Question on proof concerning : A bounded, monotone sequence converges

Question

Proposition : A bounded, monotone sequence converges.

Proof : Suppose $\{a_{n}\}$ is a non decreasing sequence that is bounded above by some number call it $M$. Since $\{a_{n}\}$ is bounded, it follows that it has a least upper bound (here lets call it $\sup(a_{n})$).

Suppose $\sup(a_{n}) = S$. Then we know $\forall \epsilon > 0$, $S - \epsilon$ can not be a bound for $\{a_{n}\}$ since $S$ is the least upper bound. It also means that $\forall N \in \mathbb{N}$, $S - \epsilon < a_{N} \leq S$. Okay here is where my question comes into play.

At this point in the proof, are we saying that since the sequence is bounded by $S$, I can find some arbitrary open interval around $S$ where some part of the sequence is contained within this interval? Is this always true? Also, $N$ is that point where the remainder of this sequence is in the $\epsilon$ interval? In other words, when I say $a_{N}$, does that mean it is the start of the sequence where $a_{n}$ is in my $\epsilon$ interval?

Lastly, we are looking to show that $|a_{n} - S| < \epsilon$. Does showing $a_{N}$ and the terms after that being contained in $\epsilon$ do so? Sorry if this last question is stupid. I always get a little confused when we switch from $a_{n}$ to $a_{N}$.

Thanks!

Answer 1

We know that for all $\varepsilon > 0$, $S-\varepsilon$ is not a bound for the sequence, so there exists $N$ such that $S - \varepsilon < a_N \leq S$. You have written $\forall N \in \mathbb{N}$, which isn't strictly true.

However, since $\{a_n\}$ is non-decreasing, we also know that $a_n \geq a_N\;\forall \, n \geq N$. Therefore, $S - \varepsilon < a_N \leq a_n \leq S\;\forall \, n \geq N$. In other words, $|S - a_n| < \varepsilon \; \forall \, n \geq N$, so $a_n \rightarrow S$.