Class Notes for MATH 454.

Transcription

1 Class Notes for MATH 454. by S. W. Drury Copyright c , by S. W. Drury.

2 Contents 1 Metric and Topological Spaces Metric Spaces Normed Spaces Some Norms on Euclidean Space Inner Product Spaces The Inequalities of Hölder and Minkowski Examples of Metric Spaces Topology of Metric Spaces Neighbourhoods and Open Sets Convergent Sequences Continuity Compositions of Functions Product Spaces and Mappings The Diagonal Mapping and Pointwise Combinations Bases and Subbases in Toplogical Spaces General Product Spaces Interior and Closure Limits in Metric Spaces Distance to a Subset Separability Relative Topologies Uniform Continuity Subsequences Completeness Boundedness and Uniform Convergence Subsets and Products of Complete Spaces

3 3.3 Contraction Mappings Extension by Uniform Continuity Completions Extension of Continuous Functions Baire s Theorem Complete Normed Spaces Compactness Compact Subsets The Finite Intersection Property Compactness in Metric Spaces equivalent formulations Preservation of Compactness by Continuous Mappings Compactness and Uniform Continuity Compactness and Uniform Convergence The Lebesgue Number Equicontinuous Sets Tychonov s Product Theorem Duals of Banach spaces and the weak star topology The Stone Weierstrass Theorem Connectedness Connected Subsets Connectivity of the Real Line Connected Components Preservation of Connectedness by Continuous Mappings Path Connectedness Measuring sets on the Line Fields and σ-fields Extending the notion of length first steps Extending premeasures from fields to σ-fields Borel sets and Lebesgue sets Uniqueness of the Extension Completions of Measure Spaces Approximating sets in Lebesgue measure A non-measurable set

4 7 Additional material and some Proofs Metric Spaces Geometry of Norms More on Topology Countability in topological spaces Relative topologies in topological spaces Uniform spaces Baire Category Theorem for compact Hausdorff topological spaces A Metric Space Miscellany Premetrics Operator Norms Continuous Linear Forms Equivalent Metrics The Abstract Cantor Set Other Formulations of Compactness Compactness and Separation Hausdorff Distance Supplemental Material, Measure Theory The Free Group on Two Generators Measure Theory Proofs Monotone Classes The Inner Outer Measure approach to measurability Index 182 1

5 1 Metric and Topological Spaces 1.1 Metric Spaces In this section we introduce the concept of a metric space. A metric space is simply a set together with a distance function which measures the distance between any two points of the space. Starting from the distance function it is possible to introduce all the concepts we dealt with so far such as convergent sequences, continuity and limits etc. Thus in order to have the concept of convergence in a certain set of objects (5 5 real matrices for example), it suffices to have a concept of distance between any two such objects. A metric space (X, d) is a set X together with a distance function or metric d : X X R + satisfying the following properties. DEFINITION d(x, x) = 0 x X. x, y X, d(x, y) = 0 x = y. d(x, y) = d(y, x) x, y X. d(x, z) d(x, y) + d(y, z) x, y, z X. The fourth axiom for a distance function is called the triangle inequality. It is easy to derive the extended triangle inequality d(x 1, x n ) d(x 1, x 2 ) + d(x 2, x 3, ) + + d(x n 1, x n ) x 1,..., x n X (1.1) directly from the axioms. 2

6 Sometimes we will abuse notation and say that X is a metric space when the intended distance function is understood. The real line R is a metric space with the distance function d(x, y) = x y. A simple construction allows us to build new metric spaces out of existing ones. Let X be a metric space and let Y X. Then the restriction of the distance function of X to the subset Y Y of X X is a distance function on Y. Sometimes this is called the restriction metric or the relative metric. If the four axioms listed above hold for all points of X then a fortiori they hold for all points of Y. Thus every subset of a metric space is again a metric space in its own right. We can construct more interesting examples from vector spaces. 1.2 Normed Spaces We start by introducing the concept of a norm. This generalization of the absolute value on R (or C) to the framework of vector spaces is central to modern analysis. The zero element of a vector space V (over R or C) will be denoted 0 V. For an element v of the vector space V the norm of v (denoted v ) is to be thought of as the distance from 0 V to v, or as the size or length of v. In the case of the absolute value on the field of scalars, there is really only one possible candidate, but in vector spaces of more than one dimension a wealth of possibilities arises. DEFINITION A norm on a vector space V over R or C is a mapping v v from V to R + with the following properties. 0 V = 0. v V, v = 0 v = 0 V. tv = t v t a scalar, v V. v 1 + v 2 v 1 + v 2 v 1, v 2 V. The last of these conditions is called the subadditivity inequality. There are really two definitions here, that of a real norm applicable to real vector spaces and that of a complex norm applicable to complex vector spaces. However, every complex vector space can also be considered as a real vector space one simply 3

7 forgets how to multiply vectors by complex scalars that are not real scalars. This process is called realification. In such a situation, the two definitions are different. For instance, x + iy = max( x, 2 y ) (x, y R) defines a perfectly good real norm on C considered as a real vector space. On the other hand, the only complex norms on C have the form for some t > 0. The inequality x + iy = t(x 2 + y 2 ) 1 2 t 1 v 1 + t 2 v t n v n t 1 v 1 + t 2 v t n v n holds for scalars t 1,..., t n and elements v 1,..., v n of V. It is an immediate consequence of the definition. If is a norm on V and t > 0 then v = t v defines a new norm on V. We note that in the case of a norm there is often no natural way to normalize it. On the other hand, an absolute value is normalized so that 1 = 1, possible since the field of scalars contains a distinguished element Some Norms on Euclidean Space Because of the central role of R n as a vector space it is worth looking at some of the norms that are commonly defined on this space. EXAMPLE On R n we may define a norm by (x 1,..., x n ) = n max j=1 x j. (1.2) EXAMPLE Another norm on R n is given by (x 1,..., x n ) 1 = n x j. j=1 4

