Finding a Real Number using epslion

Question

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?

Answer 1

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 $$|b| \leqslant |a-b| + |a|\implies |a|-|b|\leqslant|a-b|.$$

Answer 2

I would argue by cases based on the value of $x$. If $x \ge 2$, you are to show $x-1\le \epsilon \implies x-2 \ge 1-\epsilon$ Add $1$ to each side and use $\epsilon \gt 0$ and you are there. You have two more cases, $x \in [1,2)$ and $x \lt 1$