Finite, countable and uncountable sets

Finite, countable and uncountable sets
Definition 1 (Injective function) Let f : S → T be a function. We say that f is injective if f(x) =f(y) ⇒
x = y.
Remark 2 f is injective iff x 6= y ⇒ f(x) 6= f(y).
Definition 3 (Onto function) Let f : S → T be a function. We say that f is onto iff ∀y ∈ T ,thereexists
some x ∈ S such that f(x) =y.
Definition 4 (Bijection) Let f : S → T be a function. We say that
onto.
f is a bijection iff f is injective and
Definition 5 (Cardinality of a set) Let S be a finite set, S = {s 1 ,s 2 , ..., s n }. Then the cardinality of S is
|S| = n.
Proposition 6 Suppose that S and T are finite. Then |S| = |T | iff there exists a bijection f : S → T .
Proof. "⇒"
Suppose that |S| = |T | = n, S, T are finite sets. Then S should be of the type S = {s 1 ,s 2 , ..., s n } and
T = {t 1 ,t 2 , ..., t n }. Consider the function f : S → T , s i → t i . This function is a bijection.
"⇐"
Suppose that f : S → T is a bijection. We need to show that |S| = |T |. Suppose not. Then we have 2
alternatives:
(a) |S| > |T |, whereS = {s 1 ,s 2 , ..., s n ,s n+1 , ..., s m } and T = {t 1 ,t 2 , ..., t n }. Then f(s 1 ) 6= f(s 2 ) 6= ... 6=
f(s n ),sincef is injective. But f(s n+1 )=t j = f(s i ) for some i ∈ {1, ..., n}, which is a contradiction with
the injectivity of f.
(b) |S| < |T |, whereS = {s 1 ,s 2 , ..., s m } and T = {t 1 ,t 2 ,...,t m ,t m+1 , ..., t n }. Then ∀i ∈ {1, ..., m},
t i = f(s j ) (∀j ∈ {1, ..., m}), else the function is not onto. But then t m+1 6= f(s j ) for any j ∈ {1, ..., m}.
Then f is not onto, which is a contradiction.
Example 7 Consider a set A = {a 1 ,a 2 , ..., a n }.ThenthepowersetofA is defined as
Then, |P(A)| =2 n .
P(A) ={B|B is a subset of A} .
Proof. Consider the set X = {0, 1} n , X = {(x 1 , ..., x n )|x i ∈ {0, 1}}. (A typical set has the form X =
{0, 1, 1, 0,...,1}). Then |X| =2 n ½. Now we find a bijection between P(A) and X, f : P(A) → X, B →
1 if ai ∈ B
f(B) =(x 1 ,x 2 , ..., x n ) with x i =
0 if a i /∈ B .
(For example, consider A = {1, 2, 3, 4}. If B = {1, 2, 3} then f(B) = {1, 1, 1, 0}; if B = ∅, then
f(B) ={0, 0, 0, 0}.)
(a) f is injective.
Consider B 1 ,B 2 ∈ P(A). Suppose that f(B 1 )=f(B 2 ). We need to show that B 1 = B 2 . Note that
a i ∈ B 1 ⇔ [f(B 1 )] i =1⇔ [f(B 2 )] i =1(since f(B 1 )=f (B 2 )) ⇔ a i ∈ B 2 .ThenB 1 = B 2 .
(b) f is onto.
We need to show that ∀x ∈ X, there exists some B ∈ P(A) such that f(B) = x. Consider x =
(x 1 ,x 2 , ..., x n ) ∈ X, andB = {a i |x i =1}. Thenf (B) =x.
Since f is injective and onto, then f is bijective, and therefore |P(A)| = |X| =2 n .
Definition 8 For infinite sets, we say that |A| = |B| iff there exists a bijection f : A → B.
Definition 9 (Denumerable and countable sets) If |A| = |N| then we say that A is denumerable. If |A| = |N|
or A is finite, then A is countable.
Proposition 10 |N| = |Z|.
1

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

Proof. It is enough to show that [0, 1] is not countable. Suppose that [0, 1] is countable. Then its elements
can be put in a sequence:
x 1 = 0.x 11 x 12 x 13 ...
x 2 = 0.x 21 x 22 x 23 ...
x 3 = 0.x 31 x 32 x 33 ...
x 4 = 0.x 41 x 42 x 43 ...
.
where x ij are the digits of the number. Now we construct a number ¯x such that it is not in the list.
Consider ¯x =0.¯x 1¯x 2¯x 3 ... defined as:
½ 5 if xii =4
¯x i =
4 otherwise .
Then ¯x 6= x n ∀n ∈ N since ¯x n 6= x nn . Therefore, no bijection between N and [0, 1] can be established,
and R is not countable.
Theorem 15 Any family of disjoint nonempty open intervals in R is countable.
Proof. Let {(a λ ,b λ )} λ∈Λ
be a family of open sets. Then there exists x λ in (a λ ,b λ ) such that x λ ∈ Q.
Consider f : Λ → Q, λ → x λ . We now claim that f is injective. To see that this is so, suppose that
λ 1 6= λ 2 , and without loss of generality, suppose that b λ1

Consider ¯x ∈ {0, 1} N defined by ¯x =(¯x 1 , ¯x 2 , ...), ¯x i =
¯x n 6= x nn . It follows that A is not countable.
½ 1 if xii =0
0 if x ii =1 .Then ¯x 6= x n ∀n ∈ N since
Remark 19 There is a bijection between P ({a 1 , ..., a n }) and {0, 1} N .
between P(N) and {0, 1} N .ThenP(N) is not countable.
In addition, there is a bijection
Theorem 20 If f : R → R is a nondecreasing function, then the set of points where f is discontinuous is
at most countable.
Proof. Consider the family {x λ } λ∈Λ
, the set of points where f is discontinuous. We need to show that Λ is
countable. Note that if x is discontinuous at x λ ,then
lim f(x) 6= lim f(x)
x→x − λ
x→x + λ
Since f is nondecreasing,
lim f(x) < lim f(x)
x→x − λ
x→x + λ
It is possible to find Q λ ∈ Q such that
lim f(x) x λ2 or
x λ1 x λ2 ⇒ Q λ1 >Q λ2 ⇒ g(λ 1 ) >g(λ 2 ). In the second case, x λ1