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Chapter 9
The Topology of Metric Spaces
9.1 Abstract Topological Spaces
We undertake a study of metric spaces because we wish to study, among other things the set of continuous functions defined on Rn, and Rn^ is a simple instance of a metric space, which we shall shortly define. The topology of a space is of particular interest to us, because the topology of the space and the set of continuous function defined on that space are intimately connected, as we shall soon see.
Definition 9.1 Let X be a set, and C a collection of subsets of X. Then (X, C) is called a topological space, and the elements of C are called the open sets of X, provided the following hold:
- ∅ ∈ C.
- X ∈ C.
- If X, Y ∈ C then X ∩ Y ∈ C.
- The union of any number (i.e. finite or infinite number) of elements of C is again an element of C. 1
2 CHAPTER 9. THE TOPOLOGY OF METRIC SPACES
Note that by induction, (3) implies that the intersection of any finite number of elements of C is an element of C, or said another way: C is closed under finite intersection, and from (4), C is closed under arbitrary union. Often, having exhibited the topological space (X, C), we will often refer to “an open set O in the topological space X”, understanding that that means O ∈ C.
Example 1 Let X = { 0 , 1 }, that is, a set consisting of two elements. Then if we let C = {{ 0 }, { 1 }, X, ∅}, then (X, C) is a topological space. This is true because (1) and (2) can be verified by inspection. (3) and (4) require that certain subsets of X are elements of C, but we have chosen C to be all subsets of C, which make (3) and (4) hold automatically. This reasoning generalizes to the following example:
Example 2 Let X be an arbitrary set, and C be the set of all subsets of X, including both ∅ and X. Then according to the reasoning of Example 1, (X, C) is a topological space. This is called the discrete topology for X.
Example 3 Let X be arbitrary, and let C = {∅, X}. Then (X, C) is a topological space, and the topology is called the trivial topology.
Example 4 [The Usual Topology for R^1 .] Let X = (−∞, ∞), and let C consist of all intervals of the form (a, b), the arbitrary union of such intervals, and the intersection of any finite number of elements of C. Then (X, C) is a topological space, and the open sets are just the open sets we studied in Chapter 1. To see this, first note that since (1, 2) ∈ C and (3, 4) ∈ C, ∅ = (1, 2) ∩ (3, 4) ∈ C. Further,
⋃ x∈X
(x − 1 , x + 1) ∈ C
which implies that X ∈ C. To prove (3), suppose (a, b) and (c, d) are elements of C. Then if (a, b) ∩ (c, d) = ∅, either (a, b) ⊂ (c, d) or (c, d) ⊂ (a, b) or a < c < b < d or c < a < d < b. In each of these four cases the intersection is an interval.^1 The intersection of two arbitrary (^1) In Exercise 1 you are asked to compute the intersection explicitly, for each of the four cases.
4 CHAPTER 9. THE TOPOLOGY OF METRIC SPACES
Informally, we say: The inverse image under f of every open set in Y is an open set in X. If f : X → Y is continuous we occasionally call f a mapping from X to Y.
Note that whether or not a particular function f is continuous depends upon the topologies, that is, what the open sets are, of both the domain and range. This is an important property of continuity. Later, when we specialize our study of topological spaces to Metric Spaces, we shall see that our − δ definition of continuity and the topological definition of continuity are the same. A useful way to test continuity of a function is given by the following theorem.
Theorem 9.2 f : (X, C) → (Y, C′) is continuous on X if and only if for every x ∈ X and every open set V containing f (x) there exists an open set U containing x such that f (U) ⊂ V.
Proof: If f : X → Y is continuous on X, then since the inverse of every open set in Y is open in X, for any V ∈ C′, f −^1 (V ) is open in X and provides the U. Thus the necessity of the condition is proved. Conversely, suppose that V is open in Y , and y ∈ V. Then for any x ∈ X such that f (x) = y, by hypothesis there exists an open set U in X, containing x, and such that f (U) ⊂ V. Label the above U by Ux to indicate its dependence upon x, and let
O = ⋃ {x:f (x)∈V }
Ux.
