Page 359 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Appendix 1: Solutions to Selected Exercises 345
14. (a)
b is the R-largest element of B iff b ∈ B and ∀x ∈ B(xRb)
iff b ∈ B and ∀x ∈ B(bR −1 x)
−1
iff b is the R -smallest element of B.
(b)
b is an R-maximal element of B iff b ∈ B and ¬∃x ∈ B(bRx ∧ b = x)
−1
iff b ∈ B and ¬∃x ∈ B(xR b
∧x = b)
iff b is an R-minimal element of B.

17. No. Let A = R × R, and let R ={((x, y), (x , y )) ∈ A × A | x ≤ x and

y ≤ y }. (You might want to compare this to exercise 8.) Let B ={(0, 0)}∪
({1}× R). We will leave it to you to check that R is a partial order on A,
and that (0, 0) is the only minimal element of B, but it is not a smallest
element.
21. (a) Suppose that x ∈ U and xRy. To prove that y ∈ U, we must show that
y is an upper bound for B, so suppose that b ∈ B. Since x ∈ U, x is an
upper bound for B,so bRx. But we also have xRy, so by transitivity
of R we can conclude that bRy. Since b was arbitrary, this shows that
y is an upper bound for B.
(b) Suppose b ∈ B. To prove that b is a lower bound for U, let x be an
arbitrary element of U. Then by definition of U, x is an upper bound
for B,so bRx. Since x was arbitrary, this shows that b is a lower bound
for U.
(c) Hint: Suppose x is the greatest lower bound of U. First use part (b) to
show that x is an upper bound for B, and therefore x ∈ U. Then use the
fact that x is a lower bound for U to show that x is the smallest element
of U – in other words, it is the least upper bound of B.

Section 4.5
1. (a) Reflexive closure: {(a, a), (a, b), (b, c), (c, b), (b, b), (c, c)}.
Symmetric closure: {(a, a), (a, b), (b, c), (c, b), (b, a)}.
Transitive closure: {(a, a), (a, b), (b, c), (c, b), (a, c), (b, b), (c, c)}.
(b) Reflexive closure: {(x, y) ∈ R × R | x ≤ y}.
Symmetric closure: {(x, y) ∈ R × R | x = y}.
Transitive closure: R.