# subsequence definition question..

• Jan 26th 2009, 10:08 AM
transgalactic
subsequence definition question..
• Jan 26th 2009, 01:54 PM
ThePerfectHacker
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 AM
transgalactic
$\displaystyle \limsup X_n = \lim X_n$

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

• Jan 28th 2009, 02:55 PM
ThePerfectHacker
Quote:

Originally Posted by transgalactic
$\displaystyle \limsup X_n = \lim X_n$

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

The sequence $\displaystyle X_n$ is convergent.
Therefore, its limit matches its limit superior.
Look at the theorems you learned.
• Jan 28th 2009, 09:22 PM
transgalactic
Xn is not convergent
its bounded
• Jan 28th 2009, 09:28 PM
Isomorphism
Quote:

Originally Posted by transgalactic
Xn is not convergent
its bounded

Quote:

Originally Posted by ThePerfectHacker
Let $\displaystyle X_n = x_{r_n}$.

According to your work $\displaystyle X_n$ is convergent. TPH's $\displaystyle X_n$ is not the same as your $\displaystyle X_n$. He is talking about the subsequence which is convergent...
• Jan 28th 2009, 11:44 PM
transgalactic
$\displaystyle \liminf x_n \leq \limsup x_n \leq \limsup X_n = \lim X_n$

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 AM
ThePerfectHacker
Quote:

Originally Posted by transgalactic
why
$\displaystyle limsupx_n \leq \limsup X_n$

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