Cardinality
Equivalent Sets
Definition: We say that two sets are equivalent
(sometimes called equipotent or equipollent), denoted by
A ~ B iff there exists a bijection f: A → B.
When two sets are equivalent, we say that they have the
same cardinality.
Finite Sets
We define some special sets of natural numbers:
ℕ1 = {1} ℕ2 = {1,2} ℕ3 = {1,2,3}, ..., ℕm = {1,2,...,m}
These sets are sometimes called initial segments.
Definition: A set A is finite iff A = ∅ or A ~ ℕm for some
m ∈ℕ.
A set is infinite iff it is not finite.
We say that ∅ has cardinality 0.
If A ~ ℕm we say that A has cardinality m.
Finite Cardinalities
When two finite sets are of the same cardinality, say of
cardinality k, then by definition, there is a bijection
between them, and from each of them onto ℕk.
Since a bijection sets up a one-to-one pairing of the
elements in the domain and codomain, it is easy to see
that all the sets of cardinality k, must have the same
number of elements, namely k.
Indeed, for any set that has k elements we can set up a
bijection between that set and ℕk. So, for finite sets, all
the sets in the same cardinality have the same number of
elements.
This is why we often refer to a cardinality as a cardinal
number.
ℕ
Since the definition of an infinite set is a "negation", we
should expect that most proofs about them will use
contradiction methods.
Theorem: ℕ is an infinite set.
Pf: BWOC assume that ℕ is a finite set.
Then there exists a bijection f: ℕk → ℕ for some k.
Let n = f(1) + f(2) + ... + f(k) + 1.
Then n is a natural number (being the sum of natural numbers)
and n > f(i) for any i.
So n is in the codomain of f, but since it can not equal any f(i), it
is not in the Rng(f). So, f is not onto →←
Denumerable Sets
Definition: A set is denumerable iff it is of the same
cardinality as ℕ.
The cardinality of the denumerable sets is denoted ℵ0 which is
read as "aleph naught" or "aleph null". (ℵ is the first letter of the
Hebrew alphabet.)
One may be tempted to say, in analogy with finite sets, that all
denumerable sets have the same number of elements, or all
denumerable sets have ℵ0 elements. As our next example will
show, you probably should avoid this temptation.
ℕ~2ℕ
Let 2ℕ denote the set of all even natural numbers.
Theorem: 2ℕ is denumerable.
Pf: To prove this we must find a bijection f: ℕ → 2ℕ.
Our candidate will be f(x) = 2x with domain ℕ.
This f is one-to-one: Suppose f(x) = f(y). Then 2x = 2y, so x=y.
This f is onto: Let y be an arbitrary element of the codomain 2ℕ
Since y is an even natural number, y = 2k for some k in ℕ.
Thus, f(k) = 2k = y, and f is onto.
f is a bijection, and so, ℕ~2ℕ i.e., 2ℕ is denumerable.
So, the number of even natural numbers is the same as the
number of all natural numbers ?????
So, what gives?
On the one hand, our intuition tells us that this last statement
can't be correct. Every even natural number is in ℕ, but ℕ
also has odd natural numbers ... so ℕ must have more
elements than 2ℕ !
But, the bijection between these sets, pairs up the elements
exactly, so there are just as many elements in one set as in
the other!
Something has to give ... these are contradictory results.
So, what gives?
Realize that intuition is built up from experience. Our direct
experience with sets is limited to finite sets ... so our intuition
is usually ok for these. But we have no direct experience
with infinite sets (for instance, you've never counted the
elements in one), and our intuition leads us astray when it
comes to infinite sets!
Denumerable + 1 = Denumerable
Theorem: If A is denumerable and x ∉ A then A ∪ {x} is
denumerable.
Pf: Since A is denumerable there exists a bijection g: ℕ →
A.
Now define f: ℕ → A ∪ {x} by:
f(1) = x
f(2) = g(1)
f(3) = g(2)
.... and in general f(n) = g(n-1) for n > 1.
f is one-to-one since no two images of f can be the same
(no two images of g are the same, and none of them is x)
f is onto because g is onto A and x is certainly in Rng(f).
So, f is a bijection.
Listable Sets?
This is a Cherowitzo special definition – you will not find
this anywhere in the literature.
A denumerable set is one whose elements can written in a
list (an infinite list) where all the elements appear
somewhere and no element appears twice. If you can
create such a list of elements of the set, then you can
define a function whose arguments are the elements of the
set and whose values are the positions in the list where the
elements appear. This function is a bijection between the
set and ℕ ... thus proving that the set is denumerable.
Thus, you can prove that a set is denumerable by creating
this list.
So, maybe, denumerable sets should be called listable
sets.
Union of Denumerable Sets
Theorem: If A and B are disjoint denumerable sets then
A ∪ B is denumerable.
Pf: Since A is denumerable we can list its elements as:
a1, a2, a3, ...
Since B is denumerable we can also list its elements as:
b1, b2, b3, ....
We can now form a list of the elements of A ∪ B this way:
a1, b1, a2, b2, a3, b3, .....
