Or maybe perhaps someone can provide a different/simpler proof for theorem 1?

Results 1 to 6 of 6

- Jan 17th 2010, 12:43 AM #1

- Joined
- Jan 2009
- Posts
- 404

- Jan 17th 2010, 02:42 PM #2

- Joined
- Jan 2009
- Posts
- 404

- Jan 17th 2010, 03:20 PM #3
Theorem 1: It can easily be seen that if is the set of all sub sequential limits of then and . But, since exists and equals we know that every subsequential limit converges to so that

Theorem 2: Using the same idea as above we see that since for every we see that and so every subsequential limit converges to . The conclusion follows.

- Jan 17th 2010, 06:43 PM #4

- Joined
- Jan 2009
- Posts
- 404

Somehow my textbook presents lim sup in a different way. It defines the lim sup as simply the limit of the seqeunce of supremums of the tails, and nothing related to subsequences and subsequential limits. (so I admit there is some gap between your proof and my background)

Is there an alternative way to prove theorem 1 using more basic definitions/principles? (e.g. epsilon-limit proofs, upper bounds, least upper bounds, etc.)

Thank you!

- Jan 17th 2010, 09:04 PM #5

- Jan 18th 2010, 01:32 PM #6

- Joined
- Jan 2009
- Posts
- 404

|a_n -a| < ε for all n≥N

=> a-ε < a_n < a+ε for all n≥N

=> in particular,

a-ε < a_N < a+ε........................(1)

sup a_n ≥ a_n for all n≥N (by definition of an upper bound)

n≥N

=> in particular,

sup a_n ≥ a_N............................(2)

n≥N

(1) and (2) show that we will*always*have

sup a_n > a-ε

n≥N

But how to do it for the "upper" bound a+ε?