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Problem 10.3.5. Suppose limn→∞ ¤n = c. Prove that lim→ xn = subsequence (ng) of (xn). ▼ Hint. First prove that n. > k. c for any

Problem 10.3.5. Suppose limn→∞ ¤n = c. Prove that lim→ xn = subsequence (ng) of (xn). ▼ Hint. First prove that n. > k. c for any

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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real math analysis please hep me solve problem 10.3.5

10.3 The Bolzano-Weierstrass Theorem Once we introduced the Nested Interval Property (Axiom 10.1.1), the Intermediate Value Theorem (Theorem 10.2.1) followed pretty readily. The proof of Extreme Value Theorem (Theorem 10.4.8) takes a bit more work. First we need to show that a function that satisfies the conditions of the EVT is bounded. Theorem 10.3.1. A continuous function defined on a closed, bounded interval must be bounded. That is, let f be a continuous function defined on [a,b]. Then there exists a positive real number B such that |f (x)| < B for all x E [a, b]. Sketch of Alleged Proof: Let's assume, for contradiction, that there is no such bound B. This says that for any positive integer n, there must exist x, E [a, b] such that |f(xn)| > n. (Otherwise n would be a bound for f.) IF the sequence (xn) converged to something in [a, b], say c, then we would have our contradiction. Indeed, we would have lim, 00 *n = c. By the continuity of f at c and Theorem 9.2.1 of Chapter 9, we would have lim, 00 f(xn)= f(c). This would say that the sequence (f(xn)) converges, so by Lemma 7.2.7 of Chapter 7, it must be bounded. This would provide our contradiction, as we had |f(xn)| > n, for all positive integers n. QED? This would all work well except for one little problem. The way it was constructed, there is no reason to expect the sequence (xn) to converge to anything and we can't make such an assumption. That is why we emphasized the IF above. Fortunately, this idea can be salvaged. While it is true that the sequence (xn) may not converge, part of it will. We will need the following definition. Definition 10.3.2. Let (nx) be a strictly increasing sequence of positive integers; that is, n1 < n2 < n3 < …. If (xn)is a sequence, then = (xn1, Xn2, Xng, ...) is called a {subsequence} of (xn). The idea is that a subsequence of a sequence is a part of the sequence, (xn), which is itself a sequence. However, it is a little more restrictive. We can choose any term in our sequence to be part of the subsequence, but once we choose that term, we can't go backwards. This is where the condition n1 < n2 < n3 < ·… comes in. For example, suppose we started our subsequence with the term x100- We could not choose our next term to be æ99. The subscript of the next term would have to be greater than 100. In fact, the thing about a subsequence is that it is all in the subscripts; we are really choosing a subsequence (ng) of the seguence of subscripts (n) in (xn).
Transcribed Image Text:10.3 The Bolzano-Weierstrass Theorem Once we introduced the Nested Interval Property (Axiom 10.1.1), the Intermediate Value Theorem (Theorem 10.2.1) followed pretty readily. The proof of Extreme Value Theorem (Theorem 10.4.8) takes a bit more work. First we need to show that a function that satisfies the conditions of the EVT is bounded. Theorem 10.3.1. A continuous function defined on a closed, bounded interval must be bounded. That is, let f be a continuous function defined on [a,b]. Then there exists a positive real number B such that |f (x)| < B for all x E [a, b]. Sketch of Alleged Proof: Let's assume, for contradiction, that there is no such bound B. This says that for any positive integer n, there must exist x, E [a, b] such that |f(xn)| > n. (Otherwise n would be a bound for f.) IF the sequence (xn) converged to something in [a, b], say c, then we would have our contradiction. Indeed, we would have lim, 00 *n = c. By the continuity of f at c and Theorem 9.2.1 of Chapter 9, we would have lim, 00 f(xn)= f(c). This would say that the sequence (f(xn)) converges, so by Lemma 7.2.7 of Chapter 7, it must be bounded. This would provide our contradiction, as we had |f(xn)| > n, for all positive integers n. QED? This would all work well except for one little problem. The way it was constructed, there is no reason to expect the sequence (xn) to converge to anything and we can't make such an assumption. That is why we emphasized the IF above. Fortunately, this idea can be salvaged. While it is true that the sequence (xn) may not converge, part of it will. We will need the following definition. Definition 10.3.2. Let (nx) be a strictly increasing sequence of positive integers; that is, n1 < n2 < n3 < …. If (xn)is a sequence, then = (xn1, Xn2, Xng, ...) is called a {subsequence} of (xn). The idea is that a subsequence of a sequence is a part of the sequence, (xn), which is itself a sequence. However, it is a little more restrictive. We can choose any term in our sequence to be part of the subsequence, but once we choose that term, we can't go backwards. This is where the condition n1 < n2 < n3 < ·… comes in. For example, suppose we started our subsequence with the term x100- We could not choose our next term to be æ99. The subscript of the next term would have to be greater than 100. In fact, the thing about a subsequence is that it is all in the subscripts; we are really choosing a subsequence (ng) of the seguence of subscripts (n) in (xn).
Problem 10.3.5. Suppose lim,→∞ xn = c. Prove that limk→00 Xng = c for any subsequence (xng) of (xn). Hint. First prove that n. > k. A very important theorem about subsequences was introduced by Bernhard Bolzano and, later, independently proven by Karl Weierstrass. Basically, this theorem says that any bounded sequence of real numbers has a convergent subsequence.
Transcribed Image Text:Problem 10.3.5. Suppose lim,→∞ xn = c. Prove that limk→00 Xng = c for any subsequence (xng) of (xn). Hint. First prove that n. > k. A very important theorem about subsequences was introduced by Bernhard Bolzano and, later, independently proven by Karl Weierstrass. Basically, this theorem says that any bounded sequence of real numbers has a convergent subsequence.
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