Continuation of sequence of supremums question

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In summary, the conversation discusses the justification for the convergence of the sequence ##A_n##, which is decreasing and bounded from below by the point it converges to. By definition, ##A_n## is the supremum of the set containing all ##a_k## where ##k\geq n+1## and the set containing ##a_n##. It is also shown that ##A_n\geq \lambda##, where ##\lambda## is the limit of ##A_n##. The conversation also discusses the proof for the convergence of a subsequence of ##A_n##, which involves choosing a sequence of points that are closer to ##A_n## than the previous one. This leads to a subsequence that decreases
  • #1
Eclair_de_XII
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TL;DR Summary
Let ##\{a_n\}\subset \mathbb{R}##. Let ##A_k\equiv \sup\{a_n:n\geq k\}## and suppose that ##A_k\rightarrow \lambda##. Prove:
(a) There is a sequence ##B_n## with the properties:
(i) ##B_n<A_n## for all ##n##
(ii) ##B_n## is an increasing sequence
(iii) ##B_n\rightarrow \lambda##
(b) There are infinitely many ##k## such that ##a_n\in [B_n,A_n]##
(c) There exists a subsequence ##\{a_{n_k}\}## that converges to ##\lambda##
(d) The limit of a subsequence ##\{a_{n_k}\}## is at most ##\lambda##
First, it must be justified that ##A_n## is decreasing and is bounded from below by the point it converges to. See the other topic

(1) ##A_n## is decreasing
By definition, ##A_n## is the supremum of the set containing all ##a_k## where ##k\geq n+1## and the set containing ##a_n##. Hence, it must be an upper bound for ##A_{n+1}##. By definition, ##A_n\geq A_{n+1}##.

(2) ##A_n\geq \lambda##
If not, then choose ##k## such that ##A_k<\lambda-\epsilon## for some ##\epsilon>0##. Since ##A_n## is decreasing, it follows that for ##n\geq k##, no matter how large the interval of convergence is, ##A_n## must lie outside of it. This contradicts the fact that it converges. Hence, ##A_n\geq \lambda##.

Note: This is just a rephrasing of Mr. Office Shredder's post from that topic I mentioned.

===(a)===
Define ##B=2\lambda-A_n##.
==(i)
This follows from the fact that ##A_n## is bounded from below by ##\lambda##
##A_n\geq \lambda##
##-A_n\leq -\lambda##
##2\lambda-A_n\leq 2\lambda-\lambda##
##2\lambda-A_n\leq \lambda##
##B_n\leq \lambda\leq A_n##

==(ii)
This follows from the fact that ##A_n## is decreasing.
##A_{n+1}\leq A_n##
##-A_{n+1}\geq -A_n##
##2\lambda-A_{n+1}\geq 2\lambda-A_n##
##B_{n+1}\geq B_n##

==(iii)
Finally, if ##A_n\rightarrow \lambda##, then for any ##\epsilon>0##, there is some integer ##N>0## such that if ##n\geq N##:

##|A_n-\lambda|<\epsilon##
##-\epsilon<\lambda-A_n<\epsilon##
##\lambda-\epsilon<2\lambda-A_n<\lambda+\epsilon##
##\lambda-\epsilon<B_n<\lambda+\epsilon##
##-\epsilon<B_n-\lambda<\epsilon##
##|B_n-\lambda|<\epsilon##

===(b)===
Let ##\epsilon>0## and let ##A_n## be fixed. By definition, ##A_n\geq a_k## whenever ##k\geq n##. Hence, there are infinitely many points of this form below ##A_n##. If we can show that ##a_k\geq B_n##, for ##k\geq n##, we will have our result.

Suppose that it is not. Then ##B_n## must be an upper bound for ##\{a_k:k\geq n\}##. Since ##B_n<A_n##, this contradicts the fact that ##A_n## is the least upper bound. Hence, ##a_k\geq B## for ##k\geq n##.

