Definition 2.1. A metric (or distance function) defined on a non-empty set X is a function d: X X R that satisfies: For all x, y, and z in X :

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1 MATH 337 Metric Spaces Dr. Neal, WKU Let X be a non-empty set. The elements of X shall be called points. We shall define the general means of determining the distance between two points. Throughout we shall define concepts prove properties in general, and then apply them specifically to the real line R and to the Euclidean space R n. Definition 2.1. A metric (or distance function) defined on a non-empty set X is a function d: X X R that satisfies: For all x, y, and z in X : (i) d (x, x ) = 0 (ii) d (x, y ) > 0 when x y ; (iii) d (x, y ) = d( y, x) ; (iv) d (x, y ) d(x,z) + d(z, y). With such a function d defined, X is called a metric space. Example 2.1. (a) Euclidean space R n is a metric space with d(u, v) = v u = (v 1 u 1 ) (v n u n ) 2, where u = {u 1,..., u n } and v = {v 1,..., v n } are vectors in R n. Properties (i), (ii), and (iii) of a metric are clearly satisfied, and we have previously proven Property (iv) as a consequence of the Triangle Inequality: For all u, v, w in R n, we have d (u,v) = v u = (v w) + (w u) v w + w u = d(w,v ) + d(u, w) = d(u, w) + d (w, v). (b) A special case of the metric in (a) is the absolute value function which defines a metric on the real line R: For x, y R, we let d (x, y ) = x y. This metric gives the standard way of measuring the distance between two points on the real line. Henceforth, let X be a metric space with distance function d. Definition 2.2. A subset E of X is called bounded if there exists a real number N and a point z X such that d (x, z) N for all x E. Example 2.2. (a) In R, let E = ( 3, 15). (b) In R 3, let S be the sphere of radius 3 that is centered at the point (4, 5, 6). We assert that both E and S are bounded.

2 Indeed, for E = ( 3, 15), let z be the midpoint of the interval: z = ( ) / 2 = 6. Then every point in E is within N = 9 of point z = 6; i.e., for all x ( 3, 15), d (x, 6) = 6 x 9. Thus, E is bounded around point 6 within a distance of N = 9. For Part (b), we have in R 3 S = {(x, y, z) : (x 4) 2 + ( y 5) 2 + (z 6) 2 = 3 2 } { } = (x, y, z) : ( x 4) 2 + (y 5) 2 + (z 6) 2 = 3 = {(x, y, z) : d ((x, y,z), (4, 5, 6) ) = 3}. Thus, S is bounded around point (4, 5, 6) within a distance of N = 3. The following proposition gives more useful characterizations of boundedness on the real line R : Proposition 2.1. A subset E of the real line is bounded if and only if there exists a real number c such that x c for all x E. Proof. Assume E is bounded on R and let z, N R be such that x z N for all x E. Let c = N + z. Then for all x E, x = x z + z x z + z N + z = c. Conversly, suppose x c for all x E. Then d (x, 0) = x 0 c for all x E and E is bounded on R. For example, with E = ( 3, 15) R, it is easier to say that x 15 for all x E, rather than to use the formal metric space definition of bounded as we did in Example 2.2 (a) by saying d (x, 6) = 6 x 9 for all x E. Definition 2.3. Let E be a subset of a metric space X and let ε > 0. (a) An ε -neighborhood about a point x, denoted N ε (x ), is the set of all points in X that have a distance to x that is less than ε : N ε (x ) = { y X : d (x, y ) < ε}. The value ε is called the radius of the neighborhood. (b) A point x is called an interior point of E if there is an ε -neighborhood about x such that N ε (x ) E. (c) Set E is called open if every point of E is an interior point of E. (d) A point p is called a limit point of E if every ε -neighborhood about p contains a point from E that is different from p. (e) Set E is called closed if E contains all of its limit points. (f) Set E is called perfect if E is closed and every point in E is a limit point of E. (g) Set E is called dense if every point in X that is not in E is then a limit point of E.

