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Cauchy Sequences 1 The Definition. I will introduce the main idea by contrasting three sequences of rational numbers. In each case, the universal set of numbers will be the set Q of rational numbers; all irrational numbers are temporarily banished. n−1 X 1 k The sequence (xn ). Let xn = . Since 2 k=0 n 1 − 12 xn = , 1 − 12 a routine calculation shows that (xn ) is convergent: lim xn = 2. n→∞ The sequence (yn ). Let yn = n−1 X k=0 1 . As you well know, k! lim yn = lim n→∞ n→∞ n−1 X k=0 1 k! ! = ∞ X 1 = e, k! k=0 so that (yn ) is trying to converge to e. However, because e was temporarily banished along with all the other irrational numbers, (yn ) is divergent, because the set Q has an infinitesimal hole where e = lim yn ought to be. n→∞ n X 1 j√ k , where bxc := min ({z ∈ Z : x ≥ z}). The following comk k=1 putation shows that (zn ) is unbounded: The sequence (zn ). Let zn = n X 1 j√ k k k=1 = 1 1 1 1 + √ + √ + · · · + √ b nc 2 3 1 1 1 because each denominator has weakly increased −→ ≥ 1 + √ + √ + · · · + √ (1) n 2 3 1 1 1 1 1 because all terms in (1) are ≥ √ −→ ≥ √ + √ + √ + · · · + √ (2) n n n n n √ n because (2) is a sum of n identical terms −→ = √ = n. n Thus, (zn ) is divergent, not because Q has a hole where lim zn ought to be, but rather n→∞ because (zn ) is inherently incapable of approaching a limit. 1 There thus seem to be two different sorts of divergent sequences—those that are capable of having limits1 and those that aren’t—and we need a method for telling them apart.2 That is what the Definition 1 accomplishes. The definition is simply a precise statement of this informal observation: In a convergent sequence, as the terms get closer and closer to the limit, the are also getting closer and closer to each other . The corresponding precise statement is this: Definition 1 A sequence (xn ) is called Cauchy sequence iff it satisfies this condition: ∀ > 0 ∃n0 ∈ N ∀k ≥ n0 ∀m ≥ n0 |xk − xm | < (3) Observe that Definition 1 is couched completely in terms of the sequence itself (see footnote#2). 2 The Main Theorems. The first thing to establish is that convergent sequences are indeed Cauchy sequences. Theorem 1 Every convergent sequence satisfies condition (3). Proof. Let (xn ) be convergent; say, lim xn = a. For any given > 0, choose n0 such that |xn − a| < for all n→∞ 2 n ≥ n0 . For any k ≥ n0 and any m ≥ n0 , then, we have |xk − xm | = |(xk − a) − (xm − a)| ≤ |(xk − a)| + |(xm − a)| < + = . 2 2 The converse to Theorem 1 is not true, at least in Q: in Exercise 1, you will show that in Q, the divergent sequence (yn ) is also a Cauchy sequence. One can think of a divergent Cauchy sequence like (yn ) as a “hole-detector,” because, so to speak, it converges to a hole—to a position where there ought to be a number but there em isn’t a number. Fortunately, in the case of R, the existence of lub’s and glb’s ensure that, in contrast to Q, there are no holes of this type: in R, a Cauchy sequence will never detect a hole, because—this is the content of Theorem 2—every Cauchy sequence converges to a number, not to a hole. Theorem 2 If (xn ) is a Cauchy sequence in R, then there is a real number a such that lim xn = a. n→∞ Proof. The first step is to establish the following claim. Claim #1: (xn ) is bounded. Proof of claim #1. Choose any value of (say, = 1); by (3), there is an integer n0 such that, for every k ≥ n0 and every m ≥ n0 , |xk − xm | < 1. In particular: by taking m = n0 , we get that |xk − xn0 | < 1 for any k ≥ n0 . This gives, for k ≥ n0 : |xk | = |(xk − xn0 ) + xn0 | ≤ |xk − xn0 | + |xn0 | < 1 + |xn0 |. It follows that |xn | ≤ B for all n, where B = max |x1 | , |x2 | , . . . , |xn0 −1 | , |xn0 | + 1 . 