8 EXAMPLE The Euclidean norm on R n is given by ( n ) 1 2 (x 1,..., x n ) 2 = x j 2. This is the standard norm, representing the standard Euclidean distance to 0. The symbol 0 will be used to denote the zero vector of R n or C n. These examples can be generalized by defining in case 1 p < ( n ) 1 p (x 1,..., x n ) p = x j p. In case that p = we use (1.2) to define. It is true that p is a norm we will prove this in section (1.5). 1.4 Inner Product Spaces Inner product spaces play a very central role in analysis. They have many applications. For example the physics of Quantum Mechanics is based on inner product spaces. DEFINITION A real inner product space is a real vector space V together with an inner product. An inner product is a mapping from V V to R denoted by and satisfying the following properties j=1 j=1 (v 1, v 2 ) v 1, v 2 w, t 1 v 1 + t 2 v 2 = t 1 w, v 1 + t 2 w, v 2 w, v 1, v 2 V, t 1, t 2 R. v 1, v 2 = v 2, v 1 v 1, v 2 V. v, v 0 v V. If v V and v, v = 0, then v = 0 V. The symmetry and the linearity in the second variable implies that the inner product is also linear in the first variable. t 1 v 1 + t 2 v 2, w = t 1 v 1, w + t 2 v 2, w w, v 1, v 2 V, t 1, t 2 R. 5

9 EXAMPLE The standard inner product on R n is given by n x, y = x j y j j=1 The most general inner product on R n is given by x, y = n j=1 k=1 n p j,k x j y k where the n n real matrix P = (p j,k ) is a positive definite matrix. This means that P is a symmetric matrix. We have n j=1 k=1 for every vector (x 1,..., x n ) of R n. The circumstance n j=1 k=1 only occurs when x 1 = 0,..., x n = 0. n p j,k x j x k 0 n p j,k x j x k = 0 In the complex case, the definition is slightly more complicated. DEFINITION A complex inner product space is a complex vector space V together with a complex inner product, that is a mapping from V V to C denoted (v 1, v 2 ) v 1, v 2 and satisfying the following properties w, t 1 v 1 + t 2 v 2 = t 1 w, v 1 + t 2 w, v 2 w, v 1, v 2 V, t 1, t 2 C. v 1, v 2 = v 2, v 1 v 1, v 2 V. v, v 0 v V. 6

10 If v V and v, v = 0, then v = 0 V. It will be noted that a complex inner product is linear in its second variable and conjugate linear in its first variable. t 1 v 1 + t 2 v 2, w = t 1 v 1, w + t 2 v 2, w w, v 1, v 2 V, t 1, t 2 C. EXAMPLE The standard inner product on C n is given by n x, y = x j y j j=1 The most general inner product on C n is given by x, y = n j=1 k=1 n p j,k x j y k where the n n complex matrix P = (p j,k ) is a positive definite matrix. This means that P is a hermitian matrix, in other words p jk = p kj. We have n j=1 k=1 for every vector (x 1,..., x n ) of C n. The circumstance n j=1 k=1 only occurs when x 1 = 0,..., x n = 0. n p j,k x j x k 0 n p j,k x j x k = 0 7

11 DEFINITION Let V be an inner product space. Then we define the associated norm. v = ( v, v ) 1 2 (1.3) It is not immediately clear from the definition that the associated norm satisfies the subadditivity condition. Towards this, we establish the abstract Cauchy- Schwarz inequality. PROPOSITION 1 (CAUCHY-SCHWARZ INEQUALITY) Let V be an inner product space and u, v V. Then holds. Link to proof. PROPOSITION 2 Link to proof. PROPOSITION 3 The norm arising from an inner product satisfies the parallelogram law u, v u v (1.4) In an inner product space (1.3) defines a norm. u + v 2 + u v 2 = 2 u v 2 for all u and v in the inner product space. Link to proof. The converse is also true every norm that satisfies the parallelogram law arises from an inner product. This is not easy to prove. Link to proof. Note however that the key idea involves the concept of polarization. For real inner product spaces a suitable polarization identity is u, v = 1 ( u + v 2 u v 2). 4 More generally if ϕ is a symmetric n-linear form on a vector space, a polarization identity expresses ϕ(v 1,..., v n ) in terms of ψ(v) = ϕ(v,..., v) (i.e. the values of ϕ on the diagonal). For example in n = 3, 1 24 (ψ(v 1 + v 2 + v 3 ) ψ(v 1 v 2 + v 3 ) ψ(v 1 + v 2 v 3 ) + ψ(v 1 v 2 v 3 )) is a polarization expression for ϕ(v 1, v 2, v 3 ), but there are many others. 8

12 1.5 The Inequalities of Hölder and Minkowski Let 1 p. Then we define p = the conjugate index of p. In case p 1 p = 1 we take p =, and in case p = we take p = 1. We have p 1 p + 1 p = 1, so that the relationship between index and conjugate index is symmetric. Note that if p = 2 then p = 2. PROPOSITION 4 (HÖLDER S INEQUALITY) For x, y C n we have n x j y j x p y p (1.5) j=1 If p = 1 or p =, Hölder s Inequality is easy to verify. In the general case we use the following lemma. LEMMA 5 Let x 0 and y 0. Let 1 < p < and let p be the conjugate index of p, so that 1 < p <. Then xy 1 p xp + 1 p yp. (1.6) Link to proof. Proof of Hölder s Inequality. We first suppose that x p = 1 and y p = 1. Then, by multiple applications of Lemma 5 we have n x j y j j=1 n x j y j j=1 n j=1 1 p x j p + 1 p y j p = 1 p x p p + 1 p y p p = 1 p + 1 p = 1. (1.7) 9

13 For the general case, we first observe that if x p = 0, then x = 0 and the result is straightforward. We may assume that x p > 0 and similarly that y p > 0. Then, applying (1.7) with x replaced by x 1 p x and y replaced by y 1 p y, we obtain n x j y j 1. x p y p j=1 Finally, multiplying by x p y p yields Hölder s inequality. The next result implies that the quantities p are actually norms for 1 p. They do not satisfy the triangle inequality if p < 1. THEOREM 6 (MINKOWSKI S INEQUALITY) Let 1 p and x, y R n. Then holds. x + y p x p + y p (1.8) Proof. The result is easy if p = 1 or if p =. We therefore suppose that 1 < p <. We have n x + y p p = x j + y j p = = j=1 n x j + y j x j + y j p 1 j=1 n ( x j + y j ) x j + y j p 1 j=1 n x j x j + y j p 1 + j=1 n y j x j + y j p 1 (1.9) The key is to apply Hölder s inequality to each of the two sums in (1.9). We have j=1 j=1 ( n n ) 1 ( p n x j x j + y j p 1 x j p x j + y j p (p 1) j=1 j=1 ) 1 p = x p x + y p 1 p. (1.10) 10