3 Notes: (i) Due to the axioms of a metric d, we need only consider ε -neighborhoods for ε > 0. (For all ε 0, we would have N ε (x ) = for all x.) (ii) Clearly, every ε - neighborhood is bounded since d(x, y) < ε ε for all y N ε (x). Example 2.3. (a) On the real line, N ε (x ) = {y R : y x < ε} = (x ε, x + ε ). (b) Consider E = [a, b ] on the real line, where a,b R, with a < b. Then a and b are not interior points of E because every ε -neighborhood (a ε, a + ε ) about a and every ε -neighborhood (b ε, b + ε ) about b will go beyond the interval [a, b ]. However, any point x such that a < x < b is an interior point of [a, b ]. Indeed, for any such x, we can find two other real numbers y and z such that a < y < x < z < b. Then let ε = min (z x, x y ). We then have (x ε, x + ε ) ( y,z) [a, b]. (c) From Part (b), the interval [a, b ] is not open on the real line. However, the interval (a, b) is open because every point of (a, b) is an interior point of (a, b). (d) On the real line, the point b is a limit point of the open interval (a, b) because every ε -neighborhood (b ε, b + ε ) about b contains a point from (a, b) between b ε and b. The point a is also a limit point of (a, b). Every interior point x of (a, b) is a limit point of (a, b) because every ε -neighborhood (x ε, x + ε ) about x contains a point from (a, b) that is different from x. (e) We assert that [a, b ] is closed on the real line. From (d), we know that every point in [a, b ] is a limit point of [a, b ]. But suppose x is outside of [a, b ]. If x < a, then there exists a real number z such that x < z < a. If we let ε = z x, then (x ε, x + ε ) = (x ε, z), which is an ε -neighborhood about x that does not contain a point of [a, b ]. That is, x is not a limit point of [a, b ] if x < a. Likewise, x is not a limit point of [a, b ] if x > b. Thus [a, b ] contains all of its limit points; hence, [a, b ] is closed. (f) From (d) and (e), we see that [a, b ] is perfect because [a, b ] is closed and every point of [a, b ] is a limit point of [a, b ]. (g) We assert that the rational numbers Q are dense on the real line R. To prove this, we must show that every x R is either in Q or a is limit point of Q. Therefore, we need only consider an irrational x. But for such an x and any ε -neighborhood (x ε, x + ε ), there exists a rational number z such that x ε < z < x + ε. Thus x is a limit point of Q. Hence, Q is dense in R. Similarly, the irrationals are dense in R. (h) On the real line, [a, b ] is bounded but Q is not. For [a, b ], let z = (a + b) / 2 be the midpoint of the interval and let M = (b a) / 2, which is half of the interval length. Then for all x [a, b ], d (x, z) x z M. Now suppose Q is bounded. Then there exists c R such that x c for all x Q. That is, all rational numbers would be between c and c, which is a contradiction because there are rational numbers between c and c +1.

4 Theorem 2.1. Every ε -neighborhood is an open set. Proof. Let N ε (x ) be an ε -neighborhood about point x. Let y N ε (x ). We must show that y is an interior point of N ε (x ) by finding a neighborhood about y that is contained in N ε (x ). Because d (y, x ) < ε, let the radius about y be = ε d( y, x) > 0. We claim N (y) is contained in N ε (x ). ε x z y ε Let z N (y). Then by Axiom (iii) of a metric, we have d (z, x ) d(z, y) + d(y, x ) < + d(y, x ) = ε d (y, x ) + d( y, x) = ε. Hence, z N ε (x ) and therefore N (y) N ε (x ). Theorem 2.2. Let E be a non-empty finite subset of X. Then E is bounded and E has no limit points. Proof. Let E = {x 1,..., x n }. Because E, we can choose a point x 1 E. Now let M = max{d( x 1, x i ) : 1 i n}. Then d (x 1, x i ) M for all x i E ; hence, E is bounded. Now let x be any point in X. Let ε = 1 2 min{d( x, x i ) : x i E, x i x}. Then for any x i E, with x i x, we have d (x, x i ) > ε. Thus, N ε (x ) contains no point from E except possibly x. Thus x cannot be a limit point of E. Corollary 2.1. A finite set is closed. Proof. Let E be the set of all limit points of the finite set E. By Theorem 2.2, E has no limit points; thus, E = E. So E contains all its limit points which makes E closed. Theorem 2.3. If p is a limit point of a set E, then every ε -neighborhood of p contains infinitely many points of E.

