B Showing that a sequence of supremums of a sequence has these two properties

[Eclair_de_XII]

TL;DR

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.

 

[Office_Shredder]

For step 1, your proof is wrong because in general $A_n$ does not have to be equal to any of the $a_i$s. For example if $a_i=1+1/i$ for all i, then $A_n=1$ for all n.Your proof really shouldn't involve any complicated inequalities. $A_n$ is an upper bound of the set that $A_{n+1}$ is the supremum of. Why?

 

[Eclair_de_XII]

in general $A_i$ does not have to be equal to any of the $a_i$s.

Oh, I had overlooked that possibility.

For example if $a_i=1+1/i$ for all i, then $A_n=1$ for all n.

Surely, you mean $a_i=1-1/i$?

Why?

Because $A_n$ is an upper bound for the set containing $a_k$ for $k\geq n+1$ in addition to the set containing $a_n$?

 

[PeroK]

Try to write down a simple informal statement of why (1) and (2) hold. Once you have that, you can formalise a proof.

 

[Office_Shredder]

Surely, you mean $a_i=1-1/i$?

I did indeed.

Because $A_n$ is an upper bound for the set containing $a_k$ for $k\geq n+1$ in addition to the set containing $a_n$?

Yes. Since $A_n$ is an upper bound for a set which $A_{n+1}$ is the supremum, it must be at least as large as $A_{n+1}$ by definition of the supremum. That gives you part 1.

For part 2, your proof is overly complicated and starts by assuming $A_n$ is unbounded (though I don't think it's necessary for the rest of your proof). Try something simpler: if $A_k$ is decreasing, and there is some $n$ for which $A_n < \lambda-\epsilon$, then every other $A_k$ for $k>n$ must be below $A_n$ which gives you a contradiction to the fact that $\lambda$ is the limit. You don't need to prove that it has some other limit, just that it doesn't match the statement you were given.