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316 Infinite Sets
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1. Define h : A × B → Z × Z by the formula
h(a, b) = ( f (a), g(b)).
As in the proof of part 1 Theorem 7.1.2, it is not hard to show that h is one-
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to-one. Since Z × Z is denumerable, we can let j : Z × Z → Z be
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a one-to-one, onto function. Then by Theorem 5.2.5, j ◦ h : A × B → Z +
is one-to-one, so by Theorem 7.1.5, A × B is countable.
2. Define h : A ∪ B → Z as follows:
f (x) if x ∈ A
h(x) =
−g(x) if x /∈ A.
We claim now that h is one-to-one. To see why, suppose that h(x 1 ) = h(x 2 ),
for some x 1 and x 2 in A ∪ B.If h(x 1 ) = h(x 2 ) > 0, then according to the
definition of h, we must have x 1 ∈ A, x 2 ∈ A, and f (x 1 ) = h(x 1 ) = h(x 2 ) =
f (x 2 ). But then since f is one-to-one, x 1 = x 2 . Similarly, if h(x 1 ) = h(x 2 ) ≤
0, then we must have g(x 1 ) =−h(x 1 ) =−h(x 2 ) = g(x 2 ), and then since g
is one-to-one, x 1 = x 2 . Thus, h is one-to-one.
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Since Z is denumerable, we can let j : Z → Z be a one-to-one, onto
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function. As in part 1, we then find that j ◦ h : A ∪ B → Z is one-to-one,
so A ∪ B is countable.
As our next theorem shows, part 2 of Theorem 7.2.1 can be extended to
unions of more than two sets.
Theorem 7.2.2. The union of countably many countable sets is countable. In
other words, if F is a family of sets, F is countable, and also every element of
F is countable, then ∪F is countable.
Proof. We will assume first that ∅ /∈ F. At the end of the proof we will discuss
the case ∅ ∈ F.
If F = ∅, then of course ∪F = ∅, which is countable. Now suppose F =
∅. Then, as described after the proof of Theorem 7.1.5, since F is count-
able and nonempty we can write the elements of F in a list, indexed by the
positive integers. In other words, we can say that F ={A 1 , A 2 , A 3 ,...}. Sim-
ilarly, every element of F is countable and nonempty (since ∅ /∈ F ), so for
each positive integer i the elements of A i can be written in a list. Thus we
can write
1
1
1
A 1 ={a , a , a ,...},
1 2 3
2
2
2
A 2 ={a , a , a ,...},
1 2 3

