## 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:

"Check that the relation defined in Example 7 is an order relation: Define if , or if and ."

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

-------

SOLUTION

First we notice that the order relation is defined on .

Comparability: Choosing , so that , either or (or both), and the elements are comparable. Supposing , then either or (or both). The elements are comparable. If , then we picked the elements contrary to our assumption. Thus comparability holds.

Non-reflexivity: we want to show that for no the statement holds. Suppose it holds for some element . This means which will contradict the identity property of equality. So . Then, checking the second criterion, , we conclude that this is also a contradiction. Non-reflexivity must hold.

Transitivity. If , we want to show that if and , this implies . There are four cases. In the first case, and implies . Since the reals are transitive axiomatically, . In the second case and . We can substitute and ascertain that . In the third case, and . Substituting again we obtain . Finally, and . This implies , but by transitivity of the reals, . Transitivity must hold.