Download Cardinality

Cardinality
Equivalent Sets
Definition: We say that two sets are equivalent (sometimes called
equipotent), denoted by A ~ B iff there exists a bijection f: A → B.
This is an equivalence relation on the class of all sets:
A ~ A for all sets A. (IA: A → A is a bijection for all sets A)
If A ~ B then B ~ A. (If f:A → B is a bijection, f :B → A is also)
If A ~ B and B ~ C then A ~ C. (If f:A → B is a bijection and
g:B → C is a bijection, then g⊙f: A → C is a bijection)
-1
The equivalence classes under this relation are called cardinalities.
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 ∅ is of cardinality 0.
If A ~ ℕm we say that A is of cardinality m.
This makes sense since A and ℕm are in the same equivalence
class, i.e., "are of the same cardinality".
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. (Also known as countably infinite.)
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.
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 sets, then A×B 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 not on the list (by
th
construction) →← The construction is: the i digit is chosen to be
th
th
anything other than 0, 9 or the digit in i place 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 5 6 7 9 ...
0.2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 9 5 ...
0.3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 9 8 5 ...
0.1 2 3 4 5 6 7 8 9 5 4 3 2 1 6 7 8 9 1 2 3 4 5 6 7 9 ...
0.4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 9 7 7 7 ...
0.9 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 ...
0.5 6 5 6 5 6 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 ...
⋮
⋱
We construct a number not on the list as follows:
0.3 2 7 1 1 1 2 .......
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
a
1
b
c
ℝ is equivalent to any open interval.
(
)
→
→
Order of Cardinal Numbers
We define an ordering on the cardinal numbers in the following
way:
For any set A we denote the cardinal number (cardinality) of A
by |A|.
|A| = |B| iff A ~ B, otherwise |A| ≠ |Β|.
|Α| ≤ |B| iff there exists an injection f: A → B.
|A| < |B| iff |Α| ≤ |B| and |A| ≠ |Β|.
With these definitions, it is easy to see that "=" for cardinal
numbers is an equivalence relation.
Also, the relation "≤" is reflexive and transitive.
We will see later that it is also antisymmetric.
Cardinal Numbers
The finite cardinal numbers are 1, 2, 3, ..., etc.
The inclusion map, f: ℕk → ℕm given by f(x) = x is an injection
iff k ≤ m. This means that | ℕk | ≤ |ℕm | iff k ≤ m. By the pigeonhole principle, |ℕk| < |ℕm| iff k < m.
The same map f: ℕk → ℕ shows that | ℕk | ≤ ℵ0 for all k. Since
ℕ is an infinite set we have | ℕk | < ℵ0
Similarly f: ℕ → ℝ shows that ℵ0 ≤ c. Since ℝ is uncountable
we have ℵ0 < c.
Cantor's Theorem
Theorem: For any set A, |A| < |P (A) |.
Pf: For x ∈ A, the function f: A → P (A) given by f(x) = {x} is an
injection, so |A| ≤ |P (A)|.
Now assume that |A| = |P (A)|, that is, there exists a bijection
g:A → P (A). Note that g(a) is a subset of A.
Define S = {a ∈ A: a ∉ g(a)}.
S is a subset of A and since g is onto there must exist some c in
A so that g(c) = S.
If c ∈ S, then by the definition of S, c ∉ g(c) = S. →←
If c ∉ S, then by the definition of S, c ∈ g(c), so c ∈ S. →←
Our assumption has led to a contradiction, so it must be false.
Thus, |A| ≠ |P (A)| so |A| < |P (A)|.
Consequences
Cantor's theorem immediately implies that there are infinitely
many infinite cardinal numbers because we can repeatedly use it to
show that:
ℵ0 < |P (ℕ)| < |P ( P ( ℕ) ) | < |P ( P ( P ( ℕ))) | < ...
We shall see a little later where c fits in this string of inequalities.
It is also immediate that there can be no largest cardinal number.
Schröder-Bernstein Theorem
Theorem: If |A| ≤ |B| and |B| ≤ |A| then |A| = |B|.
The proof of this result is fairly long and complicated. I will not
present it, but I do encourage you to look at it in the text.
Example: The closed interval [0,1] has cardinality c.
Let the cardinality of [0,1] be |A|.
The inclusion map f: (0,1) → [0,1] shows that c ≤ |A|.
The inclusion map g: [0,1] → (-1,2) shows that |A| ≤ c.
Thus, by the Schröder-Bernstein Theorem, |A| = c.
It follows that any closed or half-open interval has cardinality c.
c
We can use the Schröder-Bernstein Theorem to prove that
c = | P (ℕ) |
Pf: We first define a function from f: P (ℕ) → (0,1) as follows:
Let A be a subset of ℕ. f(A) = 0.a1a2a3 ... where
{
ai = 3
5
}
when i∈ A .
when i ∉ A
It is easy to see that f is an injection.
Now define g: (0,1) → P (ℕ) as follows:
Each real number in (0,1) can be represented by a binary decimal.
As with the base 10 decimals, the representation is not unique.
(0.00111111...) = (0.010000 ...) We chose the form that has the
repeated 0 ending. Let g(0.b1b2b3....) = {i: bi = 1}.
g is also an injection, so by Schröder-Bernstein (0,1) ~ P (ℕ).
The Axiom of Choice
The following axiom can not be proved or disproved (i.e., it is
independent ) of the axioms of the Zermelo-Fraenkel set theory.
Axiom of Choice: Given any collection of non-empty sets A, there
is a function F (called a choice function) which selects an element
from each set in A.
For any finite collection A, this axiom is provable, but for infinite
sets this can't be done.
A well known example which shows the subtlety of this axiom is:
If a shoe store had an infinite number of pairs of shoes and socks,
then choosing one shoe from each pair does not require the axiom
but choosing one sock from each pair does.
The Comparability Theorem
There are several statements that are equivalent to the axiom of
choice:
(1) For every collection A of nonempty, disjoint sets, there is a
set consisting of exactly one member of each set in A.
(2) Every set can be well ordered.
(3) For any two cardinal numbers, |A| and |B|, one and only one
of the following must hold:
|A| < |B|
|A| = |B|
|B| < |A|.
(This is known as the comparability theorem)
Banach-Tarski Paradox
Most mathematicians are comfortable with the axiom of choice,
however, by using it one can prove:
Banach-Tarski Theorem: Any solid sphere can be decomposed
into five pieces that can be reassembled into two solid spheres with
the same radius as the original.
Repeated use of this construction can lead to a pea being
decomposed into a finite number of pieces which can be
reassembled to form enough peas to fill a volume the size of the
sun!!!!!
Results like this lead many to be wary of the axiom of choice, so
proofs that use it often make this use explicit.
Theorem 5.36
Theorem: Every infinite set has a denumerable subset.
Pf: Suppose A is an infinite set. Since A is not finite, A ≠∅. Choose
an element a1 in A. Now, A – {a1}, is also infinite, so we may
choose an element a2 from it (a1 and a2 are of course different
elements of A). We continue in this way to chose elements. By the
Axiom of Choice, we have constructed a set B = {a1,a2,...,an,...} of
elements of A which are all distinct. The function f: ℕ → B given
by f(n) = an is a bijection, showing that B is a denumerable subset of
A.
The Continuum Hypothesis
We have already seen that ℵ0 < c.
We can now ask the question, is there a cardinal number strictly
between these two?
Cantor conjectured that the answer is no, and this is known as the
continuum hypothesis.
The statement that there is no cardinal number strictly between
|A| and |P (A)| for any infinite set A is known as the generalized
continuum hypothesis. (The continuum hypothesis is the case
where A = ℕ.)