topological space (R, Z), which possesses the following properties: (1) it is a (T1) space; (2) it is separable; (3) it is not a Hausdorff space; (4) it does not have a countable base. PROOF. All four assertions are fairly straightforward: (1) the Zariski topology on the real line is (T1), because the complement to any point is open; (2) it is separable, since it is weaker than the standard topology in R; (3) it is not Hausdorff, because any two nonempty open sets have nonempty intersection; (4) it does not have a countable base, because the intersection of all the sets in any countable collection of open sets is nonemply and thus the complement to any point in that intersection does not contain any element from that collection. The definition of Zariski topology on R (Definition 1.4.6) can be straightforwardly generalized to R n for any n 2, and the assertions of the proposition above remain true. However, this definition is not the natural one, because it generalizes the wrong form of the notion of Zariski topology. The correct form of that notion originally appeared in algebraic geometry (which studies zero sets of polynomials) and simply says that closed sets in the Zariski topology on R are sets of zeros of polynomials p(x) R[x]. This motivates the following definitions. DEFINITION The Zariski topology is defined in Euclidean space R n by stipulating that the sets of zeros of all polynomials are closed; on the unit sphere S n R n+1 by taking for closed sets the sets of zeros of homogeneous polynomials in n + 1 variables; on the real and complex projective spaces RP (n) and CP (n) (Example 1.3.7, Example 1.3.8) via zero sets of homogeneous polynomials in n + 1 real and complex variables respectively.

18 22 1. BASIC TOPOLOGY EXERCISE Verify that the above definitions supply each of the sets R n, S n, RP (n), and CP (n) with the structure of a topological space satisfying the assertions of Proposition Compactness The fundamental notion of compactness, familiar from the elementary real analysis course for subsets of the real line R or of Euclidean space R n, is defined below in the most general topological situation Types of compactness. A family of open sets {O α } T, α A is called an open cover of a topological space X if X = α A O α, and is a finite open cover if A is finite. DEFINITION The space (X, T ) is called compact if every open cover of X has a finite subcover; sequentially compact if every sequence has a convergent subsequence; σ compact if it is the union of a countable family of compact sets. locally compact if every point has an open neighborhood whose closure is compact in the induced topology. It is known from elementary real analysis that for subsets of a R n compactness and sequential compactness are equivalent. This fact naturally generalizes to metric spaces (see Proposition ). PROPOSITION Any closed subset of a compact set is compact. PROOF. If K is compact, C K is closed, and Γ is an open cover for C, then Γ 0 := Γ {K C} is an open cover for K, hence Γ 0 contains a finite subcover Γ {K C} for K; therefore Γ is a finite subcover (of Γ) for C. PROPOSITION Any compact subset of a Hausdorff space is closed. PROOF. Let X be Hausdorff and let C X be compact. Fix a point x X C and for each y C take neighborhoods U y of y and V y of x such that U y V y =. Then y C U y C is a cover of C and has a finite subcover {U xi 0 i n}. Hence N x := n i=0 V y i is a neighborhood of x disjoint from C. Thus X C = is open and therefore C is closed. x X C PROPOSITION Any compact Hausdorff space is normal. N x

19 1.5. COMPACTNESS 23 PROOF. First we show that a closed set K and a point p / K can be separated by open sets. For x K there are open sets O x, U x such that x O x, p U x and O x U x =. Since K is compact, there is a finite subcover O := n i=1 O x i K, and U := n i=1 U x i is an open set containing p disjoint from O. Now suppose K, L are closed sets. For p L, consider open disjoint sets O p K, U p p. By the compactness of L, there is a finite subcover U := m j=1 U p j L, and so O := m j=1 O p j K is an open set disjoint from U L. DEFINITION A collection of sets is said to have the finite intersection property if every finite subcollection has nonempty intersection. PROPOSITION Any collection of compact sets with the finite intersection property has a nonempty intersection. PROOF. It suffices to show that in a compact space every collection of closed sets with the finite intersection property has nonempty intersection. Arguing by contradiction, suppose there is a collection of closed subsets in a compact space K with empty intersection. Then their complements form an open cover of K. Since it has a finite subcover, the finite intersection property does not hold. EXERCISE Show that if the compactness assumption in the previous proposition is omitted, then its assertion is no longer true. EXERCISE Prove that a subset of R or of R n is compact iff it is closed and bounded Compactifications of non-compact spaces. DEFINITION A compact topological space K is called a compactification of a Hausdorff space (X, T ) if K contains a dense subset homeomorphic to X. The simplest example of compactification is the following. DEFINITION The one-point compactification of a noncompact Hausdorff space (X, T ) is ˆX := (X { }, S), where S := T {(X { }) K K X compact}. EXERCISE Show that the one-point compactification of a Hausdorff space X is a compact (T1) space with X as a dense subset. Find a necessary and sufficient condition on X which makes the one-point compactification Hausdorff. EXERCISE Describe the one-point compactification of R n.

