This is a question from a book that I am reading:
Prove that a monotone non-decreasing sequence of real numbers which is bounded above converges to a limit $a$, and that limit is the least upper bound of the set $\{a_1, a_2, …\}$. (Similarly, prove the same with non-increasing sequences bounded below, ie $\lim_{n\rightarrow \infty} a_n =$ greatest lower bound.)
My issue with this is that this is perfectly intuitive. It makes complete sense, but I just have no idea how to formulate this in a mathematically formal way.
My attempt goes like this: Let $\{a_n\}$ be bounded above and monotone non-decreasing. Then, $a_n \leq a_{n+1} \leq M$, where $M$ denotes an upper bound. From here, it is clear that there exists some least upper bound of this sequence $\leq M$, denoted $a^*$. For this part, I'm not sure if I can just say that this is clear, even though it does seem obvious. I assume that I could argue by contrapositive, and if I assumed there was no least upper bound then there is no way that this sequence is bounded above.
Then, $a_n \leq a^* \leq M$ for all $n \in \mathbb{N}$. This leads to $0 \leq a^* – a_n \leq M – a_n$. At this point, it looks pretty close to being the definition of the limit. Can I choose $\epsilon = 1$, and note that since $a_n \leq a^*$ for any $n$, then certainly for any sufficiently large $n>N$, $a^* – a_n = |a_n – a^*| < \epsilon = 1$, implying that $a^*$ is in fact our limit.
Best Answer
Given $\varepsilon>0$ there exists $N$ such that $a^*-\varepsilon<a_N\leq a^*$.
Let any $n>N$. As the sequence $\{a_n\}$ is monotone non-decreasing, $a_n\geq a_N>a^*-\varepsilon$. But, for all $n$, $a_n\leq a^*$. So, in particular, this means that for all $n>N$, $a^*-\varepsilon<a_n\leq a^*<a^*+\varepsilon$ implying for all $n>N$, $|a_n-a^*|<\varepsilon$.
This is exactly the definition of limit of a sequence, which proves the problem.