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# Is that p holds for at most a finite number of

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is that P holds for at most a finite number of integers, hence it ceases to hold from some point onwards. Let us return to the exercise. Because L + cannot be in the first set, there is N 1 N such that a n L + for all n N 1 . Similarly, there will exist N 2 N such that a n L - for all n N 2 . Taking n = max( N 1 , N 2 ) (or anything larger), we get L - a n L + , thus L - L 2 . Since is arbitrary, we conclude that L - L 0; i.e., L L . We proved (or noticed) in the process something which is actually a useful property of these limit analogues, namely: LU: Let > 0 be given. There exists N N such that a n L + for all n N . LI: Let > 0 be given. There exists N N such that a n L - for all n N . To see that we have equality if and only the sequence converges, assume first that { a n } converges to a limit a . Let us establish here something that could be useful later on: Given a sequence of real numbers { a n } , x R , the following two state- ments are equivalent: (a) a n > x for an infinite number of indices n . (b) The sequence has a subsequence all of whose terms are > x . The equivalence should be sort of obvious. Or fully obvious. Or close to obvious. If the first statement holds, we can determine integers n 1 , n 2 , . . . such that 1 n 1 < n 2 < · · · inductively as follows. We define n 1 as the first element of the set of n N such that a n > x . Assuming n k defined, we define n k +1 as being the first element of the necessarily non-empty set of all n N such that n > n k and a n > x . Then { a n k } is a subsequence all of whose terms are > x . Conversely, if there is such a subsequence, there obviously is an infinite number of indices n for which a n > x . 3

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Assume now, with the sequence converging to a , that x is in the first set; i.e., (in our new formulation), there is a subsequence { a n k } of { a n } such that a n k > x for all k N . Since this sequence also converges to a we conclude that a x . That is, a is an upper limit of the first set, hence a L . In exactly the same way, mutatis mutandis , one sees that a L . Since L L , we conclude that L = L = a . Conversely, assume L = L . We know from what we just did that if the sequence has a limit, that limit equals the common value of L, L , so let us try to prove that the sequence converges to this common value. It’s either that, or nothing. By properties LU , LI above, given > 0 we can find N 1 , N 2 such that a n L + for all n N 1 and a n L - for all n N 2 . But since now L = L , this means that for n max( N 1 , N 2 ) we have L - = L - a n L + ; i.e., | a n - L | ≤ . The exercise is over. Some comments and extensions. The reason why the lim sup and the lim inf are important is that they always exist, specially if we extend the reals by adding ±∞ to the set, as detailed in Homework n that lets infinity into your life. If we do this, so that the l. u. b. of a set that isn’t bounded above is , the g. l. b. of a set that is not bounded below is -∞ , and l. u. b. = -∞ , g. l. b. = , then one can prove the following additional properties of the lim sup and of the lim inf: (a) { a n } is bounded above if and only if lim sup n →∞ a n < .
• Fall '09
• Schonbek
• Supremum, Limit of a sequence, Limit superior and limit inferior, lim sup

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