My Question: I am not sure about the very last inequality in the proof below; namely, where did we get $\mid a_{n}-a_{N}\mid$ and $\mid a_{N}-b\mid$? I see that $\mid a_{n}-a_{N}\mid<\epsilon/2$ and that $\mid a_{N}-b\mid\leqslant\epsilon/2$, which validates the last inequality, but i just don't see how we chose those two pieces. You can probably skip 2/3 or more of what I have written below to answer my question, I just wrote everything so that it would be clear where I am coming from. Also, I don't see why $\mid a_{n}-b\mid\leqslant\mid a_{n}-a_{N}\mid+\mid a_{N}-b\mid$ is true. Thanks for your time.
Notation let the sequence $a_1, a_2, ..., a_n=(a_n$) for $n\in \mathbb{N}$
Definition: The Cauchy condition states that if a sequence $a_{n}$ is convergent then the following condition must be true as well:
$$ (\forall\epsilon\in\mathbb{R}_{+})(\exists N\in\mathbb{N})(n,m\geqslant N\implies\mid a_{n}-a_{m}\mid<\epsilon)$$
Theorem: $\mathbb{R}$ is complete with respects to the Cauchy sequences in the sense that if $(a_{n})$ is a sequence of real numbers obeying the Cauchy condition then $(a_{n})$ converges to a limit in $\mathbb{R}$.
Proof. Let the set $A$ be the set of all the real numbers comprising the sequence $(a_{n})$, $$ A=\{x\in\mathbb{R}:(\exists n\in\mathbb{N})(a_n=x)\}$$
We first observe that the set $A$ is bounded, by taking $\epsilon=1$ in the Caunchy condition which implies that there must exist a $N_{1}\in\mathbb{N}$ such that for all $n,m\geqslant N_{1}$ , $\mid a_{n}-a_{m}\mid<1$. Now, clearly that the finite set $A^*=\{a_{1},a_{2},...,a_{N_{1}},a_{N_{1}}-1,a_{N_{1}}+1\}$ is bounded, since all finite sets are bounded. Let us say that all the elements of $A^*$ belong to to the interval $[-M,M]$. Since we require that for all $n,m\geqslant N_{1}$, $\mid a_{n}-a_{m}\mid<1$ must be true, it follows that for all $n\geqslant N_{1}$, the inequality $ \mid a_{n}-a_{N_{1}}\mid<1$ is also true. The latter inequality suggest that $A$ is also contained in the interval [-M,M], hence $A$ is bounded.
Next we consider the set, $$S=\{s\in[-M,M]:\exists\mbox{ infinitely many }n\in\mathbb{N},\mbox{ for which }a_{n}\geqslant s\}$$
That is that $a_{n}\geqslant s$ occurs infinitely many times in the sequence $(a_{n})$. Clearly $-M\in S$ and $S$ is bounded above by $M$. According to the least upper bound property of $\mathbb{R}$ we know that $S$ must have a least upper bound, i.e. , $\mbox{l.u.b.}(S)=b$ where $b\in\mathbb{R}$. Thus, we claim that the Caunchy sequence $(a_{n})$ converges to the limit $b$.
Given that for all $\epsilon>0$ we must show that there exists an $N$ such that for all $n\geqslant N$, $\mid a_{n}-b\mid<\epsilon$. The Cauchy condition provides us with a $N_{2}$ such that
$$m,n\geqslant N_{2}\implies\mid a_{n}-a_{m}\mid<\frac{\epsilon}{2}$$
All of the element of $S$ are less than or equal to $b$ so, $b+\epsilon/2\notin S$ . Thus if the element $b+\epsilon/2$ does occur in $(a_{n})$ then it could only occur finitely many times, and thus there must exist a $N_{3}\geqslant N_{2}$ such that $$n\geqslant N_{3}\implies a_{n}\leqslant b+\frac{\epsilon}{2}$$ Since $\mbox{l.u.b.}(S)=b$ , the number $b-\epsilon/2$ cannot be be an upper bound for $S$; thus there exists an element $s\in S$ such that $s>b-\epsilon/2$ which implies that $a_{n}\geqslant s>b-\epsilon/2$ occurs infinitely often. In particular, there exists a $N\geqslant N_{3}$ such that $a_{N}>b-\epsilon/2$ . Since $N\geqslant N_{3}$ , we have $a_{N}\leqslant b+\epsilon/2$ and so we have that $$a_{N}\in(b-\epsilon/2,b+\epsilon/2]$$ Moreover, since $N\geqslant N_{3}\geqslant N_{2}$ $$\mid a_{n}-b\mid\leqslant\mid a_{n}-a_{N}\mid+\mid a_{N}-b\mid<\epsilon$$ which verifies convergence.