Then O is open because it is the union of open sets, and clearly^2 O = f −^1 (V ).
Corollary 9.3 Let f : R^1 → R^1 be any function where R^1 = (−∞, ∞) with the usual topology (see Example 4), that is, the open sets are open intervals (a, b) and their arbitrary unions. Then in R^1 , f is continuous in the − δ sense if and only if f is continuous in the topological sense. (^2) Provide the details. See Exercise 2.
9.1. ABSTRACT TOPOLOGICAL SPACES 5
Proof: Suppose f is − δ continuous, and x 0 ∈ R^1. Let V be an open set in R^1 , and suppose v = f (x 0 ) ∈ V. Since V is open, it is the union of open intervals, and hence v ∈ (a, b) for some a, b. Then let = min{v − a, b − v}, and note that (v − , v + ) ⊂ (a, b). Since f is assumed continuous in the − δ sense, there exists a δ > 0 such that |x − x 0 | < δ implies |f (x) − f (x 0 )|^ < ^ implies^ x^ ∈^ V.^ Letting^ U^ = (x 0 −^ δ, x 0 +^ δ) in Theorem 9.2 shows that^ f is continuous in the topological sense. Conversely, suppose f is continuous in the topological sense. Let x be arbitrary, and y = f (x). Let > 0. Then since V = (y − , y + ) is an open set in the range, by hypothesis there exists an open set U, containing x, in the domain of f , and such that f (U) ⊂ V. But U open implies there exists an interval (a, b), containing x, which is contained in U. That is, z ∈ (a, b) implies f (z) ∈ V. Now let δ = min{x − a, b − x} and observe that if |z − x| < δ, then z ∈ (a, b) and hence f (z) ∈ V. But f (z) ∈ V implies |y − f (z)| < . Thus, f is − δ continuous at x.
Exercise 1 Compute the four intersections in Example 1.
The following two exercise indicate the degree with which continuity is connected to the topology of the spaces involved:
Exercise 2 In the proof of Theorem 9.2, why is O = f −^1 (V )?
The following exercises show the intimate connection between topology and continuity.
Exercise 3 Let X be an arbitrary set with the discrete topology: C is the set of all subsets of X. Let (Y, C′) be an arbitrary topological space. Then f : (X, C) → (Y, C′) is continuous, for any f!
Exercise 4 Let Y be an arbitrary set endowed with the trivial topology: C′^ = {∅, Y }, and (X, C) an arbitrary topological space. Then f : (X, C) → (Y, C′) is continuous, for any f.
Exercise 5 Let X = (−∞, ∞) and the open sets of X be half-open, half-closed intervals of the form [a, b), for a < b, and their arbitrary unions. Let Y = (−∞, ∞), and the open sets
9.2. METRIC SPACES: 7
The major theorem relating closed sets and open sets is the following:
Theorem 9.4 A set A in a topological space (X, C) is closed if and only if its complement, Ac, is open.
Proof: Suppose A is closed, and x ∈ Ac. Then since A contains all its limit points, x is not a limit point of A, that is, there exists an open set O containing x, such that O ∩ A = ∅. Then x ∈ O ⊂ Ac, and by Theorem 9.1 Ac^ is open. Conversely, suppose Ac^ is open. If x ∈ Ac, then since Ac^ ∩ A = ∅, x cannot be a limit point of A. Therefore, all limit points of A are contained in A, that is, A is closed.
Theorem 9.5 The closure of A is closed, for any set A.
Proof: We prove that the complement is open, which in light of Theorem 9.4, is equivalent. Suppose x ∈ (A)c^ = Ac^ ∩ (A′)c. x ∈ A′^ ⇒ x ∈ O, O ∩ A = ∅, O some open set. If O ∩ A′^ = ∅, then there exists some limit point a′^ of A, a′^ ∈ O. But a′^ a limit point of A means that every open set containing a′^ has non-empty intersection with A. But O is an open set that contains a′, and does not intersect A, a contradiction. Hence O ∩ A′^ = ∅. But then x ∈ O ⊂ (A)c, and hence (A)c^ is open.
