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                                   320                       Infinite Sets
                                   be an arbitrary positive integer and prove D  = f (n). Now recall that we chose
                                   D carefully so that we would be able to prove D  = f (n), and the reasoning
                                   behind this choice hinged on whether or not n ∈ f (n). Perhaps the easiest way
                                   to write the proof is to consider the two cases n ∈ f (n) and n /∈ f (n) separately.
                                   In each case, applying the definition of D easily leads to the conclusion that
                                   D  = f (n).


                                                    +
                                                             +
                                   Proof. Suppose f : Z → P (Z ). We will show that f cannot be onto by
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                                   finding a set D ∈ P (Z ) such that D /∈ Ran( f ). Let D ={n ∈ Z | n /∈ f (n)}.
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                                   Clearly D ⊆ Z ,so D ∈ P (Z ). Now let n be an arbitrary positive integer.
                                   We consider two cases.
                                     Case 1. n ∈ f (n). Since D ={n ∈ Z | n /∈ f (n)}, we can conclude that
                                                                   +
                                   n /∈ D. But then since n ∈ f (n) and n /∈ D, it follows that D  = f (n).
                                     Case 2. n /∈ f (n). Then by the definition of D, n ∈ D. Since n ∈ D and
                                   n /∈ f (n), D  = f (n).
                                     Since these cases are exhaustive, this shows that ∀n ∈ Z (D  = f (n)), so
                                                                                   +
                                   D /∈ Ran( f ). Since f was arbitrary, this shows that there is no onto function
                                   f : Z → P (Z ). Clearly P (Z )  = ∅, so by Theorem 7.1.5, P (Z )is
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                                   uncountable.
                                     The method used in the proof of Theorem 7.2.5 is called diagonaliza-
                                   tion because of the diagonal arrangement of the boxed answers in Figure 1.
                                   Diagonalization is a powerful technique that can be used to prove many the-
                                   orems, including our next theorem. However, rather than doing another di-
                                   agonalization argument, we’ll simply apply Theorem 7.2.5 to prove the next
                                   theorem.


                                   Theorem 7.2.6. R is uncountable.
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                                   Proof. We will define a function f : P (Z ) → R and show that f is
                                   one-to-one. If R were countable, then there would be a one-to-one function
                                                                                            +
                                   g : R → Z . But then g ◦ f would be a one-to-one function from P (Z )to
                                           +
                                   Z and therefore P (Z ) would be countable, contradicting Cantor’s theorem.
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                                   Thus, this will show that R is uncountable.
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                                     To define f, suppose A ∈ P (Z ). Then f (A) will be a real number between
                                   0 and 1 that we will specify by giving its decimal expansion. For each positive
                                               th
                                   integer n, the n digit of f (A) will be the number d n defined as follows:

                                                               3  if n /∈ A
                                                         d n =
                                                               7  if n ∈ A.
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