Page 180 - HOW TO PROVE IT: A Structured Approach, Second Edition

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166 Relations
Case 1.(x, y) ∈ A × B. Then x ∈ A and y ∈ B, so clearly x ∈ A ∪ C and
y ∈ B ∪ D. Therefore (x, y) ∈ (A ∪ C) × (B ∪ D).
Case 2.(x, y) ∈ C × D. A similar argument shows that (x, y) ∈ (A ∪ C) ×
(B ∪ D).
Since (x, y) was an arbitrary element of (A × B) ∪ (C × D), it follows that
(A × B) ∪ (C × D) ⊆ (A ∪ C) × (B ∪ D).

Proof of 5. Suppose A×∅ = ∅. Then A×∅ has at least one element, and by
the definition of Cartesian product this element must be an ordered pair (x, y)
for some x ∈ A and y ∈ ∅. But this is impossible, because ∅ has no elements.
Thus, A×∅ = ∅. The proof that ∅×A = ∅ is similar.

Commentary. Statement 4 says that one set is a subset of another, and the proof
follows the usual pattern for statements of this form: We start with an arbitrary
element of the first set and then prove that it’s an element of the second. It is
clear that the arbitrary element of the first set must be an ordered pair, so we
have written it as an ordered pair from the beginning.
Thus, for the rest of the proof we have (x, y) ∈ (A × B) ∪ (C × D)asa
given, and the goal is to prove that (x, y) ∈ (A ∪ C) × (B ∪ D). The given
means (x, y) ∈ A × B ∨ (x, y) ∈ C × D, so proof by cases is an appropriate
strategy. In each case it is easy to prove the goal.
Statement 5 means A×∅= ∅ ∧ ∅×A = ∅, so we treat this as two goals
and prove A×∅ = ∅ and ∅×A=∅ separately. To say that a set equals the
empty set is actually a negative statement, although it may not look like it on
the surface, because it means that the set does not have any elements. Thus,
it is not surprising that the proof that A×∅ = ∅ proceeds by contradiction.
The assumption that A×∅ = ∅ means ∃p(p ∈ A×∅), so our next step is
to introduce a name for an element of A×∅. Once again, it is clear that the
new object being introduced in the proof is an ordered pair, so we have written
it as an ordered pair (x, y) from the beginning. Writing out the meaning of
(x, y) ∈ A×∅ leads immediately to a contradiction.
The proof that ∅×A = ∅ is similar, but simply saying this doesn’t prove
it. Thus, the claim in the proof that this part of the proof is similar is really
an indication that the second half of the proof is being left as an exercise. You
should work through the details of this proof in your head (or if necessary write
them out on paper) to make sure that a proof similar to the proof in the first
half will really work.
Because the order of the coordinates in an ordered pair matters, A × B
and B × A mean different things. Does it ever happen that A × B = B × A?
Well, one way this could happen is if A = B. Clearly if A = B then A × B =
A × A = B × A. Are there any other possibilities?