 #1
Eclair_de_XII
 1,067
 90
 TL;DR Summary

Let ##\{a_n\}## be a sequence in ##\mathbb{R}##. Let ##A_k:=\sup\{a_n:n\geq k\}## and suppose that ##\{A_k\}## converges to some real number ##\lambda##. Show that:
(1) ##A_k## is a decreasing sequence
(2) ##A_k\geq \lambda## for all ##k##
===(1)===
Let ##n\in \mathbb{N}##. Express ##A_n## and ##A_{n+1}## as:
##A_n=\sup\{a_n,a_{n+1},\ldots\}##
##A_{n+1}=\sup\{a_{n+1},\ldots\}##
Suppose for some ##m\geq {n+1}##, ##a_m=A_{n+1}##. By definition, ##a_m\geq a_k## for ##k\geq {n+1}##.
If ##a_n<a_m##, then ##a_m\geq a_k## for ##k\geq n+1## and ##k=n##. Hence, ##a_m=A_n##.
Now suppose that ##a_n\geq a_m##. Then ##a_n\geq a_m \geq a_k## for ##k\geq {n+1}##. Hence, ##A_n=a_n\geq a_m = A_{n+1}##
===(2)===
Fact: A decreasing sequence of real numbers bounded from below must converge to its infimum.
We prove the second fact by proving that if ##A_k## is not bounded from below, it cannot converge to ##\lambda##.
Assume ##\{A_k\}## is not bounded from below. Then for all real numbers, particularly, ##\lambda##, there is an integer ##m## such that ##A_m<\lambda##. Choose ##\epsilon=A_m\lambda## and let ##N\in \mathbb{N}##. Then whenever ##n\geq N##:
\begin{align*}
A_n\lambda&\geq&A_n\lambda\\
&\geq&A_m\lambda\\
&=&\epsilon
\end{align*}
if ##N>m##. If ##m\geq N##, choose ##n=m##:
\begin{align*}
A_m\lambda&\geq&A_m\lambda\\
&\geq&A_m\lambda\\
&=&\epsilon
\end{align*}
Hence, ##A_k## must be bounded from below. Since it is decreasing, it must converge and it converges to its infimum. It also converges to ##\lambda##. Any convergent sequence cannot converge to two different numbers, which means that ##\lambda## is the infimum.
Let ##n\in \mathbb{N}##. Express ##A_n## and ##A_{n+1}## as:
##A_n=\sup\{a_n,a_{n+1},\ldots\}##
##A_{n+1}=\sup\{a_{n+1},\ldots\}##
Suppose for some ##m\geq {n+1}##, ##a_m=A_{n+1}##. By definition, ##a_m\geq a_k## for ##k\geq {n+1}##.
If ##a_n<a_m##, then ##a_m\geq a_k## for ##k\geq n+1## and ##k=n##. Hence, ##a_m=A_n##.
Now suppose that ##a_n\geq a_m##. Then ##a_n\geq a_m \geq a_k## for ##k\geq {n+1}##. Hence, ##A_n=a_n\geq a_m = A_{n+1}##
===(2)===
Fact: A decreasing sequence of real numbers bounded from below must converge to its infimum.
We prove the second fact by proving that if ##A_k## is not bounded from below, it cannot converge to ##\lambda##.
Assume ##\{A_k\}## is not bounded from below. Then for all real numbers, particularly, ##\lambda##, there is an integer ##m## such that ##A_m<\lambda##. Choose ##\epsilon=A_m\lambda## and let ##N\in \mathbb{N}##. Then whenever ##n\geq N##:
\begin{align*}
A_n\lambda&\geq&A_n\lambda\\
&\geq&A_m\lambda\\
&=&\epsilon
\end{align*}
if ##N>m##. If ##m\geq N##, choose ##n=m##:
\begin{align*}
A_m\lambda&\geq&A_m\lambda\\
&\geq&A_m\lambda\\
&=&\epsilon
\end{align*}
Hence, ##A_k## must be bounded from below. Since it is decreasing, it must converge and it converges to its infimum. It also converges to ##\lambda##. Any convergent sequence cannot converge to two different numbers, which means that ##\lambda## is the infimum.