9.2 Metric Spaces:
9.2.1 Definition of a Metric:
We think of a metric as a way of measuring distance between points in a topological space. A metric has certain properties, which we elaborate below. If X is a set and d(x, y) is a metric on X, then the pair (X, d) is called a metric space.
Definition 9.7 A metric d on a space X is a function d : X × X → [0, ∞) with the following properties:
- d(x, y) ≥ 0 for all x, y ∈ X.
8 CHAPTER 9. THE TOPOLOGY OF METRIC SPACES
- d(x, y) = 0 if and only if x = y.
- d(x, y) = d(y, x).
- d(x, z) ≤ d(x, y) + d(y, z) for all x, y, z ∈ X. (Triangle inequality)
Note: We shall often write d(x, y) as ||x − y|| or occasionally as |x − y|, when doing so will cause no confusion.
Example 6 X = R^1 , and d(x, y) = |x − y|, the usual absolute value on R^1. This is the “euclidean metric” on R^1. It is easy to check that it satisfies all four properties of a metric.
Example 7 X is an arbitrary non-empty set, and
d(x, y) =
1 if x = y 0 if x = y.
It is not difficult^3 to verify that this is a metric! In this metric, all points are “far apart.”
Example 8 X = C[0, 1], the set of continuous functions on [0, 1], and the metric d is the “sup-norm”:
d(f, g) = (^0) ≤supx≤ 1 |f (x) − g(x)| = ||f − g||[0,1].
To verify that this is a metric, properties (1) and (3) are immediate. Property (2) holds because
||f − g||A = 0 ⇐⇒ sup x∈A |f (x) − g(x)| = 0 ⇐⇒ f (x) = g(x) for all x ∈ A ⇐⇒ f (x) ≡ g(x).
Property (4) was proved in Chapter **.
Exercise 10 Prove that the distance function of Example 7 above is a metric.
(^3) See Exercise 10.
10 CHAPTER 9. THE TOPOLOGY OF METRIC SPACES
- If {Oα : α ∈ A} is a family of sets in C indexed by some index set A, then ⋃ α∈A
Oα ∈ C.
Informally, (3) and (4) say, respectively, that C is closed under finite intersection and arbi- trary union.
Exercise 11 Prove Theorem 9.6.
Theorem 9.7 (The ball in metric space is an open set.) Let (X, d) be a metric space. Then for any x ∈ X and any r > 0 , the ball B(x, r) is open.
Proof: Let x ∈ X and r > 0. Let y ∈ B(x, r), and let r 1 = r − ||x − y||. Since ||x − y|| < r, r 1 > 0. We claim that B(y, r 1 ) ⊂ B(x, r). To see this, let z ∈ B(y, r 1 ). Then ||z − y|| < r 1 = r − ||x − y||, so ||x − y|| + ||y − z|| < r. By Triangle inequality for metric spaces, ||x − z|| ≤ ||x − y|| + ||y − z|| < r. But then z ∈ B(x, r). Thus, B(y, r 1 ) ⊂ B(x, r), so by Theorem 9.1, B(x, r) is open.
Continuous Functions on a Metric Space
We remarked earlier, that our notion of a continuous function from R^1 to R^1 in terms of − δ would carry over to metric space, and now we are in a position to state and prove the theorem:
Theorem 9.8 Suppose f : (X, d) → (Y, d′) is a function from one metric space to another. Then f is continuous in the topological sense if and only if for every x ∈ X and > 0 there exists a δ > 0 such that f (B(x, δ)) ⊂ B(f (x), ).
