Chapter 4
Natural numbers
In this chapter we will construct the natural numbers using sets. we will then define addition,
multiplication as set operations and note they have the properties we expect.
To create some perspective it is good to know how numbers will be defined. Zero is always
taken to be the empty set. For the remaining numbers there are twi conventions. Either
1 = {∅}, 2 = {{∅}}, 3 = {{{∅}}}, . . . and in general n = {n − 1}. Or 1 = {∅}, 2 = {∅, {∅}},
3 = {∅, {∅}, {∅, {∅}}}, . . . and in general n = n − 1 ∪ {n − 1}.
In both cases the set n lies in the nth level of the hierarchy Vn = P (n) (∅). What is not clear
is that the set of all such n exists. As we have mentioned before we will need to introduce
an extra axiom to guarantee its existence. It is customary to call this set ω. It is of extreme
importance to take one step at the time and check that we have enough tools in our disposal
to prove principles and properties that are taken for granted.
We follow the second convention. Note that in this case we have two properties that are not
usually associated with natural numbers, like 0 ∈ 1 ∈ 2 ∈ . . . and 0 ⊆ 1 ⊆ 2 ⊆ . . .
1
Inductive sets
The first task is to create the framework that will allows us to define the set n for all (infinitely
many!) natural numbers. The key definitions are of a successor and inductive set.
Definition. Let a be a set. Its successor is
a+ = a ∪ {a}.
So 1 is the successor of 0, 2 the successor of 1 etc.
Definition. A set A is inductive if
(i) ∅ ∈ A
(ii) A is closed under successor; i.e. ∀a ∈ A (a+ ∈ A) or ∀a (a ∈ A) =⇒ (a+ ∈ A).
The existence of an inductive set is not guaranteed by the axioms seen so far. One may be
tempted to combine the union, powerset and subset axioms, but this does not work as any
1
inductive set has to be infinite and any (finite) combination of the axioms can only result in a
finite set. So we need a sixth axiom.
6. Infinity Axiom There exists an inductive set.
∃A [∅ ∈ A ∧ ∀a ∈ A (a+ ∈ A)].
It is easy to check that the natural number sets defined above lie in every inductive set A as
∅ ∈ A, 1 = 0+ ∈ A and so on. It is customary to use this property to define natural numbers.
Definition. A natural number is a set that belongs to all inductive sets.
We want to take ω to be the set of natural numbers. But we need to prove this is a set first!
We apply the subset axioms, in a way similar to the proof of the existence of the intersection.
Theorem 4A. There exists a set whose elements are precisely the natural numbers.
Proof. We know from the Infinity Axiom that an inductive set A exists. We apply a subset
axiom to A to produce the set of all natural numbers. Clearly the formula that x ∈ A must
satisfy is that x lies in all inductive sets. Let us break this process down to two steps.
Let S be any set and q(S) be the formula that tests whether S is inductive. This is easy to
write down:
q(S) = (∅ ∈ S) ∧ [∀a ( (a ∈ S) =⇒ a+ ∈ S )].
A mechanical check confirms this is a formula with variable S. Note however that coding ∅ ∈ S
is long: there exists s in S such that, for all x, x does not belong to s.
The formula that identifies natural numbers (and only them) is
p(x) = ∀S (q(S) =⇒ x ∈ S).
A mechanical check confirms this is a formula with variable x. It codes ‘x is a natural number’
as it is true precisely when x belongs to all inductive sets.
So the set of natural numbers exists by an application of the subset axiom related to p to A
proves the existence of the set of natural numbers.
Now we make the crucial definition.
Definition. ω is the set of natural numbers. So
n∈ω
⇐⇒
n is a natural number
⇐⇒
n belongs to all inductive sets.
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An immediate consequence is the following.
Theorem 4B. ω is an inductive set and is a subset of all inductive sets. In particular any
inductive subset of ω coincides with ω.
