Subsequences and the Bolzano-Weierstrass Theorem

Transcription

1 Subsequences and the Bolzano-Weierstrass Theorem

2 A subsequence of a sequence x n ) n N is a particular sequence whose terms are selected among those of the mother sequence x n ) n N The study of subsequences is important because it provides vital information about the convergence of the parent sequence For example, we will see that a sequence which contains two subsequences converging to different limits is divergent An interesting result in this section Bolzano-Weierstrass Theorem) states that every bounded sequence has a convergent subsequence Plan Subsquences and properties The Bolzano-Weierstrass Theorem 1

3 Definition 1 Given a sequence x n ) n N of real numbers, let n 1 < n 2 < < n k < < be a strictly increasing sequence of natural numbers. sequence given by xn1, x n2,, x nk, ) = ) x nk is called a subsequence of x n ) n N. k N Then the This is a good place for you to pause the video for a moment. Before you proceed, read this definition as many times as possible and try your best to unpack every aspect of this definition 2

4 Example 2 Let 1 1, 1 2,, 1 ) k, be a sequence. Then ) 1 2 1, 1 2 2,, 1 2 k, is a subsequence of the given sequence. Let be a sequence. Then 2 1, 2 2,, 2 k, ) 2 2 1, 2 2 2,, 2 2 k, ) is a subsequence of the given sequence. 3

5 Let x n ) n N be a sequence such that x n = 1) n 1 n. Then x 2k ) k N is a subsequence given by x 2k = 1) 2k 1 2k = 1 2k. Also, x 2k+1 ) k N is a subsequence given by x 2k+1 = 1) 2k+1 1 2k + 1 = 1 2k

6 Let 1 1, 1 2,, 1 ) k, be a sequence. Then ) 1 cos 1), 1 cos 2),, 1 cos k), is a not subsequence of the given sequence because is not an increasing function. cos k)) k N 5

7 Test your understanding: You should pause here, and 1) find an example of a sequence 2) write down an example of a subsequence of the sequence given in 1) and 2) find an example of a sequence which is not a subsequence of the sequence given in 1) Theorem 3 If x n ) n N is a sequence of real numbers which is convergent to x then any subsequence x nk )k N of x n) n N is convergent to x. 6

8 Proof. Let ɛ > 0. By assumption, there exists a natural number N depending on ɛ such that if n > N then x n x < ɛ. I claim that since n k ) k N is an increasing sequence then it is true that n k k for every natural number k. Indeed by induction we known that n 1 1 since n 1 is a natural number. Next, suppose that Since n k ) k N is increasing then n l l for some l N, l 1. n l+1 > n l strict inequality is important here!!!) Now by the inductive hypothesis, and it follows that Since n l+1, l are natural numbers Moving forward, if k > N then n l l n l+1 > l n l+1 l + 1. n k k > N and consequently, Thus, xnk x < ɛ lim x n k k = x 7

9 Corollary 4 If x n ) n N is a sequence of real numbers admitting two subsequences which converge to different limits, then x n ) n N is divergent. 8

10 Example 5 Let us consider the following sequence a n = 1) n ). n Then a 2k = 1) 2k ) 2k and a 2k+1 = 1) 2k ) 2k + 1 Now note that lim a 2k = lim k k 2k = lim 2 + lim k k = k = 1 2k = 2 lim a 2k+1 = lim 2 1 k k 2k + 1 = lim 2) + lim k k ) 2k ) 2k + 1 = 2 Thus, a n ) n N has two subsequences converging to two different limits. As such a n ) n N is a divergent sequence. 9

11 Example 6 Let us consider the following sequence ) 2πn a n = cos ). 3 n Then and lim a 3k = lim k k cos = lim k lim a 3k+1 = lim cos k k = lim k cos = 1 2 lim k 2π3k 3 ) = k ) ) 3k ) 2π 3k + 1) ) 3 3k ) 2π ) 3 3k ) k = 1 2 Therefore, the sequence a n ) n N is a divergent sequence. 10

12 Theorem 7 Any sequence has a monotone subsequence 11

13 Proof. Let a n ) n N be a sequence. We will call a peak a term a m of the sequence such that a m a n for all n m. For the first case, let us suppose that a n ) n N has infinitely many peaks. We list those peaks by increasing order of their subscripts and clearly by construction we have a m1, a m2,, a mk, a m1 a m2 a mk Which is clearly a decreasing subsequence of a n ) n N. For the second case, let us suppose that a n ) n N has a finite number of peaks possibly none.) We list them in increasing order of their subscripts as follows a m1, a m2,, a mk. Since a mk is the last peak, then it is clear a mk+1 is not a peak. Thus, there is n 1 > m k + 1 such that a mk +1 < a n1. But a n1 is not a peak. Thus, there is n 2 > n 1 such that a n1 < a n2. We may then iterate this process indefinitely to construct an increasing sequence a n1 < a n2 < a n3 < < a nk < 12

14 Theorem 8 The Bolzano-Weierstrass Theorem) Any bounded sequence has a convergent subsequence. Proof. Let a n ) n N be a sequence which is also bounded. Now from our previous result, we know that a n ) n N has a monotone subsequence say ank )k N. Since ) a nk is a bounded sequence as a subsequence of a k N bounded sequence) then a nk must be a convergent subsequence of the )k N parent sequence a n ) n N. 13

15 Example 9 Consider the sequence with terms cosn). Since this sequence is bounded, then it has a convergent subsequence. 14