20 24 1. BASIC TOPOLOGY Other compactifications are even more important. EXAMPLE Real projective space RP (n) is a compactification of the Euclidean space R n. This follows easily form the description of RP (n) as the identification space of a (say, northern) hemisphere with pairs of opposite equatorial points identified. The open hemisphere is homeomorphic to R n and the attached set at infinity is homeomorphic to the projective space RP (n 1). EXERCISE Describe the complex projective space CP (n) (see Example 1.3.8) as a compactification of the space C n (which is of course homeomorphic to R 2n ). Specifically, identify the set of added points at infinity as a topological space. and desribe open sets which contain points at infinity Compactness under products, maps, and bijections. The following result has numerous applications in analysis, PDE, and other mathematical disciplines. THEOREM The product of any family of compact spaces is compact. PROOF. Consider an open cover C of the product of two compact topological spaces X and Y. Since any open neighborhood of any point contains the product of opens subsets in x and Y we can assume that every element of C is the product of open subsets in X and Y. Since for each x X the subset {x} Y in the induced topology is homeomorphic to Y and hence compact, one can find a finite subcollection C x C which covers {x} Y. For (x, y) X Y, denote by p 1 the projection on the first factor: p 1 (x, y) = x. Let U x = C O x p 1 (C); this is an open neighborhood of x and since the elements of O x are products, O x covers U x Y. The sets U x, x X form an open cover of X. By the compactness of X, there is a finite subcover, say {U x1,..., U xk }. Then the union of collections O x1,..., O xk form a finite open cover of X Y. For a finite number of factors, the theorem follows by induction from the associativity of the product operation and the case of two factors. The proof for an arbitrary number of factors uses some general set theory tools based on axiom of choice. PROPOSITION The image of a compact set under a continuous map is compact. PROOF. If C is compact and f : C Y continuous and surjective, then any open cover Γ of Y induces an open cover f Γ := {f 1 (O) O Γ} of C which by compactness has a finite subcover {f 1 (O i ) i = 1,..., n}. By surjectivity, {O i } n i=1 is a cover for Y.

21 1.6. CONNECTEDNESS AND PATH CONNECTEDNESS 25 A useful application of the notions of continuity, compactness, and separation is the following simple but fundamental result, sometimes referred to as invariance of domain: PROPOSITION A continuous bijection from a compact space to a Hausdorff space is a homeomorphism. PROOF. Suppose X is compact, Y Hausdorff, f : X Y bijective and continuous, and O X open. Then C := X O is closed, hence compact, and f(c) is compact, hence closed, so f(o) = Y f(c) (by bijectivity) is open. Using Proposition we obtain COROLLARY Under the assumption of Proposition spaces X and Y are normal. EXERCISE Show that for noncompact X the assertion of Proposition no longer holds Connectedness and path connectedness There are two rival formal definitions of the intuitive notion of connectedness of a topological space. The first is based on the idea that such a space consists of one piece (i.e., does not fall apart into two pieces ), the second interprets connectedness as the possibility of moving continuously from any point to any other point Definition and invariance under continuous maps. DEFINITION A topological space (X, T ) is said to be connected if X cannot be represented as the union of two nonempty disjoint open sets (or, equivalently, two nonempty disjoint closed sets); path connected if for any two points x 0, x 1 X there exists a path joining x 0 to x 1, i.e., a continuous map c: [0, 1] X such that c(i) = x i, i = {0, 1}. PROPOSITION The continuous image of a connected space X is connected. PROOF. If the image is decomposed into the union of two disjoint open sets, the preimages of theses sets which are open by continuity would give a similar decomposition for X. (1) Interval is connected (2) Any path-connected space is connected. PROPOSITION

