Finite, countable and uncountable sets

½
Proof. Define a function f : N → Z, n →
n
2
if n is even
−(n+1)
2
if n is odd .
(a) f is injective.
Suppose that f(n 1 )=f (n 2 ).
(a.1) Let f(n 1 )=f(n 2 ) ≥ 0. Then n1
2 = n2
2 ⇒ n 1 = n 2 .
(a.2) Let f(n 1 )=f(n 2 ) ≤ 0. Then −(n+1)
2
⇒ n 1 = n 2 .
(b) f is onto.
Consider any z ∈ Z.
(b.1) Let z ≥ 0, and consider n =2z, whichisanevennaturalnumber.Thenf(n) = n 2 = (2z)
2
= z.
(b.2) Let z < 0, and consider n = −(2z +1), an odd natural number. Then f(n) = −(n+1)
2
=
−(−(2z+1)+1)
2
= z.
Then |N| = |Z|.
2
= −(n+1)
Theorem 11 Any subset of a countable set is also countable.
Proof. Assigned for homework.
Theorem 12 Let S 1 ,S 2 ,... be a countable family of countable sets. Then S i≥1 S i is countable.
Proof. S m is countable for m ≥ 1. Then it can be put into a bijection with a (possibly proper) subset of N
S m = {(m, 1), (m, 2), (m, 3), (m, 4)...}.
Then:
S 1 = (1, 1), (1, 2), (1, 3), ...
S 2 = (2, 1), (2, 2), (2, 3), ...
S 3 = (3, 1), (3, 2), (3, 3), ...
.
S m = (m, 1) (m, 2) (m, 3) ...
.
And (1, 1) canbeassignedto1, (2, 1) to 2, (1, 2) to 3, (3, 1) to 4, (2, 2) to 5, andsoon,soallelements
can be covered without repeating.
Now consider f : {(m, n) |m, n ∈ N ∗ } → N ∗ , (m, n) → (m+n−1)(m+n−2)
2
+ n. Thenf is an injection, but
not necessarily onto (since some S i ’s can be finite and some elements of some S i ’s can be dropped because
of repetition). However, for = {(m, n) |m, n ∈ N ∗ } then f : {(m, n) |m, n ∈ N ∗ } → f[S] is onto. Then,
there is a bijection between S i≥1 S i and f[S]. Since f[S] ⊆ N, thenf[S] is countable, because of Theorem
11. Then S i≥1 S i is countable.
Corollary 13 |Q| = |N|.
Proof. Note that Q can be expressed as
Q = [ Q∈Z\{0}
½ p
q |p ∈ Z ¾
.
n o
Call the set p
q |p ∈ Z = B Q , the set of rational numbers with denominator q. Note that |B Q | = |Z|.
(f : B Q → Z, p q → p is a bijection). Then B Q is countable.
But note that the union is made in Q ∈ Z\{0}, which is countable. So we have that Q is a countable
union of countable sets, so Q is countable. If a set is countable, it can be the case that either |Q| = |N| or
that Q is finite. Since Q is not finite, then |Q| = |N|.
Theorem 14 R is not countable.
2