1 (claim #1) I will call such sequences convergence-ready. Furthermore, this method must be intrinsic—must depend only on the sequence itself. We do not have such a method yet, note: at the moment, only reason we know that (yn ) is capable of approaching a limit is that we already know what its limit is (in R). 2 2 We can now apply Balzano–Weierstrass: (xn ) has a convergent subsequence (xn` ); put a := lim xn` . `→∞ Claim #2: lim xn = a. n→∞ Proof of claim #2, by the –n0 definition. For a given > 0 challenge, find the n0 response as follows. 1. Choose n1 such that |xk − xm | < for all k ≥ n1 and m ≥ n1 ; 2 2. Choose `0 such that • |xn` − a| < for all ` ≥ `0 and 2 • n `0 ≥ n 1 . 3. Choose n0 to be n`0 (since n`0 ≥ n1 ). For all n ≥ n0 , then, we have |xn − a| = |(xn − xn`0 ) − (xn`0 − a)| ≤ |xn − xn`0 | + |xn`0 − a| < + = . (claim #2 and Theorem 2) 2 2 Theorem 2 can also be regarded from a slightly different perspective: it tells us that for sequences of real numbers, Definition 1 completely captures the property of convergence-readiness: in order to be certain that a sequence of real numbers is convergence-ready, we need to know only that it is Cauchy—no further information is required.3 Exercise 1 Show that (yn ) is a Cauchy sequence. Hint: it is MUCH easier to use Theorem 1 to prove this than it is to prove directly that (yn ) satisfies condition (3) of Definition 1. 3 Applications. Cauchy sequences are instrumental in two related endeavors. 1. One can detect and fill gaps in Q and other subsets of R by means of lub’s and glb’s, but this technique cannot be used anywhere else. Cauchy sequences, on the other hand, can be used to detect and fill holes in many more sets than just subsets of R. 2. Many theorems about R that are proved by using lub’s and glb’s have analogues for these more general sets, and Cauchy sequences often replace lub’s and glb’s in the proofs of the generalizations. I will briefly discuss each of these applications. 3.1 Filling Holes in Metric Spaces. The concepts “convergent sequence” and “Cauchy sequence” make sense in any set in which the distance between elements of the set can be determined. Such a set is called a metric space; here is the precise definition. Definition 2 A metric space is a nonempty set S together with a distance function d : S × S −→ [0, ∞) with the following properties. For all x, y and z in S: • d(x, x) = 0; • If x 6= y, then d(x, y) > 0; • d(x, y) = d(y, x); and • d(x, y) ≤ d(x, z) + d(z, y). 3 In other words: if a sequence of real numbers is Cauchy, then that sequence definitely can converge—because, by Theorem 2, that sequence in fact does converge (in R). 3 Examples of metric spaces abound. Here are a few examples. 1. Any nonempty subset S of R. Here, d(x, y) = |x − y|. p (a − c)2 + (b − d)2 . v u n uX 3. The set Rn . In this case, for ~x = (x1 , . . . , xn ) and ~y = (y1 , . . . , yn ), d(~x, ~y ) = k~x − ~y k = t (xi − yi )2 . 2. The set of complex numbers {a + bi : a, b ∈ R}. In this case, d(a + bi, c + di) = i=1 4. `2 : the set of all real-valued sequences (xn ) for which ∞ X i=1 v u∞ uX 2 xi converges. In this case, d ((xn ) , (yn )) = t (xi − yi )2 . i=1 (It can be proved that this sum converges.) 5. The set of continuous functions f : [0, 1] −→ R, with distance function d(f, g) = Z 1 |f (x) − g(x)| dx. 0 Definition 3 Let (xn ) be a sequence in a metric space S. • For a ∈ S, “ lim xn = a” means: n→∞ ∀ > 0 ∃n0 ∀n ≥ n0 d(xn , a) < . • Sequence (xn ) is convergent iff it has a limit. • Sequence (xn ) is a Cauchy sequence iff (4) Exercise 2 Fill the correct definition into box (4). In any metric space S, a divergent Cauchy sequence, because it “converges to a hole,” detects a hole into which S could fit another point. A metric space that has no such holes is called a complete metric space: Definition 4 A metric space S is complete iff every Cauchy sequence in S has a limit in S. Metric space examples #2, #3 and #4 above are all complete, bit example #5 is not. Filling the holes in incomplete metric spaces. There is a structure called the completion a metric space in which all holes detected by divergent Cauchy sequences are filled in.4 In the completion space, every Cauchy sequence turns out to have a limit. Thus, just as was the case for sequences of real numbers, Definition 1 completely captures the property of convergence-readiness for these sequences as well. 4 The details of how to construct a completion are too complicated to include in this handout, but ask me if you are curious! 