## 1.3 Exercise 7

After doing the same procedure for an exercise on equivalence relations, this question should pose no problem. Important is starting to visualize relations on the Cartesian product of a set. What does an order relation look like? It is upper or lower triangular (by comparability and transitivity), with the identity line deleted (because of non-reflexivity). Contrast this with what an equivalence relation looks like: a square (because of symmetry and transitivity) with an extended diagonal (the identity line by reflexivity).

"Show that the restriction of an order relation is an order relation."

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

--------

SOLUTION

Suppose there is an order relation defined on the elements of .

Comparability: Pick with . and , but importantly, either or (or both). Now suppose these same elements . Then , and . We already know that either one or the other (or both) of such elements belong to . Thus either or (or both). The restriction inherits comparability.

Non-reflexivity. Picking , it is clear that . We know, however, that . If such is picked so that it also belongs to , then . But such an element of the cross product is not in , we've established, so it cannot possibly be in the intersection of with . Thus, inherits non-reflexivity.

Transitivity. Picking different elements , it is clear that in particular the elements , and . Suppose these elements , and thus , and . Now further suppose . By transitivity of the order relation in , . This last element is in both and , and thus it is in the intersection. Transitivity of the order relation is inherited by the restriction.

We conclude that the restriction of an order relation is an order relation.