14 since p (p 1) = p and 1 = (p 1) 1. Similarly p p n j=1 y j x j + y j p 1 y p x + y p 1 p. (1.11) Combining now (1.9), (1.10) and (1.11), we obtain x + y p p ( x p + y p ) x + y p 1 p. (1.12) Now if x + y p = 0 we have the conclusion (1.8). If not, then it is legitimate to divide (1.12) by x + y p 1 p and again the conclusion follows. The p-norms are used most frequently in the cases p = 1, p = 2 and p =. The case p = 2 is special in that the 2-norm is the Euclidean norm which arises from an inner product. In particular the standard Cauchy-Schwarz inequality ( n n ) 1 ( ) 2 n 1 2 x k y k x j 2 y k 2 k=1 j=1 is just the case p = 2 of Hölder s Inequality (1.5). Link to Geometry of Norms. k=1 1.6 Examples of Metric Spaces In the previous section we discussed the concept of the norm of a vector. In a normed vector space, the expression u v represents the size of the difference u v of two vectors u and v. It can be thought of as the distance between u and v. Just as a vector space may have many possible norms, there can be many possible concepts of distance. EXAMPLE Let V be a normed vector space with norm. Then V is a metric space with the distance function d(u, v) = u v. We check that the triangle inequality is a consequence of the subadditivity of the norm. d(u, w) = u w = (u v)+(v w) u v + v w = d(u, v)+d(v, w) 11

15 EXAMPLE It follows that every subset X of a normed vector space is a metric space in the distance function induced from the norm. EXAMPLE For any set X, the standard discrete metric is the distance function { 1 if x1 x d(x 1, x 2 ) = 2, 0 if x 1 = x 2. EXERCISE For any finite set X show that d(a, B) = card((a \ B) (B \ A)) is a metric on the set of all subsets of X. EXAMPLE Let, and denote the standard inner product and Euclidean norm on R n. Let S n 1 denote the unit sphere S n 1 = {x; x R n, x = 1} then we can define the geodesic distance between two points x and y of S n 1 by d(x, y) = arccos( x, y ). (1.13) We will show that d is a metric on S n 1. This metric is of course different from the Euclidean distance x y. To verify that (1.13) is in fact a metric, at least the symmetry of the metric is evident. Suppose that x, y S n 1 and that d(x, y) = 0. Then x, y = 1 and x y 2 = x 2 2 x, y + y 2 = = 0. It follows that x = y. To establish the triangle inequality, let x, y, z S n 1, θ = arccos( x, y ) and ϕ = arccos( y, z ). Then we can write x = cos θ y + sin θ u and z = cos ϕ y + sin ϕ v where u and v are unit vectors orthogonal to y. An easy calculation now gives x, z = cos θ cos ϕ + u, v sin θ sin ϕ. Now, since 0 θ, ϕ π, we have sin θ sin ϕ 0. By the Cauchy-Schwarz Inequality (1.4), we find that u, v 1. Hence x, z cos θ cos ϕ sin θ sin ϕ = cos(θ + ϕ). Since arccos is decreasing on [ 1, 1] this immediately yields d(x, z) θ + ϕ = d(x, y) + d(y, z). 12

16 2 Topology of Metric Spaces 2.1 Neighbourhoods and Open Sets It is customary to refer to the elements of a metric space as points. In this chapter we will develop the point-set topology of metric spaces. This is done through concepts such as neighbourhoods, open sets, closed sets and sequences. Any of these concepts can be used to define more advanced concepts such as the continuity of mappings from one metric space to another. They are, as it were, languages for the further development of the subject. We study them all and most particularly the relationships between them. DEFINITION and Let (X, d) be a metric space. For t > 0 and x X, we define U(x, t) = {y; y X, d(x, y) < t} B(x, t) = {y; y X, d(x, y) t}. the open ball U(x, t) centred at x of radius t and the corresponding closed ball B(x, t). DEFINITION Let V be a subset of a metric space X and let x V. Then we say that V is a neighbourhood of x or x is an interior point of V iff there exists t > 0 such that U(x, t) V. Thus V is a neighbourhood of x iff all points sufficiently close to x lie in V. 13

17 PROPOSITION 7 If V is a neighbourbood of x and V W X. Then W is a neighbourhood of x. If V 1, V 2,..., V n are finitely many neighbourhoods of x, then n j=1v j is also a neighbourhood of x. Proof. The first statement is left as an exercise for the reader. For the second, applying the definition, we may find t 1, t 2,..., t n > 0 such that U(x, t j ) V j. It follows that n U(x, t j ) j=1 n V j. (2.1) But the left-hand side of (2.1) is just U(x, t) where t = min t j > 0. It now follows that n j=1v j is a neighbourhood of x. Link to Filters. Neighbourhoods are a local concept. We now introduce the corresponding global concept. DEFINITION Let (X, d) be a metric space and let V X. Then V is an open subset of X iff V is a neighbourhood of every point x that lies in V. EXAMPLE For all t > 0, the open ball U(x, t) is an open set. To see this, let y U(x, t), that is d(x, y) < t. We must show that U(x, t) is a neighbourhood of y. Let s = t d(x, y) > 0. We claim that U(y, s) U(x, t). To prove the claim, let z U(y, s). Then d(y, z) < s. We now find that j=1 d(x, z) d(x, y) + d(y, z) < d(x, y) + s = t, so that z U(x, t) as required. EXAMPLE In R every interval of the form ]a, b[ is an open set. Here, a and b are real and satisfy a < b. We also allow the possibilities a = and b =. 14

18 THEOREM 8 A subset of R is open (for the standard metric on R) if and only if it is a countable disjoint union of open intervals. THEOREM 9 In a metric space (X, d) we have X is an open subset of X. is an open subset of X. If V α is open for every α in some index set I, then α I V α is again open. If V j is open for j = 1,..., n, then the finite intersection n j=1v j is again open. Proof. For every x X and any t > 0, we have U(x, t) X, so X is open. On the other hand, is open because it does not have any points. Thus the condition to be checked is vacuous. To check the third statement, let x α I V α. Then there exists α I such that x V α. Since V α is open, V α is a neighbourhood of x. The result now follows from the first part of Proposition 7. Finally let x n j=1v j. Then since V j is open, it is a neighbourhood of x for j = 1,..., n. Now apply the second part of Proposition 7. DEFINITION Let X be a set. Let V be a family of open sets satisfying the four conditions of Theorem 9. Then V is a topology on X and (X, V) is a topological space. Not every topology arises from a metric. We say that a topological space X is metrizable if there exists a metric on X that gives rise to the given topology. We highlight some of the main difference between metric spaces and topological spaces. Topological spaces cannot always distinguish between points. For example, if X = {x, y} and V = {X, }, then every open subset that contains x also contains y. Every open subset that contains y also contains x. Another valid topology on this space is V = {X, {x}, }. In a metric space, given two distinct points x and y, you can always find disjoint open subsets U 1 and U 2 with x U 1, y U 2. 15

