Page 331 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Countable and Uncountable Sets 317
and, in general,
i i i
A i ={a , a , a ,...}.
1 2 3
i
+
+
Note that, by the definition of union, ∪F ={a | i ∈ Z , j ∈ Z }.
j
+
+
Now define a function f : Z × Z →∪F by the formula
i
f (i, j) = a .
j
+ + + +
Clearly f is onto. Since Z × Z is denumerable, we can let g : Z → Z ×
+
Z be a one-to-one, onto function. Then f ◦ g : Z →∪F is onto, so ∪F is
+
countable.

Finally, suppose ∅ ∈ F. Let F = F \ {∅}. Then F is also a countable

family of countable sets and ∅ /∈ F , so by the earlier reasoning, ∪F is

countable. But clearly ∪F =∪F ,so ∪F is countable too.
Another operation that preserves countability is the formation of finite se-
quences. Suppose A is a set and a 1 , a 2 ,..., a n is a list of elements of A.We
might specify the terms in this list with a function f : I n → A, where for each
th
i, f (i) = a i = the i term in the list. Such a function is called a finite sequence
of elements of A.
Definition 7.2.3. Suppose A is a set. A function f : I n → A, where n is a
natural number, is called a finite sequence of elements of A, and n is called the
length of the sequence.
Theorem 7.2.4. Suppose A is a countable set. Then the set of all finite se-
quences of elements of A is also countable.
Proof. For each n ∈ N, let S n be the set of all sequences of length n of elements
of A. We first show that for every n ∈ N, S n is countable. We proceed by
induction on n.
In the base case we assume n = 0. Note that I 0 = ∅, so a sequence of length
0 is a function f : ∅ → A, and the only such function is ∅. Thus, S 0 = {∅},
which is clearly a countable set.
For the induction step, suppose n is a natural number and S n is countable. We
must show that S n+1 is countable. Consider the function F : S n × A → S n+1
defined as follows:
F( f, a) = f ∪{(n + 1, a)} .

In other words, for any sequence f ∈ S n and any element a ∈ A, F( f, a)isthe
sequence you get by starting with f, which is a sequence a length n, and then
tacking on a as term number n + 1. You are asked in exercise 2 to verify that F