Page 332 - HOW TO PROVE IT: A Structured Approach, Second Edition

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318 Infinite Sets
is one-to-one and onto. Thus, S n × A ∼ S n+1 . But S n and A are both countable,
so by Theorem 7.2.1 S n × A is countable, and therefore S n+1 is countable.
This completes the inductive proof that for every n ∈ N, S n is countable.
Finally, note that the set of all finite sequences of elements of A is ∪ n∈N S n , and
this is countable by Theorem 7.2.2.

As an example of the use of Theorem 7.2.4, you should be able to show
that the set of all grammatical sentences of English is a denumerable set. (See
exercise 10.)
By now you may be wondering if perhaps all sets are countable! Is there any
set-theoretic operation that can be used to produce uncountable sets? We’ll see
in our next theorem that the answer is yes, the power set operation. This fact was
discovered by the German mathematician Georg Cantor (1845–1918) by means
of a famous and ingenious proof. In fact, it was Cantor who first conceived of the
idea of comparing the sizes of infinite sets. Important mathematical theorems
are often named after their discoverers, so we have identified Theorem 7.2.5 as
Cantor’s theorem. Cantor’s proof is somewhat harder than the previous proofs
in this chapter, so we’ll discuss the strategy behind the proof before presenting
the proof itself.

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Theorem 7.2.5. (Cantor’s theorem) P (Z ) is uncountable.
Scratch work

The proof is based on statement 2 of Theorem 7.1.5. We’ll show that there
is no function f : Z → P (Z ) that is onto. Clearly P (Z ) = ∅,soby
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Theorem 7.1.5 this shows that P (Z ) is not countable.
Our strategy will be to let f : Z → P (Z ) be an arbitrary function and
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prove that f is not onto. Reexpressing this negative goal as a positive statement,
we must show that ∃D[D ∈ P (Z ) ∧∀n ∈ Z (D = f (n))]. This suggests
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that we should try to find a particular set D for which we can prove both
D ∈ P (Z ) and ∀n ∈ Z (D = f (n)). This is the most difficult step in figuring
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out the proof. There is a set D that makes the proof work, but it will take some
cleverness to come up with it.
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We want to make sure that D ∈ P (Z ), or in other words D ⊆ Z ,sowe
know that we need only consider positive integers when deciding what the
elements of D should be. But this still leaves us infinitely many decisions to
make: For each positive integer n, we must decide whether or not we want n
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to be an element of D. We also need to make sure that ∀n ∈ Z (D = f (n)).
This imposes infinitely many restrictions on our choice of D: For each positive
integer n, we must make sure that D = f (n). Why not make each of our