5 Proof. Let N ε ( p) be an ε -neighborhood about p. By definition of limit point, there exists a point y 0 E such that y 0 p and y 0 N ε ( p). Let ε 1 be a smaller radius than ε chosen so that 0 < ε 1 < d( p, y 0 ). Then there exists y 1 E such that y 1 p and y 1 N ε1 ( p). Because d ( p, y 0 ) > ε 1, y 0 N ε1 ( p) ; thus, y 1 y 0. Having chosen y k, we can choose y k +1 E, such that d ( p, y k +1 ) < ε k +1 < d( p, y k ) so that y k +1 y k. We thereby obtain an infinite sequence y 0, y 1, y 2,... of distinct points from E that are in N ε ( p). ε 1 p y 1 y 0 ε Alternate Proof. Suppose N ε ( p) contains only a finite number of points y 1, y 2,..., y n from E that are different from p. Let = 1 2 min{d( p, y 1 ), d( p, y 2 ),..., d( p, y n )}. Then 0 < < d( p, y i ) < ε for 1 i n ; thus, none of the y i are in N ( p). Because < ε, we have N ( p) N ε ( p). Thus, no other points from E can be in N ( p) or else they would also be in N ε ( p). Thus, N ( p) is a neighborhood about p that does not contain any points from E distinct from p, which contradicts the fact that p is a limit point of E. Note: Theorem 2.3 also implies that a finite set has no limit points. Let E c denote the complement of E, which consists of all points in X that are not in E. Theorem 2.4. A set E is open if and only if E c is closed. Proof. Assume E is open. We must show that E c is closed by showing that it contains all of its limit points. So suppose x is a limit point of E c. If x were not in E c, then it would be in E which is open. So then x would be an interior point of E. Hence, there would be an ε -neighborhood about x that is completely contained in E, which would contradict the fact that x is a limit point of E c. So x E c and E c is closed. Now assume that E c is closed. We must show that E is open by showing that every point in E is an interior point. So let x E. Then x E c and x cannot be a limit point of E c because E c is closed. So there must be an ε -neighborhood N ε (x ) about x that does not contain any points of E c. That is, N ε (x ) E, which makes x an interior point of E. Thus, E is open.

6 Example 2.4. On the real line, (a, b) is open; thus, (a, b) c = (, a] [b, ) is closed. In fact, any ray of the form (, a] or of the form [b, ) is closed. These rays are also perfect. Because (E c ) c = E, an immediate result of Theorem 2.3 is: Corollary 2.1. A set E is closed if and only if E c is open. Example 2.5. On the real line, [a, b] is closed; thus, [a, b] c = (, a) (b, ) is open. In fact, any ray of the form (, a) or of the form (b, ) is open. Definition 2.4. The set E is the set of all limit points of E. The closure of E is the set E = E E. Example 2.6. On the real line, the closure of (a, b) is [a, b]. Indeed, from Example 2.3 (d) and (e), every point of (a, b) is a limit point of (a, b), the points a and b are also limit points of (a, b), but no other point is a limit point of (a, b). So the closure of (a, b) is (a, b) = (a, b) (a, b ) = [a, b]. On the other hand, the closure of [a, b] is [a, b]. Indeed, [a, b] is closed; thus it contains all of its limit points: [a, b ] [a, b]. Thus, [a, b] = [a, b] [a, b ] = [a, b]. The following result shows that E is the smallest closed set that contains E. Theorem 2.5. Let E be a subset of a metric space X. (a) E is closed. (b) E = E if and only if E is closed. (c) If E F X and F is closed, then E F. Proof. (a) We will show that E c is open. Let x E c. Then x is not in E and x is not a limit point of E. So there exists an ε -neighborhood N ε (x ) about x that contains no points of E. But we must also show that it contains no points of E. By Theorem 2.1, N ε (x ) is open. If y N ε (x ), then there exists a δ -neighborhood N δ (y) about y such that N δ (y) N ε (x ). So N δ (y) cannot contain any points of E ; thus, y E. So N ε (x ) does not intersect E = E E ; that is, N ε (x ) E c. Thus, E c is open and E is closed. (b) If E is closed, then E E (all limit points of E are in E ); thus, E = E E = E. On the other hand, if E = E, then E E = E E, so E E and E is closed. (c) Suppose E F X and F is closed. Let x E = E E. If x E, then x F by assumption. If x E, then x is a limit point of E which makes it a limit point of F because E F. Thus, x F because F is closed. Hence, E F.

