Let (xn) be a bounded sequence and let s = sup{xn : n e N}. Show that if s ¢ {xn : n E N}, then there is a subsequence of (xn) that converges to s. (Hint: Use Lemma 1.5.1 and use s - 1< s, s- 1/2 < s, s – 1/3 < s, ...)

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.1: Infinite Sequences And Summation Notation
Problem 72E
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LEMMA 1.5.1
An upper bound u is the supreme of S in R if and
only if V e > 0, 3 a E S such that u – e < a.
-
S
a
u = sup S
Transcribed Image Text:LEMMA 1.5.1 An upper bound u is the supreme of S in R if and only if V e > 0, 3 a E S such that u – e < a. - S a u = sup S
Let (xn) be a bounded sequence and let s =
sup{xn : n e N}. Show that if s 4 {xn :
n E N}, then there is a subsequence of (xn) that converges to s. (Hint: Use Lemma 1.5.1 and use
s - 1< s, s – 1/2 < s, s – 1/3 < s, . .
..)
|
Transcribed Image Text:Let (xn) be a bounded sequence and let s = sup{xn : n e N}. Show that if s 4 {xn : n E N}, then there is a subsequence of (xn) that converges to s. (Hint: Use Lemma 1.5.1 and use s - 1< s, s – 1/2 < s, s – 1/3 < s, . . ..) |
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