Download Contents 1 The Natural Numbers

MATH 304:
CONSTRUCTING THE REAL NUMBERS∗†
Peter Kahn
Spring 2007
Contents
1 The
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
1.10
Natural Numbers
The Peano Axioms . . . . . . . . . .
Proof by induction . . . . . . . . . .
Comparing integral systems . . . . .
Definition by Induction . . . . . . . .
Addition . . . . . . . . . . . . . . . .
Ordering the natural numbers . . . .
Variations on induction . . . . . . . .
Well-ordering . . . . . . . . . . . . .
Multiplication . . . . . . . . . . . . .
The natural numbers and cardinality
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1
2
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15
18
20
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24
Except for illustrative examples, the following sections describe a systematic construction of the real numbers using only the basic concepts of set theory. They follow
the standard progression of such constructions: natural numbers, integers, rational
numbers, and real numbers. Exercises with asterisks are considered challenging.
1
The Natural Numbers
The natural numbers N consist of the positive integers 1, 2, 3, . . ., together with the
number 0. Whereas small positive integers have been used, with varied symbolic
representation, since prehistoric times, the number 0 is a relatively late addition
∗
The construction of the real numbers described in Chapter 3 of Robert Stoll’s 1963 book Set
Theory and Logic, W.H. Freeman and Company, San Francisco, London, inspired the development
in these notes. There will be occasional references to Stoll’s book.
† c
°May 21, 2007.
1
(estimated at around 300 B.C.), hence slightly less “natural” than the others. Thus,
some mathematicians exclude it from N. This doesn’t change the theory in any
significant way, just the form of certain definitions and proofs. It turns out to be
slightly more convenient to include 0 in N, so this is what we (and most contemporary
mathematicians) do.
There are a number of different approaches to constructing the natural numbers.
Our point of view will be that the entire set N = {0, 1, 2, 3, . . .} is given as a member of
our universe, together with one crucial property of N. We codify this property in the
form of an axiom, which we shall use to analyze N and to derive new properties. In this
approach, we are reminded of the dictum of the distinguished German mathematician
Leopold Kronecker (1823-1891):
God created the integers, all else is the work of man.
In our derivations of new properties of N, we shall insist on using only the basic
axiom, together with the rules of symbolic logic and introductory set theory as presented in the earlier chapters. Of course, once we have proved a result, we may use it
as well. By deriving all the usual features of N in this way, not only do we place them
on a solid logical foundation but we also validate our choice of initial assumptions.
1.1
The Peano Axioms
To focus on key properties of N, we make the following definitions:
Definition 1. An integral system < X, x0 , s > consists of a set X, a distinguished
element x0 ∈ X, called the initial element, and a function s : X → X, called the
successor function, satisfying:
2
a. s is an injection.
b. x0 6∈ R(s).
c. Suppose that Y is a subset of X containing x0 such that, for every x ∈ Y we
must have s(x) ∈ Y . Then Y = X.
Property (c) is known as the Principle of Induction (for < X, x0 , s >).
The stated property of Y (namely, that if x ∈ Y , then s(x) ∈ Y ) is sometimes
described by the statement “Y is closed under s.” It is also sometimes abbreviated
by writing s(Y ) ⊆ Y , where s(Y ) is shorthand for {z : z = s(x), for some x ∈ Y }.
The Principle of Induction states that there is only one subset of X that contains x0
and is closed under s: namely X itself. We’ll see shortly how this translates into the
usual method of proof by induction.
Let us now define the function σ : N → N by the rule σ(n) = n + 1. We shall
assume
The Axiom of Integrality: < N, 0, σ > is an integral system.
The basic
the natural numbers; its three features, as given in Definition 1, are part of our axiom
for the
concept of N. A nearly identical presentation of these features (written as five items) natural
numbers
The Axiom of Integrality is a fundamental assumption that we are making about
is known as the Peano Axioms, named after the Italian mathematician G. Peano,
who published them in 1889.
In the definition of the successor function σ, we used the notation n + 1 because it
is the common way that we write the successor of n. We do not mean by this that
we have yet defined or have available a notion of addition of natural numbers, and
3
that n + 1 is the result of adding 1 to n. Addition is a secondary concept that we
shall define later. To emphasize this, we shall for the time being use the notation
σ(n) = n0 , avoiding n + 1 until later.
1.2
Proof by induction
The student may recognize the Principle of Induction from other courses, perhaps
notably beginning calculus. Anticipating the way we are going to be using this principle for proofs later in these notes, let us formulate the general idea here in the
terminology and symbolism of the Symbolic Logic notes and then reformulate it in
more familiar terms.
Proofs by induction involve statements of the form (∀n)P (n), where P is a predicate
involving the single variable n, which is assumed to range over N. Let T (P ) be the
truth set of P . By this we mean the set of all specializations a of the variable n for
which P (a) is true. To prove the given statement, we must show that T (P ) = N.
According to the Principle of Induction for < N, σ, 0 >, this can be accomplished
³¡
by showing two things: namely, that (i) 0 ∈ T (P ), and that (ii) (∀k ∈ N) k ∈
¢
¡
¢´
T (P ) ⇒ k 0 ∈ T (P ) .
Reformulating this, we get the familiar version of the induction principle: namely,
to prove that (∀n)P (n) is true, first prove that P (0) is true, and then prove
¡
¢
(∀k ∈ N) P (k) is true ⇒ P (k 0 ) is true .
