Introduction to Topology Randall R. Holmes Auburn University Typeset by AMS-TEX 1 Chapter 1. Metric Spaces 1. De¯nition and Examples. As the course progresses we will need to review some basic notions about sets and functions. We begin with a little set theory. Let S be a set. For A; B ⊆ S, put A [ B := fs 2 S j s 2 A or s 2 Bg A \ B := fs 2 S j s 2 A and s 2 Bg S ¡ A := fs 2 S j s2 = Ag 1.1 Theorem. Let A; B ⊆ S. Then S ¡ (A [ B) = (S ¡ A) \ (S ¡ B). Exercise 1. Let A ⊆ S. Prove that S ¡ (S ¡ A) = A. Exercise 2. Let A; B ⊆ S. Prove that S ¡ (A \ B) = (S ¡ A) [ (S ¡ B). (Hint: Either prove this directly as in the proof of Theorem 1.1, or just use the statement of Theorem 1.1 together with Exercise 1.) Exercise 3. Let A; B; C ⊆ S. Prove that A M C ⊆ (A M B) [ (B M C), where A M B := (A [ B) ¡ (A \ B). 1.2 De¯nition. A metric space is a pair (X; d) where X is a non-empty set, and d is a function d : X £ X ! R such that for all x; y; z 2 X (1) d(x; y) ¸ 0, (2) d(x; y) = 0 if and only if x = y, (3) d(x; y) = d(y; x), and (4) d(x; z) · d(x; y) + d(y; z) (\triangle inequality"). In the de¯nition, d is called the distance function (or metric) and X is called the underlying set. 1.3 Example. For x; y 2 R, set d(x; y) = jx ¡ yj. Then (R; d) is a metric space. ½ 0 x = y; 1.4 Example. Let X be a non-empty set. For x; y 2 X, set d(x; y) = . Then 1 x 6= y: (X; d) is a metric space. (d is called the discrete metric on X.) 1.5 Example. Let X be the set of all continuous functions f :[a; b] ! R. For f; g 2 X, set R b d(f; g) = a jf(t) ¡ g(t)j dt. Then (X; d) is a metric space. 1.6 Example. Let p be a ¯xed prime number. For m; n 2 Z set ½ 0 m = n; d(m; n) = p¡t m 6= n; where m ¡ n = ptk with k an integer that is not divisible by p. Then (Z; d) is a metric space. 2 Exercise 4. Let X be the collection of the interiors of those rectangles in R2 having sides parallel to the coordinate axes. For A; B 2 X, let d(A; B) denote the area of A M B. Prove that (X; d) is a metric space. (Hint: Use Exercise 3.) For sets X1;:::;Xn, de¯ne Yn Xi := f(x1; : : : ; xn) j xi 2 Xig: i=1 This set is called the Cartesian product of X1;:::;Xn. Sometimes it is written X1£¢ ¢ ¢£Xn. Exercise 5. Let Rn := R £ ¢ ¢ ¢ £ R (n factors). For x; y 2 Rn; c 2 R set x + y = (x1 + y1; : : : ; xn + yn); cx = (cx1; : : : ; cxn); x ¢ y = x1y1 + ¢ ¢ ¢ + xnyn; 2 2 1=2 kxk = (x1 + ¢ ¢ ¢ + xn) Prove the following: a. x ¢ (y + z) = x ¢ y + x ¢ z: b. (cx) ¢ y = c(x ¢ y): ° °2 c. x ¢ y · jjxjj jjyjj (Hint: ° kykx ¡ kxky ° ¸ 0. Use the fact that kzk2 = z ¢ z together with (a) and (b) to expand the left hand side.) d. jjx + yjj · jjxjj + jjyjj (Hint: Square both sides and use (c).) £ ¤ n Pn 2 1=2 n Exercise 6. For x; y 2 R , set d(x; y) = i=1(xi ¡ yi) . Prove that (R ; d) is a metric space. (The function d is called the Euclidean metric on Rn.) (Hint: For the triangle inequality, note that d(x; z) = kx ¡ zk. Use Exercise 5(d).) Yn 1.7 Theorem. Let (X1; d1);:::; (Xn; dn) be metric spaces and let X = Xi. For i=1 x; y 2 X, set d(x; y) = maxfdi(xi; yi) j 1 · i · ng. Then (X; d) is a metric space. n n 1.8 Corollary. For x; y 2 R , set ½(x; y) = maxfjxi ¡ yij j 1 · i · ng. Then (R ; ½) is a metric space. The function ½ is called the square metric on Rn. 2. Continuous Functions. 2.1 De¯nition. Let (X; d) and (Y; d0) be metric spaces. A function f : X ! Y is continuous at the point a 2 X if for each ² > 0, there exists a ± > 0 such that whenever x 2 X satis¯es d(x; a) < ±; 3 then f(x) satis¯es d0(f(x); f(a)) < ²: The function f is continuous if it is continuous at each point of X. In the case X = Y = R (usual metric), we have d(x; a) = jx ¡ aj and d0(f(x); f(a)) = jf(x) ¡ f(a)j, so this de¯nition of continuity agrees with the usual de¯nition. 