===(c)===
Let ##\epsilon>0##. Then there is ##N\in \mathbb{N}## such that if ##n\geq N##:

##-\epsilon+\lambda<A_n<\epsilon+\lambda##

Choose ##\epsilon_1\in (0,\epsilon)##. Then there is ##n_1\geq n## such that ##a_{n_1}\in (A_n-\epsilon_1,A_n]##. Now choose an ##\epsilon_2>0## smaller than ##\epsilon_1## such that ##a_{n_1}\notin (A_n-\epsilon_2,A_n]##. For example, choose ##\epsilon_2\in (0,A_n-a_{n_1})##. Then there must be some ##n_2\geq n## such that ##a_{n_2}\in (A_n-\epsilon_2,A_n]##.

With each step, we choose a point that is closer to ##A_n## than the point chosen before it. Hence, for ##n_k\geq n##:

##\lambda-\epsilon\geq A_n-\epsilon<A_n-\epsilon_k<a_{n_k}\leq A_n<\lambda+\epsilon##
##\lambda-\epsilon<a_{n_k}<\lambda+\epsilon##
##-\epsilon<a_{n_k}-\lambda<\epsilon##
##|a_{n_k}-\lambda|<\epsilon##

===(d)===
Let some subsequence ##a_{n_k}## converge to ##\gamma\in\mathbb{R}##. Suppose that ##\gamma>\lambda##. Choose ##m\in\mathbb{N}## such that ##\gamma>A_m>\lambda##. Set ##\epsilon=\gamma-A_m##. Let ##N\in\mathbb{N}##. Then whenever ##n\geq N##, specifically, when ##n\geq m##:

\begin{align*}
|\gamma-a_n|&=&\gamma-a_n\\
&\geq&\gamma-A_m\\
&=&\epsilon
\end{align*}
 
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  • #2
Your proof for b is wrong. Consider the sequence ##a_k=0## if k is odd, and when ##k=2m## is even, ##a_{2m}=1+1/m##. Then ##A_{2m}=A_{2m-1}=1+1/m##, ##\lambda=1##, ##B_{2m}=B_{2m-1}=1-1/m##, but it is certainly not true that ##a_k\geq B_n## for ##k\geq n##, since when ##k## is odd ##a_k## is zero. Just because ##B_n## is larger than some points in a set doesn't mean it's larger than all the points in a set.

I thought your choice for ##B_n## is fairly clever and I agree with all your proofs for (a).
 
  • #3
What about if I used the fact that ##A_n## is a supremum for the subsequence ##\{a_k:k\geq n\}##?

For any positive ##\epsilon_1##, there must exist some ##n_1\geq n## such that ##a_{n_1}\in (A_n-\epsilon_1,A_n]##. In particular, choose ##\epsilon=A_n-B_n##. Now choose ##\epsilon_2\in(0,\epsilon_1)## such that ##(A_n-\epsilon_2,A_n]## excludes the ##a_{n_1}##. Now there is some ##n_2\geq n## such that ##a_{n_2}\in(A_n-\epsilon_2,A_n]##. Repeat this process in order construct a subsequence contained in ##[B_n,A_n] ## that converges to ##A_n##.
 
  • #4
Yes, that looks like what you want.

One thing to keep in mind for this, while your choice for ##B_n## is clever, it's also unnecessary. Any increasing sequence that converges to ##\lambda## will also work here.For part (c) this sentence stood out.

Eclair_de_XII said:
With each step, we choose a point that is closer to ##A_n## than the point chosen before it. Hence, for ##n_k\geq n##:

##\lambda-\epsilon\geq A_n-\epsilon<A_n-\epsilon_k<a_{n_k}\leq A_n<\lambda+\epsilon##
##\lambda-\epsilon<a_{n_k}<\lambda+\epsilon##
##-\epsilon<a_{n_k}-\lambda<\epsilon##
##|a_{n_k}-\lambda|<\epsilon##

The sentence you wrote obviously disproves what you are trying to do. It sounds like your sequence ##a_{n_k}## converges to ##A_n##, but that means it doesn't converge to ##\lambda## unless they happen to coincide.

You need to actually look at a sequence of ##A_k##s and for each one pick a ##a_{n_k}## that is really close to ##A_k## in order to get convergence to ##\lambda##.
 
