Page 338 - HOW TO PROVE IT: A Structured Approach, Second Edition

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324 Infinite Sets
have A 3 ⊆ X. Continuing in this way we can define sets A n for every positive
integer n, and for every n we must have A n ⊆ X. As you will see, letting
X =∪ n∈Z A n works. In the following proof, we actually do not mention the
+
sets W and Z.

Proof. Suppose A B and B A. Then we can choose one-to-one func-
tions f : A → B and g : B → A. Let R = Ran(g) ⊆ A. Then if we think of
g as a function from B to R, it is one-to-one and onto, so by Theorem 5.3.4,
g −1 : R → B.
We now define a sequence of sets A 1 , A 2 , A 3 ,... by recursion as follows:

A 1 = A \ R;
+
for every n ∈ Z , A n+1 = g( f (A n )) ={g( f (a)) | a ∈ A n }.
Let X =∪ n∈Z A n and Y = A \ X. Of course, every element of A is in either
+
X or Y, but not both. Now define h : A → B as follows:

f (a) if a ∈ X
h(a) = −1
g (a) if a ∈ Y.
Note that for every a ∈ A,if a /∈ R then a ∈ A 1 ⊆ X. Thus, if a ∈ Y then
−1
a ∈ R,so g (a) is defined. Therefore this definition makes sense.
We will show that h is one-to-one and onto, which will establish that A ∼ B.
To see that h is one-to-one, suppose a 1 ∈ A, a 2 ∈ A, and h(a 1 ) = h(a 2 ).
Case 1. a 1 ∈ X. Suppose a 2 ∈ Y. Then according to the definition of
−1
h, h(a 1 ) = f (a 1 ) and h(a 2 ) = g (a 2 ). Thus, the equation h(a 1 ) = h(a 2 )
−1 −1
means f (a 1 ) = g (a 2 ), so g( f (a 1 )) = g(g (a 2 )) = a 2 . Since a 1 ∈ X =
∪ n∈Z A n , we can choose some n ∈ Z + such that a 1 ∈ A n . But then a 2 =
+
g( f (a 1 )) ∈ g( f (A n )) = A n+1 ,so a 2 ∈ X, contradicting our assumption that
a 2 ∈ Y.
Thus, a 2 /∈ Y,so a 2 ∈ X. This means that h(a 2 ) = f (a 2 ), so from the equa-
tion h(a 1 ) = h(a 2 )weget f (a 1 ) = f (a 2 ). But f is one-to-one, so it follows that
a 1 = a 2 .
Case 2. a 1 ∈ Y. As in case 1, if a 2 ∈ X, then we can derive a contradiction,
−1
so we must have a 2 ∈ Y. Thus, the equation h(a 1 ) = h(a 2 ) means g (a 1 ) =
−1
−1
−1
g (a 2 ). Therefore, a 1 = g(g (a 1 )) = g(g (a 2 )) = a 2 .
In both cases we have a 1 = a 2 ,so h is one-to-one.
To see that h is onto, suppose b ∈ B. Then g(b) ∈ A, so either g(b) ∈ X or
g(b) ∈ Y.
Case 1. g(b) ∈ X. Choose n such that g(b) ∈ A n . Note that g(b) ∈ Ran(g) =
R and A 1 = A \ R,so g(b) /∈ A 1 . Thus, n > 1, so A n = g( f (A n−1 )), and there-
fore we can choose some a ∈ A n−1 such that g( f (a)) = g(b). But then since