Having trouble understanding a cardinality-related argument when proving that all Cauchy sequences of reals numbers converge to a real limit. Came across it on CC Pugh's Real Mathematical Analysis, page 18.
Suppose $(a_n) : n \in \Bbb N $ is a Cauchy sequence;
First the set $A = \{ x \ |\ x = a_n \; \text{for some} \; n \in \Bbb N \}$ is proved to be bounded and is considered to be contained in $[-M, M]$ for $M\in \Bbb N$. Then the following set is considered;
$$S = \{s \in [-M, M] \ | \ \text{there are infinitely many $n$ in $\Bbb N$ for which $a_n \ge s $} \}$$
The set is bounded from above and is non-empty rendering the existence of $\sup S = b$ in $\Bbb R$. Then the author proceeds to show that $b$ is the limit to which $(a_n)$ converges.
Since $(a_n)$ is a Cauchy sequence $\exists N \in \Bbb N$ such that $m, n \ge N \implies |a_m - a_n| \lt \frac \epsilon 2$ where $\epsilon$ is an arbitrary positive quantity.
Here my issues begin..
$ (b + \frac \epsilon 2) \ \notin \ S$ since $s \ \in S \implies s \le b$. Therefore the author says $a_n$ exceeds $(b + \frac \epsilon 2)$ only finitely often. This apparently suggests $ \exists \ N_1(\ge N) \in \Bbb N $ such that $$ n \ge N_1 \implies a_n \lt (b + \frac \epsilon 2) $$
This is my sketchy idea of why this is true: There are countably infinite elements in $(a_n)$ and hence there are a similar number of elements $(a_n)$ which are less than $(b + \frac \epsilon 2)$. These must be close to each other since this is a Cauchy sequence.
But I am having trouble convincing myself in terms of a rigorous argument. Especially why $N_1 \ge N$??? Furthermore since $\exists \ s \in S$ such that $s \gt (b - \frac \epsilon 2)$ There are infinitely many $n$ for which $a_n \ge s \gt (b - \frac \epsilon 2)$. This apparently justifies the existence of $r \ge N_1$ such that $a_r \gt (b - \frac \epsilon 2)$??? How come??
Any help would be appreciated...