The cardinality of a set is roughly the number of elements in a set. This poses few difficulties with finite sets, but infinite sets require some care.
I can tell that two sets have the same number of elements by trying to pair the elements up. Consider the sets
They have the same number of elements because I can pair the elements of the first set with the elements of the second:
This kind of pairing is called a bijection or a one-to-one correspondence; it's easy to understand with finite sets, but I need to be more careful if I'm going to use the same idea with infinite sets. I'll begin by reviewing the some definitions and results about functions.
Definition. Let X and Y be sets and let be a function.
1. f is injective (or
one-to-one) if implies
.
2. f is surjective (or
onto) if for all , there is an
such that
.
3. f is bijective (or a one-to-one correspondence) if it is injective and surjective.
Definition. Let S and T be sets, and let be a function from S to T. A function
is called the inverse of f if
I proved the following result earlier.
Theorem. Let S and T be sets, and let be a function. f is invertible if and only if f is
bijective.
Example. Let
Define by
Show that f is bijective.
I'll construct an inverse for f. The inverse should "undo" the effect of f:
As you can see, I need to define
I've constructed so that
for all
. To be complete, I should check
that it works the other way, too:
So really is the inverse of f, and f is a
bijection. (For that matter,
is a bijection as
well, because the inverse of
is f.)
Notice that this function is also a bijection from S to T:
If there is one bijection from a set to another set, there are many
(unless both sets have a single element).
I introduced bijections in order to be able to define what it means for two sets to have the same number of elements. The number of elements in a set is called the cardinality of the set.
Definition. (a) Let S and T be sets. S and T have the same cardinality if there is a bijection f from S to T.
Notation: means that S and T have the same
cardinality.
(b) A set S is finite if it is empty, or if
there is a bijection for some
integer
. A set which is not finite is infinite.
(c) If S is a nonempty finite set and there is a bijection for some integer
, I'll say that S has cardinality
n or that S has n elements. In this case,
I'll write
.
At this point, there is an apparently silly issue that needs to be
resolved: Could a finite set be bijective with both and
(say)? Of course, everyday
experience says that this is impossible. However, mathematicians
always take the point of view that if something is really
obvious, then it ought to be easy to justify.
Actually, this particular point isn't that simple to justify --- try to prove it yourself! --- but it's true, and I'll omit the proof.
Example. Prove that the set of natural numbers
has the same
cardinality as the set
of positive even
integers.
Define by
This function has an inverse
given by
Note that since , m is even, so m is divisible by 2 and
is actually a positive integer.
Here's the proof that f and are inverses:
This situation looks a little strange. E is contained in
, but I've just shown that the two sets "have
the same number of elements". The only reason this looks funny
is that it contradicts your real world experience --- which only
deals with finite objects. In fact, it's
characteristic of infinite sets that they have the same
number of elements as some of their proper subsets.
Informally, a set has the same cardinality as the natural numbers if the elements of an infinite set can be listed:
In fact, to define listable precisely, you'd end up saying "the set has the same cardinality as the natural numbers". But this is a good picture to keep in mind. I'll show that the real numbers, for instance, can't be arranged in a list in this way.
The next part of this discussion points out that the notion of cardinality behaves the way "the number of things in a set" ought to behave.
Proposition. Let S, T, and U be sets.
(a) The identity function given by
is a bijection.
(b) The inverse of a bijection is a bijection.
(c) If and
are bijections, then the
composite
is a
bijection.
Proof. (a) The identity function has an inverse, namely itself. Therefore, the identity function is a bijection.
(b) If is a bijection, then by definition it has
an inverse
. To be inverses means that
But these equation also say that f is the inverse of , so it follows that
is a bijection.
(c) Suppose that and
are bijections.
Let
and
be their
respective inverses. I'll prove that
is the
inverse of
.
This proves that is the inverse of
, so
is a bijection.
Corollary. Let S, T, and U be sets.
(a) (Reflexivity) .
(b) (Symmetry) If , then
.
