This exercise provides the steps necessary to prove that every partial orderingis in a
Chapter 3, Problem 19(choose chapter or problem)
This exercise provides the steps necessary to prove that every partial orderingis in a sense the same as the set inclusion relation on a collection of subsetsof a set. Let A be a set with a partial order R. For each let a A, iff iffT = {(m, n): m, n , m n and n = 5} {(5, m): m }S = {(m, n): m, n , m n and m = 5} {(m, 5): m }m nm n,m V n iff. Let Then is a subset of andthus may be partially ordered by .(a) Show that if then(b) Show that if Sa Sb, then(c) Show that for every an immediate predecessor of b A, b in A correspondsa R b.a R b, Sa Sb.Sa = {x A: x R a} = {Sa: a A}. (A)to an immediate predecessor of in .(d) Show that if and x is the least upper bound for B, then is theleast upper bound forProofs to
Unfortunately, we don't have that question answered yet. But you can get it answered in just 5 hours by Logging in or Becoming a subscriber.
Becoming a subscriber
Or look for another answer