Proof. ∅ ∈ ω as ∅ is a natural number. Next we show n ∈ ω =⇒ n+ ∈ ω.
n∈ω
=⇒
n belongs to all inductive sets
=⇒
n+ belongs to all inductive sets
=⇒
n+ ∈ ω.
Any n ∈ ω lies in every inductive set A, so ω ⊆ A.
The last observation of the theorem above is the principle of induction in disguise. Say we want
to prove that all natural numbers satisfy a property expressed by the formula p. It is enough
to consider the subset
T = {n ∈ ω : p(n)}.
If we can show that T is inductive, then it is an inductive subset of ω and so has to be ω. In
other words all natural numbers satisfy p. An example is the following.
Theorem 4C. Every non-zero natural number is the successor of some natural number.
Proof. We let
T = {∅} ∪ {n ∈ ω : ∃m ∈ ω (n = m+ )}.
T is inductive because ∅ ∈ T and
n∈T
=⇒
n = m+ for some m ∈ ω
=⇒
n+ = (m+ )+ for some m ∈ ω
m+ ∈ω
=⇒
n+ ∈ T.
It is furthermore a subset of ω so it must equal ω. Thus ∅ 6= n ∈ ω is the successor of some
m ∈ ω.
From now one we will start using natural numbers to denote the corresponding sets. So ∅ = 0.
3
2
Peano’s postulates and transitive sets
Peano showed that a small number of axioms are enough to construct the natural numbers. He
attributed them to Dedekind and so they are naturally known as ‘Peano’s postulates’. Years
later Peano’s axioms were proved from the six axioms we have discussed so far. We follow this
approach and show that ω as constructed satisfies these postulates.
A key concept is that of a Peano system. This is a triplet hN, S, ei of a set N , a function
S : N 7→ N and an element e ∈ N , which satisfies three postulates:
(i) e ∈
/ ran(S)
(ii) S is injective
(iii) Any T ⊆ N that contains e and is closed under S is in fact N .
Here T is closed under S means that x ∈ T =⇒ S(x) ∈ T or equivalently S[T ] ⊆ T.
The first two conditions ensure that N consists of e, S(e), S (2) (e), . . . . The last condition is an
inductive principle for the system. Picture from p. 70 in the book.
We have nearly shown that ω becomes a Peano system if we take S = σ : ω 7→ ω to be the
“successor function” given by σ(n) = n+ and e = 0.
Theorem 4D. The triplet hω, σ, 0i is a Peano system, where
σ = {hn, n+ i : n ∈ ω}.
Proof. Everything is in order: 0 ∈ ω, σ : ω → ω and 0 ∈
/ σ(ω). Condition (iii) above is just the
induction principle: any T ⊆ ω that contains 0 and is closed under σ is an inductive subset of
ω and so equals ω.
To prove that ω is injective we need the notion of transitive sets, not to be confused with
transitive relations.
Definition. A set A is transitive if any of the following equivalent conditions holds:
(i) a ∈ A =⇒ a ⊆ A
(ii) x ∈ a ∈ A =⇒ x ∈ A
(iii) A ⊆ P(A)
S
(iv) A ⊆ A.
Transitive sets are defined because they capture some of the properties of natural numbers. To
prove that all natural numbers are transitive we need an intermediate result about unions of
successors of transitive sets.
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Theorem 4E. Let a be a transitive set. Then
[
(a+ ) = a.
Proof. We rely of part (iv) of the definition: a is transitive =⇒
[
(a+ )
S
a ⊆ a.
[
(a ∪ {a})
[
[
a ∪ {a}
[
a∪a
=
=
=
(iv)
= a.
We can now deduce that any natural number is a transitive set.
Theorem 4F. Every natural number is a transitive set.
Proof. We do this by induction. Let
T = {n ∈ ω : n is transitive} = {n ∈ ω : ∀x (x ∈ n =⇒ x ⊆ n)}.
