a 2n = . On the other hand, the subsequence a 2n+1 =

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1 Math 316, Intro to Analysis subsequences. This is another note pack which should last us two days. Recall one of our arguments about why a n = ( 1) n diverges. Consider the subsequence a n = It converges to. On the other hand, the subsequence a n+1 = converges to. If (a n ) had a limit L then the subsequences (a n ) and (a n+1 ) would have the same limits: 1 = L and 1 = L. But this is impossible since Today we make this reasoning explicit. starting with the notion of a subsequence. Definition 1. Let f : N N be a strictly increasing sequence on natural numbers. (That is, if n < m then f(n) f(m)). Let a : N R be a sequence of real numbers. Then a f is called a subsequence of a. The idea is to build a new sequence by skipping some (possibly many) of the entries of the original sequence, but respecting the order of the original sequence, and never repeating the same term in the sequence. Notation: We often write (a n ) instead of a(n) similarly we will write our sequence of natural numbers as k n instead of f(n). Thus, the standard notation for a subsequence is a kn where k 1 < k < k 3 <... is a strictly increasing sequence. The book often uses the notation n k, so that k suddenly becomes the independent variable. I will instead use k n, so that n is still the independent variable. Example: Let a n = ( 1) n and b n = 1 n. Get closed expressions for the following subsequences and decide if they converge. (1) a n+1 = () a n = (3) a n+1 = (4) a n = (5) b n+1 = (6) b n = (7) b n+1 = (8) b n = 1

2 Theorem. Let (a n ) be any sequence and a R be any number. a n a if and only if every subsequence of (a n ) converges to a The idea of the proof is that ( =), a n is a subsequence of a n and that (= ) if a n a and a kn is a subsequence then as n gets big so does k n, but as the subscript in a n gets big a n should get close to a. Proof of Theorem. Let a n be a sequence of real numbers. ( =) Suppose that for every subsequence (a kn ) of (a n ), a kn a. Notice that (a n ) is a subsequence of itself (Take k n = n). Thus, by assumption a kn = a n converges of a, as we claimed. (= ) Now suppose that a n a then for all ɛ > 0 there is an N such that if n > N then a n a < ɛ. Now, let k n be any strictly increasing sequence of natural numbers. Claim 1: k n n for all n. We will prove this claim soon. (by induction) Suppose that n > N, but then k n n > N (By Claim 1) so that a kn a < ɛ. This completes the proof. Proof of Claim 1. The proof is by induction. Base Case. Show that k 1 1 inductive step Let k N. Assume that k n n. Thus, by the principle of mathematical induction k n n for all n N Use this theorem to prove the following (which we have previously proven directly) Corollary 3. a n = ( 1) n does not converge. Cool. We ve have some divergent sequences with convergent subsequences. In fact these convergent subsequences can be used to detect divergence! In order to study this phenomenon more we make some notation: Definition 4. If (a n ) is a sequence and a R then a is a sub-sequential limit of (a n ) if there is a subsequence a kn which converges to a. Exercise 5. Make a guess for what the set of all sub-sequential limits of a n = ( 1) n is.

3 Theorem 6. For a sequence (a n ) and a R, a is a sub-sequential limit of (a n ) if and only if for all ɛ > 0 there exists infinitely many n N with a n a < ɛ. Proof. ( =) Suppose that a is a sub-sequential limit. Then there is a subsequence a kn with a kn a. Consider any ɛ > 0. Since a kn a there is a N such that if n < N then 3 How large is the set {k n : n > N}? Say a few words completing the proof. (= ) Suppose that for every ɛ there are infinitely many n N with a n a < ɛ. We need to build a subsequence with a kn a. We will do so using the following inductive argument: Claim : There exists a strictly increasing sequence n 1 < n <... of natural numbers such that a nk a < 1 k for all k N. Proof of Claim. Our proof will proceed inductively. We will generate a first term in the subsequence. Then after assuming that we have generated the first n terms in the subsequence we will get the n + 1 th term. Base case: By assumption there exist infinitely many n such that a n a < 1. Let k 1 be one of these. I ll include a first inductive step even though it is not strictly necessary in order to provide motivation. First inductive step: By assumption there are infinitely many n with a n a < 1. {1,,... k 1 } is a finite set. If you start with an infinite set and remove a finite number of elements the set is still infinite. Thus, there are infinitely many n N with n k 1 and a n a 1. Let k N be one such number. inductive step: Fix n N. Assume that that we have found k 1 < k < < k n with a kj a < 1 j for j = 1,,..., n. By assumption, there are infinitely many m N such that a m a < 1 n+1. The set {1,,... k n} is finite, so there are infinitely many natural numbers m with Let k n+1 be one such number. Iterating this process, we have a sub-sequence a kn such that for all n, a kn a < 1 n. Show that a nk a.