19 Sequence techniques are not viable in topological spaces. A new concept, that of a net, is needed to replace the sequence concept. Some metric space concepts such as uniform continuity and completeness cannot be defined on topological spaces. Explictly, you can find two metrics on the same space and with the same topology (i.e. the same open subsets) for which one is complete and the other not. Similarly for uniformly continuous functions. EXAMPLE Let s resurrect the neighbourhood concept for topological spaces. So, if X is a topological space and x X, we will say that V is a neighbourhood of x iff there exists U an open subset of X such that x U V. Now we need to know that the metric space definition of open is still good in this context. Claim V open V is a neighbourhood of each of its points. If V is open and x V, then taking U = V above, we see that V is a neighbourhood of x. Claim V is a neighbourhood of each of its points V open. For every x V there exists an open subset U x such that x U x V. But then V {x} U x V x V x V Hence V = x V U x is a union of open sets and hence open. Link to Discrete topology and metrics. 2.2 Convergent Sequences A sequence x 1, x 2, x 3,... of points of a set X is really a mapping from N to X. Normally, we denote such a sequence by (x n ). For x X the sequence given by x n = x is called the constant sequence with value x. DEFINITION Let X be a metric space. Let (x n ) be a sequence in X. Then (x n ) converges to x X iff for every ɛ > 0 there exists N N such that d(x n, x) < ɛ for all n > N. In this case, we write x n x or x n n x. 16

20 Sometimes, we say that x is the limit of (x n ). Proposition 10 below justifies the use of the indefinite article. To say that (x n ) is a convergent sequence is to say that there exists some x X such that (x n ) converges to x. EXAMPLE sequence Perhaps the most familiar example of a convergent sequence is the x n = 1 n in R. This sequence converges to 0. To see this, let ɛ > 0 be given. Then choose a natural number N so large that N > ɛ 1. It is easy to see that n > N 1 n < ɛ Hence x n 0. PROPOSITION 10 Let (x n ) be a convergent sequence in a metric space X. Then the limit is unique. Proof. Suppose that x and y are both limits of the sequence (x n ). We will show that x = y. If not, then d(x, y) > 0. Let us choose ɛ = 1 d(x, y). Then there exist 2 natural numbers N x and N y such that n > N x d(x n, x) < ɛ, n > N y d(x n, y) < ɛ. Choose now n = max(n x, N y ) + 1 so that both n > N x and n > N y. It now follows that 2ɛ = d(x, y) d(x, x n ) + d(x n, y) < ɛ + ɛ a contradiction. PROPOSITION 11 Let X be a metric space and let (x n ) be a sequence in X. Let x X. The following conditions are equivalent to the convergence of (x n ) to x. For every neighbourhood V of x in X, there exists N N such that n > N x n V. (2.2) 17

21 The sequence (d(x n, x)) converges to 0 in R. We leave the details of the proof to the reader. The first item here is significant because it leads to the concept of the tail of a sequence. The sequence (t n ) defined by t k = x N+k is called the Nth tail sequence of (x n ). The set of points T N = {x n ; n > N} is the Nth tail set. The condition (2.2) can be rewritten as T N V. Sequences provide one of the key tools for understanding metric spaces. They lead naturally to the concept of closed subsets of a metric space. DEFINITION Let X be a metric space. Then a subset A X is said to be closed iff whenever (x n ) is a sequence in A (that is x n A n N) converging to a limit x in X, then x A. The link between closed subsets and open subsets is contained in the following result. THEOREM 12 open. In a metric space X, a subset A is closed if and only if X \ A is It follows from this Theorem that U is open in X iff X \ U is closed. Proof. First suppose that A is closed. We must show that X \A is open. Towards this, let x X \ A. We claim that there exists ɛ > 0 such that U(x, ɛ) X \ A. Suppose not. Then taking for each n N, ɛ n = 1 we find that there exists n x n A U(x, 1 ). But now (x n n) is a sequence of elements of A converging to x. Since A is closed x A. But this is a contradiction. For the converse assertion, suppose that X \ A is open. We will show that A is closed. Let (x n ) be a sequence in A converging to some x X. If x X \ A then since X \ A is open, there exists ɛ > 0 such that U(x, ɛ) X \ A. (2.3) But since (x n ) converges to x, there exists N N such that x n U(x, ɛ) for n > N. Choose n = N +1. Then we find that x n A U(x, ɛ) which contradicts (2.3). In a topological space X, we define the concept of closed set as follows: E is closed if and only if X \E is open. If E is closed, x n E and x n x as n, then x E. But the converse is false, namely there may exist a subset E such that the limit of every convergent sequence in E lies in E but E is not closed. To 18

22 make this work one needs to replace the sequence concept with the concept of a net. Link to Nets. We have the following corollary of Theorems 9. COROLLARY 13 In a topological space X we have X is a closed subset of X. is a closed subset of X. If A α is closed for every α in some index set I, then α I A α is again closed. If A j is closed for j = 1,..., n, then the finite union n j=1a j is again closed. EXAMPLE In a metric space every singleton is closed. To see this we remark that a sequence in a singleton is necessarily a constant sequence and hence convergent to its constant value. EXAMPLE Combining the previous example with the last assertion of Corollary 13, we see that in a metric space, every finite subset is closed. EXAMPLE Let (x n ) be a sequence converging to x. Then the set {x n ; n N} {x} is a closed subset. EXAMPLE In R, the intervals [a, b], [a, [ and ], b] are closed subsets. EXAMPLE A more complicated example of a closed subset of R is the Cantor set. There are several ways of describing the Cantor set. Let E 0 = [0, 1]. To obtain E 1 from E 0 we remove the middle third of E 0. Thus E 1 = [0, 1] [ 2, 1]. 3 3 To obtain E 2 from E 1 we remove the middle thirds from both the constituent intervals of E 1. Thus E 2 = [0, 1 9 ] [ 2 9, 1 3 ] [ 2 3, 7 9 ] [ 8 9, 1]. 19