The first part is often called the initial step of the induction proof, and the second
is called the inductive step. The hypothesis in the inductive step is often called the
4
Familiar
version
of
induction
inductive hypothesis or induction hypothesis.
We present one example of this method that should be familiar to someone who has
studied calculus. The example is intended as an illustration, not as part of the logical
development we are presenting. So, we shall be assuming the standard properties of
the natural numbers as they are used in arithmetic, elementary algebra, and calculus
courses.
For any natural number n, let P (n) be the predicate 1+3+. . .+(2n+1) = (n+1)2 .
We use induction to prove (∀n)P (n). P (0) is the assertion 1 = (0 + 1)2 , which is
¡
¢
obviously true. We must now show that (∀k) P (k) ⇒ P (k + 1) . So, choose any
natural number k and assume P (k), i.e., assume 1 + 3 + . . . + (2k + 1) = (k + 1)2 . We
must prove P (k+1). Begin by writing P (k+1): 1+3+. . .+(2(k+1)+1) = ((k+1)+1)2 .
¡
¢
The left-hand side of this, can be written as 1 + 3 + . . . + (2k + 1) + (2(k + 1) + 1).
Familiar
Since P (k) is true, by the induction assumption, we may replace the part in large
example
¡
¢
2
parentheses by (k + 1)2 , obtaining (k + 1)2 + 2(k + 1) + 1, which equals (k + 1) + 1 . of an
induction
proof
This calculation shows that
¡
¢2
1 + 3 + . . . + (2(k + 1) + 1) = (k + 1) + 1 ,
which is just the assertion P (k + 1). Therefore induction step is complete. The
principle of induction now implies that P (n) is true for all n, so the desired result is
proved.
We now return to the logical development and illustrate method of proof by induction by proving the following simple but useful statement:
Proposition 1. Every natural number is either equal to 0 or is the successor of
5
another natural number.
This proposition may seem obvious to us, but it is not contained directly in the
definition we have given, so it needs to be proved.
Proof. The predicate in the statement of the proposition is n ∈ {0} ∪ R(σ), so its
truth set is obviously {0} ∪ R(σ). Let us call this X for short. We show that X
Another
example
0
is true by definition. Now choose any k ∈ X. Since k = σ(k), we obviously have of an
induction
0
0
k ∈ R(σ). But R(σ) ⊆ X, by definition, so k ∈ X. By the principle of induction, proof
satisfies the hypotheses of the induction principle. The first property, that 0 ∈ X,
X = N, as desired.
We remark that the Axiom of Integrality asserts in part that σ is injective. Together
with the foregoing proposition, this means that every non-zero natural number n can
be written in the form m0 , with m uniquely determined by n. We may call m the
predecessor of n.
Exercise 1. Using the standard facts about natural numbers that you have learned
in earlier courses, prove the following by induction on n:
a. 2 + 4 + . . . + (2n) = n(n + 1).
¡
¢2
b. 13 + 23 + . . . + n3 = n(n + 1)/2 .
1.3
Comparing integral systems
We now wish to compare different integral systems, and for this, we use the following
definition:
6
Definition 2. Let < X, x0 , s > and < Y, y0 , t > be two integral systems. An integral
isomorphism of integral systems < X, x0 , s >' < Y, y0 , t > is a function f : X → Y
such that
a. f is a bijection.
b. f (x0 ) = y0 .
c. (∀x ∈ X), f (s(x)) = t(f (x)).
If an integral isomorphism exists, we say that < X, x0 , s > is integrally isomorphic to
< Y, y0 , t >.
An integral isomorphism in this sense, simply establishes a bijection between elements of X and Y in such a way as to match initial elements and preserve the
successor relation. Thus, isomorphic integral systems are just copies of one another.
Exercise 2. Prove that this notion of integral isomorphism is an equivalence relation.
Exercise 3. Prove that for every integral system < X, x0 , s >, there is at most one
integral isomorphism < N, 0, σ >'< X, x0 , s >. (Hint: Suppose that f and g are
integral isomorphisms, and prove by induction that (∀n) f (n) = g(n).) The existence
of at least one isomorphism requires the notion of definition by induction, which is
discussed below.
Exercise 4. In this exercise you may use the standard properties of numbers (real,
rational, natural, etc.). For each of the following examples, state which is an integral
system and which isn’t. In either case, justify your answer.
7
a. < Z, 0, σ >, where Z denotes the set of all (i.e., positive, negative and zero)
integers, and 0 and σ are as before.
b. < Q+ , 0, σ >, where Q+ is the set of non-negative rational numbers, and 0 and
σ are as before.
c. < N(10), 10, σ >, where N(10) is the set of natural numbers ≥ 10, and σ is as
before.
d. < N− , 0, σ >, where N− denotes the non-positive integers, and 0 and σ are as
before.
1.4
Definition by Induction
The principle of induction is used not only to prove statements of the sort already
indicated, but it is also used to define functions that have N as their domain.
For example, assuming for the moment that we have full use of the system of real
numbers R, choose any real number a. We might try to define an “exponential”
Two
function f : N → R as follows: f (0) = 1; f (n0 ) = af (n). Indeed, for the first few familiar
values of n, we get f (0) = 1, f (1) = af (0) = a, f (2) = a · a = a2 , f (3) = a · f (2) = examples
of
a · a = a , . . ., etc. For those familiar with computer science, this is just the standard definition
by
recursive definition of an . The definition of f at the successor of n is defined in terms induction
2
3
of its value at n.
