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

In summary, (1) and (2) state that for a decreasing sequence of real numbers bounded from below, the limit must converge to its infimum. The proof involves showing that if the sequence is not bounded from below, it cannot converge to its infimum. This is done by showing that if ##A_n## is not bounded from below, then for all real numbers ##\lambda##, there is an integer ##m## such that ##A_m<-|\lambda|##. From this, it can be shown that ##A_k## must be bounded from below, and therefore must converge to its infimum.
  • #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.
 
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  • #2
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?
 
  • #3
Office_Shredder said:
in general ##A_i## does not have to be equal to any of the ##a_i##s.

Oh, I had overlooked that possibility.

Office_Shredder said:
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##?

Office_Shredder said:
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##?
 
  • #4
Try to write down a simple informal statement of why (1) and (2) hold. Once you have that, you can formalise a proof.
 
  • #5
Eclair_de_XII said:
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.
 
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