Foundations and Proof
Notes by Dr. Lynne H. Walling
and Dr. Steffi Zegowitz
September 2016
9
9.1
Uncountable Sets and Power Sets
Uncountable Sets
Definition 9.1. A set X is called uncountable if it is infinite but not countable.
We want to show that R is uncountable. To do this we will show that the interval
(0, 1) = {x ∈ R : 0 < x < 1 } is uncountable. As an exercise, one can show that there
exists a bijection between the interval (0, 1) and R.
Now, assume that every real number between 0 and 1 has a decimal expansion of the
form
X
0.a1 a2 a3 · · · =
ak 10−k
k∈N
where ak is an integer such that 0 ≤ ak ≤ 9 for each k ∈ N. Note that
X
0.999 · · · =
9 · 10−k
k∈N
=9
1/10
1 − 1/10
=1
since k∈N 10−k is a convergent geometric series. Consequently, if there exists some N ∈ N
such that aN 6= 9 and an = 9 for all n ∈ N with n > N , we have that
P
0.a1 a2 a3 · · · = 0.a1 a2 · · · aN −1 bN
where bN = aN + 1. We will assume the result that for every α ∈ R with 0 < α < 1, there
is a unique way to write α as 0.a1 a2 a3 · · · where ak is an integer such that 0 ≤ ak ≤ 9 for
each k ∈ N, and that for all N ∈ N, there exists an n ∈ N such that n ≤ n and an 6= 9.
We have the following theorem.
Theorem 9.2. The interval (0, 1) = {x ∈ R : 0 < x < 1} is uncountable.
Proof. (Cantor’s diagonalisation argument). We know that the interval (0, 1) is infinite,
since f : N → (0, 1) defined by f (k) = 10−k , for all k ∈ N, is easily shown to be injective.
To the contrary, suppose that (0, 1) is countable. Hence, we can enumerate the elements
of (0, 1) as α1 , α2 , α3 , . . .. We write each αk as a decimal expansion as described above,
that is
αk = 0.ak1 ak2 ak3 · · ·
1
9 Uncountable Sets and Power Sets
2
where aki is an integer such that 0 ≤ ak ≤ 9 for each k ∈ N. Further, we assume that, for
all N ∈ N, there exists an n ∈ N such that n ≤ N and an 6= 9. Now, for each k ∈ N, set
(
1 if akk 6= 1,
bk =
2 if akk = 1,
and set β = 0.b1 b2 b3 · · · . Then β ∈ R with 0 < β < 1 and for all N ∈ N, there exists
an n ∈ N such that n ≤ N and bn 6= 9. Hence by assumption, β = αm for some m ∈ N.
But bm 6= amm , which contradicts the uniqueness of the representation of β as a decimal
expansion not ending in an infinite sequence of 9s. Hence, we must have that (0, 1) is
uncountable.
Remark. Suppose that we have m ∈ N where am is an integer such that 0 ≤ am ≤ 9 (not
all 0) for each m, and
α = 0.a1 a2 · · · am a1 a2 · · · am a1 a2 · · · am · · · = 0.a1 a2 · · · am .
Then α is a rational number. This follows since if b =
m
P
ak · 10m−k , then b ∈ N and
k=1
α=
X
b · 10−mn
n∈N
10−m
=b
1 − 10−m
b
= m
.
10 − 1
Note that the map g : (0, 1) → R given by g(x) = x, for all x ∈ (0, 1) is injective, so
|(0, 1)| ≤ |R|. Since |N| < |(0, 1)|, we have that |N| < |R|, that is R is uncountable. The
following corollary, whose proof is left as an exercise, shows that |(0, 1)| = |R|. This is
another way to argue that R is uncountable.
Corollary 9.3. There exists a bijection between the interval (0, 1) and R.
9.2
Power Sets
Definition 9.4. Let A be a set and define P(A) = {C : C ⊆ A }. Then P(A) is called
the power set of A.
Example. We have P(∅) = {∅}, so |P(∅)| = 1.
Example. We have that P({1, 2}) = {∅, {1}, {2}, {1, 2}}, so |P({1, 2})| = 4 = 22 .
Example. We have that P({1, 2, 3}) = {∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}.
So |P({1, 2, 3})| = 8 = 23 .
Note that, for any nonempty set X, we know that ∅, X are distinct subsets of X.
Hence, we have that |P(X)| ≥ 2. We have the following results, whose proofs are left as
an exercise.
Theorem 9.5. Suppose that A is a finite set with |A| = n for some n ∈ N0 . Then
|P(A)| = 2n .
9 Uncountable Sets and Power Sets
3
Proposition 9.6. Let A, B be sets. Then
(i) A ⊆ B if and only if P(A) ⊆ P(B).
(ii) P(A) ∪ P(B) ⊆ P(A ∪ B).
(iii) P(A) ∩ P(B) = P(A ∩ B).
We have the following theorem.
Theorem 9.7 (Cantor’s Theorem). Let X be a set. Then |X| < |P(X)|.
Proof. Suppose X = ∅. Then |X| = 0 < 1 = |P(X)|. So suppose X is nonempty, and
define f : X → P(X) by f (x) = {x}, for all x ∈ X. We will show that f is injective.
So suppose that x, x0 ∈ X are such that f (x) = f (x0 ). Then {x} = {x0 }. Hence, we have
that x1 = x2 . Therefore, f is injective, so |X| ≤ |P(X)|.
Now, we want to show that there exists no bijection between X and P(X). To the
contrary, suppose there exists such a bijection g : X → P(X). Define
A = {x ∈ X : x 6∈ g(x)}.
Then A is a subset of X, so A ∈ P(X). Further, since g is bijective, there exists some
z ∈ X such that g(z) = A. By the definition of g, z ∈ A if and only if z 6∈ g(z) = A,
which is a contradiction since z ∈ A if and only if z 6∈ A. Therefore, there does not exist
a bijective function g : X → P(X). It follows that |X| < |P(X)|.
9 Uncountable Sets and Power Sets
4
Exercises
9.1. Prove that there is a bijective map from (0, 1) onto R. (Suggestion: In Section 1
of the notes, we constructed a bijective map between the closed intervals [a, b] and
[c, d]. Mimic this construction to define a map from the open interval (0, 1) to the
open interval (−1, 1), and prove this map is bijective. Then use the result of Exercise
1.4 to prove there is a bijective map from (0, 1) to R.)
9.2. Suppose A is a finite set with |A| = n for some n ∈ Z with n ≥ 0. Use induction
to show that |P(A)| = 2n . (Suggestion: For the induction step, suppose A is a
nonempty set, and fix an element u ∈ A. Let B = A r {u}. Argue that there is a
bijection between {C : C ⊆ B } and {D : D ⊆ A, u ∈ D }. Then show that this
means that P(A) = 2P(B), and use your induction hypothesis to conclude that
|P(A)| = 2k+1 where |A| = k + 1.)
9.3. Let A, B be sets. Prove the following.
(a) (A ⊆ B) ⇐⇒ (P(A) ⊆ P(B)).
(b) P(A) ∪ P(B) ⊆ P(A ∪ B).
(c) P(A) ∩ P(B) = P(A ∩ B).