4 3.2 Absolute Convergence. The Calculus II theorem known as Absolute Convergence Test, which says that ∞ X t=0 xt converges if if ∞ X |xt | t=0 does, is an example of a theorem about R that can be generalized to many complete metric spaces,5 , provided one changes the proof. The Calculus II proof of this theorem does not work in the general case, because the Calculus II proof depends upon a trick that is specific to R, but the Cauchy-sequence-based proof below works perfectly in the more general case.6 n X xt Sn := Theorem 3 For a real-valued sequence (xn ), put t=1 . If (Tn ) converges, then (Sn ) must also n X |xt | Tn := t=1 converge. Proof. The first step is to expand |Tk − Tm | for any k ≥ m: k m X X |Tk − Tm | = |xt | − |xt | t=1 t=1 k X = |xt | t=m+1 = k X |xt |. (5) t=m+1 Let > 0 be given. Since (Tn ) converges, (Tn ) is a Cauchy sequence; therefore, there is an “-response integer” n0 such that, for all k ≥ m ≥ n0 , |Tk − Tm | < ; or, by (5), there is an integer n0 such that k ≥ m ≥ n0 =⇒ k X |xt | < . (6) t=m+1 But, for all k ≥ m ≥ n0 , we also have k m X X |Sk − Sm | = xt − xt t=1 t=1 k X = xt t=m+1 by the Triangle Inequality −→ ≤ k X |xt | t=m+1 by (6) −→ < . Thus, the “-response integer” n0 that works for (Tn ) also works for (Sn ), so that (Sn ) is also a Cauchy sequence. By Theorem 2, then, lim Sn exists. n→∞ 4 A Different Proof of Theorem 2. The following proof of Theorem 2 is slightly more complex than the one on p.2, but this proof is motivated by a simple, clear idea, namely the mental picture of a gradually-tightening Cauchy sequence homing in on a position. 5 including metric space examples #2, #3 and #4 In this handout, I am applying this proof technique only to the original Calculus II theorem. Again, I will be happy to show you how the generalizations are proved; ask me if you are curious. 6 5 Setting the stage. Notation.7 For a sequence (xn ), I will put • Sn = {xn , xn+1 , xn+2 , . . .}, • yn := lub(Sn ), and • zn := glb(Sn ). One new definition. Definition 5 Let S be a bounded nonempty set of real numbers. The diameter of S is the number diam(S) := lub ({|x − y| : x, y ∈ S}) . Preliminary exercises. Exercise 3 Prove that sequence (xn ) is a Cauchy sequence if and only if lim diam(Sn ) = 0. n→∞ Exercise 4 Prove that diam(Sn ) = yn − zn . Alternate proof of Theorem 2. Let (xn ) be a Cauchy sequence. Because Sn ⊇ Sn+1 for all n, we have zn ≤ zn+1 ≤ yn+1 ≤ yn (n = 1, 2, 3, . . .), so that ([z1 , y1 ] , [z2 , y2 ] , [z3 , y3 ] , . . .) is a nested sequence of closed intervals. Furthermore, because (xn ) is Cauchy, by Exercises 3 and 4, we have lim (yn − zn ) n→∞ = by Exercise 4 −→ = by Exercise 3 −→ = lim n→∞ diam(Sn ) 0. By the Nested Interval Theorem,8 then, the intersection of the intervals {[xn , yn ]} contains exactly one number a: ∞ \ [yn , zn ] = {a}. (7) n=1 To complete the proof, I will show that lim xn = a. Let > 0 be given. We have that lim (yn − zn ) = 0, so we n→∞ n→∞ can choose an integer n0 such that (yn0 − zn0 ) < . Now, for any n ≥ n0 , we have that xn is in Sn0 , so that zn0 = inf (Sn0 ) ≤ xn ≤ sup (Sn0 ) = yn0 . (8) In addition, by (7), we have that a ∈ [zn0 , yn0 ], so that zn 0 ≤ a ≤ y n 0 . Clearly it follows from (8) and (9) that |xn − a| < . 7 8 I used this notation in the handout on lim sup. Theorem 2.5.9 on p.57. 6 (9)