subsequence definition question..

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January 26th 2009, 10:08 AM
transgalactic

subsequence definition question..

http://img141.imageshack.us/img141/7...3651xp3.th.gif
January 26th 2009, 01:54 PM
ThePerfectHacker

Let $X_n = x_{r_n}$ .
Now, $\liminf x_n \leq \limsup x_n \leq \limsup X_n = \lim X_n$

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

The rest of them are similar.
January 28th 2009, 12:20 AM
transgalactic

$ \limsup X_n = \lim X_n $

this expression is not correct
X_n is bounded
the limit of the sequence doesnt equal the limsup
January 28th 2009, 02:55 PM
ThePerfectHacker

Quote:

Originally Posted by transgalactic

$ \limsup X_n = \lim X_n $

this expression is not correct
X_n is bounded
the limit of the sequence doesnt equal the limsup

The sequence $X_n$ is convergent.
Therefore, its limit matches its limit superior.
Look at the theorems you learned.
January 28th 2009, 09:22 PM
transgalactic

Xn is not convergent
its bounded
January 28th 2009, 09:28 PM
Isomorphism

Quote:

Originally Posted by transgalactic

Xn is not convergent
its bounded

Answer:

Quote:

Originally Posted by ThePerfectHacker

Let $X_n = x_{r_n}$ .

According to your work $X_n$ is convergent. TPH's $X_n$ is not the same as your $X_n$ . He is talking about the subsequence which is convergent...
January 28th 2009, 11:44 PM
transgalactic

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

why
$ 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
January 29th 2009, 08:27 AM
ThePerfectHacker

Quote:

Originally Posted by transgalactic

why
$ limsupx_n \leq \limsup X_n $

If $A,B$ are non-empty bounded subsets with $A\subseteq B$ then $\sup (B) \leq \sup (A)$ .

Now, $\{ x_{r_k} | k\geq n\} \subseteq \{ x_k | k\geq n\}$ .
Therefore, $\sup \{ x_k | k\geq n\} \leq \sup \{ x_{r_k} | k\geq n\}$ .
Thus, $\lim \sup \{x_k |k\geq n\} \leq \lim \sup \{ x_{r_k}| k\geq n\}$ .
This gives, $\limsup x_n \leq \limsup X_n$

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