# Analysis subsequential limits

• Sep 20th 2010, 10:47 AM
newuser
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)

• Sep 20th 2010, 02:50 PM
emakarov
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?
• Sep 20th 2010, 03:23 PM
bubble86
Can only help you if I can understand your question. Explain or state the definition of what Sn and SL(Sn) are.
• Sep 20th 2010, 03:49 PM
newuser
Sn is sequence of real numbers

SL(Sn) is the set of all limits of all convergent subsequences of the sequence Sn
• Sep 20th 2010, 04:11 PM
emakarov
Yes, but what is sup Sn? Is it a constant sequence where each element is equal to the supremum of the sequence $\displaystyle \{S_n\}_{n=0}^\infty$? Or is it a supremum of an individual number $\displaystyle S_n$? Or maybe, what I think is likely, it is $\displaystyle \sup_{m\ge n}S_m$, as in here, so the whole expression is $\displaystyle \displaystyle\limsup_{n\to\infty}S_n := \lim_{n\to\infty}(\sup_{m\geq n}S_m)$?
• Sep 20th 2010, 05:11 PM
newuser
Quote:

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 $\displaystyle \{S_n\}_{n=0}^\infty$? Or is it a supremum of an individual number $\displaystyle S_n$? Or maybe, what I think is likely, it is $\displaystyle \sup_{m\ge n}S_m$, as in here, so the whole expression is $\displaystyle \displaystyle\limsup_{n\to\infty}S_n := \lim_{n\to\infty}(\sup_{m\geq n}S_m)$?

It is the last definition that you gave with the link
• Sep 21st 2010, 03:27 AM
emakarov
Let's prove that there exists a subsequence $\displaystyle S_{n_1}, S_{n_2},\dots$ whose limit is $\displaystyle \lim_{n\to\infty}(\sup_{m\geq n}S_m)$. Let $\displaystyle T_n=\sup_{m\geq n}S_m$ and let $\displaystyle a=\lim_{n\to\infty}T_n$. We can ensure that for every $\displaystyle i$ and every $\displaystyle j>i$, $\displaystyle |S_{n_j}-a|<1/i$, i.e., the entire tail of the subsequence starting from $\displaystyle S_{n_i}$ lies in the $\displaystyle 1/i$-neighborhood of $\displaystyle a$.

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

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

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

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