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Cauchy sequences. Definition: A sequence (xn ) is said to be a Cauchy sequence if given any ε > 0, there exists K ∈ N such that |xn − xm | < ε for all n, m ≥ K. Thus, a sequence is not a Cauchy sequence if there exists ε > 0 and a subsequence (xnk : k ∈ N) with |xnk − xnk+1 | ≥ ε for all k ∈ N. 3.5.5 A sequence converges if and only if it is a Cauchy sequence. Proof. If a sequence converges, it is easy to see that it must be a Cauchy sequence (exercise). Next, if it is a Cauchy sequence, then it is bounded (why) and hence it has a convergent subsequence say (xnk ) by the Bolzano-Weierstrass theorem. Suppose lim(xnk : k ∈ N) = x, it is easy to check that lim(xn ) = x (exercise). Example If (sn ) is a sequence such that |sn − sn+1 | ≤ rn for all n ∈ N, 0 ≤ r < 1, then the sequence converges. In particular, if sn = x1 + x2 + · · · + xn where |xn | ≤ rn , then (sn ) is a Cauchy sequence and hence converges. 3.5.6 (a) x1 = 1, x2 = 2 and xn+2 = (xn+1 + xn )/2 for all n. 1 (−1)n (b) sn = −1 + + · · · + . 2! n! (c) |xn+1 − xn | ≤ r|xn − xn−1 | for all n ≥ 2. Example 3.5.10. Solution of x3 − 7x + 2 = 0. Define xn+1 = 17 (x3n + 2) for suitable choice of x1 . Indeed, one can also define u = sup{x : x3 − 7x + 2 < 0}. 1 2 Construction of Real numbers by Cauchy Sequence. Recall that we have constructed the set of rational numbers. However, we have yet shown that there exists a set of real numbers, i.e., a complete ordered field. We will now construct this set using rational numbers. Since we have yet to define real numbers. We will say (xn ) is a Cauchy sequence of rational numbers if given any positive rational number r, there exists K ∈ N such that |xn − xm | < r for all n, m ≥ K. Let S be the collection of Cauchy sequences of rational numbers. We say (xn ) ∼ (yn ) if given any positive rational number ε, there exists K ∈ N such that |xn − yn | < ε for all n ∈ K. We then show that this defines an equivalence relation on S. Exercise: (xn ) ∼ (xn+K : n ∈ N). We will now define R as the set of equivalence classes in S and let us identify any rational number r with the equivalence class that contains the sequence (r). Note that if (a) ∼ (b) where a, b ∈ Q, then a = b; see Homework/Exercise. We will now show that R is a field. 3 The only more difficult part is to show that each nonzero equivalence class has a multiplicative inverse. First, we will need a simple observation. Fact: If (xn ) is not a Cauchy sequence that is equivalence to 0, then there exists M, K ∈ N such that |xn | > 1/M for all n ≥ K. It is then clear that (xn ) is equivalence to a sequence (yn ) of rational numbers with absolute value more than 1/M . Using the above sequence, we now claim that the sequence (1/yn ) is Cauchy and clearly it is the multiplicative inverse of (xn ). We next show that R is an ordered field. Similar to the previous argument, we will take the positive real numbers to be the set of equivalence classes (yn ) with yn > r for all n for some positive rational number r. Finally, we show that it is complete. In general, let ((y(k)n : n ∈ N) : k ∈ N) be a Cauchy sequence of Cauchy sequences of rational numbers. Then note that lim(y(k) : k ∈ N) = (zn : n ∈ N) for some Cauchy sequence of rational numbers (zn ). Hence every Cauchy sequence converges. For more details, see the Appendix at the end of the note. 4 Now suppose S is a nonempty set of equivalence classes of Cauchy sequences that is bounded above. For each fixed n ∈ N, consider the set ( recall that we have identified constant sequences (r) with rational numbers r) Tn = {k ∈ Z : k/2n is an upper bound of S}. First, let us show that it is nonempty. Let X be an upper bound of S. We may assume that X > 0. Now, note that 1/X > 0 and there exists rational number r = m/K, m, K ∈ N