Page 323 - HOW TO PROVE IT: A Structured Approach, Second Edition

P1: PIG/KNL P2: IWV/
0521861241c07 CB996/Velleman October 20, 2005 1:12 0 521 86124 1 Char Count= 0

Equinumerous Sets 309
Since f and g are both one-to-one, it follows that a 1 = a 2 and c 1 = c 2 ,so
(a 1 , c 1 ) = (a 2 , c 2 ).
To see that h is onto, suppose (b, d) ∈ B × D. Then since f and g are both
onto, we can choose a ∈ A and c ∈ C such that f (a) = b and g(c) = d.
Therefore h(a, c) = ( f (a), g(c)) = (b, d), as required. Thus h is one-to-one
and onto, so A × C ∼ B × D.
2. Suppose A and C are disjoint and B and D are disjoint. You are asked in
exercise 13 to show that f ∪ g is a one-to-one, onto function from A ∪ C
to B ∪ D,so A ∪ C ∼ B ∪ D.

It is not hard to show that ∼ is reflexive, symmetric, and transitive, so it is
an equivalence relation. In other words, we have the following theorem:

Theorem 7.1.3. For any sets A, B, and C:
1. A ∼ A.
2. If A ∼ B then B ∼ A.
3. If A ∼ B and B ∼ C then A ∼ C.

Proof.

1. The identity function i A is a one-to-one, onto function from A to A.
2. Suppose A ∼ B. Then we can choose some function f : A → B that is
one-to-one and onto. By Theorem 5.3.4, f −1 is a function from B to A.
But now note that ( f −1 −1 = f , which is a function from A to B,soby
)
Theorem 5.3.4. again, f −1 is also one-to-one and onto. Therefore B ∼ A.
3. Suppose A ∼ B and B ∼ C. Then we can choose one-to-one, onto functions
f : A → B and g : B → C. By Theorem 5.2.5, g ◦ f : A → C is one-to-
one and onto, so A ∼ C.

Theorems 7.1.2 and 7.1.3 are often helpful in showing that sets are equinu-
merous. For example, we showed earlier that Z × Z ∼ Z and Z ∼ Z,
+
+
+
+
+
so by part 3 of Theorem 7.1.3 it follows that Z × Z ∼ Z. Part 2 tells us
+
that we need not distinguish between the statements “A is equinumerous with
B” and “B is equinumerous with A”, because they are equivalent. For ex-
+ + + +
ample, we already know that Z × Z ∼ Z , so we can also write Z ∼
+
Z × Z . By part 1 of Theorem 7.1.2, Z × Z ∼ Z × Z, so we also have
+
+
+
Z ∼ Z × Z.
+
We have now found three sets, Z, Z × Z , and Z × Z, that are equinumer-
+
+
+
ous with Z . Such sets are especially important and have a special name.