23 E E E Figure 2.1: The sets E 0, E 1 and E 2. Continuing in this way, we find that E k is a union of 2 k closed intervals of length 3 k. The Cantor set E is now defined as E = E k. k=0 By Corollary 13 it is clear that E is a closed subset of R. The sculptor Rodin once said that to make a sculpture one starts with a block of marble and removes everything that is unimportant. This is the approach that we have just taken in building the Cantor set. The second way of constructing the Cantor set works by building the set from the inside out. Let us define K = { ω k 3 k ; ω k {0, 2}, k = 1, 2,...}. k=1 A moment s thought shows us that the points n k=1 ω k3 k given by the 2 n choices of ω k for k = 1, 2,..., n are precisely the left hand endpoints of the 2 n constituent subintervals of E n. Also a straightforward estimate on the tail sum 0 k=n+1 ω k 3 k k=n k 3 n, shows that each sum k=1 ω k3 k lies in E n for each n N. It follows that K E. 20

24 For the reverse inclusion, suppose that x E. Then for every n N, let x n be the left hand endpoint of the subinterval of E n to which x belongs. Then We write x x n 3 n. (2.4) n x n = ω k 3 k (2.5) k=1 where ω k takes one or other of the values 0 and 2. It is not difficult to see that the values of ω k do not depend on the value n under consideration. Indeed, suppose that (2.5) holds for a specific value of n. Then x [x n, x n + 3 n ]. At the next step, we look to see whether x lies in the left hand third or the right hand third of this interval. This determines x n+1 by x n+1 = x n + ω n+1 3 (n+1) where ω n+1 = 0 if it is the left hand interval and ω n+1 = 2 if it is the right hand interval. The values of ω k for k = 1, 2,..., n are not affected by this choice. It now follows from (2.5) and (2.4) that x = ω k 3 k k=1 so that x K as required. 2.3 Continuity The primary purpose of the preceding sections is to define the concept of continuity of mappings. This concept is the mainspring of mathematical analysis. DEFINITION Let X and Y be metric spaces. Let f : X Y. Let x X. Then f is continuous at x iff for all ɛ > 0, there exists δ > 0 such that z U(x, δ) f(z) U(f(x), ɛ). (2.6) The combination suggests the role of the devil s advocate type of argument. Let us illustrate this with an example. 21

25 EXAMPLE The mapping f : R R given by f(x) = x 2 is continuous at x = 1. To prove this, we suppose that the devil s advocate provides us with a number ɛ > 0 chosen cunningly small. We have to reply with a number δ > 0 (depending on ɛ) such that (2.6) holds. In the present context, we choose so that for x 1 < δ we have δ = min( 1 ɛ, 1) 4 x 2 1 x 1 x + 1 < ( 1 4 ɛ)(3) < ɛ since x 1 < δ and x + 1 = (x 1) + 2 x < 3. EXAMPLE Continuity at a point a single point that is, does not have much strength. Consider the function f : R R given by { 0 if x R \ Q, f(x) = x if x Q. This function is continuous at 0 but at no other point of R. EXAMPLE given by An interesting contrast is provided by the function g : R R { 0 if x R \ Q or if x = 0, g(x) = 1 if x = p where p Z \ {0}, q N are coprime. q q The function g is continuous at x iff x is zero or irrational. To see this, we first observe that if x Q \ {0}, then g(x) 0 but there are irrational numbers z as close as we like to x which satisfy g(z) = 0. Thus g is not continuous at the points of Q \ {0}. On the other hand, if x R \ Q or x = 0, we can establish continuity of g at x by an epsilon delta argument. We agree that whatever ɛ > 0 we will always choose δ < 1. Then the number of points z in the interval ]x δ, x + δ[ where g(z) ɛ is finite because such a z is necessarily a rational number that can be expressed in the form p q where 1 q < ɛ 1. With only finitely many points to avoid, it is now easy to find δ > 0 such that z x < δ = g(z) g(x) = g(z) < ɛ. There are various other ways of formulating continuity at a point. 22

26 THEOREM 14 Let X and Y be metric spaces. Let f : X Y. Let x X. Then the following statements are equivalent. f is continuous at x. For every neighbourhood V of f(x) in Y, f 1 (V ) is a neighbourhood of x in X. For every sequence (x n ) in X converging to x, the sequence (f(x n )) converges to f(x) in Y. Proof. We show that the first statement implies the second. Let f be continuous at x and suppose that V is a neighbourhood of f(x) in Y. Then there exists ɛ > 0 such that U(f(x), ɛ) V in Y. By definition of continuity at a point, there exists δ > 0 such that z U(x, δ) f(z) U(f(x), ɛ) f(z) V z f 1 (V ). Hence f 1 (V ) is a neighbourhood of x in X. Next, we assume the second statement and establish the third. Let (x n ) be a sequence in X converging to x. Let ɛ > 0. Then U(f(x), ɛ) is a neighbourhood of f(x) in Y. By hypothesis, f 1 (U(f(x), ɛ)) is a neighbourhood of x in X. By the first part of Proposition 11 there exists N N such that But this is equivalent to n > N x n f 1 (U(f(x), ɛ)). n > N f(x n ) U(f(x), ɛ). Thus (f(x n )) converges to f(x) in Y. Finally we show that the third statement implies the first. We argue by contradiction. Suppose that f is not continuous at x. Then there exists ɛ > 0 such that for all δ > 0, there exists z X with d(x, z) < δ, but d(f(x), f(z)) ɛ. We take choice δ = 1 n for n = 1, 2,... in sequence. We find that there exist x n in X with d(x, x n ) < 1 n, but d(f(x), f(x n)) ɛ. But now, the sequence (x n ) converges to x in X while the sequence (f(x n )) does not converge to f(x) in Y. 23

