Page 179 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Ordered Pairs and Cartesian Products 165
Proof of 1. Let p be an arbitrary element of A × (B ∩ C). Then by the defi-
nition of Cartesian product, p must be an ordered pair whose first coordinate
is an element of A and second coordinate is an element of B ∩ C. In other
words, p = (x, y) for some x ∈ A and y ∈ B ∩ C. Since y ∈ B ∩ C, y ∈ B
and y ∈ C. Since x ∈ A and y ∈ B, p = (x, y) ∈ A × B, and similarly p ∈
A × C. Thus, p ∈ (A × B) ∩ (A × C). Since p was an arbitrary element of
A × (B ∩ C), it follows that A × (B ∩ C) ⊆ (A × B) ∩ (A × C).
Now let p be an arbitrary element of (A × B) ∩ (A × C). Then p ∈ A × B,
so p = (x, y) for some x ∈ A and y ∈ B. Also, (x, y) = p ∈ A × C,so
y ∈ C. Since y ∈ B and y ∈ C, y ∈ B ∩ C. Thus, p = (x, y) ∈ A × (B ∩ C).
Since p was an arbitrary element of (A × B) ∩ (A × C) we can con-
clude that (A × B) ∩ (A × C) ⊆ A × (B ∩ C), so A × (B ∩ C) = (A × B) ∩
(A × C).

Commentary. Before continuing with the proofs of the other parts, we give a
brief commentary on the proof just given. Statement 1 is an equation between
two sets, so as we saw in Example 3.4.4, there are two natural approaches
we could take to prove it. We could prove ∀p[p ∈ A × (B ∩ C) ↔ p ∈
(A × B) ∩ (A × C)] or we could prove both A × (B ∩ C) ⊆ (A × B) ∩ (A ×
C) and (A × B) ∩ (A × C) ⊆ A × (B ∩ C). In this proof, we have taken the
second approach. The first paragraph gives the proof that A × (B ∩ C) ⊆
(A × B) ∩ (A × C) and the second gives the proof that (A × B) ∩ (A × C) ⊆
A × (B ∩ C).
In the first of these proofs we take the usual approach of letting p be an arbi-
trary element of A × (B ∩ C) and then proving p ∈ (A × B) ∩ (A × C). Be-
cause p ∈ A × (B ∩ C) means ∃x∃y(x ∈ A ∧ y ∈ B ∩ C ∧ p = (x, y)), we
immediately introduce the variables x and y by existential instantiation. The
rest of the proof involves simply working out the definitions of the set theory op-
erations involved. The proof of the opposite inclusion in the second paragraph
is similar.
Note that in both parts of this proof we introduced an arbitrary object p
that turned out to be an ordered pair, and we were therefore able to say that
p = (x, y)forsomeobjectsxandy.InmostproofsinvolvingCartesianproducts
mathematicians suppress this step. If it is clear from the beginning that an object
will turn out to be an ordered pair, it is usually just called (x, y) from the outset.
We will follow this practice in our proofs.
We leave the proofs of statements 2 and 3 as exercises (see exercise 5).

Proof of 4. Let (x, y) be an arbitrary element of (A × B) ∪ (C × D). Then
either (x, y) ∈ A × B or (x, y) ∈ C × D.