(c) (Transitivity) If and
, then
.
In other words, having the same cardinality is an equivalence relation.
Proof. (a) By the lemma, the identity function
is a bijection, so
.
(b) If , then there is a bijection
. By the lemma,
is a
bijection. Therefore,
.
(c) If and
, then there are
bijections
and
. By the lemma,
is a bijection, so
.
Example. Prove that the interval has the same cardinality as
.
First, notice that the open interval has the same cardinality as the real line.
To prove this, I have to construct a bijection
. It's easy: just define
To show that f is bijective, I have to show that it has an inverse;
the inverse is .
Now I know that and
have the same
cardinality. Next, I'll show that
and
have the
same cardinality.
The idea is to multiply by to stretch
to
. Then I subtract
to shift
to
.
All together, I define by
First, if , then
, so
. This shows that g takes inputs in
and produces
outputs in
.
To show that g is bijective, I have to produce an inverse. The standard "swap the x's and y's" procedure works; you get
Here's the proof that g and are inverses:
Therefore, g is a bijection, so and
have the
same cardinality. By transitivity,
and
have the same cardinality.
Example. Prove that has the same cardinality as
.
Define by
Note that if
Therefore, f does map to
.
I claim that . If
, then
, so
maps
to
. Moreover,
Thus, f is a bijection.
Define by
If , then
. Therefore, g does
map
to
.
I claim that . If
, then
, so
maps
to
. Moreover,
Therefore, g is a bijection.
With the bijections f and g, I have , so
and
have the same
cardinality.
Example. Prove that the intervals and
have the same cardinality by
constructing a bijection from one to the other.
It's important that both of these intervals are closed
intervals. If both were open --- say and
--- we can still take the approach
we'll take in this example. We would have some difficulty, however,
if the intervals were (say)
and
. We'll see how to handle that kind of situation
later.
The interval has length 8 and the interval
has length 4. Since
, I'll
define a bijection
by "scaling up by
a factor of 2". So I start this way:
As it stands, this doesn't work, because , and I'd like 0 to go to -3 in
. I fix this by subtracting 3:
First, I need to show that f actually takes to
. Suppose
. Then
Hence, . Therefore, it's valid to write
.
As usual, I'll show f is bijective by constructing an inverse . You can do this by working backward on
scratch paper, or by doing a scaling argument like the one I used to
construct f. Either way, I get
As I did with f, I need show that g takes its supposed domain into its supposed codomain
. Suppose
. Then
Thus, . Therefore, it's valid to write
.
Finally, I need to show f and g are inverses.
Hence, f and g are inverses. Therefore, f and g are bijections. This
proves that and
have the same
cardinality.
In many situations, it's difficult to show that two sets have the same cardinality by actually constructing a bijection between them. The theorem that follows gives an indirect way to show that two sets have the same cardinality.
Theorem. (Schröder-Bernstein) Let S and T be
sets. Suppose there are injective functions and
. Then S and T have the same
cardinality.
The proof of the Schröder-Bernstein theorem is a little tricky, so I won't do it here.
The Schröder-Bernstein theorem says that if S has the same cardinality as a subset of T, and T has the same cardinality as a subset of S, then S and T must have the same cardinality.
It is a powerful tool for showing that sets have the same cardinality. Here are some examples.
Example. Show that the open interval and the closed interval
have the same
cardinality.
The open interval is a subset of the closed
interval
. In this situation, there is an
"obvious" injective function
,
namely the function
for all
. (f is called an inclusion
map.) If
, then
, so f is injective.
Next, I'll construct an injective function . The idea is to find a "copy" of
in
, then do some scaling and
translation to map
onto the copy. I'll use the
interval
as my target in
. The target has length 0.5, so I'll multiply by 0.5
to shrink
to
. Next, I'll add
0.25 to shift
to
. All
together, I get
First, if , then
, so
. This proves that g is a function from
to
.
