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Ordering Relations 189
22. Consider the following putative theorem:
Theorem? Suppose R is a relation on A. If R is symmetric and transitive,
then R is reflexive.

Is the following proof correct? If so, what proof strategies does it use? If
not, can it be fixed? Is the theorem correct?

Proof. Let x be an arbitrary element of A. Let y be any element of A
such that xRy. Since R is symmetric, it follows that yRx. But then by
transitivity, since xRy and yRx we can conclude that xRx. Since x was
arbitrary, we have shown that ∀x ∈ A(xRx), so R is reflexive.

∗ 23. This problem was suggested by Prof. William Zwicker of Union College.
Suppose A is a set, and F ⊆ P (A). Let R ={(a, b) ∈ A × A | for ev-
ery X ⊆ A \{a, b},if X ∪{a}∈ F then X ∪{b}∈ F}. Show that R is
transitive.

4.4. Ordering Relations

Consider the relation L ={(x, y) ∈ R × R | x ≤ y}. You should be able to
check for yourself that it is reflexive and transitive, but not symmetric. It fails
to be symmetric in a rather extreme way because there are many pairs (x, y)
such that xLy is true but yLx is false. In fact, the only way xLy and yLx can
both be true is if x ≤ y and y ≤ x, and thus x = y. We therefore say that L is
antisymmetric. Here is the general definition.

Definition 4.4.1. Suppose R is a relation on a set A. Then R is said to be
antisymmetric if ∀x ∈ A∀y ∈ A((xRy ∧ yRx) → x = y).

We have already seen a relation with many of the same properties as L. Look
again at the relation S defined in part 1 of Example 4.3.1. Recall that in that
example we let A ={1, 2}, B = P (A), and S ={(x, y) ∈ B × B | x ⊆ y}.
Thus, if x and y are elements of B, then xSy means x ⊆ y. We checked in
the last section that S is reflexive and transitive, but not symmetric. In fact, S
is also antisymmetric, because for any sets x and y,if x ⊆ y and y ⊆ x then
x = y. You may find it useful to look back at Figure 3 in the last section, which
shows the directed graph representing S.
Intuitively, L and S are both relations that have something to do with com-
paring the sizes of two objects. Each of the statements x ≤ y and x ⊆ y can