Proof: Suppose f is continuous in the topological sense. Let x ∈ X and > 0. Let V = B(f (x), ). By Theorem 9.2, since V is open in Y , there exists a U open in X such that
9.2. METRIC SPACES: 11
x ∈ U and f (U) ⊂ V. Since U is open in X and x ∈ U there exists a ball centered at x and contained in U. Suppose the radius of this ball is δ. That is, B(x, δ) ⊂ U. But then
f (B(x, δ)) ⊂ f (U) ⊂ V = B(f (x), ),
which proves the theorem in the forward direction.^4 Conversely, suppose O is an open set in (Y, d′), and let x ∈ f −^1 (O). Since O is open, there exists a ball B(f (x), ) ⊂ O for some > 0. By assumption, there exists a δ > 0 such that f (B(x, δ)) ⊂ B(f (x), ) ⊂ O. So, B(x, δ) ⊂ f −^1 (O), which shows that f −^1 (O) is open in (X, d).
Exercise 12 Let X be arbitrary and the metric d be that of Example 7 above:
d(x, y) =
1 if x = y 0 if x = y.
Determine the family C of open sets of (X, d).
Definition 9.11 Let (X, d) be a metric space. Let {xn} be a sequence of elements of X. We say the sequence {xn} converges to x, and write
n^ lim→∞ xn^ =^ x
or xn → x
if and only if
n^ lim→∞ d(xn, x) = lim n→∞ ||xn^ −^ x||^ = 0,
that is, ||xn − x|| → 0. (^4) It is interesting to note that the translation of this last statement is: “for all y, if ||y − x|| < δ then ||f (y) − f (x)|| < .”
9.2. METRIC SPACES: 13
Exercise 18 Let (R^1 , d) be the real numbers with the “usual” distance metric. Determine all the limit points of the set of rationals, Q.
Exercise 19 Prove that in R^2 there exists a countable family B of open balls which form a basis for all the open sets in the topology of the space: every open set O in the topology of R^2 can be written as the (necessarily countable) union of sets from B. Hint: prove that the set of all balls whose centers have rational coordinates, and whose radii are rational, meet the requirements. A topological space with such a countable family B of open sets is called second count- able. (Note that your proof would generalize to Rn^ as well, and hence Rn^ is second countable as well.)
Theorem 9.9 If A and B are closed, then A ∪ B is closed. Hence the union of any finite number of closed sets is closed.
Exercise 20 Prove Theorem 9.9.
Theorem 9.10 If A is an index set^5 , and {Gα : α ∈ A} is a family of closed sets, then the intersection ⋂ α∈A
Gα
is closed. Restated: the intersection of an arbitrary number of closed sets is closed.
Proof: (^) { ⋂ α∈A
Gα
}c = ⋃ α∈A
(Gα)c,
which, by Theorem 9.4, is open. QED.
Definition 9.13 The interior of A, denoted A^0 is defined as follows:
A^0 = {a ∈ A : B(x, ) ⊂ A for some > 0 }. (^5) Just think of A as a set of indices, such as (in the simple case) A = { 1 , 2 , 3 ,... }.
14 CHAPTER 9. THE TOPOLOGY OF METRIC SPACES
Exercise 21 In R^1 with the usual topology, what is the interior of [0, 1]? What is the interior of Q?
Definition 9.14 The exterior of A is defined as the interior of Ac.
Definition 9.15 The boundary of A is the set of points x ∈ X which lie in neither A^0 nor the exterior of A. It is denoted ∂A.
Theorem 9.11 The boundary of A is the set of x ∈ X for which every open set containing x contains both points of A and points of Ac.
Proof: See Exercise 23.
Example 9 In R^2 , let D be the “closed” unit disc^6 ,
D = {(x, y) : x^2 + y^2 ≤ 1 }.
Then D^0 is the “open” disc B((0, 0), 1):
D^0 = {(x, y) : x^2 + y^2 < 1 }.
Also, the exterior of D, (Dc)^0 = {(x, y) : x^2 + y^2 > 1 }.
