2 Metric Spaces Definitions Exotic Examples... 3

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1 Contents 1 Vector Spaces and Norms 1 2 Metric Spaces Definitions Exotic Examples Topologies Open Sets Examples Bases Metric Space Topology Continuity 6 5 Additional Topological Notions 8 1 Vector Spaces and Norms Recall that a real vector space is a set V equipped with with two operations, vector addition and scalar multiplication. Formally, these are maps from V V V and from R V V, respectively. The two operations must satisfy the following properties (for all x, y V and for all r, s R): (i) (V, +) is an abelian group (ii) r(x + y) = rx + ry (iii) (r + s)x = rx + sx (iv) r(sx) = (rs)x (v) 1x = x In the above definition we can replace R with any other field such as the complex numbers C or even the binary numbers Z 2 to obtain a complex vector space, a Z 2 vector space, and so on. We won t consider any of these, so from here on vector space will mean a real vector space. Examples. The following are all vector spaces. 1. R n with the familiar addition and scalar multiplication of n-tuples (x 1, x 2,..., x n ). 2. Mat n m (R), the set of n m matrices with real coefficients. 3. C([0, 1]), the set of continuous, real valued functions on the interval [0, 1]. 4. R[x], polynomials in one real variable x with coefficients in R. If V is a vector space, then a norm on V is a function : V R such that (i) x 0 (ii) x = 0 if and only if x = 0 1

2 (iii) rx = r x (iv) x + y x + y (the Triangle Inequality) A vector space equipped with a norm is called a normed space. Euclidean space R n is a familiar normed space with norm x = (x 1, x 2,..., x n ) = ( x x 2 ) 1/2 n which is the length of x in R n. Not every vector space is necessarily a normed space, and it s possible that different normed spaces can have the same underlying vector space. 2 Metric Spaces 2.1 Definitions A metric space is a set X together with a metric or distance function d: X X R such that, for all x, y, z X: (i) d(x, y) = d(y, x) (ii) d(x, y) 0 (iii) d(x, y) = 0 if and only if x = y (iv) d(x, y) d(x, z) + d(z, y) (the Triangle Inequality) Metric spaces and distance functions are related to normed spaces. If V is a normed space, we can turn V into a metric space by defining d(x, y) = y x In R n, if x = (x 1, x 2,..., x n ) and y = (y 1, y 2,..., y n ), then ( n ) 1/2 d(x, y) = (y i x i ) 2 i=1 When n = 1 this reduces to the ordinary absolute value d(x, y) = y x and when n = 2 to the familiar distance function in the plane (really just the Pythagorean Theorem): d(x, y) = (y 1 x 1 ) 2 + (y 2 x 2 ) 2 2

3 2.2 Exotic Examples You might think that introducing a new definition and notation d is unnecessary, seeing that we have the norm available. Why not just keep writing y x when we want to calculate a distance? If we were only working in Euclidean space R n you d be right. But not every metric space is a vector space. Vector spaces have algebraic structure i.e. addition and scalar multiplication that tell us how to combine vectors algebraically in V to get new vectors. A metric space is just a set, whose elements we call points, equipped with a distance function. It doesn t make sense to add points together. Even when V is a vector space the situation is not simple. We have seen that a norm on V gives rise to a metric d by setting d(x, y) = y x. Is the converse true? Given a metric d on a vector space V, is it possible to define a norm by defining the length of a vector to be the distance to 0, i.e. x = d(x, 0)? For some metrics this works, but for others it does not. Consider the following examples: Example If p is an integer, p 1, define a distance function on R 2 by The distance from x to 0 defines the p-norm: d(x, y) = ( x 2 x 1 p + y 2 y 1 p ) 1/p x p = ( x 1 p + x 2 p ) 1/p Example When p = 2 the p-norm is the same as the Euclidean norm. Example When p = 1 the distance and norm are given by d(x, y) = x 2 x 1 + y 2 y 1 and x 1 = x 1 + x 2 The 1-norm is also called the taxicab norm, and the associated metric the taxicab distance. Example Define a distance function on R 2 by d(x, y) = max( x 2 x 1, y 2 y 1 ) This metric gives rise to a norm called the infinity norm, x = max( x 1, x 2 ). Example Define a distance function on R 2 by { 1 if x y d(x, y) = 0 if x = y This distance function does not define a norm on R 2. Each of these three examples can be extended in an obvious way to n dimensional space R n. Problems 1. Verify that the 1-norm and the associated distance function satisfy the definitions. 2. Verify that the -norm and the associated distance function satisfy the definitions. 3