As another example, we might try to define the function g : N → R as follows:
g(0) = 1; g(n0 ) = n0 g(n). The first few values of this function are 1, 1, 2, 6, 24, 120, . . . .
Most students have seen this function before: it is just n!, the factorial function. Here,
8
the definition of g at the successor of n depends both on n itself and on the value of
g at n.
These definitions work fine if one is interested only in finitely many values of the
function—and this is all that comes into play in computer science or combinatorics.
By themselves, these rules for defining (or, more precisely, attempting to define) the
function, do not actually give a function with domain N. They give only a finite
sequence of function values.
This gap is filled by the following theorem, which shows that the examples just
given do indeed determine functions with domain N. It also gives a general criterion
for making valid definitions “by induction.” The theorem can be proved using the
principle of induction, but the proof is fairly lengthy, so we shall omit it. A proof can
be found in the first two chapters of Stoll’s book. We shall use the result stated by
the theorem, however.
Theorem 1. Suppose that A is a given set with a fixed element a0 selected in A.
Suppose further that we have functions hn : A → A, one for each n ∈ N. Then, there
exists a unique function k : N → A satisfying (i) k(0) = a0 , and (ii) k(n0 ) = hn (k(n)).
Notice that the theorem requires us to have available the element a0 and the stated
functions hn . It then gives us the desired function k. The functions hn are either
given as part of the background information or can be defined by some method other
than induction.
Let us illustrate this by showing how it applies to the two examples above.
For the first example, let A be the set of real numbers R, and let a be a real number
as before. For each n, let hn : R → R be given by the formula hn (x) = ax. (In this
case, all these functions hn are equal.) Notice that the definition of hn does not use
9
induction. Then, the theorem tells us that there exists a unique function k such that
k(0) = a and k(n0 ) = hn (k(n)) = ak(n). This is precisely the exponential function f
of the first example.
For the second example, let A = N and now choose a = 1. Define hn by the formula
hn (x) = n0 x. The reader should convince himeself/herself that this definition does
not use induction. In this case, all the functions hn are different from each other.
The theorem now tells us that there exists a unique function k : N → N satisfying
k(0) = 1 and k(n0 ) = n0 k(n). This is precisely the factorial function g of the second
example.
Exercise 5. Use the method of definition by induction to prove that for each integral
system, < X, x0 , s >, there exists at least one integral isomorphism f :< N, 0, σ >→<
X, x0 , s >.
Exercises 3 and 5 tell us that every integral system is a copy of < N, 0, σ > in a
unique way, or, to put it another way, there is only one integral isomorphism class
of integral systems, and every representative is isomorphic to < N, 0, σ > in a unique
Uniqueness
way. Because of this , we say that the properties in Definition 1 characterize (or of
determine, or define) < N, 0, σ > Henceforth, we write < N, 0, σ > more briefly as N natural
numbers
We now use this method of definition by induction to define addition of natural
numbers.
1.5
Addition
We have started with the most primitive features that lie at the heart of our concept
of the natural numbers, encoded as The Integrality Axiom (or Peano’s Axioms). No10
tice that, except for the operation of succession, (i.e., counting), we are not assuming
anything about operations with the natural numbers (such as addition and multiplication). In particular, we are in a pre-arithmetic mode with respect to N. We now
remedy this by defining the operation of addition , +, for N. Induction can then be
used to verify that this operation has all the familiar properties. The definition and
derivation of the properties of multiplication are similar; we do this later.
Choose any m ∈ N. We shall use induction to define a function am : N → N, and
we shall denote the function value am (n) by m + n. To define am we make use of
Theorem 1, and this requires us to define functions hn : N → N. In this case, the
definition is easy: we define each hn to be the successor function σ. We must also
choose a fixed member of N, which in this case, we choose to equal m. Then Theorem
1 tells us that there exists a unique function N → N, which we shall denote am such
that
a. am (0) = m.
b. am (n0 ) = (am (n))0 .
Definition 3. For any m and n in N, we denote am (n) by m + n and call it the sum
of m and n ( or the result of adding m to n, or m plus n).
In terms of the + notation, the two defining features of am may be written as
follows:
a. m + 0 = m.
b. m + n0 = (m + n)0 .
11
We are left with the important task of justifying this definition, which amounts to
showing that it gives us the operation of addition with which we are all familiar, i.e.,
that it has all the familiar properties. Those that are not immediate consequences of
the definition are all proved by induction. We sketch a few proofs by induction for
illustration.
Lemma 1. Choose any natural numbers `, m, and n. Then: (i)
m+0 = m = 0+m;
(ii) m + 1 = m0 = 1 + m; (iii) ` + (m + n) = (` + m) + n.
Proof. (i) m + 0 = am (0) = m, by definition, proving the first equality. The second
equality, namely, m = a0 (m) must be proved by induction on m. By definition,
a0 (0) = 0, so the assertion is true for m = 0. Now assume it is true for m = k:
that is, k = a0 (k). Evaluate a0 (k 0 ) = (a0 (k))0 = k 0 , the first equality following from
property (b) of a0 and the second from the induction hypothesis. This completes the
proof of (i).
