1. ## Analysis subsequential limits

I need to prove

i) lim n to infinity sup Sn is an element of SL(Sn)
ii) same thing but replace sup with inf

I know SL(Sn) is the set of all limits of all convergent subsequences of Sn.

Also, lim n to infinity sup Sn can be defined as sup SL(Sn)

So it may be easier to show that Sup SL(Sn) is an element of SL(Sn)

2. Sorry, not a specialist in this area, but what is Sn? Is it just a sequence of real numbers S1, S2, ...? If so, then what is sup Sn?

3. Can only help you if I can understand your question. Explain or state the definition of what Sn and SL(Sn) are.

4. Sn is sequence of real numbers

SL(Sn) is the set of all limits of all convergent subsequences of the sequence Sn

5. Yes, but what is sup Sn? Is it a constant sequence where each element is equal to the supremum of the sequence $\{S_n\}_{n=0}^\infty$? Or is it a supremum of an individual number $S_n$? Or maybe, what I think is likely, it is $\sup_{m\ge n}S_m$, as in here, so the whole expression is $\displaystyle\limsup_{n\to\infty}S_n := \lim_{n\to\infty}(\sup_{m\geq n}S_m)$?

6. Originally Posted by emakarov
Yes, but what is sup Sn? Is it a constant sequence where each element is equal to the supremum of the sequence $\{S_n\}_{n=0}^\infty$? Or is it a supremum of an individual number $S_n$? Or maybe, what I think is likely, it is $\sup_{m\ge n}S_m$, as in here, so the whole expression is $\displaystyle\limsup_{n\to\infty}S_n := \lim_{n\to\infty}(\sup_{m\geq n}S_m)$?

It is the last definition that you gave with the link

7. Let's prove that there exists a subsequence $S_{n_1}, S_{n_2},\dots$ whose limit is $\lim_{n\to\infty}(\sup_{m\geq n}S_m)$. Let $T_n=\sup_{m\geq n}S_m$ and let $a=\lim_{n\to\infty}T_n$. We can ensure that for every $i$ and every $j>i$, $|S_{n_j}-a|<1/i$, i.e., the entire tail of the subsequence starting from $S_{n_i}$ lies in the $1/i$-neighborhood of $a$.

Indeed, suppose $i$ is given and $S_{n_1}, \dots, S_{n_{i-1}}$ have already been chosen. From the definition of $a$, there is a point $m$ such that

for all $n\ge m$, $|T_n-a|<1/(2i)$ (*)

Without loss of generality, we can assume that $m>n_{i-1}$; otherwise, choose a new $m$ as $n_{i-1}+1$; then (*) still holds. In particular, $|T_{m}-a|<\1/(2i)$, i.e., $|(\sup_{m\ge m}S_m)-a|<1/(2i)$. From the definition of sup, there exists an index, which we'll call $n_i$, such that $n_i\ge m>n_{i-1}$ and $|S_{n_i}-\sup_{m\ge m}S_m|<1/(2i)$. Thus, $|S_{n_i}-a|<1/i$.

TL;DR: We choose a subsequence whose limit is the limit superior of $\{S_n\}$ (call it $a$). Since $a$ is a limit, sups of tails of $\{S_n\}$ come close to $a$. Since those are sups, there are individual elements $S_n$ that come close to $a$.