It is enough to prove that the set T is inductive. The first condition is easy: 0 ∈ T as there
are not elements to worry about. For the second we must prove that n ∈ T =⇒ n+ ∈ T . For
S
this we use part (iv) of the definition once again: n ∈ T =⇒
n ⊆ n. We establish the same
condition for n+ .
[
(n+ )
Th. 4E
=
n
⊆
n+ .
We are done!
Next we finish off the proof of Theorem 4D.
Last part of the proof of Theorem 4D. We have to show that σ(n) = σ(m) =⇒ n = m.
σ(n) = σ(m)
=⇒
n+ = m+
[
[
(n+ ) = (m+ )
=⇒
n = m.
=⇒
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In the last implication we applied Theorem 4F to get that n+ and m+ are transitive sets and
then Theorem 4E to get the conclusion.
The last result for this section is that ω is a transitive set itself.
Theorem 4G. ω is a transitive set.
Proof. We rely of part (i) of the definition: ω is transitive if n ∈ ω =⇒ n ⊆ ω. We prove this
by induction. We let
T = {n ∈ ω : n ⊆ ω}
and prove it is an inductive set.
∅ ∈ T as ∅ ⊆ ω. As for the “inductive step”, n ∈ T =⇒ n ⊆ ω. We also have {n} ⊆ ω (for
all sets n and ω that satisfy n ∈ ω). So we conclude n+ = n ∪ {n} ⊆ ω ⇐⇒ n+ ∈ T and we
are done.
3
Recursion on ω
Recursion is a notion very similar to induction. I am guilty of not distinguishing between the two
as much as I should. Recursion is explained by examples. The function h(n) = an , n ∈ ω, a ∈ R
is defined recursively by h(0) = 1 and h(n + 1) = ah(n). A set can also be defined recursively.
The most famous example are the Fibonacci numbers defined by: F0 = F1 = 1 and for n > 1
Fn = Fn−1 + Fn−2 . Note however that in this case Fn depends on the two preceding Fibonacci
numbers.
In general we say that a function h with domain ω is defined recursively if h(0) is given and
h(n+ ) = F (h(n)), for some function F : ω 7→ ω. The example given above F : R 7→ R is given
by F (t) = at.
A fundamental results that has many applications is that the recursive definition does produce
a function. More precisely given a function F one can use it to define a function recursively.
Even more precisely:
Theorem (Recursion on ω). Let A be a set, a ∈ A and F : A 7→ A be a function. There exists
a unique function h : ω 7→ A that satisfies
(i) h(0) = a
(ii) h(n+ ) = F (h(n)).
6
The proof is long and convoluted and is left as an exercise in HWK5. The rough idea is to
consider functions that approximate h in the sense that they agree with what h “ought” to be
in their domain and then take their union.
Leaving the proof aside, let us quickly check that recursive definitions are not possible even on
Z, which is very similar to ω.
To see why let F : Z 7→ Z be given by F (t) = t2 + 1. Attempting to define h recursively by
h(n + 1) = F (h(n)) gives h(n + 1) = h(n)2 + 1. This is problematic as h(n + 1) > 0 and also
h(n + 1) ≥ h(n) + 1 > h(n) > h(n − 1) > . . . .
If we take F : Z 7→ Z given by
(
F (t) =
t+1 , t<0
,
t
, t≥0
Then there re infinitely many h that satisfy h(n + 1) = F (h(n)) see the book on p. 76.
The moral reason for all this is that Z has no starting point where the recursion can start.
The first application of the recursion theorem is that any Peano system is isomorphic to hω, σ, 0i.
Theorem 4H. Let hN, S, ei be a Peano system. There is a bijection h : ω 7→ N that preserves
the successor operation
h(σ(n)) = S(h(n))
and the zero element
h(0) = e.
Proof. By the Recursion Theorem there is a function h : ω 7→ N such that h(0) = e and
h(n+ ) = S(h(n)). We have to show it is injective and surjective.