4 3. If V is a normed vector space, define the unit circle of V to be the set of vectors of norm one: unit circle = {x V x = 1} Describe the unit circles on the normed spaces (R 2, 1 ), (R 2, ), (R 3, 1 ) and (R 3, ) 4. Verify that the metric of Example satisfies the definition. Why can t we use this metric to define a norm on R 2? Which property of the definition goes wrong? 3 Topologies Our goal is to give the proper definition of continuous function, which you will see is extraordinarily simple once we ve developed some technical baggage, mainly in the form of definitions. 3.1 Open Sets Given a set X, a topology T on X is a collection of subsets of X such that: (i) X T (ii) T (iii) If {U α α I} is a family of sets in T, then the union U = α I U α is also in T. (iv) Finite intersections of sets in T are in T The elements of the set X are called points and the sets in T are called the open sets for the topology. You should not make any advance assumptions about the exact nature of the open sets. The only requirement is that the family of sets satisfies the axioms above. There are many strange and arbitrary constructions some quite useful where the open sets are completely anti-intuitive. The phrase arbitrary union in (iii) above deserves some clarification: By an arbitrary collection of open sets, we mean that it is possible that the collection may uncountably infinite. In this case it would be inapprpriate to use the notation i U i for the infinite union, because this implies that the sets are countable, i.e. that they can be denumerated U 1, U 2, U 3 etc. Instead we use an index set I which may be uncountable, and for each α I there corresponds exactly one U α. 3.2 Examples Before moving on to metric spaces, and R n in particular, here are several examples of strange and unusual topologies. The Discrete Topology. Let X be a set and let P(X) be the power set of X, the set of all subsets of X. The discrete topology is obtained by setting T = P(X), i.e. by declaring every subset to be an open set. The name discrete refers to the fact that every point of X is itself an open set. 4

5 The Indiscrete Topology. If X is a set, then the indiscrete topology consists of the subsets T = {, X}. So the discrete topology on X has as many open sets as possible, and the indiscrete topology has only two (which is a few as possible). Sierpinski Space. Let X be a set with two elements, X = {a, b}. Then the set of subsets T = {, {a}, X} defines a topology on X. Zariski Topology. Let X be the real numbers R. If A is a finite set of points in R, then we declare the complement R \ A to be a open set. The collection of all such sets, together with the empty set and R, define a topology on R which is called the Zariski topology. 3.3 Bases A subset B T is called a basis for the topology if every set in T can be written as a union of sets in B. Theorem Let B T. Then the following are equivalent: (i) B is a basis for T. (ii) For every open set U, and every x U, there exists a B B such that x B U. Proof. (i) (ii) Assume B is a basis, and let U T, x U. Since U = B α for some collection of basic open sets, x B α for at least one these. (ii) (i) Let U T. By assumption there exists a B x B with x B x U for every x U. I claim U = x U B x. Note that U x U B x, since every x U is contained in one of the B x s. Conversely, U x U B x, since each B x U. Theorem Suppose B is a basis for T, and A is any subset of X. Then A is open if and only if for every x A, there exists a B B such that x B A. Proof. ( ) Theorem 3.3.1(ii) ( ) Suppose for every x A, there exists a B x B such that x B x A. Then A = x A B x, by the same reasoning as in the previous theorem. Since basic sets are open by definition, A is a union of open sets; therefore A is open. Suppose you are handed a collection of sets B, and you would like to know if there exists a topology T on X such that B is a basis for T. This is not possible for every collection B, but there is a technical condition which I won t go into here that will insure such a topology exists. When it does, we say that B generates T. 3.4 Metric Space Topology Let (X, d) be a metric space. If x X, we define the open ball of radius r, centered at x to be the set B r (x) = {y X d(x, y) < r} The phrase open ball is justified, because the collection of all open balls B = {B r (x) x X, r R, r > 0} generates a topology on X, called the metric space topology. The open balls in a metric space give an easy test to see whether a given set is open: 5

6 Theorem In a metric space X, a set U is open if and only if for every x U, there exists an r > 0 such that B r (x) U. We will mainly be interested in the special case where X is Euclidean space R n (i.e. the set R n topologized by the Euclidean metric): Theorem In R n a set U is open if and only if for every x U, there exists an r > 0 such that B r (x) U. These two results are essentially a restatement of Theorem 3.3.2, but with a subtle difference: In Theorem 3.3.2, we are given a point x in U, and need to find a basic open set such that x B U. In Theorem we re not looking for just any open ball; the ball has to be centered at x. In any case the proofs are identical. Note that the textbook uses Theorem as the definition of an open set. Problems 1. Prove that Zariski topology on R satisfies the definition of a topology. 2. Assume, as in the textbook, that an open set in R n is defined by the property in Theorem Prove that the collection of all such open sets satisfies the formal definition of a topology. 4 Continuity Definition Let X and Y be topological spaces. A function f : X Y is continuous if f 1 (W ) is open in X for every open set W in Y. Definition Suppose f : X Y is a function with f(x 0 ) = y 0. We say f is continuous at x 0 if for every open neighborhood W (y 0 ), there exists a neighborhood U(x 0 ) such that f(u) W. If the topologies of X and Y are generated by bases B X and B Y, then we can replace U and W in both definitions above by basic open sets. In particular, when X and Y are metric spaces, the definition for continuity at a point reduces to a more familiar ɛ-δ form. Here s how the definition would look for maps from R n to R m : Definition Suppose f : R n R m is a function with f(x 0 ) = y 0. We say f is continuous at x 0 if for every ɛ > 0, there exists a δ > 0 such that f(b δ (x 0 )) B ɛ (y 0 ). Recall that the way continuity was presented in calculus: we first defined continuity at a point, and then declared f to be continuous on an open interval I if f was continuous at each point in I. This calculus definition is compatible with Definition and Definition because of the following. Theorem Let X and Y be topological spaces, f : X Y a function. Then the following are equivalent: (i) f is continuous. 6

