181
Chapter 11 Partially Ordered Sets and Boolean Algebras
Exercises for Section 11.1. Partially Ordered Sets
1. (a) R = {(a, b)}
2. (a) Yes.
(b) R = {(a, b), (b, a)} (c) R = ∅ (d) R = {(a, b), (b, a), (c, d)}
(b) Yes.
3. (a) No. For example, (−2) | 2 and 2 | (−2) but 2 ̸= −2.
4. (a) Yes
(b) Yes.
5. (a) No, for example, 7 ̸ R 7;
(c) Yes.
(b) Yes.
(b) No, for example, 3 R 5 and 5 R 3 but 3 ̸= 5.
(d) No, for example, 5 R 3 and 3 R 5 but 5 ̸ R 5.
6. (a) R = {(w, w), (x, x), (y, y), (z, z)}
(b) R = {(w, w), (x, x), (y, y), (z, z), (w, x), (x, w), (x, y)}
(c) R = {(w, x), (x, y)}
(d) R = {(w, w), (x, x), (w, x), (x, w)}
(e) R = {(w, w), (x, x), (y, y), (z, z), (w, x), (x, y)}
(f) R = {(w, w), (x, x), (y, y), (z, z), (w, x), (x, w)}
(g) R = ∅
(h) R = {(w, x), (x, w)}
7. The relation R is antisymmetric and transitive but R is not reflexive since (2, 2) ∈
/ R.
8. The relation R is reflexive, antisymmetric and transitive.
9. The relation R on Z defined by a R b if a = b.
10. (a) partial order
(b) Since 1 ̸ R2 1, it follows that R2 is not reflexive. Since 1 R2 2 and 2 ̸ R2 1 but 1 ̸= 2, it
follows that R2 is not antisymmetric. Since 1 R2 2, 2 R2 1 and 1 ̸ R2 1, it follows that R2
is not transitive.
(c) Since 1 ̸ R3 1, it follows that R3 is not reflexive. Since 2 R3 3, 3 R3 2 and 2 ̸= 3, it follows
that R2 is not antisymmetric. Since 2 R3 3, 3 R3 4 and 2 ̸ R3 4, R3 is not transitive.
(d) Since 1 ̸ R4 1, it follows that R4 is not reflexive. Since 1 R4 2, 2 R4 3 and 1 ̸ R4 3, R4 is
not transitive.
(e) Since 1 ̸ R5 1, it follows that R5 is not reflexive.
11. Although R is reflexive and antisymmetric, it is not transitive. For example, a R c and c R b
but a ̸ R b.
12. Yes. R = {(2, 2), (2, 3), (2, 4), (2, 5), (3, 3), (3, 5), (4, 3), (4, 4), (4, 5), (5, 5)}.
13. Let S = {a, b, c, d, e, f, g} and
≼ =
{(a, a), (b, b), (c, c), (d, a), (d, b), (d, c), (d, d),
(e, a), (e, b), (e, c), (e, e), (f, a), (f, b), (f, c), (f, d),
(f, e), (f, f ), (g, a), (g, b), (g, c), (g, d), (g, e), (g, g)}
182
14. For each A ∈ P, A ∩ A = A ̸= ∅. Thus A ̸ R A and R is not reflexive. If A and B are distinct
elements of P, then A R B and B R A but A ̸ R A. This R is not transitive. If A R B and
B R A, then A and B are disjoint and so A ̸= B. Thus R is not antisymmetric.
15. Let S = {a, b} and R = {(a, a), (a, b), (b, b)}. Then (S, R) is a poset.
16. Proof. Let a ∈ A. Then a ∈ S. Since a ≼ a, it follows that ≼ is reflexive in (A, ≼). Assume
next that a ≼ b and b ≼ a, where a, b ∈ A. Since a, b ∈ S, it follows that a = b. Hence ≼ is
antisymmetric in (A, ≼). Now assume that a ≼ b and b ≼ c, where a, b, c, ∈ A. Thus a, b, c ∈ S
and so a ≼ c. Hence ≼ is transitive. Therefore, (A, ≼) is a poset.
17. Proof. Let (S, R) be a poset. Let a ∈ S. Since a R a, it follows that a R−1 a. Hence R−1 is
reflexive. Assume that a R−1 b and b R−1 a, where a, b ∈ S. Then b R a and a R b. Because R
is antisymmetric, b = a. Thus R−1 is antisymmetric. Next assume that a R−1 b and b R−1 c,
where a, b, c ∈ S. Then b R a and c R b. Since R is transitive, c R a. Thus a R−1 c and so
R−1 is transitive. Therefore, (S, R−1 ) is a poset.
