Prove that limited increasing or decreasing successions have a limit.

Question

Be $u_n$ an increasing sequence of numbers ${(u_1,u_2,u_3, \ldots)}$ such that for all $\ n$, $u_{n-1}\le u_n$

Prove that if $u_n\le a, a \in \mathbb R$, for all n then $\lim_{n\rightarrow\infty}u_n =b $ for some number $b\le a$.

I am using as definition of limit of a succession the following:

$\lim_{n\rightarrow\infty}u_n =b \Leftrightarrow ∀\delta \gt 0, ∃p \in \mathbb N :n\gt p \Rightarrow |u_n -b|\lt \delta$

Answer 1

This is the https://en.wikipedia.org/wiki/Monotone_convergence_theorem. Here's a copy of the proof. If you have questions you can ask in the comments:

Let $\{ a_n \}$ be such a sequence. By assumption, $\{ a_n \}$ is non-empty and bounded above. By the least upper bound property of real numbers, $c = \sup_n \{a_n\}$ exists and is finite. Now, for every $\varepsilon > 0$, there exists $N$ such that $a_N > c - \varepsilon $, since otherwise $c - \varepsilon $ is an upper bound of $\{ a_n \}$, which contradicts to the definition of $c$. Then since $\{ a_n \}$ is increasing, and $c$ is its upper bound, for every $n > N$, we have $|c - a_n| \leq |c - a_N| < \varepsilon $. Hence, by definition, the limit of $\{ a_n \}$ is $\sup_n \{a_n\}.$