Page 181 - HOW TO PROVE IT: A Structured Approach, Second Edition

P1: PIG/ P2: OYK/
0521861241c04 CB996/Velleman October 20, 2005 2:54 0 521 86124 1 Char Count= 0

Ordered Pairs and Cartesian Products 167
Here’s an incorrect proof that A × B = B × A only if A = B: The first
coordinates of the ordered pairs in A × B come from A, and the first coordinates
of the ordered pairs in B × A come from B. But if A × B = B × A, then the
first coordinates in these two sets must be the same, so A = B.
This is a good example of why it’s important to stick to the rules of proof-
writing we’ve studied rather than allowing yourself to be convinced by any rea-
soning that looks plausible. The informal reasoning in the preceding paragraph
is incorrect, and we can find the error by trying to reformulate this reasoning
as a formal proof. Suppose A × B = B × A. To prove that A = B we could
let x be arbitrary and then try to prove x ∈ A → x ∈ B and x ∈ B → x ∈ A.
For the first of these we assume x ∈ A and try to prove x ∈ B. Now the incor-
rect proof suggests that we should try to show that x is the first coordinate of
some ordered pair in A × B and then use the fact that A × B = B × A.We
could do this by trying to find some object y ∈ B and then forming the ordered
pair (x, y). Then we would have (x, y) ∈ A × B and A × B = B × A, and it
would follow that (x, y) ∈ B × A and therefore x ∈ B. But how can we find
an object y ∈ B? We don’t have any given information about B, other than the
fact that A × B = B × A. In fact, B could be the empty set! This is the flaw in
the proof. If B = ∅, then it will be impossible to choose y ∈ B, and the proof
will fall apart. For similar reasons, the other half of the proof won’t work if
A = ∅.
Not only have we found the flaw in the proof, but we can now figure out
what to do about it. We must take into account the possibility that A or B might
be the empty set.

Theorem 4.1.4. Suppose A and B are sets. Then A × B = B × A iff either
A = ∅, B = ∅,or A = B.
Proof. (→) Suppose A × B = B × A. If either A = ∅ or B = ∅, then there
is nothing more to prove, so suppose A = ∅ and B = ∅. We will show that
A = B. Let x be arbitrary, and suppose x ∈ A. Since B = ∅ we can choose
some y ∈ B. Then (x, y) ∈ A × B = B × A,so x ∈ B.
Now suppose x ∈ B. Since A = ∅ we can choose some z ∈ A. Therefore
(x, z) ∈ B × A = A × B,so x ∈ A. Thus A = B, as required.
(←) Suppose either A = ∅, B = ∅,or A = B.
Case 1. A = ∅. Then A × B = ∅×B =∅ = B×∅ = B × A.
Case 2. B = ∅. Similar to case 1.
Case 3. A = B. Then A × B = A × A = B × A.

Commentary. Of course, the statement to be proven is an iff statement, so
we prove both directions separately. For the → direction, our goal is A =
∅∨B = ∅∨A = B, which could be written as (A = ∅∨B = ∅) ∨A = B,