Chapter 6 Completeness Lecture 18 Recall from Definition 2.22 that a Cauchy sequence in (X, d) is a sequence whose terms get closer and closer together, without any limit being specified. In the Euclidean line, every Cauchy sequence converges; but there are many Cauchy sequences in Q R that have no limit in Q. This distinction prompts the following definition. Definition 6.1. A metric space (X, d) is complete if every Cauchy sequence tends to a limit in X. Examples 6.2. 1. The real numbers are complete; this follows from the completeness axiom, which asserts that every bounded above subset S R has a least upper bound, or supremum. In fact R is the unique complete ordered field. 2. Any open ball B 1 (0) in Euclidean n-space is not complete. For the sequence ((1 (1/n), 0,..., 0) : n 1) converges to (1, 0,..., 0) in R n, and is therefore Cauchy by Proposition 2.23; but (1, 0,..., 0) / B 1 (0), so ((1 (1/n), 0,..., 0) has no limit in B 1 (0), by Theorem 2.18. 3. The vertex set V of any path connected graph is complete with respect to the edge metric, because every Cauchy sequence is eventually constant. A particular case of Example 6.2.2 is the open interval ( 1, 1) R. Example 6.2.1 generalises to higher dimensions. Proposition 6.3. Euclidean n-space R n is complete, for every n 1. Proof. Let (x k = (x 1,k,..., x n,k ) : k 1) be a Cauchy sequence in R n, and consider the sequence (x j,k : k 1) R for any fixed 1 j n. Since 42
x j,k x j,l d 2 (x k, x l ), it follows that (x j,k : k 1) is a Cauchy sequence, and therefore tends to a limit a j because R is complete. In other words, there exists N j such that x j,k a j < ɛ/n 1/2 for every k N j. It therefore suffices to confirm that x k a := (a 1,..., a n ) in R n, as k ; this is true because k > max 1 j n N j implies that d 2 (x k, a) = ( n x j,k a j 2) 1/2 ( < n(ɛ 2 /n) ) 1/2 = ɛ. j=1 A similar proof shows that the cartesian product of any finite number of complete metric spaces is complete. It is useful to understand a little more about Cauchy sequences. Lemma 6.4. Suppose that a Cauchy sequence (x n : n 1) in X has a convergent subsequence (x nr : r 1), and that lim r x nr = x; then (x n ) is also convergent, and lim n x n = x. Proof. Since (x n ) is Cauchy, for any ɛ > 0 there exists N such that m, n N implies d(x m, x n ) < ɛ/2. Since x nr x, there exists R such that r R implies d(x nr, x) < ɛ/2. So for n N, choose r R such that n r N; then d(x n, x) d(x n, x nr ) + d(x nr, x) < ɛ/2 + ɛ/2 = ɛ by the triangle inequality, and x n x as required. There are several simple ways to recognise complete metric spaces. Proposition 6.5. A closed subspace Y of a complete metric space X is itself complete. Proof. Suppose that (y n ) is Cauchy in Y. Since X is complete, y n x for some x X; but Y is closed, so x Y by Theorem 2.19. Without the completeness of X, Proposition 6.5 does not necessarily hold. For example, let X be an open interval (0, b) in the Euclidean line, and let Y be the interval (0, a] X for any 0 < a < b. Then Y is closed in X, because its complement (a, b) is open; but Y is not complete, because the Cauchy sequence (1/n : n 1) is not convergent in Y. Proposition 6.6. Any compact metric space X is complete. Proof. Let (x n ) be Cauchy in X. Since X is sequentially compact by Theorem 5.17, there exists a convergent subsequence x nr x in X, as r. Thus x n x as n by Lemma 6.4, and X is complete. The converse of Proposition 6.6 is not generally true; for example, the Euclidean line is complete, but not compact. 43
