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 PrototypePHX March 7th, 2013 09:01 AM

Order relation

I have the following relation:
$R= \{(x,y) \in \mathbb{R} \times \mathbb{R} | |x| < |y| \vee x= y\}$
Show that R is partially ordered. Is it totally ordered?

My solution:
It's not a total order since $\forall x \in \mathbb{R}: ( \neg(|-x| < |x|) \wedge -x \neq x)$
A relation is partially ordered iff it is reflexive, antisymmetric and transitive.

Reflexitivity: $|x|=< |x| \vee x= x$.
This is a tautology, hence R is reflexive.

Symmetry: $(|x| < |y| \vee x = y) \wedge (|y| < |x| \vee y = x)
\Leftrightarrow x = y \vee (|x| < |y| \wedge |y| < |x|)
\Leftrightarrow x = y$

This statement is true iff x = y. That means R is antisymmetric.

Transitivity: $(|x| < |y| \vee x = y) \wedge (|y| < |z| \vee y = z)
\text{case 1:} x = y \Rightarrow x=y \wedge (|y| < |z| \vee y = z)
\Leftrightarrow x=z \vee |x| < |z|.
\text{case 2:} x \neq y \Rightarrow |x| < |y| \wedge (|y| < |z| \vee y = z)
\Leftrightarrow |x| < |z| \vee x = z$

Thus R is transitive.
Since R is reflexive, antisymmetric and transitive, it is partially ordered.

However I'm very unsure on the transitive part. Thanks for help.

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