(ii) Evaluate m + 1 = am (1) = am (00 ) = (am (0))0 = m0 . (The reader should fill in the
reasons for each equality.) The second equality, namely, m0 = a1 (m), is proved by
induction. When m = 0, both sides reduce to 1, so the assertion is true for m = 0.
Assume it for m = k, i.e., k 0 = a1 (k), and then evaluate a1 (k 0 ) = (a1 (k))0 = (k 0 )0 .
Switching the order of the equality, we obtain (k 0 )0 = a1 (k 0 ). But this is precisely the
second equality in assertion (ii) of the lemma, with m = k 0 , as desired. Therefore,
the induction, hence the proof of (ii), is complete.
(iii) Let P (n) be the predicate
(∀`)(∀m)(` + (m + n) = (` + m) + n).
12
We prove (∀n)P (n) by induction on n. P (0) is the statement (∀`)(∀m)(` + (m + 0) =
(` + m) + 0). Both sides reduce to ` + m, by what has already been proved, so P (0) is
true. Now assume P (k) for some k. We evaluate the left-hand side of P (k 0 ) for any
selected ` and m:
` + (m + k 0 ) = ` + (m + k)0
¡
¢0
= ` + (m + k)
¡
¢0
= (` + m) + k
= (` + m) + k 0 .
Again, we leave it to the reader to supply reasons for each step. The resulting equality
is precisely P (k 0 ), so the induction proof of (iii) is complete.
The reader may have noticed that statement (i) of the lemma is the familiar additive
identity rule for arithmetic and that the third statement of the lemma is the familiar
associative law for addition. Usually, these are stated as axioms or properties that we
assume. The lemma shows, however, that these can be proved from more rudimentary
concepts.
The associative law is needed because addition is defined to be a binary operation
on natural numbers. Once more than two numbers are to be added, we must decide
how to group them so that we can do the addition two numbers at a time. In the
case of three numbers, there are two ways of grouping the numbers (without changing
their order)—or, as we may say, there are two ways of associating the numbers (into
pairs). The associative law tells us that these two ways give equal sums.
Exercise 6. Show that there are five ways of associating four numbers into pairs,
13
and then prove, using the associative law, that all five ways produce the same sum.
How many ways of associating five numbers are there? List them.
This exercise shows that the associative law can be extended to the case of four
numbers, and similar arguments extend the law to five numbers, etc. Indeed, there is
a general inductive argument (which we do not go into in these notes) that extends
the result to any number of numbers. This means that, for addition, we do not need
to use parentheses to express arbitrarily long sums (e.g., a0 + a1 + . . . + an ), since any
association into pairs yields the same result. So, from now on we feel free to omit
parentheses in these situations.
Incidentally, earlier, we expressed some discomfort in using the notation “n + 1” to
represent the successor of n, since we had not yet defined addition. However, now, in
light of assertion (ii) of the lemma, we are able to use either n + 1 or 1 + n in place
of n0 .
Exercise 7. Use the definition of addition and the method of proof by induction to
prove the following statements:
a. (∀m)(∀n)(m + n = n + m).
the commutative law for addition
¡
¢
b. (∀k)(∀m)(∀n) (m + k = n + k) ⇒ (m = n) . the cancellation law for addition
The statements in the earlier lemma and in the exercise give the most important
features of the additive algebra of the natural numbers, so we summarize them in the
following theorem. Further algebra will come into play when we discuss multiplication
later.
14
a. x + (y + z) = (x + y) + z.
Basic
properties
of
associativity of addition addition
b. (x + z = y + z) ⇒ x = y.
cancellation for addition
c. x + 0 = 0 + x = x
identity law for addition
Theorem 2. Let x, y, z be any natural numbers. Then the following hold:
d. x + y = y + x.
1.6
commutativity of addition
Ordering the natural numbers
Although we have spoken informally about a sequential aspect to the natural numbers,
the concept of an integral system does not, by itself, contain any explicit information
about an ordering of the elements, except for the reference to one element as the
initial element. But the natural ordering of the counting numbers is one of their
basic features, so we now introduce this into the picture with a precise definition.
Definition 4. Given m, n ∈ N, we write m ≤ n (or, equivalently, n ≥ m) if and only
if there exists a k ∈ N such that m + k = n.
For later use, it is convenient to rephrase this definition in the following way: Let
m and n be any natural numbers. The equation
m + x = n has a solution in N if and only if m ≤ n.
Exercise 8. Prove the following addendum to the foregoing: If the equation has a
solution in N, then the solution is unique.
15
Exercise 9. Show that ≤ is a partial order on N.
Exercise 10. Let < be the strict ordering corresponding to ≤. Show that m < n if
and only if there exists a k ∈ N such that k 6= 0 and m + k = n (or, equivalently,
k > 0 and m + k = n).
Exercise 11. Let m, n, and p be any natural numbers. Prove that m ≤ n ⇐⇒
m + p ≤ n + p.
Exercise 12. Suppose that m, n, p and q are natural numbers such that m+n = p+q.
Prove that if m < p, then n > q.
Exercise 13. * Let m be any natural number. Prove by induction on n, that no
natural number n satisfies m < n < m + 1.
Again, this is an example of one of those so-called “obvious” statements that is not
so obvious to prove. Another way of stating this fact is the following: For all m and
n, n < m + 1 if and only if n ≤ m. Convince yourself that this is, indeed, equivalent
to the assertion in the preceding exercise.