  • #5
Office_Shredder said:
You need to actually look at a sequence of ##A_k## s and for each one pick a ##a_{n_k}## that is really close to ##A_k## in order to get convergence to ##\lambda##.

I figure I could choose some ##A_k## at random, set ##\epsilon=A_k-\lambda##, then choose an ##n_1>k## such that ##a_{n_1}\in (A_k-\epsilon,A_k]##. From then on, I'd pick some ##k_1>k## such that ##a_{n_1}>A_{k_1}##. Set ##\epsilon=A_{k_1}-\lambda##. Then choose ##a_{n_2}\in (A_{k_1}-\epsilon,A_{k_1}]##. Now find a ##k_2>k_1## such that ##A_{k_2}<a_{n_2}##. And so on.

Basically, I'm just choosing some ##A_k## at random, choosing some ##a_{n_k}## between ##\lambda## and ##A_k##, then choosing an element of the sequence of supremums smaller than the chosen ##a_{n_k}##. In this way, I obtain a subsequence of ##\{A_n\}## and a subsequence ##\{a_{n_k}\}## that decreases alongside ##A_n##. It sounds like it would converge by the Sandwich Theorem but this book didn't cover that yet, so it wouldn't be valid to use it, I should think.
 
  • #6
Yeah I think you can just say it like this, without relying on any theorems.

For every ##k##, pick ##A_{n_k}## such that ##|\lambda - A_{n_k}| < 1/(2k)##, and then pick ##a_{n_k}## such that ##|A_{n_k}-a_{n_k}| < 1/(2k)##. Then by the triangle inequality ##|\lambda - a_{n_k}| < 1/k##. So ##a_{n_k}## converges to ##\lambda##.

Your proof for d doesn't look complete, I'm not sure what the conclusion is there. You also just drop the || from your first inequality but you don't actually know what the sign of ##\gamma-a_n## is, so probably you should think about it a bit more.
 
  • #7
Office_Shredder said:
but you don't actually know what the sign of ##\gamma-a_n## is, so probably you should think about it a bit more.

I disagree. By definition, ##A_m\geq a_n## for all ##n\geq m##. By hypothesis, ##\gamma>A_m\geq a_n##. My conclusion is that if there is a subsequence ##a_{n_k}## that converges to some number, then that number cannot exceed ##\lambda##.
 
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  • #8
Oh you're right, the proof works. You shouldn't even mention ##N##, just delete all references to it.
 
  • #9
Right-o, right-o. In my defense, I was trying to recite the negation of the epsilon-delta definition of the convergence of sequences of real numbers. As always, thank you for your valuable input, besides.
 
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  • #10
If you really want to hammer it home I would write something like

Let ##N=m,##. Then for all ##n\geq N ##, we have ##|\gamma - a_n| \geq \epsilon##
 
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Related to Continuation of sequence of supremums question

1. What is the definition of a sequence of supremums?

A sequence of supremums is a set of numbers that are the least upper bounds of a given set of numbers. In other words, each number in the sequence is greater than or equal to every number in the set.

2. How is a sequence of supremums related to the concept of limits?

A sequence of supremums is closely related to the concept of limits. In fact, the limit of a sequence of supremums is equal to the supremum of the set of limits. This means that as the sequence of supremums approaches a certain value, the supremum of the set also approaches that value.

3. Can a sequence of supremums have an infinite number of terms?

Yes, a sequence of supremums can have an infinite number of terms. This is because a set can have an infinite number of elements, and each element can have a supremum.

4. How is the continuity of a function related to a sequence of supremums?

The continuity of a function is closely related to a sequence of supremums. A function is continuous at a point if and only if the supremum of the function at that point is equal to the limit of the sequence of supremums as it approaches that point.

5. What is the importance of studying the continuation of sequence of supremums?

Studying the continuation of sequence of supremums is important in many areas of mathematics, such as analysis and topology. It helps us understand the behavior of sets and functions and allows us to make predictions about their limits and continuity. It also has practical applications in fields such as physics and engineering.

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