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Unformatted text preview: HW2 ST520  Hannig Adam Labadorf ST520  Hannig September 9, 2007 HW2 1. lim A k is defined as the union of all intersections from some n → ∞ . We see in the set that, from k = 2 , all sets contain { 1 } . Likewise, from k = 4 , all sets contain { 1 , 2 } and from k = 6 all sets contain { 1 , 2 , 3 } . Following this pattern, we find that lim A k contains all positive whole numbers. Thus, lim A k = { ω : ω ∈ Z + > } lim A k is defined as the set of outcomes that occurs in infinitely many sets. We see that for even values of k all positive integers occur infinitely. Likewise, for all odd values of k greater than 1, all negative integers occur infinitely. Thus, we say that lim A k is equal to all positive and negative integers except 0. lim A k = { ω : ω ∈ Z ,ω 6 = 0 } 2. As n → ∞ , we see that the expression [ 1 6 + 1 n +1 , 1 n +2 ) approaches the interval [ 1 6 , 0) . The in tervals include 0 when n = 5 ( [ 1 6 + 1 6 , 1 7 ) = [0 , 1 7 ) ) and continue to do so to infinity. Therefore, the upper limit of the range in both lim A k and lim A k is 0 inclusive since 0 occurs in every set from some n and in infinitely many sets. Because the lower bound of the interval never actu ally reaches 1 6 the lower bound of both limits approaches 1 6 but never reaches it. Therefore, since there are infinitely many values approaching 1 6 and infinitely many sequential sets that approach 1 6 , the lower bound on the limitss is simply 1 6 . Since lim A k = lim A k we say that lim A k exists and is equal to...
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This note was uploaded on 06/16/2009 for the course STAT 520 taught by Professor Hannig,j. during the Spring '09 term at Colorado State.
 Spring '09
 Hannig,J.
 Statistics, Probability

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