Notes on metric spaces and topology. Math 309: Topics in geometry. Dale Rolfsen. University of British Columbia

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1 Notes on metric spaces and topology Math 309: Topics in geometry Dale Rolfsen University of British Columbia Let X be a set; we ll generally refer to its elements as points. A distance function, or metric on X is a function which satisfies d(x, y) = d(y, x) d(x, y) 0, and = 0 x = y d(x, z) d(x, y) + d(y, z) d : X X R (symmetry) (positive definite) (triangle inequality). The distance function is basic to both geometry and topology. In geometry, an isometry between metric spaces (X, d X ) and (Y, d Y ) is a bijective function f : X Y satisfying d Y (f(x), f(y)) = d X (x, y). A note on the jargon, a function f : X Y taking a set X (the domain) to a set Y (the target) is injective (or one-to-one) if f(x 1 ) f(x 2 ) unless x 1 = x 2, surjective (or onto) if for every y Y, there is an x X such that f(x) = y, bijective if both injective and surjective. If f is a bijective function, then it has an inverse function, f 1 : Y X, which of course satisfies f 1 (y) = x if and only if f(x) = y. If U X, then f(u) denotes the set {f(x) Y x U}. The range of f is f(x), a subset of Y, which is all of Y if f is surjective. If V Y, f 1 (V ) is the set {x X f(x) V }. This notation is used even if f does not have an inverse function f 1. 1

2 In topology, one allows equivalences which are not as rigid as in goemetry. A topological equivalence X Y is only required to be continuous and have a continuous inverse. To make this precise, we obviously need to define continuity. The definition one learned in calculus (slightly generalized to metric spaces) is that f : X Y is continuous if for each x X and ɛ > 0 there exists δ > 0 such that y X and d X (x, y) < δ d Y (f(x), f(y)) < ɛ. There is a less tedious way of stating the same thing, using the idea of open set, which is a fundamental ingredient to topology. In a metric space (X, d), the ɛ-neighbourhood of a point x X is defined as N ɛ (x) = {y X : d(x, y) < ɛ}. Define a subset S X to be open if for every x S there exists ɛ > 0 such that N ɛ (x) S. That is, every point of S has some ɛ-neighbourhood completely inside S. A set S is said to be closed if its complement X \ S is open. Examples: with X = R and the usual metric d(x, y) = x y, if a < b: the open interval (a, b) is open and the closed interval [a, b] is a closed set, the set Z of integers is closed, the set Q of rational numbers is neither open nor closed. A moot point is that the empty set is regarded as being open, as the condition for every x... is vacuously satisfied. Problem T1: Let T denote the collection of all open subsets of X, as defined using a given metric d. Then: (i) T and X T, (ii) if U T and V T then U V T and (iii) if {U α } is any collection with U α T, then α U α T. 2

3 The more abstract notion of a topological space is simply a pair (X, T ) consisting of a set X of points and a collection T of subsets of X satisfying (i) to (iii) above. If U T, then U is said to be an open set (in that topology). If the underlying topology T is understood, then we may simplify notation by calling X a topological space. Problem T2: Suppose (X, d X ) and (Y, d Y ) are metric spaces and f : X Y is a function. Then f is continuous (as defined above) if and only if f 1 (U) is open in X whenever U Y is an open set in Y. Thus we say that a function is continuous provided preimages of open sets are open. If X, Y, Z are topological spaces (with topologies not necessarily coming from a metric) and f : X Y and g : Y Z are continuous functions, then it is clear from the definition that the composite g f : X Z is also continuous. A bijective function f : X Y, between topological spaces is called a homeomorphism, or topological equivalence if both f and its inverse are continuous. This means that f also provides a bijection between the open sets of the two spaces, and the spaces are regarded as the same topologically. A coffee cup and doughnut are famous examples. New spaces from old: subspaces: If (X, T ) is a topological space, and Y X, then Y has the subspace topology T Y := {U Y U T }. products: Suppose (X 1, T 1 ) and (X 2, T 2 ) are topological spaces. The product X 1 X 2 := {(x 1, x 2 ) x i X i } is the set of ordered pairs, and is given the product topology, which is the smallest topology for which U 1 U 2 is open, whenever U i is open in X i. In other words, a set V X 1 X 2 is open if for every point x = (x 1, x 2 ) in V, there are open sets U 1 T 1 and U 2 T 2 with x U 1 U 2 V. A product of several spaces X 1,..., X n, denoted X 1 X n, and its 3