Proposition 2. ≤ is a linear order on N.
Proof. (a) A linear order is a partial order for which any two elements are comparable (cf. Set Theory notes). So, we have to show that for any m and n in N, either
m ≤ n or n ≤ m. We prove this by induction on m. So, let P (m) be the assertion
³
(∀n)
(m ≤ n)
∨
´
(n ≤ m) .
Since, for any n, 0 + n = n, it follows that 0 ≤ n, and so P (0) is true.
16
Now assume P (m) is true for m = k. We prove P (k 0 ) by first choosing any n. We
must show k 0 ≤ n or n ≤ k 0 . By the induction hypothesis, there are two cases, k ≤ n
or n ≤ k: (i) Suppose n ≤ k . Since k 0 = k + 1, we have k ≤ k 0 , so, by the transitivity
of ≤, n ≤ k 0 . So P (k 0 ) holds in this case. (ii) Now suppose that k ≤ n. Case (i)
includes the case n = k, so we may as well assume k < n. We shall show that k 0 ≤ n.
Since k < n, it cannot be the case that n = 0. Therefore, n has a unique predecessor
p, i.e., n = p + 1. By Exercise 13, the strict inequality k < n = p + 1 is equivalent to
k ≤ p, i.e., p = k + `, for some ` ∈ N. Therefore, n = p + 1 = k + 1 + `, which implies
k 0 ≤ n, as desired. This completes the induction.
Often the fact that ≤ is a linear order is phrased in terms of <, which we then also Strict
linear
call a (strict) linear ordering. In terms of <, the condition is phrased as follows: For ordering
any natural numbers m and n, exactly one of the following holds: m < n, m = n, or
n < m.
Exercise 14. Prove the following, for all m, n, p ∈ N:
a. m < n
∧
n<p
⇒
m < p ; This is just the transitivity of <. Since
the transitivity of ≤ has already been proved, you may use that for this proof.
b. n = 0
∨
n > 0.
c. ¬(n < n).
d. m < n
⇒
σ(m) = n
∨
σ(m) < n.
17
1.7
Variations on induction
Using the linear ordering of N, it is now easy to formulate two variations of the usual
method of induction.
Variation 1: This method does not have a commonly used name since it amounts
to little more than a re-labeling. We might call it shifted induction, however, when
we refer to it. This kind of induction arises when the predicate one would like to
verify is not usefully defined for some small values of n. Perhaps P (n) is defined only
for n ≥ some fixed number n0 which depends on the particular problem. We can
Shifted
induction
still do induction in this case, except that our initial value is no longer 0 but is n0 .
¡
The situation we are facing then is that P (n0 ) is known to be true and (∀k) P (k) ⇒
¢
P (k + 1) is known to be true, provided k ≥ n0 . We would then like to be able to
conclude that P (n) is true for all n ≥ n0 . Essentially, our entire frame of reference is
shifted over by n0 units.
This variant of induction is easy to derive from the standard induction principle
via the following re-labeling trick. Define a new predicate Q by the formula Q(n) =
P (n0 + n). Since, by definition, k ≥ n0 if and only if there is a natural number m such
that k = n0 + m, k ranges over all natural number ≥ n0 precisely when m ranges over
all natural numbers. Therefore, the situation we confront can now be reformulated as
¡
¢
follows: Q(0) is true, and (∀m) Q(m) ⇒ Q(m + 1) is true. So, ordinary induction
implies that Q(n) is true for all natural numbers n, and, hence, P (n) is true for all
natural numbers n ≥ n0 .
Variation 2: This method is sometimes called strong induction or complete induction
Strong
even though it is equivalent to ordinary induction. In this form of induction, we still induction
18
have the same initial step. But the induction step is replaced by the following
³
´
¡
¢
(∀k) (∀i) (i ≤ k) ⇒ P (i) ⇒ P (k + 1) .
That is, if, for all k, P (1), P (2), . . . , P (k) are all true, then P (k + 1) is true. The
initial step plus the induction step then imply the usual conclusion: (∀n)P (n) is true.
Exercise 15. Given a predicate P = P (n), formulate a predicate Q such that by
applying ordinary induction to Q, you get strong induction for P .
Of course, Variation 1 can be combined with strong induction: i.e., you don’t have
to start at 0 in strong induction.
Here is an example showing how strong induction, although logically equivalent to
ordinary induction, is a convenient tool in proofs. We shall be referring to standard
multiplicative properties of natural numbers, which have not yet been established in
our logical development, but we are doing so only for illustration.
Proposition 3. Every natural number n > 1 can be written as a product of primes.
Proof. This proof is just an illustrative sketch. Since there are no natural numbers
n satisfying 1 < n < 2, we may start the strong induction at n = 2. When n = 2,
the proposition is immediate, since 2 itself is a prime, so it is a one-fold product of
primes. Suppose the proposition is true for all natural numbers i ≤ k, and consider
Example
of
it can be written as a product of two natural numbers k + 1 = ab, with neither a proof
by
nor b equal to k + 1 or 1. Therefore, both a and b are < k + 1. Since there are strong
induction
the natural number k + 1. If it is prime, we are done. If not, then by definition,
no natural numbers strictly between k and k + 1, both a and b are ≤ k, hence the
19
strong induction hypothesis may be applied to them. They are thus both products
of primes, and so, clearly their product is a product of primes.