Clearly, every element of A ∪ B appears exactly once
on this list, so A ∪ B is denumerable. More formally, we can
define the bijection f from A ∪ B onto ℕ by:
2k −1 when x=a k
f  x =
.
2k
when x=bk
{
}
ℤ is denumerable
We can apply the theorems we have just proved to obtain
this result.
First, notice that -ℕ is denumerable. (Consider f(x) = -x ).
Since 0 ∉ ℕ we have that A = ℕ ∪ {0} is denumerable.
Then ℤ = ℕ ∪ {0} ∪ -ℕ = A ∪ -ℕ is denumerable.
ℚ is denumerable
From the last chapter we know that ℕ×ℕ is denumerable since
n-1
the function f: ℕ×ℕℕ given by f(n,m) = 2 (2m – 1) is a
bijection.
Theorem: If A and B are denumerable sets, then A×B is
denumerable.
Pf: Since A and B are denumerable, there exist bijections f:Aℕ
and g:Bℕ. Now consider the function h:A×Bℕ×ℕ given by
h(a,b) = (f(a), g(b)). We show that h is a bijection. Suppose that
h(a,b) = h(c,d) ⇒ (f(a),g(b)) = (f(c),g(d)) ⇒ f(a) = f(c) and g(b) =
g(d). Since f is an injection a = c, and since g is an injection b = d.
Now, let (c,d) be an arbitrary element of ℕ×ℕ. Since f is onto,
there is an a ∈ ℕ with f(a) = c, and since g is onto there is a b ∈ ℕ
with g(b) = d. So, h(a,b) = (f(a),g(b)) = (c,d), so h is a bijection.
The composition of h with f above shows that A×B is denumerable.
ℚ is denumerable
Since ℤ is denumerable, the last theorem shows that ℤ×ℤ
is denumerable.
Theorem: Any infinite subset of a denumerable set is
denumerable.
Pf: Since the original set is denumerable, its elements can
be put into a list. By passing through this list and removing
any element which is not in the subset, we obtain a list of the
elements of the subset.
By identifying each fraction p/q with the ordered pair (p,q) in
ℤ×ℤ we see that the set of fractions is denumerable.
By identifying each rational number with the fraction in reduced
form that represents it, we see that ℚ is denumerable.
Countable Sets
Definition: A countable set is a set which is either finite or
denumerable.
In most theorems involving denumerable sets the term
denumerable can be replaced by countable. Proofs involve
extending the proofs for denumerable sets by checking the
cases when one or more of the sets involved are finite.
Thus:
Theorem: If A is countable and x ∉ A then A ∪ {x} is
countable.
Theorem: If A and B are disjoint countable sets then A ∪ B
is countable.
Theorem: If A and B are countable, then AxB is countable.
Theorem: Any subset of a countable set is countable.
Countable Union
Theorem: Let A be a countable family of countable sets.
Then ∪A is countable.
This theorem while plausible can not be proved from the
generally accepted axioms of set theory (Zermelo-Fraenkel
axioms). To prove it we need an additional axiom known as
the Axiom of Choice. We will postpone this proof until we
have talked about this axiom.
The plausibility of the result comes from the fact that we can
prove all the various cases except that of a denumerable
family of disjoint denumerable sets without appealing to the
Axiom of Choice.
Uncountable Sets
A set which is not countable is called uncountable.
Theorem: The set of real numbers (0,1) is an uncountable set.
Before proving this result we need to say a few things
about the decimal representation of real numbers.
First of all, contrary to what you learned in elementary
school, there is no such thing as a "terminating" decimal
number.
¼ = 0.25 is not a valid representation of this real number, rather
= 0.250000000000 ... a repeating decimal with 0 as the
repeating portion.
Secondly, decimal representations are not unique! Some
numbers have more than one decimal representation.
¼ = 0.2499999999999999 ... . However, the only numbers with
more than one representation are those having a repeating 0 or a
repeating 9.
Uncountable Sets
Theorem: The set of real numbers (0,1) is an uncountable
set.
Pf: BWOC suppose that this set of reals is countable. We
may then list all the elements of the set, one above the other
as below. We can now find a real number in the set which is
th
not on the list (by construction) →← The construction is: the i
th
digit is chosen to be anything other than 0, 9 or the digit in i place
th
of the i number of the list.
Example of the Construction
0.1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4
0.2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5
0.3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6
0.1 2 3 4 5 6 7 8 9 5 4 3 2 1 6 7 8 9 1 2 3 4
0.4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7
0.9 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2
0.5 6 5 6 5 6 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7
⋮
⋱
We construct a number not on the list as follows:
0. 8 5 2 1 1 2 8
.......
...
...
...
...
...
...
...
c
The cardinality of the set (0,1) is denoted by c which stands
for "the continuum".
There are many sets equivalent to (0,1). While we can
prove the following equivalences algebraically (by writing
out a specific bijection), we will give some geometric
arguments that show that the bijections exist without
explicitly writing them out.
Any open interval of real numbers, (a,b) is equivalent to
(0,1).
0
1
a
b
c
ℝ is equivalent to any open interval.
(
)
→
→