Why is the set R = $\{(x,y) \in \mathbb{R} \times \mathbb{R} | |x| < |y| \bigvee x=y) \}$ a partial order?

Question

For a set to be a total order it must have 4 properties:

1. Reflexive
2. Antisymmetry
3. Transitive
4. $\forall x \in A\forall y \in A((x,y) \in R \bigvee (y,x) \in R)$, where A is a set and R is a relation on that set.

I have checked the first three and found that R is a partial order on $\mathbb{R}$; when I check this fourth property I get that:

$$ \forall x \in \mathbb{R} \forall y \in \mathbb{R} [|x| < |y| \bigvee |y| < |x| \bigvee x=y] $$ I cannot find a counter example to this last statement. It seems that any pair of real numbers (x,y) will satisfy one of the three properties of the previous statement - but that would make this a total order, not a partial order.

Answer 1

HINT: the absolute value function isn't injective . . .