The point here is that since we are dealing with a multiplicative property, simply
knowing the proposition for k may not be adequate for proving it for k + 1, since
these two numbers are not related multiplicatively in a nice way. However, knowing
the proposition for all natural numbers ≤ k gives us the flexibility we need for the
proof. This concludes our discussion of strong induction.
1.8
Well-ordering
Finally, here is another basic and useful fact about N, which plays a role in proofs
later on.
Proposition 4. Every non-empty subset S of N has a least element (i.e., an element
that is minimal among all elements in S).
Proof. Let S be any subset of N. The proof will use the contrapositive method.
Assume that S has no minimal element, and let T be the set difference N \ S. Let
¡
¢
P (n) be the predicate (∀k) (k ≤ n) ⇒ (k ∈ T ) . This asserts that the set of all
natural numbers less than or equal to n is a subset of T .
Clearly P (0) is true, since it asserts simply that 0 ∈ T . If 0 were not in T , it
would be in the complementary set S, and it would clearly be a minimal member of
S, contradicting our assumption about S.
Exercise 16. Prove the induction step (∀m)P (m) ⇒ P (m + 1).
The exercise concludes the induction proof that (∀n)P (n) is true. But this means
that T = N, hence S = ∅.
20
Thus, assuming that S has no minimal element, the above proof shows that it must
be empty, which is precisely the contrapositive of the statement of the proposition.
This property of N is known as the well-ordering principle. We sometimes say, The
natural
numbers
simply, the natural numbers are well-ordered.
are
Let us be more precise about the definition of well-ordering. We say a linear ordering wellordered
on a set S (or a strict linear ordering on S) is a well-ordering provided that every
non-empty subset of S has a minimal element. So, to prove that a relation on S is
a well-ordering, first, one has to prove that it is a linear ordering (or a strict linear
ordering), and then one has to prove that it satisfies this extra minimality property.
The following exercise shows that there are well-ordered sets that do not satisfy the
Axiom of Integrality (i.e., induction does not hold).
Exercise 17. * Consider the cartesian product N × N = N2 . Using the ordering
Example
of a
wellnamely, we write (a, b) < (c, d) if and only if (a < c) or ((a = c) ∧ (b < d)).
ordered
set
a. Show that this defines a strict linear order on N2 , and then show that this linear for
which
order well-orders N2 .
induction
fails
of N just defined, we now endow N2 with the associated (strict) lexicographic order :
b. Define a function τ : N2 → N2 as follows: For any (a, b) ∈ N2 , let T (a, b) be
the set of all (c, d) > (a, b). Since N2 is well-ordered, T (a, b) has a minimal
element, say (m, n). Define τ (a, b) = (m, n). Prove that τ is injective and
that (0, 0) 6∈ R(τ ). From the definition, it should be clear that τ is a kind of
“successor function.” (In fact, it is not hard to prove that τ (a, b) = (a, b + 1),
but, unless you prove this, do not use this fact for this part of the exercise.)
21
c. Show that < N2 , τ, (0, 0) > fails to be an integral system. That is, show why
the induction principle fails. (Here, you may want to prove and use the formula
for τ given above.)
1.9
Multiplication
We proceed similarly to the case of addition, using the method of induction to define
multiplication.
Choose any natural number m. We shall define a function bm : N → N which is
supposed to represent multiplication by m. To do this by the method of induction, we
need to define functions hn : N → N, which we define as follows: hn (x) = x + m, for
each n. Note that we use addition in this definition, but that’s okay because addition
has already been defined. The method of induction also requires us to choose a
particular element in N: our choice is the natural number 0.
Then, Theorem 1 tells us that there exists a unique function bm : N → N satisfying
bm (0) = 0, and
bm (n0 ) = bm (n) + m.
The first few values of bm are:
22
bm (0) = 0
bm (1) = m
bm (2) = m + m
bm (3) = (m + m) + m
and so on. Clearly, bm represents repeated addition of m, which coincides with our
intuitive understanding of multiplication.
We write bm (n) as mn and call it the product of m by n or m times n. (Occasionally,
we interpose a dot, as in a · b, if this makes the product more readable.)
In terms of the product notation that we just introduced, the key defining properties
of bm can be written as
m · 0 = 0, and
m · n0 = mn + m.
We now list the basic properties of multiplication as a theorem.
Theorem 3. For any natural numbers `, m, and n, the following are true:
a. (`m)n = `(mn).
the associative law for multiplication
Basic
the commutative law for multiplication properties
of
multithe multiplicative identity law plication
b. `m = m`.
c. ` · 1 = ` = 1 · `.
23
d. 0 · m = 0 = m · 0.
e. Suppose that ` 6= 0. Then, we have (` · m = ` · n) ⇒ m = n.
the cancellation law for multiplication
f. Suppose that ` 6= 0. Then we have (` · m ≤ ` · n) ⇐⇒ m ≤ n.
g. `(m + n) = `m + `n and (` + m)n = `n + mn.
the distributive laws
Except for the second equality in item (d), which is true by definition, all of the
statements in this theorem can be proved by the method of induction. The proofs are
not very different from those already done for the facts about addition. However, the
first equality in (d) can also be proved directly (i.e., without induction) , provided
one has first proved the second distributive law. The proof is a little trick involving
the evaluation of 0 + 0 · m as follows: 0 + 0 · m = 0 · m = (0 + 0)m = 0 · m + 0 · m. Thus,
we get 0 + 0 · m = 0 · m + 0 · m, which, by additive cancellation, yields 0 = 0 · m. We
give this proof because a nearly identical proof can be used in more general contexts,
which we’ll encounter later.