Examples 6.7. 1. Any finite metric space is compact, and therefore complete. In particular, every subset S V of the vertices of a finite path-connected graph Γ is complete with respect to the edge metric. 2. Because Euclidean n-space is complete, it follows from Proposition 6.5 that any closed ball B r (x) R n is complete. Alternatively, B r (x) is compact by Theorem 5.20, and is therefore complete by Proposition 6.6. Complete subspaces have certain simple properties. Proposition 6.8. Any complete subspace Y X is closed in X. Proof. Let x Y. By Theorem 2.19, there exists a sequence (y n ) in Y such that y n x as n. Thus (y n ) is Cauchy, by Proposition 2.23, and its limit lies in Y because Y is complete. Thus x Y, so Y is closed. Example 6.9. Let Y be the subspace P[0, 1] C[0, 1], and recall from Example 2.21 that Y is not closed. So it is not complete in C[0, 1], by Proposition 6.8. Lecture 19 In our final lecture, we introduce one of the most beautiful theorems of mathematics, which has numerous applications beyond the scope of this course. Definition 6.10. Given any metric space (X, d), a self-map f : X X is a contraction whenever there exists a constant 0 < K < 1 such that (6.11) d(f(x), f(y)) Kd(x, y) for all x, y X. Lemma 6.12. Any contraction f : X X is continuous. Proof. If x X and y B ɛ (x), then d(f(x), f(y)) < Kɛ < ɛ, by (6.11). So f(b ɛ (x)) B ɛ (f(x)) and f is continuous at x by (4.2) with δ = ɛ. A fixed point of a self-map f : X X is a point x X for which f(x) = x. Theorem 6.13 (The Contraction Mapping Theorem). Let X be complete, and f : X X a contraction; then f has a unique fixed point. 44
Proof. To prove that such an f has at least one fixed point, choose x 1 X at random, and define x n+1 = f(x n ) for every n > 1. Then d(x r, x r 1 ) Kd(x r 1, x r 2 )... K r 2 d(x 2, x 1 ) for any r > 2, because f is a contraction. So the triangle inequality gives d(x m, x n ) d(x m, x m 1 ) + + d(x r, x r 1 ) + + d(x n+1, x n ) (K m 2 + + K r 2 + + K n 1 )d(x 2, x 1 ) = K n 1 (K m n 1 + + K r n 1 + + 1)d(x 2, x 1 ) = ( K n 1 (1 K m n )/(1 K) ) d(x 2, x 1 ) < ( K n 1 /(1 K) ) d(x 2, x 1 ) for any m n > 1. But K n 1 0 as n. So d(x m, x n ) < ɛ whenever m n N for suitably large N, and (x n ) is Cauchy. Since X is complete, it follows that x n p for some p X. But f is continuous by Lemma 6.12, so f(x n ) f(p) and f(x n ) = x n+1 p as n. Thus f(p) = p, as required. To prove uniqueness, suppose that f(q) = q; then d(p, q) = d(f(p), f(q)) Kd(p, q). Since K < 1, the only possibility is d(p, q) = 0, whence p = q. Amongst the simpler applications of the Contraction Mapping Theorem are certain familiar approximation pocedures. Example 6.14. Consider the map g : R R of the Euclidean line, defined by g(x) = 1 ( x + 2 ). 2 x This restricts to a self-map of the interval [1, 2], because 1 x 2 1/2 x/2 1 and 1/2 1/x 1, so 1 g(x) 2. If x, y [1, 2] and x y, then y x (y x)/xy = 1/x 1/y 0 because xy 1; so (y x)/2 1/x 1/y (y x)/2 (y x)/2. Hence g(x) g(y) y x /2, and g is a contraction with K = 1/2. Moreover, [1, 2] is complete by Proposition 6.5, because it is a closed subspace of the complete space R. So Theorem 6.13 confirms that g has a unique fixed point p [1, 2]. The fixed point must satisfy Therefore p 2 = 2, and p = + 2. g(p) = (p + 2/p)/2 = p. 45
One of the more remarkable aspects of the proof of the Contraction Mapping Theorem is that x 1 may be chosen at random. In other words, the sequence (x 1, f(x 1 ), f 2 (x 1 ),... ) converges to the fixed point p for every x 1 in X! If Example 6.14 is used to construct an approximation to 2, any x 1 [1, 2] may be selected. Thus x 1 = 1 yields (1, 3/2, 17/12, 577/408,... ) = (1, 1.5, 1.416666666, 1.414215686,... ) whereas x 1 = 7/5 gives (7/5, 99/70, 19601/13860,... ) = (1.4, 1.414285714, 1.414213564,... ) to 9 decimal places. Not bad, when the correct answer is 1.414213562! 46