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344 Appendix 1: Solutions to Selected Exercises
20. Suppose R is transitive, and suppose (X, Y) ∈ S and (Y, Z) ∈ S.Toprove
that (X, Z) ∈ S we must show that ∀x ∈ X∀z ∈ Z(xRz), so let x ∈ X and
z ∈ Z be arbitrary. Since Y ∈ B, Y = ∅, so we can choose y ∈ Y. Since
(X, Y) ∈ S and (Y, Z) ∈ S, by the definition of S we have xRy and yRz.
But then since R is transitive, xRz, as required. The empty set had to
be excluded from B so that we could come up with y ∈ Y in this proof.
(Can you find a counterexample if the empty set is not excluded?)
23. Hint: Suppose aRb and bRc.Toprove aRc, suppose that X ⊆ A \{a, c}
and X ∪{a}∈ F; you must prove that X ∪{c}∈ F. To do this, you may
find it helpful to consider two cases: b /∈ X or b ∈ X. In the second of

these cases, try working with the sets X = (X ∪{a}) \{b} and X =
(X ∪{c}) \{b}.

Section 4.4

1. (a) Partial order, but not total order.
(b) Not a partial order.
(c) Partial order, but not total order.
4. (→) Suppose that R is both antisymmetric and symmetric. Suppose that
(x, y) ∈ R. Then since R is symmetric, (y, x) ∈ R, and since R is anti-
symmetric, it follows that x = y. Therefore (x, y) ∈ i A . Since (x, y)was
arbitrary, this shows that R ⊆ i A .
(←) Suppose that R ⊆ i A . Suppose (x, y) ∈ R. Then (x, y) ∈ i A ,so
x = y, and therefore (y, x) = (x, y) ∈ R. This shows that R is symmetric.
To see that R is antisymmetric, suppose that (x, y) ∈ R and (y, x) ∈ R.
Then (x, y) ∈ i A ,so x = y.
8. To see that T is reflexive, consider an arbitrary (a, b) ∈ A × B. Since R
and S are both reflexive, we have aRa and bSb. By the definition of T,
it follows that (a, b)T (a, b). To see that T is antisymmetric, suppose that
(a, b)T (a , b ) and (a , b )T (a, b). Then aRa and a Ra, so since R is anti-

symmetric, a = a . Similarly, bSb and b Sb, so since S is antisymmetric,

we also have b = b . Thus (a, b) = (a , b ), as required. Finally, to see

that T is transitive, suppose that (a, b)T (a , b ) and (a , b )T (a , b ). Then

aRa and a Ra , so since R is transitive, aRa . Similarly, bSb and b Sb ,

so bSb , and therefore (a, b)T (a , b ).

Even if both R and S are total orders, T need not be a total order.
11. The minimal elements of B are the prime numbers. B has no smallest
element.