Math 3000
Additional Homework Problems on Order Relations
1. Which of these relations (R) on the given set (S) are antisymmetric?
(a) S = {1,2,3,4,5} R = {(1,3), (1,1), (2,4), (3,2), (5.4), (4,2)}
Not antisymmetric (2,4) & (4,2)
(b) S = {1,2,3,4,5} R = {(1,4), (1,2), (2,3), (3,4), (5,2), (4,2), (1,3)} Antisymmetric
(c) S = ℤ, x R y iff x2 = y2. Not antisymmetric 1 R -1 & -1 R 1
(d) S = ℝ, x R y iff x ≤ 2y. Not antisymmetric 2 R 3 & 3 R 2
2. Show that the relation R on ℕ given by a R b iff b = 2ka for some integer k ≥ 0 is a partial order.
For any natural number n, n R n since n = 20n. Thus, R is reflexive. If n R m and m R n, then n =
2km and m = 2sn. We would then have n = 2k+sn, so k + s = 0 and since these integers are nonnegative we must have k = s = 0, so n = m and R is antisymmetric. If n R m and m R p then n =
2km and m = 2sp. So, n = 2k+sp and we have n R p. R is thus transitive and so it is a partial order.
3. Define the relation on ℝ x ℝ by (a,b) R (x,y) iff a ≤ x and b ≤ y. Prove that R is a partial order on
ℝ x ℝ.
For all (a,b) in ℝ x ℝ, we have (a,b) R (a,b) since a ≤ a and b ≤ b. If (a,b) R (x,y) and (x,y) R (a,b)
then a ≤ x and b ≤ y, as well as, x ≤ a and y ≤ b. Since, a ≤ x and x ≤ a we have a = x. Similarly, b =
y and we have (a,b) = (x,y). Thus, R is antisymmetric. If (a,b) R (x,y) and (x,y) R (w,z), then a ≤ x
and x ≤ w, so a ≤ w. Similarly, b ≤ y and y ≤ z, so b ≤ z. Thus, (a,b) R (w,z) and R is transitive.
Therefore, R is a partial order.
4. For a poset A with partial order R and B ⊆ A, write out what it means for an element a A to
not be a least upper bound of B.
Either a is not an upper bound of B or there exists an upper bound c of B with c R a.
5. In the following “proofs” assign a grade of A if the argument is correct, a grade of C if there is a
minor problem with the proof or an F if there is a major problem with the proof.
(a) Claim: Let A be a set with a partial order R. If C ⊆ B ⊆ A and sup(C) and sup(B) exist, then
sup(C) ⊆ sup(B).
“Proof” sup(B) is an upper bound for B. Therefore, sup(B) is an upper bound for C. Thus
sup(C) ⊆ sup(B).
A
(b) Claim: Let A be a set with a partial order R. If B ⊆ A, u is an upper bound for B, and u B,
then sup(B) exists and u = sup(B).
“Proof” Since u B, u R sup(B). Since u is an upper bound, sup(B) R u. Thus, u = sup(B).
C Although this proves that u = sup(B) if sup(B) exists, it does not prove that sup(B) exists.
(c) Claim: For A, B ⊆ ℝ with the usual ≤ order, sup (A ∪ B) = sup(A) + sup(B).
“Proof” If x A ∪ B, then x A or x B. Thus, x ≤ sup(A) or x ≤ sup(B). In either case, x ≤
sup(A) + sup(B). Therefore, sup(A ∪ B) ≤ sup(A) + sup(B). On the other hand, both A ⊆ A ∪ B
and B ⊆ A ∪ B, so by part (a) of this exercise, sup(A) ≤ sup (A ∪ B) and sup(B) ≤ sup(A ∪ B).
Therefore, sup(A) + sup(B) ≤ sup(A ∪ B).
F The claim is false. The last line of the "proof" does not follow from the previous line.