Prove that if x > limsup s_n, then x is not the limit of any subsequence

[fmam3]

Homework Statement

Directly from the definition, for a sequence $$(s_n)_{n \in \mathbb{N}} \subseteq \mathbb{R}$$ prove that if $$x > \limsup s_n$$, then $$x$$ is not the limit of any subsequence of $$(s_n)$$. (i.e. Do not use the fact that $$\limsup s_n$$ is the supremum of the set of subsequential limits.)

Homework Equations

I have been told by my instructor that my proof will fail due to problems with inequalities --- but I fail to see where it would fail; i.e. are there any errors where $$>$$ should be $$\ge$$ or vice-versa?

The Attempt at a Solution

Please see the attachment.

Thanks all!

 

[CaffeineJunky]

This should fall into place by using both facts. Try using a contradiction. I have a rough proof below:

WARNING: A POSSIBLE SOLUTION IS HERE. DO NOT LOOK UNLESS YOU WANT TO.

Suppose x > lim sup (S_n) and some subsequence (S_K(n)) has x as a limit, where K(n) is a strictly increasing function of natural numbers. If x is the limit of (S_K(n)), then for every epsilon there exists a M such that for all m > M, |S_K(m) - x| < epsilon. Since (S_K(n)) contains only terms in (S_n) and all terms of (S_n) are smaller than x, then every term of (S_K(n)) is smaller than x. Hence for any m > M, |S_K(m) - x| > 0. Let T = lim inf (|S_K(m) - x|) for all m > M. If T = 0, then x = lim sup (S_K(n)) which is less than or equal to (S_n), a contradiction. Hence T is non-zero. Since T is non-zero and positive, take epsilon to be smaller than T, say epsilon = T / 2. |S_K(m) - x| >= T > epsilon for all m > M. Hence x is not the limit of (S_K(m)), a contradiction.

 

[fmam3]

This should fall into place by using both facts. Try using a contradiction. I have a rough proof below:

WARNING: A POSSIBLE SOLUTION IS HERE. DO NOT LOOK UNLESS YOU WANT TO.

Suppose x > lim sup (S_n) and some subsequence (S_K(n)) has x as a limit, where K(n) is a strictly increasing function of natural numbers. If x is the limit of (S_K(n)), then for every epsilon there exists a M such that for all m > M, |S_K(m) - x| < epsilon. Since (S_K(n)) contains only terms in (S_n) and all terms of (S_n) are smaller than x, then every term of (S_K(n)) is smaller than x. Hence for any m > M, |S_K(m) - x| > 0. Let T = lim inf (|S_K(m) - x|) for all m > M. If T = 0, then x = lim sup (S_K(n)) which is less than or equal to (S_n), a contradiction. Hence T is non-zero. Since T is non-zero and positive, take epsilon to be smaller than T, say epsilon = T / 2. |S_K(m) - x| >= T > epsilon for all m > M. Hence x is not the limit of (S_K(m)), a contradiction.

@CaffeineJunky

Thanks for your proof! May I ask some follow up questions --- just for the sake of knowing the motivations, etc.

You wrote let $$T = \liminf (|S_{K(m)} - x|)$$ for all $$m > M$$ and showed that we must have $$T > 0$$ and then set $$\varepsilon = T / 2$$. I have two questions:

(1) What lead your motivation for constructing $$T = \liminf (|S_{K(m)} - x|)$$? I just want to know what "sparked" you to think of using the limit inferior for this proof, as I'd never thought of introducing this when I wrote my attempted proof (see attached).

(2) May I confirm my understanding of your last second to last sentence? You wrote that if $$T > 0$$, then we have $$|S_{K(m)} - x| \geq T$$.

That is, if in general we have a sequence, say, $$(a_n)$$ in $$\mathbb{R}$$, and if $$\lim a_n = a$$, where $$a > 0$$, then we can conclude that there exists some $$N$$ such that for all $$n > N, a_n > 0$$; that is, it is not necessarily true that we have $$a_n > a$$ for all $$n > N$$. But in this case, since $$T$$ is also the limit inferior, we must have that $$|S_{K(m)} - x| \geq T > 0$$ by properties of the infimum. Is this understanding correct?

Thanks!

 

[CaffeineJunky]

I'm not sure if the proof is 100% correct. I mean, I did word it poorly and things do need to be fixed up. but I believe the general idea is correct.

1. It just seemed to fit in order to find a value of epsilon: Take the smallest element of the set and take a smaller element to be your epsilon. Then no matter what you wouldn't be able to produce a M such that the limit inequality is less than the chosen number, which ends up contradicting the fact that every epsilon had such an M. (Although in this case, the number M isn't really as important as a well-chosen epsilon.)

Keep in mind that the goal is to show that there is no convergent subsequence whose limit is x. I had to use the negation of the definition in order to prove this to be the case, since we have little else to work with.

2. I believe your assertion is correct. It is certainly the case that A_n doesn't always has to be greater than a. Think of something such as sin(n) * exp(-n) + 1. This constantly oscillates but it still converges to 1 as n becomes arbitrarily large, because at some point it has to be within a small enough neighborhood of 1 for sufficiently large n. If it doesn't, then intuitively convergence cannot happen. (Remember, only the ultimate behavior of a sequence matters when dealing with convergence.)

P.S. I am unable to view your attachment since it is pending approval.

 

[fmam3]

Thanks for the reply!

I'm new to this forum and I was not aware that attachments need to be "approved" --- for future postings, I'll just directly type my message rather than attaching it.

Thank you again for your proof and clarifications!