$ \forall \epsilon \in \mathbb{R}^{*}_{+} , | x -y | < \epsilon \iff x = y $

Question

Prove that for all $ \epsilon > 0, \epsilon \in \mathbb{R} $ for every $ x, y \in \mathbb{R} $ if $ | x - y | < \epsilon \iff x = y$

(this question has similar ones in, but this one has the full proof, not only the hint).

Answer 1

$( \Rightarrow ) $ I will prove by contraposition. Our affirmative states that:

$$ \forall \epsilon > 0 \; ( | x - y | < \epsilon ) \implies x = y $$

The contrapositive is

$$ \lnot ( x = y ) \implies \lnot \forall \epsilon > 0 \; ( | x - y | < \epsilon ) \\ x \neq y \implies \exists \epsilon > 0 \; \lnot ( | x - y | < \epsilon ) \\ x \neq y \implies \exists \epsilon > 0 \; ( | x - y | \ge \epsilon ) $$

Therefore, if we prove the contrapositive, we have proved the affirmative. We need to find an $ \epsilon \le | x - y |$.

Let $ \epsilon = \frac{|x-y|}{2} $, since $ x \neq y $, without loss of generality, let $ x > y $, that implies $ x - y > 0 $, since we have $ | x - y | = x - y > 0 \implies | x - y | > 0 $.

Therefore $ | x - y | \ge \epsilon = \frac{|x-y|}{2} \implies 2 |x-y| \ge |x-y| $. That is true for all positive values, since we have proved our contraposition our initial hypothesis is valid.

$ ( \Leftarrow ) $ Direct proof. Let $ x = y \implies x - y = 0 \implies | x - y | = | 0 | < \epsilon \;\;\; \square$