Finally, the boundary of D,
∂D = {(x, y) : x^2 +^ y^2 = 1}. Proof: Suppose (x, y) ∈ D. If x^2 +y^2 < 1 (so that (x, y) ∈ B(0, 1)), by Theorem 9.7 (“The ball is open”), there exists^7 an^ >^ 0 such that^ B((x, y), )^ ⊂^ B(0,^ 1).^ Hence^ B(0,^ 1)^ ⊂^ D^0. On the other hand, if x^2 + y^2 = 1 and 0 < < 1 /2, ( (^) x 1 − ,^
y 1 −
) ∈ {B(0, 1)}c^ ∩ B((x, y), 2 ) (^6) Exercise 16 justifies the name. (^7) Challenge: can you construct ?
16 CHAPTER 9. THE TOPOLOGY OF METRIC SPACES
Exercise 25 Prove that in a metric space, the distance function is continuous: Let (X, d) be a metric space, and a ∈ X. Define f (x) = d(a, x) = ||x − a||. Then prove that f is continuous: X → R^1.
Example 10 [Closed Graph Theorem] Let f be a continuous real-valued function defined on some interval [a, b]. Then in R^2 , the graph of f is a closed set. Proof: We show that the complement is open. Let (x 0 , y 0 ) be in the complement of the graph of f. Then y 0 = f (x 0 ), and since f is continuous on [a.b], there exists a δ > 0 such that |x − x 0 | < δ implies |f (x) − f (x 0 )| < |y^0 −^2 f (x^0 )|. Then the set { (x, y) : |x − x 0 | < δ and |y − y 0 | < |y^0 −^2 f^ (x^0 )|
}
contains no points of the graph of f. This set is a rectangle centered at (x 0 , y 0 ). Thus there is a ball centered at (x 0 , y 0 ) contained in the rectangle, which therefore is in the complement of the graph of f , and hence the graph of f is closed.
Euclidean Space: Rn
Let Rn^ = R^1 × R^1 ×... × R^1 , where we take the Cartesian product of R^1 with itself n times. The objects x ∈ Rn^ are n-tuples: x = (x 1 , x 2 ,... , xn) where each xi ∈ R^1. The Euclidean distance between points in Rn^ is given by
d(x, y) =
( (^) ∑n i=
(xi − yi)^2
) 1 / 2 .
It can be verified that this distance function is a norm. Triangle inequality follows from an inequality which we won’t prove here, called the Cauchy-Schwartz inequality.
Complete Metric Spaces
Definition 9.17 We say a metric space (X, d) is complete if every Cauchy sequence (in the metric d) converges to some element of the space.
9.2. METRIC SPACES: 17
Compactness; Compact Sets
Definition 9.18 Let A be a set in a topological space X. A cover of A is a family F of subsets of X with the property that the union of the sets in F contains A:
A ⊂ ⋃ B∈F
B.
A cover is called an open cover if every set B in the family F is an open set.
Definition 9.19 Suppose F is a cover of the set A. A subcover of F is a family F ′^ ⊂ F , which is also a cover of A. If, in addition, F ′^ contains only finitely many sets, we call it a finite subcover.
Definition 9.20 A set A in a topological space (X, C) is called compact if every open cover of A has a finite subcover. If X itself is compact, we say X is a compact space.
Example 11 N is not compact in R^1. To see this, let On = (n − 1 / 4 , n + 1/4), for n = 1, 2 ,.... Then N ⊂ ∪∞ n=1On, since n ∈ On for every n ∈ N. But if N were compact, finitely many of the On would cover N. But each On contains only one integer! Hence the natural numbers would be finite, a contradiction.
Example 12 A finite set in R^1 is compact.
Suppose the finite set E = {x 1 ,... , xk} is covered by ∪αOα. Then xi ∈ Oαi , 1 ≤ i ≤ k, and hence the finite collection {Oα 1 ,... , Oαk } covers E.
Theorem 9.12 If f is a continuous mapping from the topological space (X, C) to the topo- logical space (Y, C′) and A ⊂ X is compact, then f (A) is compact in (Y, C′).