7 (ii) f is continuous at x for every x X. One of the problems with the way continuity is defined in calculus is that the definitions don t work for arbitrary domain sets D. You may recall that we went to a lot of trouble to define continuity from the left and from the right when D was a closed interval. But what if D = Q? Or what if D is a single point? The situation is even more complicated for maps f : D R n R m. In order to talk about the continuity of such a map, we need to declare what an open set is in D. That is, we need to topologize D. Definition Suppose X is a topological space, and D is a subset of X. The we can give D a topology called the subspace topology by defining the open sets of D to be the sets of the form D U, for U an open set in X. Such a set is called relatively open. This definition is really a theorem as well, because we need to prove that the open sets in D are in fact a topology on D. Theorem If D X and B X is a basis for X, then B D = {B D B B X } is a basis for the subspace topology of D. As an example, consider the case where D = [0, 1], the closed unit interval in R. In the subspace topology, half open intervals of the form (a, 1] are open in D. Furthermore, if we are given a function f : D R, then f is continuous at the closed endpoint x = 1 so long as the inverse image of an open neighborhood of f(1) is an open neighborhood of 1, in the subspace topology on D. We can use basic open sets and rephrase this in terms of ɛ s and δ s: f is continuous at 1 if for every ɛ > 0, there exists a δ > 0, such that f((1 δ, 1 ]) (f(x) ɛ, f(x) + ɛ). This is the same as the definition you used to call continuity from the left. Problems 1. Let f : R R be the piecewise defined function { x + 1 if x 1 f(x) = 0 if x < 1 We know from calculus that this function is not continuous, but lets experimant with the definitions to see exactly what goes wrong. a. Let W be the open interval ( 1, 1) in the codomain of f. What is f 1 (W )? Is this set open or closed? b. What if W = ( 3, 2)? c. What if W = ( 1, 3)? d. What if W = (1, 3)? 2. Let f : R R be given by Is f continuous at x = 0? Explain. f(x) = { x if x Q 0 if x / Q 7

8 3. Let g : [0, ) R be given by g(x) = { x if x > 0 2 if x = 0 Explain why g is not continuous at x = D = {x 1,..., x n } be a finite collection of points in R n. What are the open sets of D in the subspace topology? Explain why a map from D to R is always continuous. 5. If f : X Y is a map of topological spaces, and the topology on X is the discrete topology, is f necessarily continuous? 5 Additional Topological Notions Let X be a topological space. The complements of the open sets in X are called the closed sets of X. In general a set may be neither open nor closed. Note that X itself and the empty set are both open and closed at the same time. These may not be the only subsets with this property. In the discrete topology, for example, every set is both open and closed. In some books a set which is both open and closed is called clopen. Theorem Let X be a topological space. Then: (i) Arbitrary intersections of closed sets are closed. (ii) Finite unions of closed sets are closed. Proof. Apply De Morgan s Laws to the corresponding statements for infinite unions and finite intersections of open sets. Let A be a subset of X. The interior of A, denoted int A, is the union of all open sets contained in A. Clearly A is open if and only if A = int A. The closure of A, denoted cl A, is the intersection of all closed sets which contain A. A is closed if and only if A = cl A. Example In any space the closure of the empty set is the empty set. Example In Euclidean space R n, the closure of an open ball B r (x) is the closed ball B r (x) = {y d(x, y) r} In other metric spaces this may be false. In general all you can say is cl B r (x) B r (x). It s possible for the interior of A to be the empty set, and for the closure to be all of X. A set with this second property, i.e. a set such that cl A = X, is called dense. Example The rational numbers Q are dense in R. Example In the Zariski topology on R, every non-empty open set is dense. 8

9 Let A be a subset of a topological space X. A point x X is a limit point (aka cluster point or accumulation point) of A if every open neighborhood of x contains at least one point of A besides possibly x itself. The derived set of A, denoted A, is the set of all limit points of A in X. Note that a limit point of A may or may not be an element of A (i.e. A A in general). If X is a metric space then you can substitute open ball of radius r in place of open neighborhood in the above definition. A point x is an isolated point of X if the set {x} is open. If x A X, then x is an isolated point of A if x is isolated in the subspace topology. For example, if A is a subset of R n, then x A is isolated if there is an open ball B r (x) such that B r (x) A = {x}. Theorem A set A is closed if and only if A contains all of its limit points. The boundary of a set A, denoted A, is the set of points in the closure of A which are not in the interior, i.e. A = cl A \ int A. Theorem Let X be a topological space, A a subset of X. Then the following are equivalent: (i) x A (ii) x cl(a) cl(x \ A) Problems 1. If A B and x is a limit point of A, must x also be a limit point of B? 2. True or False: cl(a B) = cl A cl B. 3. True or False: A A. 4. Prove that if X is a topological space with the discrete topology, then X has no limit points. 5. Prove Theorem