27 If X and Y are topological spaces f : X Y and x X, we say that f is continuous at x if and only if f 1 (V ) is a neighbourhood of x in X for every neighbourhood V of f(x) in Y. We next build the global version of continuity from the concept of continuity at a point. DEFINITION Let X and Y be metric spaces and let f : X Y. Then the mapping f is continuous iff f is continuous at every point x of X. There are also many possible reformulations of global continuity. THEOREM 15 Let X and Y be metric spaces. Let f : X Y. Then the following statements are equivalent to the continuity of f. For every open set U in Y, f 1 (U) is open in X. For every closed set A in Y, f 1 (A) is closed in X. For every convergent sequence (x n ) in X with limit x, the sequence (f(x n )) converges to f(x) in Y. THEOREM 16 Let X and Y be topological spaces. Let f : X Y. Then the following statements are equivalent to the continuity of f. For every open set U in Y, f 1 (U) is open in X. For every closed set A in Y, f 1 (A) is closed in X. Proofs. Let f be continuous. We check that the first statement holds. Let x f 1 (U). Then f(x) U. Since U is open in Y, U is a neighbourhood of f(x). Hence, f 1 (U) is a neighbourhood of x. We have just shown that f 1 (U) is a neighbourhood of each of its points. Hence f 1 (U) is open in X. For the converse, we assume that the first statement holds. Let x be an arbitrary point of X. We must show that f is continuous at x. Let V be a neighbourhood of f(x) in Y. Then, there exists U open in Y such that x U V. Hence using the hypothesis, f 1 (U) is open in X. Since x f 1 (U) f 1 (V ), we see that f 1 (V ) is a neighbourhood of x in X. 24

28 The second statement is clearly equivalent to the first. For instance if A is closed in Y, then Y \ A is an open subset. Then X \ f 1 (A) = f 1 (Y \ A) is open in X and it follows that f 1 (A) is closed in X. The converse entirely similar. In the metric space context, the third statement is equivalent directly from the definition. One very useful condition that implies continuity is the Lipschitz condition. DEFINITION Let X and Y be metric spaces. Let f : X Y. Then f is a Lipschitz map iff there is a constant C with 0 < C < such that d Y (f(x 1 ), f(x 2 )) Cd X (x 1, x 2 ) x 1, x 2 X. In the special case that C = 1 we say that f is a nonexpansive mapping. In the even more restricted case that we say that f is an isometry. d Y (f(x 1 ), f(x 2 )) = d X (x 1, x 2 ) x 1, x 2 X, PROPOSITION 17 Every Lipschitz map is continuous. Proof. We work directly. Let ɛ > 0. The set δ = C 1 ɛ. Then d X (z, x) < δ implies that d Y (f(z), f(x)) Cd X (z, x) Cδ = ɛ. as required. 2.4 Compositions of Functions DEFINITION Let X, Y and Z be sets. Let f : X Y and g : Y Z be mappings. Then we can make a new mapping h : X Z by h(x) = g(f(x)). In other words, to map by h we first map by f from X to Y and then by g from Y to Z. The mapping h is called the composition or composed mapping of f and g. It is usually denoted by h = g f. Composition occurs in very many situations in mathematics. It is the primary tool for building new mappings out of old. 25

29 THEOREM 18 Let X, Y and Z be topological spaces. Let f : X Y and g : Y Z be continuous mappings. Then the composition g f is a continuous mapping from X to Z. THEOREM 19 Let X, Y and Z be topological spaces. Let f : X Y and g : Y Z be mappings. Suppose that x X, that f is continuous at x and that g is continuous at f(x). Then the composition g f is a continuous at x. Proof of Theorems 18 and 19. By Theorem 16, it will suffice to show only Theorem 19. Let V be a neighbourhood of g(f(x)) in Z. Then since g is continuous at f(x) it follows that g 1 (V ) is a neighbourhood of f(x) in Y. But now, since f is continuous at x, f 1 (g 1 (V )) is a neighbourhood of x in X. But f 1 (g 1 (V )) = (g f) 1 (V ). Since this is true for every neighbourhood of g(f(x)) = (g f)(x) in Z, it follows that g f is continuous at x. 2.5 Product Spaces and Mappings In order to discuss combinations of functions we need some additional machinery. DEFINITION Let (X, d X ) and (Y, d Y ) be metric spaces. Then we define a product metric d on the product set X Y which allows us to consider X Y as a product metric space. We do this as follows d((x 1, y 1 ), (x 2, y 2 )) = max(d X (x 1, x 2 ), d Y (y 1, y 2 )) (2.7) PROPOSITION 20 Equation (2.7) defines a bona fide metric on X Y. Proof. The first three conditions in the definition of a metric (on page 2) are obvious. It remains to check the triangle inequality. Let (x 1, y 1 ), (x 2, y 2 ) and (x 3, y 3 ) be three generic points of X Y. Then d((x 1, y 1 ), (x 3, y 3 )) is the maximum of d X (x 1, x 3 ) and d Y (y 1, y 3 ). Let us suppose without loss of generality that 26

30 d X (x 1, x 3 ) is the larger of the two quantities. Then, by the triangle inequality on X, we have d X (x 1, x 3 ) d X (x 1, x 2 ) + d X (x 2, x 3 ). (2.8) But the right hand side of (2.8) is in turn less than providing the required result. d((x 1, y 1 ), (x 2, y 2 )) + d((x 2, y 2 ), (x 3, y 3 )) With the definition out of the way, the next step is to see how it relates to other topological constructs. LEMMA 21 Let X and Y be metric spaces. Let x X and let (x n ) be a sequence in X. Let y Y and let (y n ) be a sequence in Y. Then the sequence ((x n, y n )) converges to (x, y) in X Y if and only if the sequence (x n ) converges to x in X and the sequence (y n ) converges to y in Y. Proof. First, suppose that ((x n, y n )) converges to (x, y) in X Y. We must show that (x n ) converges to x in X. (It will follow similarly that (y n ) converges to y in Y.) This amounts then to showing that the projection π : X Y X onto the first coordinate, given by π((x, y)) = x is continuous. But the definition of the product metric ensures that π is nonexpansive (see page 25) and hence is continuous. The key inequality is d X (x 1, x 2 ) d X Y ((x 1, y 1 ), (x 2, y 2 )). For the converse, we have to get our hands dirtier. Let ɛ > 0. Then there exists N such that d X (x n, x) < ɛ for n > N. Also, there exists M such that d Y (y n, y) < ɛ for n > M. It follows that for n > max(n, M) both of the above inequalities hold, so that But this is exactly equivalent to max(d X (x n, x), d Y (y n, y)) < ɛ. d X Y ((x n, y n ), (x, y)) < ɛ as required for the convergence of ((x n, y n )) to (x, y). There is a simple way to understand neighbourhoods and hence open sets in product spaces. 27

