1873_solutions

# That s is a bounded set that has a nonnegative subset

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Unformatted text preview: n , suppose that  0. We know that there are infinitely many integers n for which x  xn  x  and for each of these integers n we have x  xn  x  . Therefore the sequence x n is frequently in the interval x , x  and we conclude that x is a partial limit of x n . b. Prove that lmsup x n  nÝ lminf x n . nÝ Since lminf n Ý x n is a partial limit of x n we know from part a that lminf n Ý x n is a partial limit of the sequence x n . Now given any partial limit q of x n , we know that q, being a partial limit of x n , cannot be less than lminf n Ý x n . In other words, whenever q is a partial limit of x n we have 153 lminf x n nÝ q which gives us q Therefore lminf n Ý xn lminf x n . nÝ is the largest partial limit of the sequence xn . 4. Suppose that x n is a sequence of real numbers, that x  lmsup x n nÝ and that u  x  v. a. Prove that the sequence x n must be bounded above. The given inequality u  x  v tells us that x Ý and so x n is bounded above. b. Prove that the sequence x n must be frequently in the interval u, Ý . The interval u, Ý is a neighborhood of the partial limit x of x n . c. Prove that the sequence x n cannot be frequently in the interval v, Ý . Hint: Choose an upper bound  of x n . Now use this theorem. If x n were frequently in the interval v, Ý then it would have to have a partial limit in this interval, contradicting the fact that x is the largest partial limit of x n . 5. Suppose that x n is a sequence of real numbers, that x is a real number and that, whenever u  x  v, the sequence x n is frequently in the interval u, Ý but is not frequently in the interval v, Ý . Prove that x  lmsup x n . nÝ The fact that x n must be frequently in the interval u, v whenever u  x  v tells us that x n is frequently in every neighborhood of x and so x must be a partial limit of x n . To show that x is the largest partial limit of x n we shall show that no number larger than x can be a partial limit of x n . Suppose that x  p and choose a number v between x and p. Since the interval v, Ý is a neighborhood of p and x n is not frequently in v, Ý we conclude that p can’t be a partial limit of xn . 6. Suppose that x n is a bounded sequence of real numbers and that, for each integer n in the domain of this sequence we have defined y n  sup x m m n . Prove that the sequence y n is decreasing and that its limit is the lower limit of the sequence x n . For each positive integer n, it follows from the fact that xm m n  1 xm m n that sup x m m n sup x m m n  1 and so the sequence y n must be decreasing. Since every lower bound of x n is also a lower bound of y n , the sequence y n must be bounded below. Therefore y n is convergent. We define y to be the limit of y n . To show that y is lmsup n Ý x n we shall use Exercise 5. Suppose that u  y  v. Using the fact that y n y as n Ý we choose N such that the inequality y n  v holds whenever n N. u yn y v Since xm yN  v whenever m N we know that the sequence x n cannot b...
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## This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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