Page 333 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Countable and Uncountable Sets 319
infinitely many decisions in such a way that it guarantees that the corresponding
restriction is satisfied? In other words, for each positive integer n, we’ll make
our decision about whether or not n is an element of D in such a way that it
will guarantee that D = f (n). This isn’t hard to do. We can let n be an element
of D if n /∈ f (n), and leave n out of D if n ∈ f (n). This will guarantee that
D = f (n), because one of these sets will contain n as an element and the other
won’t. This suggests that we should let D ={n ∈ Z | n /∈ f (n)}.
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Figure 1 may help you understand the definition of the set D. For each
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m ∈ Z , f (m) is a subset of Z , which can be specified by saying, for each
positive integer n, whether or not n ∈ f (m). The answers to these questions
can be arranged in a table as shown in Figure 1. Each row of the table gives the
answers needed to specify the set f (m) for a particular value of m. The set D
can also be specified with a row of yesses and noes, as shown at the bottom of
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Figure 1. For each n ∈ Z we’ve decided to determine whether or not n ∈ D
by asking whether or not n ∈ f (n), and the answers to these questions are the
ones surrounded by boxes in Figure 1. Because n ∈ D iff n /∈ f (n), the row of
yesses and noes that specifies D can be found by reading the boxed answers
along the diagonal of Figure 1, and reversing all the answers. This is guaranteed
to be different from every row of the table in Figure 1, because for each n ∈ Z +
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it differs from row n in the n position.

Figure 1

If you found this reasoning difficult to follow, don’t worry about it. Remem-
ber, the reasoning used in choosing the set D won’t be part of the proof anyway!
After you finish reading the proof, you can go back and try reading the last two
paragraphs again.
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It should be clear that the set D we have chosen is a subset of Z ,so
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D ∈ P (Z ). Our other goal is to prove that ∀n ∈ Z (D = f (n)), so we let n