Exercise 18. Prove the statements of Theorem 3.
1.10
The natural numbers and cardinality
In this section we discuss cardinality issues related to N, making use of the material
in Section 6 of the Set Theory notes. The student should refer to that section when
needed for reminders about definitions.
We begin by defining the initial segments of the set N.
Definition. Let n be any natural number, and define the nth initial segment Nn to
be {k : (k ∈ N) ∧ (k < n)}.
24
Obviously, N0 is the empty set. Moreover, since, for any natural number m, there
are no natural numbers strictly between m and m + 1 (Exercise 13), we have N1 =
{0}, N2 = {0, 1}, N3 = {0, 1, 2}. More generally, for n > 0, we could write Nn =
{0, 1, . . . , m}, where here m is the predecessor of n. To promote ease of reading, it
is useful to give this number its usual name, namely n − 1. However, we need to
remember that subtraction has not yet been defined, so at this point, the use of n − 1
does not indicate that we are subtracting 1 from n, only that we are considering the
immediate predecessor of n. With this notation understood, we may sometimes write
Nn as
{0, 1, 2, . . . , n − 1}.
25
Definition. A set S is called finite if and only if #(S) = #(Nn ), for some natural
number n. A set that is not finite is called infinite.
The reader should have no trouble with the following exercise.
Exercise 19. Suppose that S is a set such that #(S) = #(Nn ), for some natural
number n, and suppose that x is an object not in S. Then #(S ∪ {x}) = #(Nn+1 )
We now recall the definition of the relation ≺ defined in the chapter Set Theory. If
S and T are two sets, we write #(S) ≺ #(T ) to indicate that there is an injection
S → T but no bijection S → T . In Set Theory, we cite some theorems that imply
that ≺ is a strict linear order.
We wish to compare the strict order relation < on N with the relation ≺ on the
collection of initial segments Nn . In order to keep assumptions to a minimum, we
shall use only the fact that ≺ is transitive and irreflexive in this context, not that it
is a linear order. In particular, we do not assume the Schröder-Bernstein Theorem or
the fact that any two cardinalities are comparable.
Lemma 2. Let m and n be any natural numbers such that m < n. Then there is no
injective map Nn → Nm .
Proof. The proof is by induction on m. When m = 0, the result is obvious, because
Nm is empty and Nn is not, and there is no map whatsoever from a nonempty set to
the empty set, much less an injective map.
So, assume inductively that the result holds for m = k. We show that when n >
k + 1, there is no injective map Nn → Nk+1 . This proof will proceed by contradiction.
Accordingly, suppose that n > k + 1 and there is an injective map f : Nn → Nk+1 .
Since n > k + 1 > k, we may use the inductive hypothesis to conclude that the range
26
Finite
and
infinite
sets
of f cannot be contained in Nk = {0, 1, . . . , k − 1}. That is, there must be some a
such that f (a) = k.
If a = n − 1, i.e., if f (n − 1) = k, then f maps all the other elements, i.e., the
elements of Nn−1 , injectively into Nk . But, since n > k + 1, we have n − 1 > k, so we
may use the induction hypothesis to conclude that this injective mapping Nn−1 → Nk
cannot exist. Therefore, f cannot satisfy f (n − 1) = k. This means that f (a) = k,
for some a other than n − 1; it also means, since f is injective, that f (n − 1) = b,
where b is some element of Nk+1 other than k.
We now define a new map f 0 : Nn → Nk+1 as follows: f 0 (x) = f (x), as long as x
is neither equal to a or to n − 1; f 0 (a) = b, and f 0 (n − 1) = k. That is, f 0 is almost
identical with f except that it switches the values of f at a and n − 1. Therefore,
f 0 is an injective map from Nn to Nk+1 satisfying f 0 (n − 1) = k. But we have just
shown in the preceding paragraph that such a map cannot exist. This is the desired
contradiction.
Corollary 3. For every pair of natural numbers m and n,
m<n
⇐⇒ #(Nm ) ≺ #(Nn ).
Proof. We first prove the implication in the =⇒ direction. Let any pair of natural
numbers m and n be given such that m < n. Then Nm ⊆ Nn , and so #(Nm ) ¹
#(Nn ). Lemma 2 implies that we cannot have #(Nm ) = #(Nn ), which proves that
#(Nm ) ≺ #(Nn ).
Now we prove the converse. Suppose #(Nm ) ≺ #(Nn ). Since this implies that there
is an injective map Nm → Nn , Lemma 2 implies that we cannot have m > n. But the
assumption #(Nm ) ≺ #(Nn ) also implies that there is no bijection Nm → Nn , so we
27
Comparing
<
and
≺
cannot have m = n. The only remaining possibility, then, is that m < n.
Exercise 20. For all natural numbers m and n, m = n if and only if #(Nm ) = #(Nn ).
We may conclude, therefore, that the set of natural numbers corresponds bijectively
to the set of all finite cardinalities in such a way as to preserve the strict orderings
defined on these sets. In particular, this shows that ≺ is a well-ordering on the set of
all cardinalities Nn , n ranging over N, indeed a copy of the strict linear ordering <
on N.
