How can I derive this expression related to the triangle inequality?

In my real analysis text book, there is an expression like this for proof of some limit of a sequence: $$n > N \implies \ |s_n - s | < 1$$ By using the triangular inequality, $n>N \implies \ |s_n| < |s| + 1$. I don't understand how this can be.

Let $N$ be a normed vector space over the field $\Bbb F$, where $\Bbb F = \Bbb R$ or $\Bbb F = \Bbb C$. We denote the norm of $n$ by $\Vert n \Vert$ for $n \in N$. Then for $m, n \in N$ we certainly have

$(n - m) + m = n, \tag{1}$

so by the triangle inequality

$\Vert n \Vert \le \Vert n - m \Vert + \Vert m \Vert, \tag{2}$

and reversing the roles of $m$ and $n$ we find that

$\Vert m \Vert \le \Vert n - m \Vert + \Vert n \Vert. \tag{3}$

(2) yields

$\Vert n \Vert - \Vert m \Vert \le \Vert n - m\Vert, \tag{4}$

and (3) gives

$\Vert m \Vert - \Vert n \Vert \le \Vert n - m \Vert. \tag{5}$

Taken together, (4) and (5) imply

$\vert \Vert n \Vert - \Vert m \Vert \vert \le \Vert n - m \Vert, \tag{6}$,

where $\vert c \vert$ is the ordinary absolute value for $c \in \Bbb F$. Applying (6) to the inequality

$\vert s_n - s \vert < 1 \tag{7}$

we see that

$\vert \vert s_n \vert - \vert s \vert \vert < 1 \tag{8}$

as well, whence

$-1 < \vert s_n \vert - \vert s \vert < 1, \tag{9}$

from which

$\vert s_n \vert < \vert s \vert + 1 \tag{10}$

immediately follows.

I should in all honesty say I didn't invent (6); it is found in any of about a gajazillion analysis texts, under discussions about properties of norms.

Hope this helps. Cheers, and as always,

Fiat Lux!