# That a sequence is Cauchy implies it's bounded.

I'm having some trouble following the proof in Ross's real-analysis textbook that all cauchy sequences are bounded. Here's what I'm able to follow thus far, along with the particular step I can't seem to grasp.

Proof: Let $(s_n)$ be a Cauchy sequence, so by definition we have \begin{align*} \forall \epsilon > 0, \exists N, \forall m, n > N, \left \lvert s_n - s_m \right \rvert < \epsilon. \end{align*} Let $\epsilon = 1$, and choose some $N$ such that, $\forall m, n > N$, $\left \lvert s_n - s_m \right \rvert < \epsilon$. Then, since this holds for all sufficiently large $m$, take $m = N + 1$, so we have that $\forall n > N, \left \lvert s_n - s_{N+1} \right \rvert < \epsilon$.

This is the point at which I'm confused. From here, Ross concludes that $\left \rvert s_n \right \rvert < \left \lvert s_{N+1} \right \rvert + 1$ for sufficiently large $n$. So, for sufficiently large $n$, $\left \lvert s_n \right \rvert \leq \left \lvert s_{N+1} \right \rvert + 1$, the terms $s_n$ for $n \leq N$ are bounded by the absolute value of whichever of these finite terms is largest, i.e., $M = \max\{\left \lvert s_{N+1} \right \rvert + 1, \left \lvert s_1 \right \rvert, \left \lvert s_2 \right \rvert, \ldots, \left \lvert s_N \right \rvert\}$, and $\forall n \in \mathbb{N}$, $\left \lvert s_n \right \rvert \leq M$.

I have two questions on this proof:

(a) First, from where do we get $\left \rvert s_n \right \rvert < \left \lvert s_{N+1} \right \rvert + 1$? I haven't been able to string together this inequality by using the Triangle Inequality or via any form of algebraic manipulation.

(b) After we secure this step, I understand where we get to $\left \lvert s_n \right \rvert \leq M$. How, though, does this allow us to conclude that the sequence is bounded? Wouldn't we want to prove that $s_n \geq M$? Or do we argue that, if $s_n \geq 0$, this holds, but if $s_n \leq 0$, it likewise holds by transitivty?

Remember that he chose $\epsilon = 1$, so $|s_n-s_{N+1}|<\epsilon = 1$ says that $|s_n|<|s_{N+1}| + 1$ by the reverse triangle inequality [$|a|-|b|\le |a-b|$].
For your second question, you're misunderstanding what bounded means. It means that the sequence (or set, in general) is bounded both below and above. So saying that $-M\le x\le M$ (or $|x|\le M$) tells you precisely that.