Finding a Real Number using epslion Fix a real number $x$ and $\epsilon>0$. If $|x-1|\le \epsilon$ show $|2-x|\ge 1- \epsilon$
I think we were supposed to use the triangle inequality to show this.
If we use the triangle inequality then $|x-1|\le ||x|-|1||$ so then $x\le \epsilon+1$ so then $-x\ge-\epsilon-1$ adding 2 to each side we get $2-x\ge 1-\epsilon$ 
does that make sense?

In general, if $a,b\in\mathbb C$ then $$\big||a|-|b|\big|\leqslant |a|-|b|. $$
For
$$|a| \leqslant |a-b| + |b|\implies |a|-|b|\leqslant|a-b| $$
and similarly
$$|b| \leqslant |a-b| + |a|\implies |a|-|b|\leqslant|a-b|.$$
