## 1.3 Exercise 8

Other than just teaching us to check that order relations satisfy the properties of comparability, non-reflexivity, and transitivity, these problems also give us insight into how we can order different elements of a set, in this particular case a very familiar one: the reals.  For example, the order relation defined on the reals of this problem shows the uncanny notion that we can order them as follows:

$0 < \ldots < -0.5 < 0.5 < \ldots < -1 < 1 <\ldots < -2.1234 < 2.1234 < \ldots$

"Check that the relation defined in Example 7 is an order relation: Define $xCy$ if $x^2 < y^2$, or if $x^2 = y^2$ and $x < y$."

(Taken from Topology by James R. Munkres, Second Edition, Prentice Hall, NJ, 2000. Page 29.)

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SOLUTION

First we notice that the order relation is defined on $\mathbb{R}$.

Comparability:  Choosing $x, y \in \mathbb{R}$, so that $x \neq y$, either $x^2 < y^2$ or $y^2 < x^2$ (or both), and the elements are comparable.  Supposing $x^2 = y^2$, then either $x < y$ or $x > y$ (or both).  The elements are comparable.  If $x = y$, then we picked the elements contrary to our assumption.  Thus comparability holds.

Non-reflexivity:  we want to show that for no $x \in \mathbb{R}$ the statement $(x, x) \in C$ holds.  Suppose it holds for some element $x$.  This means $x^2 < x^2$ which will contradict the identity property of equality. So $x^2 = x^2$.  Then, checking the second criterion, $x < x$, we conclude that this is also a contradiction.  Non-reflexivity must hold.

Transitivity.  If $x, y, z \in \mathbb{R}$, we want to show that if $xCy$ and $yCz$, this implies $xCz$.  There are four cases.  In the first case, $x^2 < y^2$ and $y^2 < z^2$ implies $x^2 < y^2 < z^2$.  Since the reals are transitive axiomatically, $x^2 < z^2$.  In the second case $x^2 < y^2$ and $y^2 = z^2$.  We can substitute and ascertain that $x^2 < z^2$.  In the third case, $x^2 = y^2$ and $y^2 < z^2$.  Substituting again we obtain $x^2 < z^2$.  Finally, $x^2 = y^2 \Rightarrow x < y$ and $y^2 = z^2 \Rightarrow y < z$.  This implies $x < y < z$, but by transitivity of the reals, $x < z$.  Transitivity must hold.

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