31 PROPOSITION 22 Let X and Y be metric spaces. Let x X and y Y. Let U X Y. Then the following two statements are equivalent U is a neighbourhood of (x, y). There exist V a neighbourhood of x and W a neighbourhood of y such that V W U. Proof. Suppose that the first statement holds. Then there exists t > 0 such that U X Y ((x, y), t) U. But it is easy to check that U X Y ((x, y), t) = U X (x, t) U Y (y, t). Of course, U X (x, t) is a neighbourhood of x in X and U Y (y, t) is a neighbourhood of y in Y. Conversely, let V and W be neighbourhoods of x and y in X and Y respectively. Then there exist t, s > 0 such that U X (x, t) V and U Y (y, s) W. It is then easy to verify that U X Y ((x, y), min(t, s)) U X (x, t) U Y (y, s) V W U, so that U is a neighbourhood of (x, y) as required. We effectively use Proposition 22 as the definition of the product topology. DEFINITION Let X and Y be topological spaces and let U X Y. Then we say that U is open in X Y iff for every point (x, y) of U there exist V and W open in X and Y respectively with x V, y W such that V W U. PROPOSITION 23 The above definition defines a topology on X Y. This topology is called the product topology. Proof. If U = there is nothing to prove. If U = X Y then we can take V = X and W = Y. If U α are open in X Y for α I and (x, y) U = U α, α I then there exists a specific α I shuch that (x, y) U α. Hence there exist V and W as above and such that V W U α. But afortiori V W U. Hence α I U α is open. 28

32 The trickiest part is to handle the finite intersections. Let U 1,..., U N be open N N in X Y. We need to show that U = U n is again open. Let (x, y) U n. n=1 Then there exist V n, W n as above open in X and Y respectively x V n, y W n N N such that V n W n U n. Now define V = V n and W = W n open subsets of X and Y respectively with x V and y W and V W V n W n U n for n = 1, 2,..., n. Then V W U and it follows that U is open as required. We remark that the neighbourhoods for the product topology are given essentially by the equivalence in Proposition 22. EXERCISE n=1 Let X and Y be topological spaces. If V is an open subset of X and W is an open subset of Y, show that V W is an open subset of X Y. If E is a closed subset of X and F is a closed subset of Y, show that E F is a closed subset of X Y. Hint: ( ) ( ) (X Y ) \ (E F ) = X (Y \ F ) (X \ E) Y n=1 n=1 Next, we introduce product mappings. DEFINITION Let X, Y, P and Q be sets. Let f : X P and g : Y Q. Then we define the product mapping f g : X Y P Q by (f g)(x, y) = (f(x), g(y)). PROPOSITION 24 Let X, Y, P and Q be topological spaces. Let f : X P and g : Y Q be continuous mappings. Then the product mapping f g is also continuous. Proof. We show that f g is continuous at every point (x, y) of X Y. Let U be a neighbourhood of (f(x), g(y)) in P Q. Let V and W be neighbourhoods 29

33 of f(x) in P and g(y) in Q respectively such that V W U. Then f 1 (V ) and g 1 (W ) are neighbourhoods of x in X and y in Y respectively. But (x, y) f 1 (V ) g 1 (W ) = (f g) 1 (V W ) (f g) 1 (U) and it follows that (f g) 1 (U) is a neighbourhood of (x, y) in X Y. 2.6 The Diagonal Mapping and Pointwise Combinations DEFINITION Let X be a set. Then the diagonal mapping on X is the mapping X : X X X given by X (x) = (x, x) x X. If X is a metric space it is easy to check that X is an isometry (for the definition, see page 25). In particular, X is a continuous mapping. For X a topological space, we can show that X is continuous. In fact, let x X. We show that X is continuous at x. To do this, let U be a neighbourhood of (x, x) in X X. Then by the definition of the product topology, there exist neighbourhoods V and W of x such that V W U. But then x V W = 1 (V W ) 1 (U). Since V W is a neighbourhood of x, so is 1 (U). This gives us the missing link to discuss the continuity of pointwise combinations. THEOREM 25 Let P, Q and R be topological spaces. Let µ : P Q R be a continuous mapping. Let f : X P and g : X Q also be continuous mappings. Then the combination h : X R given by h(x) = µ(f(x), g(x)) x X is also continuous. Proof. It suffices to write h = µ (f g) X and to apply Theorem 18 and Proposition 24 together with the continuity of X. There are numerous examples of Theorem 25. In effect, the examples that follow are examples of continuous binary operations. 30

34 EXAMPLE Let P = Q = R = R. Let µ(x, y) = x + y, addition on R. Then if f, g : X R are continuous so is the sum function f + g defined by (f + g)(x) = f(x) + g(x) x X. It remains to check the continuity of µ. We have µ(x 1, y 1 ) µ(x 2, y 2 ) = (x 1 x 2 ) + (y 1 y 2 ) x 1 x 2 + y 1 y 2 d R R ((x 1, y 1 ), (x 2, y 2 )) + d R R ((x 1, y 1 ), (x 2, y 2 )) = 2d R R ((x 1, y 1 ), (x 2, y 2 )), so that µ is Lipschitz with constant C = 2 and hence continuous. EXAMPLE Let P = Q = R = R. Let µ(x, y) = xy, multiplication on R. Then if f, g : X R are continuous so is the pointwise product function fg defined by (fg)(x) = f(x)g(x) x X. We check that µ is continuous at (x 1, y 1 ). Observe that so that xy x 1 y 1 = x 1 (y y 1 ) + (x x 1 )y 1 + (x x 1 )(y y 1 ) xy x 1 y 1 x 1 y y 1 + x x 1 y 1 + x x 1 y y 1 Now let ɛ > 0 be given. We choose δ = min(1, ( x 1 + y 1 + 1) 1 ɛ). Then implies that d R R ((x, y), (x 1, y 1 )) < δ xy x 1 y 1 < x 1 δ + δ y 1 + δ 2 ( x 1 + y 1 + 1)δ ɛ. This estimate establishes that µ is continuous at (x 1, y 1 ). We leave the reader to check that addition and multiplication are continuous operations in C. Two other operations on R that are continuous are max and min. We leave the reader to show that these are distance decreasing. 31