18. No. Observe that s2 ≺ s6 and s6 ≺ s3 but s2 ̸≼ s3 . Thus ≼ is not transitive.
19. The pairs of elements in (a) and in (d) are comparable.
20. (a) See Figure 40.
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Figure 40: A Hasse diagram for the poset in Exercise 20
(b) 2 and 4 are comparable; while 4 and 6 are incomparable.
21. Proof. Let (S, ≼) be a finite nonempty poset. Among the chains in S, let x1 ≺ x2 ≺ · · · ≺ xk
be one containing a maximum number of elements. There can be no element x ∈ S such that
xk ≺ x, for otherwise, x1 ≺ x2 ≺ · · · ≺ xk ≺ x is a chain with more than k elements. Hence
xk is a maximal element.
22. (a) Let a1 = 2, a2 = 3, a3 = 5, b = a1 a2 a3 , c1 = 7b, c2 = 11b, c3 = 13b, c4 = 17b, c5 = 19b
and let S consists of these 9 integers. Then (S, |) is a poset having the five maximal
elements c1 , c2 , c3 , c4 , c5 and three minimal elements a1 , a2 , a3 .
(b) For positive integers m and n, give an example of a poset having exactly m maximal
elements and exactly n minimal elements.
Let p1 , p2 , . . . , pm+n be m+n distinct primes. Let ai = pi for 1 ≤ i ≤ n, let b = a1 a2 · · · an
and let cj = pn+j b for 1 ≤ j ≤ m. Let S = {a1 , a2 , . . . , an , b, c1 , c2 , . . . , cm }. Then
(S, |) is a poset having the m maximal elements c1 , c2 , . . . , cm and n minimal elements
a1 , a2 , . . . , an .
23. Proof. Assume, to the contrary, that there exists a finite poset (S ≼) containing a unique
minimal element m such that m is not a least element. Since m is not a least element, there
exists an element a ∈ S such that m ̸≺ a. Since m is a minimal element, it is impossible for
a ≺ m. Thus a and m are incomparable. If there is no element a′ such that a′ ≺ a, then a is a
minimal element of (S, ≼), contradicting our assumption that m is the only minimal element
of (S, ≼). Hence there must be some element a′ such that a′ ≺ a. Since S is finite, there is
a chain a1 ≺ a2 ≺ · · · ≺ ak = a with a maximum number of elements. However, then a1 is a
minimal element. Because a1 ≺ a and a and m are incomparable, a1 ̸= m, which contradicts
the assumption that m is the unique minimal element of (S, ≼).
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Figure 41: A Hasse diagram for the poset in Exercise 24
24. (a) See Figure 41.
(b) 30 is the unique maximal element and 1 is the unique minimal element.
(c) 30 is the greatest element and 1 is the least element.
25. (a) See Figure 42.
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Figure 42: A Hasse diagram for the poset in Exercise 25
(b) 6, 10, 14, 15, 21 and 35 are maximal elements, while 2, 3, 5 and 7 are minimal elements.
(c) There is no greatest or least element.
26. (a) First, x is initialized as a1 . We proceed through s′ , searching for (a2 , a1 ) and then (a3 , a1 )
and so on. Only (a4 , a1 ) is found in s′ . Thus x is reassigned the element a4 . In future
searchings through s′ for an element (ai , a4 ) in s′ , no such element is found. So x = a4 is
output as a minimal element in (S, ≼).
(b) See Figure 43.
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Figure 43: A Hasse diagram for the poset in Exercise 26
27. Algorithm Find a maximal element in a finite poset.
This algorithm finds a maximal element in a finite poset (S, ≼), where s : a1 , a2 , . . . , an is a
list of the n elements of S and s′ : e1 , e2 , . . . , em is a list of the m elements (a, b) of ≼ for which
a ̸= b.
Input: Two positive integers n and m, a list s : a1 , a2 , . . . , an of the n elements of S and
a list s′ : e1 , e2 , . . . , em of the m elements (a, b) of ≼ for which a ̸= b.
Output: a maximal element y of (S, ≼).
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1. y := a1
2. for i := 2 to n do
3.
4.
for j := 1 to m do
if ej = (y, ai ) then y := ai and j := m + 1.
5. output y
28. (a) (1, 1) and (2, 2) are comparable.
(b) (1, 2) and (1, 3) are incomparable.
(c) (1, 1) ≺ (1, 2) ≺ (1, 4) ≺ (2, 1), ≺ (2, 2), ≺ (2, 4) ≺ (3, 1) ≺ (3, 2) ≺ (3, 4) ≺ (4, 1) ≺
(4, 2) ≺ (4, 4) ≺ (5, 1) ≺ (5, 2) ≺ (5, 4) ≺ (6, 1) ≺ (6, 2) ≺ (6, 4).
