- #1
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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) \forall \epsilon > 0, a_{n} < L + \epsilon holds, for almost all terms of the sequence,
b) \forall \epsilon > 0, L - \epsilon < a_{n} 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 a \in \textbf{R} for which there exists a subsequence (bn) of the sequence (an) such that \lim_{n \rightarrow \infty} b_{n} = a. Hence, for every \epsilon' > 0, there exists an element x of A such that L - \epsilon' < x. Because x belongs to A, for some \epsilon'' > 0, the interval <x - \epsilon'', x+ \epsilon''> contains almost all elements of a subsequence of the sequence (an). This is where I'm stuck and highly confused.
a) \forall \epsilon > 0, a_{n} < L + \epsilon holds, for almost all terms of the sequence,
b) \forall \epsilon > 0, L - \epsilon < a_{n} 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 a \in \textbf{R} for which there exists a subsequence (bn) of the sequence (an) such that \lim_{n \rightarrow \infty} b_{n} = a. Hence, for every \epsilon' > 0, there exists an element x of A such that L - \epsilon' < x. Because x belongs to A, for some \epsilon'' > 0, the interval <x - \epsilon'', x+ \epsilon''> contains almost all elements of a subsequence of the sequence (an). This is where I'm stuck and highly confused.