22 26 1. BASIC TOPOLOGY PROOF. (1) Any open subset X of an interval is the union of disjoint open subintervals. The complement of X contains the endpoints of those intervals and hence cannot be open. (2) Suppose X is path-connected and let x = X 0 X 1, where X 0 and X 1 are open and nonempty. Let x 0 X 0, x 1 X 1 and c: [0, 1] X is a continuous map such that c(i) = x i, i {0, 1}. By Proposition the image c([0, 1]) is a connected subset of X in induced topology which is decomposed into the union of two nonempty open subsets c([0, 1]) X 0 and c([0, 1]) X 1, a contradiction. REMARK Connected space may not be path-connected as is shown by the union of the graph of sin 1/x and {0} [ 1, 1] in R 2 (see the figure). 1 y O y = sin 1 / x x 1 FIGURE Connected but not path connected space PROPOSITION The continuous image of a path connected space X is path connected. PROOF. Let f : X Y be continuous and surjective; take any two points y 1, y 2 Y. Then by surjectivity the sets f 1 (y i ), i = 1, 2 are nonempty and we can choose points x i f 1 (y 1 ), i = 1, 2. Since X is path connected, there is a path α : [0, 1] X joining x 1 to x 2. But then the path f α joins y 1 to y Products and quotients. PROPOSITION The product of two connected topological spaces is connected. PROOF. Suppose X, Y are connected and assume that X Y = A B, where A and B are open, and A B =. Then either A = X 1 Y for some open X 1 X or there exists an x X such that {x} Y A and {x} Y B.

23 1.6. CONNECTEDNESS AND PATH CONNECTEDNESS 27 x y FIGURE Path connectedness The former case is impossible, else we would have B = (X \ X 1 ) Y and so X = X 1 (X \ X 1 ) would not be connected. In the latter case, Y = p 2 ({x} Y A) p 2 ({x} Y B) (where p 2 (x, y) = y is the projection on the second factor) that is, {x} Y is the union of two disjoint open sets, hence not connected. Obviously p 2 restricted to {x} Y is a homeomorphism onto Y, and so Y is not connected either, a contradiction. PROPOSITION The product of two path-connected topological spaces is connected. PROOF. Let (x 0, y 0 ), (x 1, y 1 ) X Y and c X, c Y are paths connecting x 0 with x 1 and y 0 with y 1 correspondingly. Then the path c: [0, 1] X Y defined by c(t) = (c X (t), c Y (t)) connects (x 0, y 0 ) with (x 1, y 1 ). The following property follows immediately from the definition of the quotient topology PROPOSITION Any quotient space of a connected topological space is connected Connected subsets and connected components. A subset of a topological space is connected (path connected) if it is a connected (path connected) space in the induced topology. A connected component of a topological space X is a maximal connected subset of X. A path connected component of X is a maximal path connected subset of X. PROPOSITION The closure of a connected subset Y X is connected.

24 28 1. BASIC TOPOLOGY PROOF. If Ȳ = Y 1 Y 2, where Y 1, Y 2 are open and Y 1 Y 2 =, then since the set Y is dense in its closure Y = (Y Y 1 ) (Y Y 2 ) with both Y Y 1 and Y Y 1 open in the induced topology and nonempty. COROLLARY Connected components are closed. PROPOSITION The union of two connected subsets Y 1, Y 2 X such that Y 1 Y 2, is connected. PROOF. We will argue by contradiction. Assume that Y 1 Y 2 is the disjoint union of of open sets Z 1 and Z 2. If Z 1 Y 1, then Y 2 = Z 2 (Z 1 Y 2 ) and hence Y 2 is not connected. Similarly, it is impossible that Z 2 Y 1. Thus Y 1 Z i, i = 1, 2 and hence Y 1 = (Y 1 Z 1 ) (Y 1 Z 2 ) and hence Y 1 is not connected Decomposition into connected components. For any topological space there is a unique decomposition into connected components and a unique decomposition into path connected components. The elements of these decompositions are equivalence classes of the following two equivalence relations respectively: (i) x is equivalent to y if there exists a connected subset Y X which contains x and y. In order to show that the equivalence classes are indeed connected components, one needs to prove that they are connected. For, if A is an equivalence class, assume that A = A 1 A 2, where A 1 and A 2 are disjoint and open. Pick x 1 A 1 and x 2 A 2 and find a closed connected set A 3 which contains both points. But then A (A 1 A 3 ) A 2, which is connected by Proposition Hence A = (A 1 A 3 ) A 2 ) and A is connected. (ii) x is equivalent to y if there exists a continuous curve c: [0, 1] X with c(0) = x, c(1) = y REMARK The closure of a path connected subset may be fail to be path connected. It is easy to construct such a subset by looking at Remark Arc connectedness. Arc connectedness is a more restrictive notion than path connectedness: a topological space X is called arc connected if, for any two distinct points x, y X there exist an arc joining them, i.e., there is an injective continuous map h : [0, 1] X such that h(0) = x and h(1) = y. It turns out, however, that arc connectedness is not a much more stronger requirement than path connectedness in fact the two notions coincide for Hausdorff spaces.

