Archive for February 11th, 2009

1.3 Exercise 8

February 11th, 2009 No comments

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.)



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.