Bartle Theorem 3.2.3 (b) Trouble applying a variation of triangular inequality (Sequence section)

Question

I have trouble understanding a proof because of applying a variation of the triangular inequality. Below images show the theorem and the proof (I highlighted the troubling section in red).

Theorem 3.2.3 (b)

Proof of 3.2.3 (b)

2.2.4 Corollary:

If $\mathtt{a, b} \in \mathbb{R}$, then $||\mathtt{a}| - |\mathtt{b}|| \le |\mathtt{a-b}|$

As you can see, the proof uses Corollary 2.2.4 and derives the following inequality:

$$-\alpha \le -|z_n - z| \le |z_n| - |z| $$

But when I apply the Corollary I get, $$-\alpha \le -|z_n - z| \le -||z_n| - |z|| $$

So I wonder how you can get $|z_n| - |z|$ from $-||z_n| - |z||$, because, provided that my application of the Corollary was correct, this is only possible when $|z_n| - |z| \lt 0$. But I don't think we can say $|z_n| - |z| \lt 0$ in general.

I'd really appreciate clarification.

Answer 1

For every $x$ you have $-|x|\le x\le |x|$, in particular $-||a|-|b||\le |a|-|b|$. So $-||z_n|-|z||\le |z_n|-|z|$, and the claimed inequality follows from your inequality.