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Informal Exercise 16. Draw a picture of a number line representing F. Draw b and c in the above proof, and indicate the sets defined by |r – b| < ɛ and |r – e| < e where e is as in the above proof. Observe that the sets do not intersect so there can be no a; simultaneously in both, which is why we got a contradiction. This explains why we chose ɛ = could have chosen ɛ = (c – b)/4, for instance, and obtained a contradiction. However, e = 2(c – b) does not work. Why not? (c - b)/2. Note, we

Informal Exercise 16. Draw a picture of a number line representing F. Draw b and c in the above proof, and indicate the sets defined by |r – b| < ɛ and |r – e| < e where e is as in the above proof. Observe that the sets do not intersect so there can be no a; simultaneously in both, which is why we got a contradiction. This explains why we chose ɛ = could have chosen ɛ = (c – b)/4, for instance, and obtained a contradiction. However, e = 2(c – b) does not work. Why not? (c - b)/2. Note, we

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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Informal Exercise 16.

Theorem 37. A convergent seguence in an ordered field has a unique limit. Proof. Suppose otherwise that (a,) is a sequence in F with two distinct limits b and c. By trichotomy we get that b<c ore<b. Assume that b<c. The case where c < b is similar. Let e = (c – b)/2. By definition of positive and Theorem 2, c - b > 0. Sinoe 2 > 0 we have 2-1 > 0. Thus the product (c – b)2-1 is positive. In other words e > 0. By definition of limit, there is a N1 € N such that la;-b| < e for all i > N1. Likewise there is a N2 €N such that a – c| < e for all i > Na. Let i be the maximum of N1 and N3. Then since i2 Ni, la; - 6| < e. $8.5. Infinite sequences and limits 189 Hence -E < a; - b< E. Using properties of ordered fields we get a; < b+e = b+ (e – b)/2 = (b+ c)/2. Next, since i2 Ng, la; – c| < E, a similar argument gives us (b+c)/2 < aj. Putting these together we conclude as < (b+ c)/2 < ai, so a, < a, by transitivity. This contradicts trichotomy. Remark 10. We often write lim a, = b when we wish to assert (1) that the sequence (a;) converges, and (2) that the unique limit of the sequence (a.) is b.
Transcribed Image Text:Theorem 37. A convergent seguence in an ordered field has a unique limit. Proof. Suppose otherwise that (a,) is a sequence in F with two distinct limits b and c. By trichotomy we get that b<c ore<b. Assume that b<c. The case where c < b is similar. Let e = (c – b)/2. By definition of positive and Theorem 2, c - b > 0. Sinoe 2 > 0 we have 2-1 > 0. Thus the product (c – b)2-1 is positive. In other words e > 0. By definition of limit, there is a N1 € N such that la;-b| < e for all i > N1. Likewise there is a N2 €N such that a – c| < e for all i > Na. Let i be the maximum of N1 and N3. Then since i2 Ni, la; - 6| < e. $8.5. Infinite sequences and limits 189 Hence -E < a; - b< E. Using properties of ordered fields we get a; < b+e = b+ (e – b)/2 = (b+ c)/2. Next, since i2 Ng, la; – c| < E, a similar argument gives us (b+c)/2 < aj. Putting these together we conclude as < (b+ c)/2 < ai, so a, < a, by transitivity. This contradicts trichotomy. Remark 10. We often write lim a, = b when we wish to assert (1) that the sequence (a;) converges, and (2) that the unique limit of the sequence (a.) is b.
Informal Exercise 16. Draw a picture of a number line representing F. Draw b and c in the above proof, and indicate the sets defined by |r – b| < e and |r – e| < e where e is as in the above proof. Observe that the sets do not intersect so there can be no a; simultaneously in both, which is why we got a contradiction. This explains why we chose e = (c – b)/2. Note, we could have chosen ɛ = (c- b)/4, for instance, and obtained a contradiction. However, e = 2(c – b) does not work. Why not? %3D
Transcribed Image Text:Informal Exercise 16. Draw a picture of a number line representing F. Draw b and c in the above proof, and indicate the sets defined by |r – b| < e and |r – e| < e where e is as in the above proof. Observe that the sets do not intersect so there can be no a; simultaneously in both, which is why we got a contradiction. This explains why we chose e = (c – b)/2. Note, we could have chosen ɛ = (c- b)/4, for instance, and obtained a contradiction. However, e = 2(c – b) does not work. Why not? %3D
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