Page 322 - HOW TO PROVE IT: A Structured Approach, Second Edition

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308 Infinite Sets
Note that the function f had to be chosen very carefully. There are many
+
other functions from Z to Z that are one-to-one but not onto, onto but not
one-to-one, or neither one-to-one nor onto, but this does not contradict our
claim that Z ∼ Z. According to Definition 7.1.1, to show that Z ∼ Z we
+
+
need only show that there is at least one function from Z to Z that is both
+
one-to-one and onto, and of course to prove this it suffices to give an example
of such a function.
+
+
Perhaps an even more surprising example is that Z × Z ∼ Z . To show
+
+
+
+
this we must come up with a one-to-one, onto function f : Z × Z → Z .
An element of the domain of this function would be an ordered pair (i, j),
where i and j are positive integers. The result of applying f to this pair should
be written f ((i, j)), but it is customary to leave out one pair of parentheses
and just write f (i, j). Exercise 12 asks you to show that the following formula
+
+
+
defines a function from Z × Z to Z that is one-to-one and onto:
(i + j − 2)(i + j − 1)
f (i, j) = + i.
2
Once again, the table of values in Figure 2 may help you understand this
example.

Figure 2

Theorem 7.1.2. Suppose A ∼ B and C ∼ D. Then:
1. A × C ∼ B × D.
2. If A and C are disjoint and B and D are disjoint, then A ∪ C ∼ B ∪ D.
Proof. Since A ∼ B and C ∼ D, we can choose functions f : A → B and
g : C → D that are one-to-one and onto.
1. Define h : A × C → B × D by the formula

h(a, c) = ( f (a), g(c)).
To see that h is one-to-one, suppose h(a 1 , c 1 ) = h(a 2 , c 2 ). This means
that ( f (a 1 ), g(c 1 )) = ( f (a 2 ), g(c 2 )), so f (a 1 ) = f (a 2 ) and g(c 1 ) = g(c 2 ).