- #1
Monochr
- 3
- 0
I am only stuck on one part of one fairly long proof.
Let {a_n} be a sequence in R. Then the following are equivalent:
(a) limsup a_n = a.
(b) For every b > a, a_n < b for all but finitely many n and for every c < a, a_n > c for infinitely many n.
assume that (b) holds. Then for every b > a, there exists m such that an < b for all n ≥ m. Hence sup_n≥m a_n ≥ b. This implies that limsup a_n ≥ b for every b > a so that limsup a_n ≥ a.
How does the limsup a_n ≥ b inequality hold? As far as I understand limsup it is a the limit of a monotonically decreasing sequence, sup_n≥m a_n, which cannot have a limit that is larger than the largest member of the sequence. But we know that b is larger. So shouldn't the limsup a_n ≥ b be a strict inequality because of this?
Homework Statement
Let {a_n} be a sequence in R. Then the following are equivalent:
(a) limsup a_n = a.
(b) For every b > a, a_n < b for all but finitely many n and for every c < a, a_n > c for infinitely many n.
assume that (b) holds. Then for every b > a, there exists m such that an < b for all n ≥ m. Hence sup_n≥m a_n ≥ b. This implies that limsup a_n ≥ b for every b > a so that limsup a_n ≥ a.
The Attempt at a Solution
How does the limsup a_n ≥ b inequality hold? As far as I understand limsup it is a the limit of a monotonically decreasing sequence, sup_n≥m a_n, which cannot have a limit that is larger than the largest member of the sequence. But we know that b is larger. So shouldn't the limsup a_n ≥ b be a strict inequality because of this?