such that 0 < m/K < 1/X. Clearly 1/X > 1/K and hence K2n ∈ Tn . We next show that Tn has a least number. For simplicity, let us first assume that the set has a nonnegative real number (an equivalence class of Cauchy sequences of rational numbers). Then it has a least element say n0 by a consequence of well ordering property of natural number. Note that in − 1 6∈ Tn and hence (in − 1)/2n is not an upper bound while in /2n is an upper bound. (Note that we have identified the number in /2n to the constant sequence (r) with r = in /2n . Define an = in /2n and check that (an ) is a Cauchy sequence of rational numbers since |an − am | ≤ 1/2n for any m ≥ n. Finally, we need to show that [(in /2n )] is the least upper bound of S. Note that ((in − 1)/2n ) ∈ [(in /2n )]. In general if X is a set with a ”distance function”, we say it is complete if and only if every Cauchy sequence in the set converges. Rn is complete. It is not an ordered field for n > 1. But it is complete, i.e., every Cauchy sequences converges. 5 Homework/Exercise: A. Show that multiplication of ”real numbers” is well defined. B. Let a, b ∈ Q. Prove that the constant sequences (a) ∼ (b) if and only if a = b. C. Show that if we define the positive real numbers as above, then the set of ”real numbers” satisfies properties 2.1.5. Sec 3.5: all questions except 1,3 and 7. (4th ed is the same as 3rd ed) Additional questions for HW4: Sec 3.5: 6,8,12. Sec 3.6: all questions except 1,6,10. (4th ed is the same as 3rd ed) Homework5: 5,7,8 in the homework on limsup and liminf below, Questions B and Sec 3.6: 1, 6, 10. Properly divergent sequences We say lim(xn ) = ∞ if given any real number α ∈ R, there exists K ∈ N such that xn > α for all n ≥ K. Note that then the sequence diverges. Similarly, we could define the meaning of lim(xn ) = −∞. Example : lim(n) = ∞ and hence lim(nk ), lim(n!), lim(rn ) = ∞ (if r > 1.) Moreover, if lim(xn+1 /xn ) = L > 1 where (xn ) is a sequence of positive real numbers, then lim(xn ) = ∞. A simple fact (3.6.4) Suppose xn ≤ yn for all n. Then lim(xn ) = ∞ =⇒ lim(yn ) = ∞. 6 Operation with infinite limits: Suppose lim(xn ) = x and lim(yn ) = l, l could be infinite while x ∈ R. Then (i) lim(xn + yn ) = x + l (x + ∞ = ∞ and x − ∞ = −∞.) (ii) lim(xn yn ) = lx if x 6= 0 (a × ∞ = ∞ if a > 0.) (iii) lim(xn /yn ) = 0 if l is infinite and yn 6= 0 for all n. Theorem 3.6.5 Let (xn ), (yn ) be sequences of positive real numbers and lim(xn /yn ) = L > 0 (L 6= ∞). Then lim(xn ) = ∞ if and only if lim(yn ) = ∞. Limit superior and limit inferior Definition Let {an )∞ n=1 be a sequence of real numbers. We define its set of cluster points as follows: S(an ) = {c ∈ R : there exists a subsequence (ank : k ∈ N) of (an ) that converges to c}. Examples (i) S((−1)n ) = {1, −1}. (ii) S( 12 , 1, 13 , 23 , 1, 14 , 42 , 34 , 1, 15 , · · · ) = [0, 1]. (iii) S(sin n) =?. Definition Let (an ) be a bounded sequence. We define lim(an ) = lim an = limsup an = sup S(an ) n→∞ n→∞ lim(an ) = lim an = liminf an = inf S(an ) n→∞ n→∞ 7 Useful Fact lim(an ) = l if and only if (i) there exists a subsequence of (an ) that converges to l and (ii) lim(ank ) ≤ l for any convergent subsequence (ank ) of (an ). Proof. It is clear that l = limn→∞ an if (i) and (ii) hold. Conversely, let limn→∞ an = l. Then (ii) follows immediately from the definition. To prove (i), by the definition of the supremum, for each k ∈ N there exists an uk ∈ S(an ) such that l ≥ uk > l − 1 . 