- #1

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- 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.