If lim sup a_n = lim inf a_n = a, then a_n->a

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January 16th 2010, 11:43 PM
kingwinner

If lim sup a_n = lim inf a_n = a, then a_n->a

http://sites.google.com/site/asdfasdf23135/ra5.JPG
January 17th 2010, 01:42 PM
kingwinner

Or maybe perhaps someone can provide a different/simpler proof for theorem 1?
January 17th 2010, 02:20 PM
Drexel28

Quote:

Originally Posted by kingwinner

http://sites.google.com/site/asdfasdf23135/ra5.JPG

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

Theorem 2: Using the same idea as above we see that since $a=\inf\text{ }S\leqslant s\leqslant a=\sup\text{ }S$ for every $s\in S$ we see that $S=\{a\}$ and so every subsequential limit converges to $a$ . The conclusion follows.
January 17th 2010, 05:43 PM
kingwinner

Quote:

Originally Posted by Drexel28

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

Theorem 2: Using the same idea as above we see that since $a=\inf\text{ }S\leqslant s\leqslant a=\sup\text{ }S$ for every $s\in S$ we see that $S=\{a\}$ and so every subsequential limit converges to $a$ . The conclusion follows.

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!
January 17th 2010, 08:04 PM
Drexel28

Quote:

Originally Posted by kingwinner

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!

I mean, it's just direct application. If $\forall\varepsilon>0$ there exists some $N\in\mathbb{N}$ such that $N\leqslant n\implies \left|a_n-a\right|<\varepsilon$ what can we say about $\left|\sup_{N\leqslant n}a_n-a\right|$ ?
January 18th 2010, 12:32 PM
kingwinner

Quote:

Originally Posted by Drexel28

I mean, it's just direct application. If $\forall\varepsilon>0$ there exists some $N\in\mathbb{N}$ such that $N\leqslant n\implies \left|a_n-a\right|<\varepsilon$ what can we say about $\left|\sup_{N\leqslant n}a_n-a\right|$ ?

|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?