# Superior limit question

1. Jul 18, 2007

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.

2. Jul 18, 2007

### NateTG

Are you confused by the concept, or are you trying to prove something?

3. Jul 18, 2007