Proof: To show f (A) is compact, suppose O is an open cover of f (A). We need to produce a finite subcover. Since each set B ∈ O is open, f −^1 (B) is open in X. Then
A ⊂ ⋃ B∈O
f −^1 (B) (why?),
9.2. METRIC SPACES: 19
Exercise 27 Let X be an arbitrary set and C be the set of all subsets of X. Let C′^ = {∅, X}. Then (X, C) is the discrete topology, while (X, C′) is the trivial topology. Prove that if X consists of at least two points, then the identity map i : (X, C′) → (X, C) is NOT continuous.
Theorem 9.15 In a metric space, a compact set is closed.
Proof: Let A be compact, and a ∈ A. We show that a is not a limit point of A, and hence A is closed. Let En =
{ x ∈ X : d(a, x) > (^) n^1
} n = 1, 2 ,...
and observe that each En is open and
A ⊂ ∪∞ n=1En.
But A compact and the En’s are nested implies that
A ⊂ En 0 ,
for some n 0. Since every point of A is at distance greater than 1/n 0 from a, B(a, 1 /n) is a ball about a which does not intersect A. Hence a is not a limit point of A.
Theorem 9.16 In the usual topology on R^1 , a closed bounded interval [a, b] is compact.
Proof: Suppose we have proven that [0, 1] is compact in the usual topology on R^1. Then the map f (x) = a + (b − a)x is continuous (for any number of reasons), and hence the image of [0, 1], namely [a, b], is compact in the usual topology on R^1. We are thus reduced to considering the case of [0, 1]. Let ⋃ α Oα be an open cover of [0, 1], and suppose there exists no finite subcover. Let
E = {x ∈ [0, 1] : the closed interval [0, x] can be covered by a finite number of the open sets Oα}
and note that 0 ∈ E. Let s = sup E. It follows from the Approximation Theorem for sup’s that for any x < s, the closed interval [0, x] can be covered by a finite number of the open sets. But s ∈ [0, 1) implies that s ∈ Oα 0 for some α 0 , and since R^1 is a metric space, s ∈ Oα 0
20 CHAPTER 9. THE TOPOLOGY OF METRIC SPACES
implies that for some > 0, (s − , s + ) ⊂ Oα 0. But then from the remarks about the Approximation Theorem, it follows that [0, s − /2] can be covered by a finite number of the open sets in the cover, which, together with Oα 0 , provide a finite subcover of [0, s + ], a contradiction.
Theorem 9.17 (Heine-Borel Theorem) A set in R^1 is compact if and only if it is closed and bounded.
Proof: We have already proved that a compact set is closed and bounded. The converse will be proved if we prove the following theorem, since a closed and bounded set in R^1 is a closed subset of an interval [a, b].
Theorem 9.18 A closed subset of a compact set is compact.
Proof: Let G be a closed subset of the compact set C. Let {Oα} be an open cover of G. Then the family {Oα}, together with Gc, form an open cover of X, and hence an open cover of C. Hence finitely many elements of the cover cover C. Since G is a subset of C, this finite subcover also covers G. Gc^ is not needed in the subcover to cover G since it is disjoint from G. Hence finitely many of the {Oα} cover G.
Exercise 28 In the proof of Theorem 9.16, what happens to the argument if s = 0? Why must s > 0?
Theorem 9.19 (Bolzano-Weierstrass) Let A be a compact subset of a metric space X. Then every sequence of elements of A has a convergent subsequence (to an element of A).
Proof: Let A be compact and {xn} be a sequence in A. Case I: Suppose there exists an a ∈ A such that for every r > 0, B(a, r) contains xn for infinitely many n. Then for each k ∈ N there exists^10 an nk so that n 1 < n 2 <... < nk and |a − xnk | < (^1) k. (^10) Provide the details. Some care needs to be taken so that {xnk } is a subsequence of {xn}.
