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Superior limit question

  1. Jul 18, 2007 #1

    radou

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    So, my book says that the real number L is a superior limit of the sequence (an) iff the following holds:

    a) [itex]\forall \epsilon > 0[/itex], [itex]a_{n} < L + \epsilon[/itex] holds, for almost all terms of the sequence,
    b) [itex]\forall \epsilon > 0[/itex], [itex]L - \epsilon < a_{n}[/itex] holds, for an infinite number of terms of the sequence.

    OK, the two facts confuse me. I know that "almost all terms of the sequence" means "all terms, except a finite number of terms". If L is a superior limit, then it is the supremum (by definition) of the set A of all real numbers [itex]a \in \textbf{R}[/itex] for which there exists a subsequence (bn) of the sequence (an) such that [itex]\lim_{n \rightarrow \infty} b_{n} = a[/itex]. Hence, for every [itex]\epsilon' > 0[/itex], there exists an element x of A such that [itex]L - \epsilon' < x[/itex]. Because x belongs to A, for some [itex]\epsilon'' > 0[/itex], the interval [itex]<x - \epsilon'', x+ \epsilon''>[/itex] contains almost all elements of a subsequence of the sequence (an). This is where I'm stuck and highly confused.
     
  2. jcsd
  3. Jul 18, 2007 #2

    NateTG

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    Are you confused by the concept, or are you trying to prove something?
     
  4. Jul 18, 2007 #3

    radou

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    The concept (eg definition) is clear to me, I just don't understand how one arrives at a) and b).
     
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