2.2 Example. Given a ¯xed c 2 Y , the constant function f : X ! Y given by f(x) = c (x 2 X) is continuous. ½ 0 x = 0; 2.3 Example. The function f : R ! R given by f(x) = is discontinuous x=jxj x 6= 0; (i.e., not continuous). 2.4 Example. Let (X; d) be a metric space. The identity function f : X ! X given by f(x) = x is continuous. 2 2.5 Example. The function f : R ! R given by f(x1; x2) = x1 + x2 is continuous where R has the usual metric and R2 has the square metric. Exercise 7. Let a; b 2 R. Prove that the linear function f : R ! R given by f(x) = ax+b is continuous. (Hint: For the case a = 0 use Example 2.2.) Exercise 8. Let (X; d) be the metric space de¯ned in Example 1.5. Prove that the R b function F : X ! R given by F (f) = a f(t) dt is continuous where R has the usual metric. Let X; Y; Z be sets and let f : X ! Y and g : Y ! Z be functions. The composition of f and g is the function g ± f : X ! Z given by (g ± f)(x) = g(f(x)). 2.6 Theorem. Let (X; d), (Y; d0), (Z; d00) be metric spaces. If f : X ! Y and g : Y ! Z are continuous, then so is g ± f : X ! Z. Exercise 9. It is shown in calculus that the function f : R ! R given by f(x) = x2 is continuous. Assuming this, prove that the function g : R ! R given by g(x) = x4 +2x2 +1 is also continuous. 3. Limit of a Sequence. 3.1 De¯nition. Let (X; d) be a metric space and let (an) = (a1; a2;::: ) be a sequence of elements of X. An element a of X is called the limit of the sequence (an) if for each ² > 0 there exists a positive integer N such that d(an; a) < ² for all n > N. In this case we say that the sequence converges to a and write limn an = a. In the case X = R (usual metric), we have d(an; a) = jan ¡ aj, so this de¯nition of limit agrees with the usual de¯nition. Exercise 10. Prove that a sequence can have at most one limit. 4 3.2 Example. Let an = 1=n 2 R (usual metric). Then limn an = 0. 3.3 Example. Let (Z; d) be the metric space of Example 1.6 relative to the ¯xed prime n number p. We have limn p = 0. Exercise 11. Let (an) and (bn) be sequences in R (usual metric) and assume that limn an = a and limn bn = b. Prove that limn(an + bn) = a + b. Exercise 12. Let (X; d) be the metric space of Example 1.5 and let fn 2 X be given by fn(x) = x=n. Prove that limn fn = z, where z denotes the zero function (z(x) = 0 8 a · x · b). 3.4 Theorem. Let (X; d) and (Y; d0) be metric spaces. A function f : X ! Y is continuous at the point a 2 X if and only if for each sequence (an) in X converging to a, we have limn f(an) = f(a). 1 + 2n2 + n4 Exercise 13. Prove that lim = 1 by using only what we have shown in n n4 this course. (Hint: Use Exercise 9, Example 3.2, and Theorem 3.4.) 4. Open Sets. Let f : X ! Y be a function. We say that f is injective if f(x1) = f(x2) implies x1 = x2 (xi 2 X), f is surjective if for each y 2 Y , there exists x 2 X such that f(x) = y, f is bijective if it is both injective and surjective. Given A ⊆ X, the subset f(A) := ff(a) j a 2 Ag of Y is called the image of A under f. Given B ⊆ Y , the subset f ¡1(B) := fx 2 X j f(x) 2 Bg of X is called the inverse image of B under f. If f is bijective, then there exists an inverse function f ¡1 : Y ! X which sends f(x) to x, and in this case f ¡1(B) coincides with the image of B under f ¡1 as the notation suggests. However, the inverse image of B under f is de¯ned even when the inverse function f ¡1 is not.