Injectivity: follows from the first two postulates on hN, S, ei (injectivity of S and e ∈
/ ran(S))
and induction on ω.
We consider the set T “on which h is injective”. Formally we set
T = {n ∈ ω : ∀n 6= m ∈ ω (h(n) 6= h(m)}.
It is enough to show that T is an inductive set, as then T = ω and consequently h(n) 6= h(m)
for all n 6= m ∈ ω.
First we prove that 0 ∈ T . We argue by contradiction. Suppose that there exists 0 6= m ∈ ω
such that f (m) = f (0). By Theorem 4C m is the successor of some p ∈ ω. Therefore
h(0) = h(m)
=⇒
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e = h(m)
=⇒
e = S(h(p)).
This is a contradiction as e ∈
/ ran(S)
Next we prove that n ∈ T =⇒ n+ ∈ T. Suppose that n ∈ T and that f (n+ ) = f (m). We
must show that n+ = m. We have just shown that m 6= 0 and so by Theorem 4C m = p+ for
some p ∈ ω. Then
h(n+ ) = h(m)
=⇒ S(h(n)) = S(h(p))
S inj.
=⇒ h(n) = h(p)
n∈T
=⇒
=⇒ n = p
=⇒
n+ = p+ = m.
Surjectivity: follows from the third (the inductive) postulate on hN, S, ei. We want to show
that ran(h) = N so it is enough to prove that ran(h) contains e and is closed under S. The
former is straightforward: e = h(0) ∈ ran(h). For the latter we must show that x ∈ ran(h) =⇒
S(x) ∈ ran(h) :
x ∈ ran(h)
=⇒
x = h(m) , for some m ∈ ω
=⇒
S(x) = S(h(m))
=⇒
S(x) = h(m+ )
=⇒
S(x) ∈ ran(h).
The book has a picture of the resulting infinite commuting diagram on p.77.
This is a good moment to recap what we have accomplished. We have shown that hω, σ, 0i is
a Peano system and in fact is one that was constructed from sets. We have also shown that
every Peano system is isomorphic (structurally equivalent) to hω, σ, 0i. This means that the
axiomatic approach coincides with the constructive approach to building the natural numbers.
4
Arithmetic on ω
Now that we have ω we must define the usual operations: addition, multiplication and exponentiation. Subtraction cannot be defined on ω as 0 − 1 ∈
/ ω. As one expects we begin with
addition, use it to define multiplication and use multiplication to define exponentiation. Once
all this is done the usual laws (commutativity, associativity and distributivity) may be proved.
We begin with addition. The aim is to produce a binary operation on ω. That is a function
8
from ω × ω into ω. The first step is not to define addition in general, but rather addition by a
(fixed) natural number.
We will call Am : ω 7→ ω the function that “maps n to n + m”. Before defining it let us see
some examples of what we expect these Am to be. A0 = Iω , A1 = σ, A2 = σ ◦ σ etc It seems
perfectly reasonable to define Am as σ (m) , the mth-iterate of σ. The problem is that this makes
no sense as m is a set!
The recursion theorem comes to the rescue. Instead of giving a formula for Am we define
it recursively. What we are after is a function Am : ω 7→ ω such that Am (0) = m and
Am (n+ ) = Am (n)+ . The recursion theorem applied to A = ω, F = σ and e = m ensures that
such a functions exists (and is unique).
Now that we have Am for any m ∈ ω we define addition
Definition. The binary operation of addition + : ω × ω 7→ ω is given by
m + n := +(hm, ni) = Am (n).
Let us see what the recursive definition give in terms of applicable properties:
(A1) m + 0 = Am (0) = m.
(A2) m + n+ = Am (n+ ) = Am (n)+ = (m + n)+ .
We can now prove what took Russell and Whitehead over 750 pages.
Theorem (An occasionally useful proposition). 1 + 1 = 2
Proof.
