1873_solutions

Subset e of s such that 4 prove that if s is any set

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: e frequently in the interval v, Ý . On the other hand, given any integer n N it follows from the fact that u  y n  sup x m m n 154 that there exists an integer m u n such that u  x m . yn y xm v Thus the set of integers m for which u  x m is unbounded above and we have shown that x n is frequently in the interval u, Ý . 7. State and prove an analogue of the preceding exercise for lower limits. Suppose that x n is a bounded sequence that that, for each n we define y n  inf x m m n . Then the sequence y n is decreasing and the limit of the sequence y n is lminf n Ý x n . This assertion can be proved by a mirror image of the proof that was used in Exercise 6 and it can also be obtained from the statement of Exercise 6, in view of Exercise 3b. 8. Given that zn  xn  yn for every positive integer n, prove that lmsup z n nÝ lmsup x n  lmsup y n . nÝ nÝ Solution: To obtain a contradiction we assume that lmsup z n  lmsup x n  lmsup y n . nÝ Now we choose a number nÝ nÝ  0 such that lmsup z n  lmsup x n  lmsup y n  . nÝ nÝ nÝ Since lmsup x n  nÝ 2  lmsup x n , nÝ there are at most finitely many integers n for which x n  lmsup x n  2 and we see, in the same way, that there are at most finitely many integers n for which y n  lmsup y n  . 2 nÝ Thus, for all but at most finitely many integers n we have  lmsup y n  z n  x n  y n lmsup x n  2 2 nÝ nÝ which is impossible since  lmsup y n   lmsup z n . lmsup x n  2 2 nÝ nÝ nÝ nÝ Alt 7: Limits of Sequences in Metric Spaces Some Exercises on Subsequences Decide whether each of the following assertions is true or false. If it is true, prove it. If it is false, illustrate this fact by giving an example. 1. If a sequence x n is eventually in a given set S then every subsequence of x n is eventually in S. The assertion is true. Suppose that x n is a sequence that is eventually in a given set S and that x n i is a subsequence of x n . Choose an integer N such that the condition x n S holds whenever n N. Now choose an integer j such that the condition n i N holds whenever i j. We see that 155 xni S whenever i j. 2. If a sequence x n is frequently in a given set S then every subsequence of x n is frequently in S. The assertion is false. If we define x n  1 n for every positive integer n then, although x n is frequently in the set 1 , the subsequence x 2n fails to be frequently in the set 1 . 3. If every subsequence of a sequence is frequently in a given set S then x n is frequently in S. The statement is obviously true because every sequence is a subsequence of itself. 4. If every subsequence of a sequence x n is frequently in a given set S then x n is eventually in S. The assertion is true. We shall show that if x n fails to be eventually in S then x n must have a subsequence that is not frequently in S. Suppose that x n is a sequence that fails to be eventually in a given set S. We know that x n is frequently in the set R S and from the preceding theorem we deduce that x n has a...
View Full Document

Ask a homework question - tutors are online