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

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• Eclair_de_XII
Eclair_de_XII
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.

Staff Emeritus
Gold Member
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##?

Homework Helper
Gold Member
2022 Award
Try to write down a simple informal statement of why (1) and (2) hold. Once you have that, you can formalise a proof.

Staff Emeritus
Gold Member
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.

Eclair_de_XII