511-Midterm-2002F-ANS - Economics 511 Professor John H....

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Economics 511 Professor John H. Nachbar Fall 2002 Midterm Answers 5. (a) Consider the sequence { 1 / 2 , 1 / 4 , 1 / 2 , 3 / 4 , 1 / 8 , 1 / 4 , 3 / 8 , 1 / 2 ,... } . The set of subsequential limits is [0 , 1]. A more sophisticated approach is to note the following theorem, which is a relative of Theorem 6 in the Sequences and Series notes. Theorem. Given a sequence { x t } , let F be the image of { x t } and let F 0 be the set of limit points of F . If x F 0 then x is a subsequential limit of { x t } . Proof. Since x is a limit point of F there is an x t 1 N 1 ( x ). Similarly, since x is a limit point of F there are infinitely many points of F in N 1 / 2 ( x ). Therefore, choose x t 2 N 1 / 2 ( x ) with t 2 > t 1 . And so on. By construction, this is a subsequence converging to x . ± It follows that if F is any countable subset of [0 , 1] with the property that F 0 = [0 , 1] then any sequence formed from all of the members of F has [0 , 1] as the set of its subsequential limits. Perhaps surprisingly, order does not matter . Nor is it necessary for any of the elements of F to appear more than once. Thus, for example, one could take any sequence for which F = Q [0 , 1].
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This note was uploaded on 05/20/2010 for the course ECON 511 taught by Professor Johnnachbar during the Fall '08 term at Washington University in St. Louis.

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511-Midterm-2002F-ANS - Economics 511 Professor John H....

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