2k But there exists a subsequence (akm ) of (an ) that converges to uk . By choosing a subsequence of the subsequence, we may assume that |akm − uk | < 1/2m for each m ∈ N. Moreover, we may assume that km < (k+1)m for k, m ∈ N (again by choosing subsequences). We now note that (akk ) is a subsequence of (an ) that converges to l. Remark By modifying the above argument, we can also show that the set of cluster points is closed, i.e., if there exists (uk ), uk ∈ S(an ) for all k such that uk → u, then u ∈ S(an ). Examples (1) limn→∞ sin(nπ/4) = 1. Proof. It is clear that sin(nπ/4) ≤ 1 for all n ∈ N and lim(sin(8n + 2)π/4) = 1. (2) Let an = 1+ 1 n if n is odd Then lim(an ) = 1 and lim(an ) = −2. if n is even. Proof. Let (ank ) be a convergent subsequence of (an ). It is clear that min{1 + 1 , 7 nk nk 7 −2 n − 2} ≤ ank ≤ max{1 + n1k , n7k − 2} and hence −2 ≤ lim(ank ) ≤ 1. Finally, since lim(a2k ) = −2 and lim(a2k+1 ) = 1, the answer follows. 8 Theorem Let (an ) be a sequence of real numbers. If it is bounded above we let Uk = sup{an : n ≥ k}. And if (an ) is bounded below we let Lk = inf{an : n ≥ k}. If (an ) is bounded, then lim(an ) = lim(Uk ) and lim(an ) = lim(Lk ). Useful properties of limit superior and limit inferior (i) lim(an ) ≤ u if and only if given any ε > 0, there exists an N > 0 such that an < u + ε for n > N (∗) Proof: ( =⇒ ) done in class using Uk = sup{an : n ≥ k}. (⇐) Suppose on the contrary that lim(an ) = a > u. Then (an ) has a subsequence (ank ) that converges to a. Take ε = (a − u)/2. Then there exists N 0 ∈ N such that |ank − a| < ε for all k ≥ N 0 . In particular, we have u + ε > ank > a − ε if k ≥ max{N, N 0 } where N is given in (∗). which is impossible as u + ε = (a + u)/2 = a − ε. (ii) (Comparison of limit superior) Let an ≤ bn for n ∈ N. Then lim(an ) ≤ lim(bn ). (iii) lim(an )+lim(bn ) ≤ lim(an +bn ) ≤ lim(an )+lim(bn ) ≤ lim(an +bn ) ≤ lim(an )+lim(bn ). (iv) lim(−an ) = − lim(an ). (v) If limn→∞ an = limn→∞ an = l, then (an ) converges to l. 9 Remark (i) If a set S is not bounded above then we write sup S = ∞, similarly we write inf S = −∞ if S is not bounded below. Clearly, if ∅ 6= A ⊂ B, then inf B ≤ inf A ≤ sup A ≤ sup B. We will usually write inf ∅ = ∞ and sup ∅ = −∞. (ii) We can extend the definition of limit superior to unbounded sequences as follows: if (an ) is not bounded, it is still possible to define limit superior and limit inferior. However, we will have to define it by sup S(an ) and inf S(an ) and include ∞ and −∞ as possible limits. Homework on limit superior and inferior (1) Given lim(an ) ≥ 0 and (bn ) is bounded, show that lim an bn = lim an lim bn . n→∞ n→∞ n→∞ (2) Show that a convergent sequence has at least a maximum or minimum element. (3) Let an > 0 for n ∈ N. If lim an+1 /an = r, show that {an }∞ n=1 converges if n→∞ r < 1. What if r ≥ 1? (4) Show that lim n→∞ ÃZ π/6 0 !1/n n sin xdx 1 = . 2 (5) Given limn→∞ an ≥ 5 with an > −3 for n ∈ N, show that 4an ≥ 5/2. n→∞ 3 + an lim (6) Show that the set of cluster points of the sequence ((−1)n ) is {−1, 1}. 10 (7) Given limn→∞ an ≥ a, show by using the definition that a1 + a2 + a3 + · · · + an ≥ a. n→∞ n lim (8) Show that lim(an )+lim(bn ) ≤ lim(an +bn ) ≤ lim(an )+lim(bn ) ≤ lim(an +bn ) ≤ lim(an )+lim(bn ). Appendix We will give a sketch to show that any Cauchy sequence of equivalence classes of Cauchy sequence of rational numbers converges to a (an equivalence class of) Cauchy sequence of rational numbers. Indeed, first we will need to choose a better representative from each equivalence class, we will choose them (by taking subsequences) such that |y(k)n − y(k)m | < 1 2k+n if m ≥ n. Moreover, we can choose a subsequence (y(mk )) of the Cauchy sequence such that |y(mk ) − y(ml )| < 1/2k+1 if l ≥ k. First note we have for all l ≥ k, |y(mk )n − y(ml )n | < 1/2k for all n ≥ 2. Indeed, for each l ≥ k, there exists N ∈ N such that |y(mk )N − y(ml )N | < 1/2k+1 . Hence |y(mk )n −y(ml )n | ≤ |y(mk )n −y(mk )N |+|y(mk )N −y(ml )N |+|y(ml )N −y(ml )n | < 1/2k for all n ≥ 2. Claim: we may take zn = y(mn )n . It is easy to see that it is a Cauchy sequence and lim([y(mk )]) = [(zn )] and hence lim([y(k)]) = [(zn )].