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.
Best Answer
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 $$\begin{gather*} a_n >a_{n_k}-\dfrac{\epsilon}{2}> a-\epsilon\\ a_n <a_{n_l}+\dfrac{\epsilon}{2}< a+\epsilon\\ \Longrightarrow |a_n-a|<\epsilon\\ \Longrightarrow \lim a_n= a. \end{gather*}$$