35 EXAMPLE One very important binary operation on a metric space is the distance function itself. Let X be a metric space, P = Q = X and R = R +. Let µ(x, y) = d(x, y). We check that µ is continuous. By the extended triangle inequality (page 2) we have and similarly d(x 2, y 2 ) d(x 2, x 1 ) + d(x 1, y 1 ) + d(y 1, y 2 ) d(x 1, y 1 ) + 2d X X ((x 1, y 1 ), (x 2, y 2 )), d(x 1, y 1 ) d(x 2, y 2 ) + 2d X X ((x 1, y 1 ), (x 2, y 2 )). We may combine these two inequalities into one as d(x 1, y 1 ) d(x 2, y 2 ) 2d X X ((x 1, y 1 ), (x 2, y 2 )). This shows that the distance function is Lipschitz with constant C = 2, and hence is continuous. Other examples of continuous binary operations are found in the context of normed spaces. Let us recall that in a normed space (V, ), the metric d V is given by d V (v 1, v 2 ) = v 1 v 2 v 1, v 2 V. We will treat only the case of real normed spaces. The complex case is similar. EXAMPLE In a normed space (V, ), the vector addition operator is continuous. Let µ(v, w) = v + w. We have µ(v 1, w 1 ) µ(v 2, w 2 ) = (v 1 v 2 ) + (w 1 w 2 ) v 1 v 2 + w 1 w 2 d V V ((v 1, w 1 ), (v 2, w 2 )) + d V V ((v 1, w 1 ), (v 2, w 2 )) = 2d V V ((v 1, w 1 ), (v 2, w 2 )), so that µ is Lipschitz with constant C = 2. While the previous example parallelled addition in R, the next is similar to multiplication in R. EXAMPLE In a normed space (V, ), the scalar multiplication operator is continuous. Thus P = R, Q = R = V, and µ : R V V is the map µ(t, v) = tv. We leave the details to the reader. 32

36 EXAMPLE Now let V be a real inner product space. Then the inner product is continuous. Thus P = Q = V, R = R and µ(v, w) = v, w. We check that µ is continuous at (v 1, w 1 ). Observe that v, w v 1, w 1 = v 1, w w 1 + v v 1, w 1 + v v 1, w w 1 so that by the Cauchy-Schwarz inequality (page 8) we have v, w v 1, w 1 v 1 w w 1 + v v 1 w 1 + v v 1 w w 1. Now let ɛ > 0 be given. We choose δ = min(1, ( v 1 + w 1 + 1) 1 ɛ). Then implies that d V V ((v, w), (v 1, w 1 )) < δ v, w v 1, w 1 < v 1 δ + δ w 1 + δ 2 ( v 1 + w 1 + 1)δ ɛ. This estimate establishes that µ is continuous at (v 1, w 1 ). 2.7 Bases and Subbases in Toplogical Spaces DEFINITION A base in a topological space X is a family B of open subsets of X such that every open subset of X can be obtained as a union of some subfamily of B. For the real line, the family of open intervals forms a base. So does the family of bounded open intervals and so does the family of open intervals that have rational endpoints. DEFINITION A basis on a set X is a family B of subsets of X such that the following axioms are satisfied: 1. For each x X there is an element B B such that x B. 2. If B 1, B 2 B and x B 1 B 2, then there exists B 3 B such that x B 3 and B 3 B 1 B 2. 33

37 EXERCISE If B is a basis on a set X, show that there is a topology on X for which B is a base. DEFINITION A subbase in a topological space X is a family S of subsets of X such that the family of all finite intersections of members of S forms a base for X. For the real line a subbase is the given by the sets ]a, [ and ], b[ as a and b run over the reals (or indeed, the rationals). DEFINITION A subbasis on a set X is any collection of subsets of X whose union is the whole of X. EXERCISE If S is a subbasis on a set X, show that the family of all finite intersections of members of S form a basis on X and hence give rise to a topology on X. 2.8 General Product Spaces Let X α be a topological space as α runs over an index set I. Then the product space α I X α is the space of all objects (x α ) α I such that x α X α for all α I. If I has two elements this reduces to the usual product of two spaces. There is a canonical mapping π α : α I X α X α which takes (x α ) α I to x α. It is called a coordinate projection. If each X α has a topology, how can we define a topology on X = α I X α. In effect, what we do is to define the coarsest topology on (x α ) α I that renders all the coordinate projections continuous. A subbasic open set in the product topology has the form π 1 α (U α ) (2.9) where U α is an arbitrary open subset of X α. Note that since we may take U α = X α, in which case πα 1 (U α ) = X, it follows that the sets (2.9) do form a subbasis. In order for π α to be continuous, each such set has to be open in X. Therefore the finite intersections of these subbasic sets, i.e. sets of the form πα 1 (U α ) (2.10) α F 34

38 where F is a finite subset of I and U α is open in X α for every α F can be taken as the basic open sets of the product topology. Finally, a subset of X is open in the product topology if and only if it is an arbitrary union of basic open sets, i.e. an arbitrary union of sets of the form (2.10). EXERCISE Show that if I is countable and each X α is a metric space, then there is a metric on α I X α that gives the product topology. To do this, starting with (X j, d j ) for j N, first define the metric ρ j (t, s) = d j(t, s) 1 + d j (t, s) on X j and show that it gives the same open subsets as d j. Then define a metric on j=1 X j by ρ(t, s) = 2 j ρ(t j, s j ) j=1 and show that the topology associated with this metric is just the product topology. Hint: To do this exercise, one needs the fact that a sequence in j=1 X j converges iff it converges coordinatewise. Coordinatewise convergence is similar to pointwise convergence. In fact, if X j = X for all j, then j=1 X j is just the space of all mappings from N to X and coordinatewise convergence is pointwise convergence. If a sequence converges in the product space, then it must also converge coordiatewise since the coordinate projections are continuous. To show the converse, suppose that a sequence (x n ) n=1 in j=1 X j converges coordinatewise to x. Let U be an open subset of the product space and suppose that x U. We need to show that the sequence lies eventually in U. As explained above, U is a union of basic open sets and x must belong to one of these. Hence we may assume without loss of generality that U is a basic open set. But then U has the form (2.10) with F finite. It s now easy to see that the sequence lies eventually in U because only finitely many coordinates need to be considered. EXERCISE Let X j = {0, 1} with the discrete topology for j N. Then j=1 X j with the product topology is called the abstract Cantor set. Show that the mapping (x j ) j=1 2 j=1 3 j x j maps the abstract Cantor set bijectively onto the standard Cantor set in [0, 1] and that the topologies match (i.e. the given map and its inverse are both continuous). 35