Show that every bounded and monotonic sequence is Cauchy.
My proof:
Suppose $(a_n)$ is a bounded and monotonic sequence.
Since $(a_n)$ is bounded, it follows that $\exists M \in \mathbb{Q}^{+}$ such that $|a_n| \leq M$.
Since $(a_n)$ is monotonic, it follows that:
- $a_{n+1} \geq a_n$ or,
- $a_{n+1} \leq a_n$
Let $\epsilon > 0$ and let $N\in\mathbb{N}^{+}$, such that $n,m > N$. Suppose without loss of generality that $n>m$, and suppose condition 1 of monotonic. It follows that: $$ |a_n - a_m| = a_n - a_m \leq a_n \leq M \leq NM $$ Now let's define $N\leq \frac{\epsilon}{M}$, therefore: $NM \leq \epsilon$.
It follows the same way for condition two, using the fact that $|a_n - a_m| = a_m - a_n \leq a_m$.
After that, I went to check if what I've concluded is indeed correct... So now consider the sequence $a_n = \frac{1}{n}$.
It follows that $|a_n| \leq 1$ and $a_{n+1} \leq a_n$, so if my proof was correct, I could pick a random $\epsilon$ and get a positive integer value for $N$ such that $N \leq \frac{\epsilon}{M}$. Suppose for example $\epsilon = 0.2$. Therefore it's impossible to get an $N$ by my definition and now I'm stuck.
I know that my definition of $N$ is incorrect and I see where it fails... I just don't know yet how to define it in a different way to get a $\leq \epsilon$ at the end, and that still makes a valid $N$ regardless of $\epsilon$.
My intuition is that I should define $N$ in terms of $N\geq \frac{\text{something}}{\epsilon}$, because for small values of $\epsilon$, makes sense that $N$ is huge.
Can someone please point out how should I define $N$? Am I in the right path?
Thanks!