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- Jan 26th 2009, 10:08 AMtransgalacticsubsequence definition question..
- Jan 26th 2009, 01:54 PMThePerfectHacker
Let $\displaystyle X_n = x_{r_n}$.

Now, $\displaystyle \liminf x_n \leq \limsup x_n \leq \limsup X_n = \lim X_n$

The reason why $\displaystyle \limsup X_n = \lim X_n$ is because $\displaystyle \{X_n\}$ is convergent.

The rest of them are similar. - Jan 28th 2009, 12:20 AMtransgalactic
$\displaystyle

\limsup X_n = \lim X_n

$

this expression is not correct

X_n is bounded

the limit of the sequence doesnt equal the limsup

- Jan 28th 2009, 02:55 PMThePerfectHacker
- Jan 28th 2009, 09:22 PMtransgalactic
Xn is not convergent

its bounded - Jan 28th 2009, 09:28 PMIsomorphism
- Jan 28th 2009, 11:44 PMtransgalactic
$\displaystyle

\liminf x_n \leq \limsup x_n \leq \limsup X_n = \lim X_n

$

why

$\displaystyle

limsupx_n \leq \limsup X_n

$

X_n is a subsequence which converges

the definition of limsup

is being the sup of the limit group of every subsequence

you cant do limsup X_n

by doing that you split this subsequence into a smaller convergent subsequences

and take the Sup of their limits

which has nothing to do with limsup x_n - Jan 29th 2009, 08:27 AMThePerfectHacker
If $\displaystyle A,B$ are non-empty bounded subsets with $\displaystyle A\subseteq B$ then $\displaystyle \sup (B) \leq \sup (A)$.

Now, $\displaystyle \{ x_{r_k} | k\geq n\} \subseteq \{ x_k | k\geq n\}$.

Therefore, $\displaystyle \sup \{ x_k | k\geq n\} \leq \sup \{ x_{r_k} | k\geq n\}$.

Thus, $\displaystyle \lim \sup \{x_k |k\geq n\} \leq \lim \sup \{ x_{r_k}| k\geq n\}$.

This gives, $\displaystyle \limsup x_n \leq \limsup X_n$