Next, I need to show that g is injective. Suppose that , I must prove that
. Now
means that
Therefore, g is injective. (In fact, g is bijective, and you could
prove injectivity by constructing --- though it would
be overdoing it a bit.)
Now I have injective functions and
. By Schröder-Bernstein,
.
Example. Prove that has the same cardinality as
.
The two sets don't "look alike" --- the first set is a single interval which is closed on both ends, while the second set consists of two open intervals. When two sets don't look alike but you think they have the same cardinality, consider using the Schröder-Bernstein theorem. I'll define injective functions going from each set into the other.
I'll describe in words how I'm getting the definitions of the functions. (Note that there are many functions you could use to do this!)
The first set is an interval of length 2, which (because of its
endpoints) won't fit in either of the intervals that make up the
second set. Since the second set's intervals don't have
endpoints, if I just slide over, its endpoints will
stick out of the ends of either
or
. So the idea is to shrink
first, then slide it inside either
or
.
If I multiply by 0.5, I get
, an interval
of length 1. This will surely fit inside
(say), and I can slide
into
by adding 2. This takes
to
. I just have to do the two steps one after the
other. So define
by
First, I have to show that this makes sense --- that is, that f
really takes into
.
Suppose
. Then
Since , obviously
, so f does map
into
.
Next, I have to show that f is injective. Suppose . Then
Hence, f is injective.
Next, I have to define an injective function . Now
occupies a total length of
, whereas the target interval
has length 2. If I multiply by
, I'll shrink
to
, which has a total length of 1. Next, I
can slide
inside
by subtracting 0.7, which should give
.
Thus, define by
I need to check that g maps into
. Let
. Then certainly
x is between 1 and 6, i.e.
. (Of course,
does not imply that
. Do you see why?) So
Since , obviously
, so g does map
into
.
Next, I have to show that g is injective. Suppose . Then
Therefore, g is injective.
Hence, and
have the same
cardinality, by the Schröder-Bernstein theorem.
I've already noted that it's easy to find finite sets of different cardinalities: for example, a set with three elements does not have the same cardinality as a set with 42 elements. I've also given examples of infinite sets which have the same cardinality. It's an important fact that not all infinite sets have the same cardinality --- there are different kinds of "infinity"! Here's some terminology which I'll used to describe the situation.
Definition. A set is
countably infinite if it has the same cardinality as the natural
numbers . An infinite set
which is not countably infinite is uncountably infinite or
uncountable.
A set is countable if it is either finite or countably infinite.
I know that some infinite sets --- the even integers, for instance --- are countably infinite. I know of other infinite sets, such as the real numbers. Is the set of real numbers countably infinite? The answer is no; the proof is due to Georg Cantor (1845--1918), and is called the diagonalization argument.
Theorem. The open interval is uncountably infinite.
Proof. I'm going to be a little informal in this proof so that the main idea isn't lost in a lot of notation.
Suppose on the contrary that is countably infinite.
Represent numbers in the interval as decimals
. (If a number ends in an infinite sequence
of 9's, rewrite it as a finite decimal --- so, for instance,
becomes 0.135.) Since
is countably
infinite by assumption, I can arrange the numbers in
in a list:
I emphasize that, by assumption, this list contains all of
the numbers in the interval .
Now go down the diagonal and make a number using the digits. In this
case, I get the number .
Take each of the digits in this number and change it to any
other digit except 9. For example, you could add 1 to each digit from
0 to 7 and change 8 or 9 to 0. This would produce the number .
(The reason you do not want to change digits to 9 is so that you don't wind up with a number that ends in an infinite sequence of 9's.)
The number differs from each of the numbers
in my list. Specifically, the
digit of
is different from the
digit in the
number on the list. This means that
is not in my list --- which is a contradiction,
because I assumed that my list contained all of the numbers
in the interval
.
Therefore, the interval must be uncountably infinite.
Since the interval has the same cardinality as
, it follows that
is uncountably
infinite as well. Notice that
(which is
countably infinite) is a subset of
. Are there any sets
which are "between"
and
in cardinality?
