I'm trying to follow a proof from Ross's Real Analysis textbook, but am finding myself unable to follow the end of the proof.
Here's a rewrite of what I believe I understand from this proof, following Ross's proof.
Proof: Let $(s_n)$ be a bounded, increasing sequence. Since it is bounded above, it contains a supremum, $u = \sup s$, by the Completeness Axiom. Letting $\lim s_n = s$, we need to show \begin{align*} \forall \epsilon > 0, \exists N \in \mathbb{N}, \forall n > N, \left \lvert s_n - s \right \rvert < \epsilon. \end{align*} So, let $\epsilon > 0$. Since $u - \epsilon < u$, $u - \epsilon$ is not an upper bound of $(s_n)$, as $u$ was the smallest such upper bound. Thus, there is some element of $(s_n)$ that is strictly greater than $u - \epsilon$. Denote this by $s_N$, i.e., $s_N > u - \epsilon$. Since $(s_n)$ is increasing, we have that $s_N \leq s_n$ for all $n > N$. So, for sufficiently large $n$, we have \begin{align*} u - \epsilon < s_N \leq s_n \leq u. \end{align*} From here, Ross reasons that $u - \epsilon < s_n \leq u$, which I understand and which follows from the above inequality chain, but he concludes that $\left \lvert s_n - u \right \rvert < \epsilon$, which is our goal. I'm having trouble seeing from where this step originates. I'm unsure, first, on how we convert a non-strict inequality into an absolute value. Second, it seems to me that we would need to show that $-\epsilon < s_n - u < \epsilon$ to conclude $\left \lvert s_n - u \right \rvert < \epsilon$.
I'd appreciate any insights people may have on this, particularly this final step.
Starting with $\mu - \epsilon < s_N \leq \mu$
Subtract $\mu$ from the equation and note that all three parts of the chained inequality are now $\leq 0$. When you take the absolute values again, you can flip the signs.