4 product topology are defined similarly. One may also define a product of infinitely many spaces, but we will not consider those at this point. A common metric space we ll be dealing with is R n = {x = (x 1,...,, x n )}, the set of ordered n-tuples x of real numbers, with the metric d(x, y) = n (x i y i ) 2. i=1 Problem T3: (a) Verify that this definition satisfies the metric conditions. (b) Show that a subset of R 2 is open relative to the metric d if and only if is open in the product topology, if we consider R 2 = R R and R has its usual topology. (c) Do the same for R n where R n is considered the product of n copies of the real line. Some common objects in R n : The origin is the point 0 = (0,..., 0) and the unit sphere is the set The unit n-cube in R n is the set S n 1 = {x R n d(0, x) = 1}. [ 1, 1] n = {(x 1,..., x n ) 1 x i 1, i} The vertices (or corners) are the points v = (v 1,..., v n ) with each v i = ±1 there are exactly 2 n vertices. Two vertices v and v are joined by an edge exactly when their coordinates disagree in exactly one place. How many edges in the n-cube? Quotients: This is an idea which makes precise the idea of attaching parts of a space, or several spaces, together. If X is a set, let P be a disjoint collection of subsets of X whose union P is all of X. Such a collection is sometimes called a partition of X, and it defines an equivalence relation where x x means that x and x belong to the same set in P. Let Y denote the set whose points are the sets in P (a.k.a. equivalence classes). There is a natural projection function p: X Y which takes x X to 4

5 the set in P containing it recall that this is a point of Y. If X has a topology, say T, then we define the quotient topology on Y to be the largest topology on Y such that p is continuous. This means that a subset V of Y is open exactly when p 1 (V ) is open in X. It is routine to check that this really defines a topology on Y. This topological space is sometimes denoted Y = X/. This quotient space is also sometimes called an identification space. In class we discussed many examples. For example, if one glues the endpoints of the interval [0, 1] together, we obtain a space homeomorphic with a circle. Here, the partition P consists of the set { 1, 1} together with the singleton sets {t} where 1 < t < 1. The proof that [ 1, 1]/ is homeomorphic with S 1 is provided by considering the function h : [ 1, 1] R 2 defined by h(t) = (cos t, sin t), this wraps the interval around S 1 in a continuous way. The inverse takes the point 1 S 1 to the pair of points { 1, 1}, which is a single point of [ 1, 1]/, and it is easy to check that open sets correspond, too. In a similar way we can build a cylinder S 1 [ 1, 1] from the square [ 1, 1] [ 1, 1] by identifying (t, 1) (t, 1) for all t [ 1, 1] Here P is the set of all such pairs, together with all other singleton sets in the square. The Möbius strip is constructed from the square [ 1, 1] [ 1, 1] by identifying (t, 1) ( t, 1) for all t [ 1, 1]. The torus (surface of a donut) is S 1 S 1, and it may be constructed from the square by the identifications (t, 1) (t, 1) and ( 1, s) (1, s). Note that in this case the four points (±1, ±1) become the same point in the quotient space. Connected spaces. Let S X, where (X, T ) is a topological space. Then S is said to be disconnected (in X) if there are open sets U and V in X such that S U and S V are nonempty and disjoint and S U V. Otherwise, S is said to be connected (in X). In particular, X itself is a connected space if and only if it is NOT the union of two disjoint nonempty open sets. Note that S is connected in X if and only if S is a connected space with the subspace topology so connectedness is an intrinsic property. Problem T4: Show that if f : X Y is a continuous mapping of topological spaces and X is connected, then the image f(x) is connected. 5

6 Note that R, with the usual metric topology, is connected. If a, b R with a < b, it s clear that any connected set S containing both a and b must contain the interval [a, b]. If there were some c [a, b] with c not in S, then the two open sets (, c) and (c, ) would show that S is disconnected. One may prove the intermediate value theorem of calculus by a connectedness argument, simply by noting that the image f([a, b]) must be connected. Intermediate Value Theorem: If a < b R and f : [a, b] R is continuous, then, given C between f(a) and f(b) there exists c [a, b] such that f(c) = C. Problem T5: Suppose A and B are connected subsets of X, and that A B is nonempty. Show that A B is connected. Is the empty set connected? 6