Mathematical Analysis I, 2004a

§9. Some Theorems on Countable Sets 19
We now put all pairs (a n ,b m )inone sequence as follows. We start with
(a 1 ,b 1 )
as the first term; then take the two pairs of rank three,
(a 1 ,b 2 ), (a 2 ,b 1 );
then the three pairs of rank four, and so on. At the (r − 1)st step, we take all
pairs of rank r, in the order indicated in (1).
Repeating this process for all ranks ad infinitum, we obtain the sequence of
pairs
(a 1 ,b 1 ), (a 1 ,b 2 ), (a 2 ,b 1 ), (a 1 ,b 3 ), (a 2 ,b 2 ), (a 3 ,b 1 ), ...,
in which u 1 =(a 1 ,b 1 ), u 2 =(a 1 ,b 2 ), etc.
By construction, this sequence contains all pairs of all ranks r, hence all pairs
that form the set A × B (for every such pair has some rank r and so it must
eventually occur in the sequence). Thus A × B can be put in a sequence. □
Corollary 3. The set R of all rational numbers 2 is countable.
Proof. Consider first the set Q of all positive rationals, i.e.,
fractions n ,withn, m ∈ N.
m
We may formally identify them with ordered pairs (n, m), i.e., with N × N.
We call n + m the rank of (n, m). As in Theorem 1, we obtain the sequence
1
1 , 1
2 , 2
1 , 1
3 , 2
2 , 3
1 , 1
4 , 2
3 , 3
2 , 4
1 , ....
By dropping reducible fractions and inserting also 0 and the negative rationals,
we put R into the sequence
0, 1, −1,
1
2 , −1 2 , 2, −2, 1
3 , −1 , 3, −3, ..., as required. □
3
Theorem 2. The union of any sequence {A n } of countable sets is countable.
Proof. As each A n is countable, we may put
A n = {a n1 ,a n2 , ..., a nm , ...}.
(The double subscripts are to distinguish the sequences representing different
sets A n .) As before, we may assume that all sequences are infinite. Now, ⋃ n A n
obviously consists of the elements of all A n combined, i.e., all a nm (n, m ∈ N).
We call n + m the rank of a nm and proceed as in Theorem 1, thus obtaining
⋃
A n = {a 11 ,a 12 ,a 21 ,a 13 ,a 22 ,a 31 , ...}.
n
2 Anumberisrational iff it is the ratio of two integers, p/q, q ≠0.