New notation: It is clearly convenient to have some more compact notation for
the cardinality #(Nn ), and, of course, we all know that the conventional terminology
and notation fits the bill: namely, instead of “#(Nn ),” we write simply “n,” and we
refer to any set with that cardinality as a “set with cardinality n.” The results just
proved provide logical justification for adopting this convention. That is, even though
the concept of n as an element of the set N is very different from the concept of the
cardinality of the set Nn , since the two correspond bijectively in an order-preserving
way, there is little cost and considerable convenience in conflating the two. In a
similar spirit, we shall use “<” instead of “≺” when talking about finite cardinalities.
That is, we shall use “m < n” to mean either that the natural number m precedes
the natural number n or that the cardinality m is strictly less than the cardinality n.
Proposition 5. Let n be any natural number, and let S be any subset of Nn . Then,
there exists a natural number m ≤ n such that
#(S) = #(Nm ).
Exercise 21. * Prove the above proposition by induction on n. The initial step of
the induction is almost obvious. We give a hint for the inductive step. Assume that
28
the result holds for n = k, and let S be a subset of Nk+1 = {0, 1, . . . , k}. If S is
already contained in Nk = {0, 1, . . . , k − 1}, then the induction hypothesis applies
immediately. If not, this means that k must be a member of S. Let T = S \ {k},
show that the induction hypothesis can be applied to T , and then use this, together
with Exercise 19 to get the desired result for S.
This proposition gives us good information about subsets of Nn , but it doesn’t tell Bounded
and
us about finite sets in general. To approach the general case, we make a definition: unbounded
subsets
we call a subset S of N bounded if there exists a natural number n such that S ⊆ Nn . of N
Otherwise, we say it is unbounded. Proposition 5 tells us that bounded sets are finite.
Exercise 22. Prove that if f : N → N is an injective map, then its range R(f ) is
unbounded. (Hint: Use Lemma 2.) Use this to conclude that N is infinite.
Proposition 6. If S is a finite subset of N, then S is bounded.
Proof. We prove the contrapositive: If S is unbounded, then S is infinite. So, suppose
that S is unbounded. We begin by defining a function f : S → S. Note that this is
not an inductive definition.
First observe that S 6= ∅, since ∅ is obviously bounded. Next, choose any n ∈ S.
We shall define f (n). Let Tn be the set
{m : (m ∈ S) ∧ (m > n)}.
If Tn = ∅, this would mean that every m ∈ S satisfied m ≤ n, i.e., S ⊆ Nn+1 ,
contradicting the assumed unboundedness of S. So Tn is non-empty. By the wellordering on N, Tn has a smallest element, say n1 .
Define f (n) = n1 . Notice that this definition insures that f (n) > n, for every n ∈ S.
29
This defines the function f : S → S. Next, we define a function g : N → S
inductively as follows: Let g(0) be the smallest element of S, say g(0) = n0 . Assuming
that g(k) has been defined, let g(k + 1) = f (g(k)). This gives an inductive definition
for g. Notice that, for every natural number n, g(n + 1) = f (g(n)) > g(n).
Exercise 23. Prove by induction that if m < n, then g(m) < g(n). (Hint: Since
m < n, we have n = m + k, for some k ≥ 1. Proceed by induction on k If k = 1, the
previous paragraph gives the desired inequality, hence proving the base case. Now do
the inductive step.)
The exercise proves that g : N → S is an injection. If S were finite, we would have
a bijection h : S → Np , for some natural number p. But then the composite h ◦ g
would be an injection N → Np . Exercise 22 shows that this is a contradiction.
The proof of Proposition 6 shows slightly more than what the proposition asserts.
Not only does it show that every unbounded subset S of N is infinite, it also shows that
#(N) ¹ #(S). Since S ⊆ N, we also have #(S) ¹ #(N). Therefore, now applying
the Schröder-Bernstein Theorem, we may conclude that any unbounded subset of
N has the same cardinality as #(N). Since every subset of N is either bounded or
unbounded, we see that the subsets of N have either finite cardinality or the same
cardinality as that of N (which is infinite).
We can state this in a slightly stronger way as follows: Let S be any set such that
there is an injective map f : S → N. Then, of course, f defines a bijection between
The
smallest
any set such that #(S) ¹ #(N). Then, either #(S) is finite or #(S) = #(N). In infinite
cardinality
ℵ0 .
other words #(N) is the smallest infinite cardinality. It is often denoted ℵ0 .
S and the subset R(f ) of N. So, the above statement can be rephrased as: Let S be
30
This leaves open one obvious question: Are there infinite cardinalities  ℵ0 ? The
answer to this is “yes,” and a proof was given via the extra-credit exercise in the
homework assignment for Week 6: specifically, for any set A it was shown that #(A) ≺
#(P(A)) (a result of G. Cantor).
It follows that ℵ0 ≺ #(P(N)). If we apply the exercise (Cantor’s Theorem) with
X = P(N), we get a strictly larger cardinality, #(P(P(N))), and so on. This shows
that there are infinitely many infinite cardinalities.
Of course, most of the mathematics we do involves more familiar sets, such as the
set of real numbers R, the set of points in the plane, etc. The above exercise does
not clarify what the cardinalities of these sets are. For example, is it possible that
ℵ0 = #(R)? Another important result of Cantor shows that the answer is “no.” We
shall go through his proof later in the course.
31