4 4 The Bolzano Weierstrass Theorem. Exercise 7. (1) The sequence a n = ( 1) n does not converge. Find a subsequence of a n which does converge. () The sequence b n = n diverges to infinity. ( i.e. For all M there exists an N such that if n > N then b n > M) Show that every subsequence of b n also diverges to infinity. What is different about these two sequences? Today we prove the following important theorem. Its proof is an application of Theorem 6 and of the nested intervals theorem. Theorem 8. If x n is a bounded sequence, then there is a subsequence x kn which converges. Proof. In order to prove the theorem we find a subsequential limit. The subsequence which converges to that limit is a convergent subsequence. We will make use of the Nested intervals theorem. The following claim (proof delayed) will give us a relevant sequence of nested intervals. Claim 3: If u is an upper bound for x n and l is lower bound then there is a sequence of nested intervals [a 1, b 1 ] [a, b ] [a 3, b 3 ] with b n a n = u l. and such that for each k n there are infinitely many n such that x n I k. We now use the claim to prove the theorem. By the nested intervals theorem there is an element of the intersection p. We will use Theorem 6 to show that p is a subsequential limit of x n. Consider any ɛ > 0. Since 1, there is a k > 0 such that 1 <. k Then k b k a k = u l k = (u l) 1 k < = ɛ Consider any q [a k, b k ]. a k p b k and a k q b k If p q then since q b k and a k p it follows that q p = q p = < ɛ

5 5 For you: If q < p then show that q p < ɛ Thus, for all q [a k, b k ] q p < ɛ. By Claim 3 there are infinitely many x n [a k, b k ]. Thus, for all ɛ > 0, there are infinitely many n with x n p < ɛ and so by Theorem 6, p is a sub-sequential limit of x n. All that remains is to prove Claim 3: If u is an upper bound for x n and l is lower bound then there is a sequence of nested intervals [a 1, b 1 ] [a, b ] [a 3, b 3 ] with b n a n = u l. n and such that for each k there are infinitely many n such that x n I k. Proof of Claim 3. We will use an inductive construction to build the intervals [a n, b n ]. Let m = u l. Since x n [l, u] for all n (and in particularly infinitely often) it follows that at least one of the following hold: (Case 1) There are infinitely many n with x n [l, m] or (Case ) There are infinitely many n with x n [m, u] If (Case 1) holds then set a 1 = l and b 1 = m. If (Case 1) does not hold, then (Case ) does, and we can set a 1 = m and b 1 = u. Either way we see that x n [a 1, b 1 ] for infinitely many n, and either (Case 1) b 1 a 1 = = = u l or (Case ) b 1 a 1 = = = u l. This completes the base case for our induction. Assume that there exists a sequence of k nested intervals [a 1, b 1 ] [a, b ] [a k, b k ] such that for all j = 1,,..., k two conditions hold: We need only build a subinterval [a k+1, b k+1 ] [a k, b k ] such that Let m k = b k+a k be the midpoint of [a k, b k ]. Since x n [a k, b k ] for infinitely many n it follows that at least one of the following hold: (Case 1) There are infinitely many n with x n [a k, m k ] or (Case ) There are infinitely many n with x n [m k, b k ]

6 6 If (Case 1) holds then set a k+1 = a k and b k+1 = m k. If (Case 1) does not hold, then (Case ) does: Set a k+1 = m k and b k=1 = b k. Either way we see that x n [a k+1, b k+1 ] for infinitely many n. In (Case 1) In (Case ) b k+1 a k+1 = = = by the inductive assumption b k+1 a k+1 = = = by the inductive assumption Thus, by the principle of mathematical induction we have a sequence of nested intervals [a 1, b 1 ] [a, b ]... such that for all k x n [a k, b k ] for infinitely many n and such that [b k a k ] u l. Thus, every sequence has a subsequential limit and so has a convergent subsequence. How many might it have? Exercise 9. If (a n ) is a convergent sequence then show that (a n ) has exactly one subsequential limit. If (a n ) is a bounded sequence but does not converge then show that (a n ) has at least two subsequential limits. This one we might work together. Think about negating the definition of a n a to find a subsequence which does not even get close to the limit point promiced by Bolzano-Weierstrass.

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