7 We conclude with some more properties of the real line. Theorem 2.6. (a) Let E be a non-empty set of real numbers that is bounded above. Let y = sup E. Then y E. Hence, if E is closed, then y E. (b) Let E be a non-empty set of real numbers that is bounded below. Let x = inf E. Then x E. Hence, if E is closed, then x E. Proof. (a) By the Least Upper Bound Property of the real line, y = sup E exists as real number. If y E, then y E. So assume y E, then consider an ε -neighborhood (y ε, y + ε ) about y. There cannot be any elements from E that are greater than y because y is an upper bound. But if the interval (y ε, y] did not contain an element from E, then y ε would be an upper bound of E which contradicts the fact that y is the least upper bound. Thus, there must be a point from E in (y ε, y]. Because y E, we must have a point from E other than y in every ε -neighborhood (y ε, y + ε ). Hence, y is a limit point of E, which means y E. The proof of (b) is similar. Theorem 2.7. (Bolzano-Weierstrass) Let E be an infinite, bounded set of real numbers. Then E has a limit point in R. Proof. Because E is bounded, there exists a closed interval [a 1, b 1 ], with a 1 < b 1, such that E [a 1, b 1 ]. Because E is infinite, at least one of the intervals [a 1, (a 1 + b 1 ) / 2] or [(a 1 + b 1 ) / 2, b 1 ] still has infinitely many points of E. Choose one such interval and label it as [a 2, b 2 ]. Then [a 1, b 1 ] [a 2, b 2 ], and b 2 a 2 = ( b 1 a 1 ) / 2. Now given [a i, b i ], we construct [a i+1, b i+1 ] such that [a i, b i ] [a i+1, b i+1 ], [a i+1, b i+1 ] has infinitely many points of E, and b i+1 a i+1 = ( b i a i ) / 2 = ( b 1 a 1 ) / 2 i. By the Nested Interval Property of R, [sup{a i }, inf{b i }] [a i, b i ]. We claim that x = sup{a i } is a limit point of E. Let (x ε, x + ε ) be an ε -neighborhood about x. We will find an interval [a i, b i ] that is completely contained in (x ε, x + ε ). Because x ε < x = sup{a i }, x ε cannot be an upper bound of {a i }, so there must exist an a i such that x ε < a i. We choose the smallest integer N such that x ε < a N and (b 1 a 1 ) / 2 N < ε. By construction, b N +1 a N +1 = ( b 1 a 1 ) / 2 N < ε. Hence, x ε < a N a N +1 b N +1 < a N +1 + ε x + ε. So (x ε, x + ε ) contains infinitely many points of E, which makes x E.

8 Exercises 1. (a) On the real line, define d (x, y ) = x y. Prove that d satisfies the axioms of a metric. (b) Explain why each of the following functions do not define metrics on the real line: d 1 ( x, y) = ( x y) 2 d 2 (x, y) = 2y x d 3 ( x, y) = x 2 y 2 2. Let X be an infinite set. Define a function d: X X R by d (x, y ) = 1 if x y. 0 if x = y (a) Prove that d satisfies the axioms of a metric. (b) Let E be any subset of X. (i) Prove that E is open. (ii) Prove that E has no limit points. (iii) Explain why E is also closed. (iv) From (i), every subset of X is open. In particular, E c is open. Therefore, what can you automatically say about E = (E c ) c? 3. Consider the natural numbers ℵ as a subset of the real line R. Prove that ℵ is (a) not bounded (b) not open (c) closed (d) not perfect. 4. Let x and y be points in a metric space X. Prove that if d (x, y) < ε for all ε > 0, then x = y. 5. Let E be the set of all limit points of a set E in a metric space X. Prove that E is closed. 6. Let Eº denote the set of all interior points of a set E. (a) Prove that Eº is open. (b) Prove that E is open if and only if E = E º. (c) If G E, and G is open, then G Eº. (d) Prove that ( E º ) c = E c.

9 More Exercises Let ( X, d) be a metric space. 7. Fix a point x X and let ε > 0. Let E = {y X : d(x, y) > ε } and F = {y X : d(x, y) ε }. (a) Prove that E is open. (b) Prove that F is closed. (c) Is N ε (x) = F? 8. Let E i sets in X. { } i= 1 be a collection of open sets in X and let { Fi } i= 1 be a collection of closed (a) Prove that (b) Prove that E i is open and that F i n n E i is open and that F i is closed. is closed, where 1 n <. (c) Give examples on the real line to show that not be closed. E i may not open and that F i may 9. Let E X. Let F = E E º. (a) Prove that F E c. (b) Prove that F is closed. (c) Prove that (E ) c = (E c )º. 10. Let E, F X. (a) Prove that Eº F º ( E F )º. Give an example on the real line to show that we may not have equality. (b) Prove that E F E F. Give an example on the real line to show that we may not have equality.