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This is Exercise 2.6.6 in Stephen Abbott, Understanding Analysis.

A sequence $(a_n)$ is $\textit{quasi-increasing}$ if for all $\epsilon >0$, there exists an $N\in\mathbb{N}$ such that whenever $n>m\geq N$ it follows that $a_n>a_m-\epsilon$

Is there an analogue of the Monotone Convergence Theorem for quasi-increasing sequences?

The amazing Brian M. Scott has provided a hint for me here https://math.stackexchange.com/a/1953323/633922 but I was unable to figure it out.

I have been stuck on this problem for quite a few days now and I get the idea is to show $(a_n)\rightarrow \sup(a_n)$ but I just do not know how to structure my proof. This is what I got so far,

Assuming $(a_n)$ is a quasi-increasing sequence that is bounded then by Axiom of Completeness it contains a $\sup(a_n)=\alpha$. Since the question is asking about monotone convergence it is natural that we want to show

$\forall \epsilon >0$, $\exists N\in\mathbb{N}$ such that if $n\geq N$ then $|a_n-\alpha|<\epsilon$. So $\alpha$ exists implies $\alpha-\epsilon$ is not an upper bound for the sequence. Now for $n>m\geq N$ we have that $\alpha-\epsilon < a_m<a_n < \alpha$ for $N$ large enough. But it is not always true that $a_n>a_m$ $(\color{red}{\text{Not sure if that is a correct assumption}})$

So I do not really know how to proceed. I feel like I need to fix my epsilons so things work out nicely but I am surely lost. I have no idea how to use $a_n>a_m-\epsilon$ to show convergence to the $\sup(a_n)$. I do notice that I can choose epsilon for $a_m-\epsilon$ such that ($a_n>a_m-\frac{\epsilon}{10}$), ($a_n>a_m-\frac{\epsilon}{100}$), but not sure how to apply it to the proof. I would really appreciate any guidance/direction/intuition.

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  • $\begingroup$ Actually, $a_n$ doesn’t converge to its sup, because (for instance) you can add finitely many very large values at the beginning and the sequence remains quasi-increasing. What $a_n$ converges to is its limsup. The simplest way to prove it, I think, is to show that a bounded quasi-increasing sequence has at most one limit point – I think you can somewhat imitate the proof of uniqueness of limits. $\endgroup$ Commented Feb 14, 2021 at 0:53
  • $\begingroup$ I see, thank you. I am wondering If I am allowed to claim it converges since that is what we are trying to show. Or should I argue by contradiction that if two limits exist they would be the same? Im having trouble understanding what $\lim\sup$ means since $\sup(a_n)$ is just a single value. $\endgroup$ Commented Feb 14, 2021 at 1:00
  • $\begingroup$ Do you know what a limit point is? (Not a limit – so it’s certainly not enough to show that any two limits should be the same. You need to prove that the limits of any two convergent subsequences must be the same). If so, the limsup is the supremum of the limit points. In other words, it’s the smallest real number $r$ such that for any $\epsilon>0$, the sequence contains finitely many terms above $r+\epsilon$. Or the limit of the decreasing sequence $\left(\sup_{p\geq n}\,{a_p}\right)_n$. $\endgroup$ Commented Feb 14, 2021 at 1:25
  • $\begingroup$ I see I looked in the book and the limit point is defined after this chapter in the topology section so I have yet to learn such a definition. So I suppose I need to use Bolzano-Weierstrass, claim there are two different limits and arrive at a contradiction? $\endgroup$ Commented Feb 14, 2021 at 1:28
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    $\begingroup$ @AlejandroRuiz see answer below. $\endgroup$ Commented Jul 20, 2021 at 1:18

1 Answer 1

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Since $(a_n)$ is bounded, it follows from Bolzano-Weiersteass Theorem that it has a convergent subsequence $(a_{n_k})$. Let $a=\lim_{k\rightarrow\infty} a_{n_k}$. Since $(a_n)$ is quasi-increasing, for any $\epsilon >0$ there exists an $N_1$ such that if $n>m\geq N_1$ then $$a_n>a_m-\dfrac{\epsilon}{2}$$ For the same $\epsilon$, since $(a_{nk})$ converges to $a$, there exists a $K$ such that $k\geq K$ implying $$|a_{n_k}-a|<\dfrac{\epsilon}{2}$$ Without loss of generality we can further assume that $n_K\geq N_1$, then letting $N=n_K$. We can see that for any $n\geq N$ we can find a $l>k\geq K$ such that $n_k\leq n\leq n_l$. Then since $(a_n)$ is quasi-increasing we have $$a_n>a_{n_k}-\dfrac{\epsilon}{2}\quad\text{and}\quad a_{n_l}>a_n-\dfrac{\epsilon}{2}$$ We have $k,l\geq K$, $|a_{n_k}-a|<\frac{\epsilon}{2}$ and $|a_{n_l}-a|<\frac{\epsilon}{2}$. Particularly, $a_{n_k} > a-\frac{\epsilon}{2}$ and $a_{n_l} > a+\frac{\epsilon}{2}$. Therefore $$\begin{gather*} a_n >a_{n_k}-\dfrac{\epsilon}{2}> a-\epsilon\\ a_n <a_{n_l}+\dfrac{\epsilon}{2}< a+\epsilon\\ \Longrightarrow |a_n-a|<\epsilon\\ \Longrightarrow \lim a_n= a. \end{gather*}$$

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  • $\begingroup$ In the last part did you mean $a_{n_l} < a+\frac{\epsilon}{2}$ less than rather than greater than $a_{n_l} > a+\frac{\epsilon}{2}$? $\endgroup$ Commented Feb 13, 2024 at 3:18

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