29. Proof. We proceed by induction. If n = 2, then the result follows from Theorem 11.19.
Assume that the result is true for k ≥ 2 totally ordered sets and let (A1 , ≼1 ), (A2 , ≼2 ), . . .,
(Ak+1 , ≼k+1 ) be k + 1 totally ordered sets. We show that the lexicographic order ≼ defined
on A1 × A2 × Ak+1 is a total order. By the induction hypothesis, the lexicographic order ≼
defined on A = A1 × A2 × · · · Ak is a total order and so (A, ≼) is a totally ordered set. Since
A1 × A2 × · · · Ak+1 = A × Ak+1 , it follows by Theorem 11.19 that (A × Ak+1 , ≼) is a totally
ordered set.
30. (a) See Figure 44.
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Figure 44: A Hasse diagram for the poset in Example 30
(b) A topological sorting of (A, ≼) is T ≺ N ≺ D ≺ G ≺ I ≺ W ≺ F .
31. By Algorithm 11.28, a4 ≺ a6 ≺ a8 ≺ a1 ≺ a2 ≺ a3 ≺ a5 ≺ a7 is a topological sorting of the
elements of S.
32. ∅ ≺ {1} ≺ {2} ≺ {3} ≺ {1, 2} ≺ {1, 3} ≺ {2, 3} ≺ X;
∅ ≺ {3} ≺ {2} ≺ {1} ≺ {2, 3} ≺ {1, 3} ≺ {1, 2} ≺ X.
33. (a) See Figure 45.
(b) 2 ≺ 4 ≺ 8 ≺ 24 ≺ 72 ≺ 40 ≺ 200 ≺ 1800;
2 ≺ 4 ≺ 8 ≺ 24 ≺ 40 ≺ 72 ≺ 200 ≺ 1800.
34. Proof. Let (S, ≼) be a poset. Since the result is true if S contains no least element, we can
assume that S contains a least element. We show that such an element is unique. Let m1 and
m2 be two least elements of S. Since m1 is a least element, m1 ≼ m2 . Since m2 is a least
element, m2 ≼ m1 . Since ≼ is antisymmetric, m1 = m2 .
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1800
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Figure 45: A Hasse diagram for the poset in Exercise 33(a)
35. Proof. Let q = r/s, where r and s are integers with 1 ≤ r < s. Let p1 , p2 , . . . , p2s−1 be
2s − 1 distinct primes. Let a = 1, bi = pi for 1 ≤ i ≤ 2r, c = b1 b2 · · · b2r and dj = p2r+j c for
1 ≤ j ≤ 2s − 2r − 1. Let S = {a, b1 , b2 , . . . , b2r , d1 , d2 , . . . , d2s−2r−1 }. Then (S, |) is a poset
with |S| = 2s. The set I = {b1 , b2 , . . . , b2r } is the set of intermediate elements of S. Then
2r
r
|I|
=
= = q.
|S|
2s
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Exercises for Section 11.2. Lattices
1. Proof. Assume, to the contrary, that there exists a poset (S, ≼) containing a nonempty subset
A having two distinct least upper bounds u1 and u2 . Since u1 is a least upper bound, u1 ≼ u2 .
Also, since u2 is a least upper bound, u2 ≼ u1 . Since ≼ is antisymmetric, u1 = u2 , contradicting
the assumption that u1 and u2 are distinct.
2. See Figure 46.
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Figure 46: A poset in Exercise 2
Let A = {x, y}. Then M1 , M2 and M3 are upper bounds for A and m1 and m2 are lower
bounds for A.
3. The poset (A, ≼) is not a lattice. The elements a2 and a3 , for example, do not have a unique
least upper bound.
4. The poset (S, ≼) is not a lattice. For example, {d, f } does not have a greatest lower bound
and so d ∧ f does not exist.
5. (a) Proof. Let d = a ∨ c and e = b ∨ c. Then b ≼ e and c ≼ e. Since a ≼ b, it follows that
a ≼ e. Thus e is an upper bound of {a, c}. Because d = a ∨ c, we have d ≼ e, that is,
a ∨ c ≼ b ∨ c.
(b) Proof. Let d = a ∧ c and e = b ∧ c. Thus d ≼ a and d ≼ c. Since a ≼ b, it follows
that d ≼ b. Thus d is a lower bound of {b, c}. Since e = b ∧ c, we have d ≼ e and so
a ∧ c ≼ b ∧ c.