The Continuum Hypothesis states that there are
no sets which are "between" and
in cardinality; it was first
stated by Cantor, who was unable to construct a proof. Kurt Gödel
[2] proved around 1940 that the Continuum Hypothesis was consistent
relative to the standard axioms of set theory. Paul Cohen [1] proved
in 1963 that the Continuum Hypothesis is undecidable: It is
independent of the standard axioms for set theory.
In other words, the question of the existence of a subset of which has cardinality different from either
or
can't be settled without adding
assumptions to standard mathematics --- and you can assume either
that such a set exists, or that it doesn't, without causing a
contradiction.
Definition. Let S be a set. The power set of S is the
set of all subsets of S.
For instance, suppose . The power set of S is
Notice that the power set includes the empty set and the set S itself.
If you're constructing a subset of a set, there are two alternatives
for each element: Either it is in the subset, or it is not. So if the
set has n elements, the two alternatives for each element give possibilities in all. Therefore, if S is finite and
, then
.
In this example, , and
.
Theorem. If S is a set, then S and do not have the same cardinality.
Proof. Suppose first that . Now
, so
. Hence,
while
, and
the result is true in this case.
Now suppose that . I'll prove the
result by contradiction. Suppose that
.
This means that there is a bijection
.
Since f is a bijection, every element of the power set --- that is,
every subset of S --- is paired up with an element of S. For example,
there must be an element for which
.
Of course, . So s is an element which is
paired up with a subset that doesn't contain it. And in general,
f takes an element of S to a subset of S, and that subset either
contains the element or it doesn't.
Here's a particular example to help you get your bearings. In the
picture below, the set is and the function
f is depicted by the arrows.
In this example, f takes b and c to subsets that contain them; f takes a and d to subsets which don't contain them.
Continuing with the proof, let
That is, T is the subset of elements of S which f takes to subsets which don't contain them. I know there is at least one such element, namely the element which f takes to the empty set.
Now f is bijective, and T is a subset of S, so there is an element
such that
. Question: Is
?
If , then by definition of T,
. This is a contradiction.
If , then
,
so
satisfies the defining condition for T --- which
means
. This is a contradiction.
Since and
both lead to
contradictions, I've actually contradicted my first assumption ---
that
. Therefore,
, as I wanted to prove.
As an example, the power set of the natural numbers has the same cardinality as
.
I showed earlier that is countably infinite, whereas
is uncountably infinite, so this confirms the theorem
in this particular case.
Proposition. If X and Y are finite, then .
Proof. Here's an informal proof. The elements
of are ordered pairs
where
and
. Since there are
choices for x and there are
choices for y, there are
such ordered pairs.
More formally, suppose
Define a function by
I'll let you verifty that it's injective and surjective, and hence, a
bijection.
For example, if S has 42 elements and T has 5 elements, then has
elements.
Interesting things happen when S and T are infinite. For example,
is countably infinite; how big is
?
Theorem.
is countably infinite.
Proof.
is the set of pairs
, where m and n are natural
numbers:
I'm going to list the pairs starting with in the order shown by the grey line. This means I'm
constructing a function
. Here it is:
Here is why this works. is the
element on the diagonal line whose elements add up to
. Previous to that, the number of element I've gone
through is
That gives .
It's a little tricky to show f is injective, so I'll omit the proof
here. There is an obvious way to make an injective function from to
:
If , then
, so
, and hence g is injective. By the
Schröder-Bernstein theorem,
and
have the same cardinality.
[1] Paul J. Cohen, Set Theory and the Continuum Hypothesis, Reading, Massachusetts: The Benjamin-Cummings Publishing Company, Inc., 1966 [ISBN 0-8053-2327].
[2] Kurt Gödel, Consistency-proof for the generalized continuum hypothesis, Proc. Nat. Acad. Sci. U.S.A., 25(1939), 220-204.
Copyright 2020 by Bruce Ikenaga