25 1.6. CONNECTEDNESS AND PATH CONNECTEDNESS 29 THEOREM A Hausdorff space is arc connected if and only if it is path connected. PROOF. Let X be a path-connected Hausdorff space, x 0, x 1 X and c: [0, 1] X a continuous map such that c(i) = x i, i = 0, 1. Notice that the image c([0, 1]) is a compact subset of X by Proposition even though we will not use that directly. We will change the map c within this image by successively cutting of superfluous pieces and rescaling what remains. Consider the point c(1/2). If it coincides with one of the endpoints x o or x 1 we define c 1 (t) as c(2t 1) or c(2t) correspondingly. Otherwise consider pairs t 0 < 1/2 < t 1 such that c(t 0 ) = c(t 1 ). The set of all such pairs is closed in the product [0, 1] [0, 1] and the function t 0 t 1 reaches maximum on that set. If this maximum is equal to zero the map c is already injective. Otherwise the maximum is positive and is reached at a pair (a 1, b 1 ). we define the map c 1 as follows c(t/2a 1 ), if 0 t a 1, c 1 (t) = c(1/2), if a 1 t b 1, c(t/2(1 b 1 ) + (1 b 1 )/2), if b 1 t 1. Notice that c 1 ([0, 1/2)) and c 1 ((1/2, 1]) are disjoint since otherwise there would exist a < a 1 < b 1 < b such that c(a ) = c(b ) contradicting maximality of the pair (a 1, b 1 ). Now we proceed by induction. We assume that a continuous map c n : [0, 1] c([0, 1]) has been constructed such that the images of intervals (k/2 n, (k + 1)/2 n ), k = 0,..., 2 n 1 are disjoint. Furthermore, while we do not exclude that c n (k/2 n ) = c n ((k + 1)/2 n ) we assume that c n (k/2 n ) c n (l/2 n ) if k l > 1. We find a k n, b k n maximizing the difference t 0 t 1 among all pairs (t 0, t 1 ) : k/2 n t 0 t 1 (k + 1)/2 n and construct the map c n+1 on each interval [k/2 n, (k+1)/2 n ] as above with c n in place of c and a k n, b k n in place of a 1, b 1 with the proper renormalization. As before special provision are made if c n is injective on one of the intervals (in this case we set c n+1 = c n ) of if the image of the midpoint coincides with that of one of the endpoints (one half is cut off that the other renormalized). EXERCISE Give an example of a path connected but not arc connected topological space.

26 30 1. BASIC TOPOLOGY 1.7. Totally disconnected spaces and Cantor sets On the opposite end from connected spaces are those spaces which do not have any connected nontrivial connected subsets at all Examples of totally disconnected spaces. DEFINITION A topological space (X, T ) is said to be totally disconnected if every point is a connected component. In other words, the only connected subsets of a totally disconnected space X are single points. Discrete topologies (all points are open) give trivial examples of totally disconnected topological spaces. Another example is the set {0, 1, 1 2, 1 3, 1 4,..., } with the topology induced from the real line. More complicated examples of compact totally disconnected space in which isolated points are dense can be easily constructed. For instance, one can consider the set of rational numbers Q R with the induced topology (which is not locally compact). The most fundamental (and famous) example of a totally disconnected set is the Cantor set, which we now define. DEFINITION The (standard middle-third) Cantor set C(1/3) is defined as follows: { x } i C(1/3); = x R : x = 3, x i i {0, 2}, i = 1, 2,.... i=1 Geometrically, the construction of the set C(1/3) may be described in the following way: we start with the closed interval [0, 1], divide it into three equal subintervals and throw out the (open) middle one, divide each of the two remaining ones into equal subintervals and throw out the open middle ones and continue this process ad infinitum. What will be left? Of course the (countable set of) endpoints of the removed intervals will remain, but there will also be a much larger (uncountable) set of remaining mysterious points, namely those which do not have the ternary digit 1 in their ternary expansion Lebesgue measure of Cantor sets. There are many different ways of constructing subsets of [0, 1] which are homeomorphic to the Cantor set C(1/3). For example, instead of throwing out the middle one third intervals at each step, one can do it on the first step and then throw out intervals of length 1 in the middle of two remaining interval and inductively 18 1 throw out the interval of length in the middle of each of 2 n intervals 2 n 3 n+1 which remain after n steps. Let us denote the resulting set Ĉ

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