Cauchy Sequences. x n = 1 ( ) 2 1 1, . As you well know, k! n 1. 1 k! = e, = k! k=0. k = k=1

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1 Cauchy Sequences 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 ) k The sequence x n ). Let x n. Since x n ) n, a routine calculation shows that x n ) is convergent: The sequence y n ). Let y n n. As you well know, k! lim y n lim n ) k! lim x n. k! e, so that y n ) is trying to converge to e. However, because e was temporarily banished along with all the other irrational numbers, y n ) is divergent, because the set Q has an infinitesimal hole where e lim y n ought to be. The sequence z n ). Let z n n k k, where x : min {z Z : x z}). The following computation shows that z n ) is unbounded: n k k n because each denominator has weakly increased n ) because all terms in ) are n because ) is a sum of n identical terms n n n. Thus, z n ) is divergent, not because Q has a hole where because z n ) is inherently incapable of approaching a limit ) n n n n lim z n ought to be, but rather

2 There thus seem to be two different sorts of divergent sequences those that are capable of having limits and those that aren t and we need a method for telling them apart. That is what the Definition 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 A sequence x n ) is called Cauchy sequence iff it satisfies this condition: ɛ > 0 n0 N k n 0 m n0 xk x m < ɛ ) 3) Observe that Definition is couched completely in terms of the sequence itself see footnote#). The Main Theorems. The first thing to establish is that convergent sequences are indeed Cauchy sequences. Theorem Every convergent sequence satisfies condition 3). Proof. Let x n ) be convergent; say, lim x n a. For any given ɛ > 0, choose n 0 such that x n a < ɛ n n 0. For any k n 0 and any m n 0, then, we have for all x k x m x k a) x m a) x k a) + x m a) < ɛ + ɛ ɛ. The converse to Theorem is not true, at least in Q: in Exercise, you will show that in Q, the divergent sequence y n ) is also a Cauchy sequence. One can think of a divergent Cauchy sequence like y n ) 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 every Cauchy sequence converges to a number, not to a hole. Theorem If x n ) is a Cauchy sequence in R, then there is a real number a such that lim x n a. Proof. The first step is to establish the following claim. Claim #: x n ) is bounded. Proof of claim #. Choose any value of ɛ say, ɛ ); by 3), there is an integer n 0 such that, for every k n 0 and every m n 0, x k x m <. In particular: by taking m n 0, we get that x k x n0 < for any k n 0. This gives, for k n 0 : x k x k x n0 ) + x n0 x k x n0 + x n0 < + x n0. It follows that x n B for all n, where ) B max x, x,..., x n0, x n0 +. claim #) 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 y n) is capable of approaching a limit is that we already know what its limit is in R).

3 We can now apply Balzano Weierstrass: x n ) has a convergent subsequence x nl ); put a : lim l x nl. Claim #: lim x n a. Proof of claim #, by the ɛ n 0 definition. For a given ɛ > 0 challenge, find the n 0 response as follows.. Choose n such that x k x m < ɛ for all k n and m n ;. Choose l 0 such that x nl a < ɛ for all l l 0 and n l0 n. 3. Choose n 0 to be n l0 since n l0 n ). For all n n 0, then, we have x n a x n x nl0 ) x nl0 a) x n x nl0 + x nl0 a < ɛ + ɛ ɛ. claim # and Theorem ) Theorem can also be regarded from a slightly different perspective: it tells us that for sequences of real numbers, Definition 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 Show that y n ) is a Cauchy sequence. Hint: it is MUCH easier to use Theorem to prove this than it is to prove directly that y n ) satisfies condition 3) of Definition. ) 3 Applications. Cauchy sequences are instrumental in two related endeavors.. 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.. 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. 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 A metric space is a nonempty set S together with a distance function with the following properties. For all x, y and z in S: dx, x) 0; If x y, then dx, y) > 0; dx, y) dy, x); and dx, y) dx, z) + dz, y). d : S S [0, ) 3 In other words: if a sequence of real numbers is Cauchy, then that sequence definitely can converge because, by Theorem, that sequence in fact does converge in R). 3

4 Examples of metric spaces abound. Here are a few examples.. Any nonempty subset S of R. Here, dx, y) x y.. The set of complex numbers {a + bi : a, b R}. In this case, da + bi, c + di) a c) + b d). 3. The set R n. In this case, for x x,..., x n ) and y y,..., y n ), d x, y) x y n x i y i ). 4. l : the set of all real-valued sequences x n ) for which It can be proved that this sum converges.) 5. The set of continuous functions f : [0, ] R, with distance function df, g) Definition 3 Let x n ) be a sequence in a metric space S. For a S, lim x n a means: x i converges. In this case, d x n ), y n )) x i y i ). i 0 i fx) gx) dx. i ɛ > 0 n0 n n0 dxn, a) < ɛ ). Sequence x n ) is convergent iff it has a limit. Sequence x n ) is a Cauchy sequence iff 4) Exercise 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 #, #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 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

5 3. Absolute Convergence. The Calculus II theorem known as Absolute Convergence Test, which says that x t converges if if t0 x t 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 S n : x t t Theorem 3 For a real-valued sequence x n ), put. If T n n ) converges, then S n ) must also T n : x t converge. Proof. The first step is to expand T k T m for any k m: m T k T m x t x t t t x t tm+ tm+ t t0 x t. 5) Let ɛ > 0 be given. Since T n ) converges, T n ) is a Cauchy sequence; therefore, there is an ɛ-response integer n 0 such that, for all k m n 0, T k T m < ɛ; or, by 5), there is an integer n 0 such that But, for all k m n 0, we also have k m n 0 tm+ S k S m by the Triangle Inequality by 6) < ɛ. x t < ɛ. 6) m x t x t t t x t tm+ tm+ Thus, the ɛ-response integer n 0 that works for T n ) also works for S n ), so that S n ) is also a Cauchy sequence. By Theorem, then, lim S n exists. 4 A Different Proof of Theorem. The following proof of Theorem is slightly more complex than the one on p., 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 #, #3 and #4 6 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. x t 5

6 Setting the stage. Notation. 7 For a sequence x n ), I will put S n {x n, x n+, x n+,...}, y n : lubs n ), and z n : glbs n ). One new definition. Definition 5 Let S be a bounded nonempty set of real numbers. The diameter of S is the number diams) : lub { x y : x, y S}). Preliminary exercises. Exercise 3 Prove that sequence x n ) is a Cauchy sequence if and only if Exercise 4 Prove that diams n ) y n z n. lim diams n) 0. Alternate proof of Theorem. Let x n ) be a Cauchy sequence. Because S n S n+ for all n, we have z n z n+ y n+ y n n,, 3,...), so that [z, y ], [z, y ], [z 3, y 3 ],...) is a nested sequence of closed intervals. Furthermore, because x n ) is Cauchy, by Exercises 3 and 4, we have lim y n z n ) by Exercise 4 lim diamsn ) ) by Exercise 3 0. By the Nested Interval Theorem, 8 then, the intersection of the intervals {[x n, y n ]} contains exactly one number a: [y n, z n ] {a}. 7) n To complete the proof, I will show that lim x n a. Let ɛ > 0 be given. We have that lim y n z n ) 0, so we can choose an integer n 0 such that Now, for any n n 0, we have that x n is in S n0, so that In addition, by 7), we have that a [z n0, y n0 ], so that Clearly it follows from 8) and 9) that x n a < ɛ. y n0 z n0 ) < ɛ. z n0 inf S n0 ) x n sup S n0 ) y n0. 8) z n0 a y n0. 9) 7 I used this notation in the handout on lim sup. 8 Theorem.5.9 on p.57. 6