1. ## subsequence definition question..

2. 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.

3. $
\limsup X_n = \lim X_n
$

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

4. 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.

5. Xn is not convergent
its bounded

6. Originally Posted by transgalactic
Xn is not convergent
its bounded
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...

7. $
\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

8. 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, $\limsup x_n \leq \limsup X_n$