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- Jan 16th 2010, 11:43 PMkingwinnerIf lim sup a_n = lim inf a_n = a, then a_n->a
- Jan 17th 2010, 01:42 PMkingwinner
Or maybe perhaps someone can provide a different/simpler proof for theorem 1?

- Jan 17th 2010, 02:20 PMDrexel28
Theorem 1: It can easily be seen that if $\displaystyle S$ is the set of all sub sequential limits of $\displaystyle \left\{a_n\right\}$ then $\displaystyle \limsup\text{ }a_n=\sup\text{ }S$ and $\displaystyle \liminf\text{ }a_n=\inf\text{ }S$. But, since $\displaystyle \lim\text{ }a_n$ exists and equals $\displaystyle a$ we know that every subsequential limit converges to $\displaystyle a$ so that $\displaystyle \limsup\text{ }a_n=\sup\text{ }S=\sup\text{ }\{a\}=a=\inf\text{ }\{a\}=\inf\text{ }S=\liminf\text{ }a_n$

Theorem 2: Using the same idea as above we see that since $\displaystyle a=\inf\text{ }S\leqslant s\leqslant a=\sup\text{ }S$ for every $\displaystyle s\in S$ we see that $\displaystyle S=\{a\}$ and so every subsequential limit converges to $\displaystyle a$. The conclusion follows. - Jan 17th 2010, 05:43 PMkingwinner
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, 08:04 PMDrexel28
I mean, it's just direct application. If $\displaystyle \forall\varepsilon>0$ there exists some $\displaystyle N\in\mathbb{N}$ such that $\displaystyle N\leqslant n\implies \left|a_n-a\right|<\varepsilon$ what can we say about $\displaystyle \left|\sup_{N\leqslant n}a_n-a\right|$?

- Jan 18th 2010, 12:32 PMkingwinner
|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?