1+1
= 1 + 0+
(A2)
= (1 + 0)+
(A1) +
= 1
=
2.
Multiplication is also defined as a binary operation. Like with addition we first define multiplication by a (fixed) natural number m. Like with addition we do this by recursion on ω. Applying
the recursion theorem to A = ω, F = Am and e = m ensures that a function Mn : ω 7→ ω
exists with the properties that Mm (0) = 0 and Mm (n+ ) = Am (Mm (n)) = m + Mm (n). [Note:
the book defines it as Mm (n) + m, the difference is not important as addition is commutative].
9
Like with addition, now that we have Mm for any m ∈ ω we define the binary operation of
multiplication.
Definition. The binary operation of multiplication · : ω × ω 7→ ω is given by
m · n := ·(hm, ni) = Mm (n).
The recursive definition gives two applicable properties:
(M1) m · 0 = 0.
(M2) m · n+ = Mm (n+ ) = m + Mm (n) = m + m · n.
Example. m · 1 = m.
m·1
=
(M2)
=
(M1)
=
(A1)
=
m · 0+
m+m·0
m+0
m.
+
Exponentiation can be defined in terms of multiplication recursively by m0 = 1 and mn =
mn · m.
The usual commutative, associative and distributive laws follow by induction from properties
(A1), (A2), (M 1) and (M 2). Some properties are left as homework. It is important to note
that properties of recursively defined functions are almost always proved by induction.
5
Ordering on ω
The final aim for this chapter is to define a liner ordering on ω, that will coincide with the
usual ordering <. It is easy to check that m < n ⇐⇒ m ∈ n. So in a way it is enough to
take “< = ∈”. The problem with this attempt is that ∈ does not define a relation on ω and
any ordering must be a relation. To fix this problem we modify ∈ to produce a relation. We
want to relate m to n if and only if m ∈ n.
Definition. Let ∈ω ⊆ ω × ω be the relation on ω defined by
∈ω = {hm, ni ∈ ω × ω : m ∈ n} .
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Implicitly we are using the fact that n ∈ ω is the set with elements precisely all the natural
numbers “smaller” than it. In other words
x ∈ n ⇐⇒ x ∈ ω ∧ x ∈ω n.
We must show that ∈ω is a linear ordering on ω. That is, ∈ω is a transitive relation that satisfies
the trichotomy law on ω.
Transitivity of ∈ω . ∈ω is a transitive relation. In other words for m, n, p ∈ ω
m ∈ω n ∧ n ∈ω p =⇒ m ∈ω p.
Proof. The transitivity of ∈ω follows from the fact that any natural number is a transitive set.
We have to show that m ∈ n ∧ n ∈ p =⇒ m ∈ p. This follows from the transitivity of p
(Theorem 4F): m ∈ n ∈ p =⇒ m ∈ p.
Trichotomy is harder and requires a couple of inductive results.
Lemma 4L. Let m, n ∈ ω. Then
m ∈ n ⇐⇒ m+ ∈ n+ .
Furthermore, no natural number is an element of itself
The second conclusion is a special instance of the Axiom of Regularity.
Proof. We establish the first claim in two steps.
⇐= : Suppose m+ ∈ n+ = n ∪ {n}. Then either m+ = n or m+ ∈ n. As m ∈ m+ we get
that either m ∈ n in which we are done; or m ∈ m+ ∈ n, in which case we are done by the
transitivity of n.
=⇒ : This implication is proved by induction. Let
T = {n ∈ ω : ∀m ∈ n (m+ ∈ n+ )}.
We must show that T is an inductive set. 0 ∈ T as 0 has no elements and the condition is true.
Next we suppose that n ∈ T and deduce that n+ ∈ T. That is, we must sow that for all m ∈ n+ ,
m+ ∈ n++ . There are two possibilities for m : either m = n or m ∈ n. In the former case we
are done as m+ = n+ ∈ n++ . In the latter case we are also done by the inductive hypothesis as
n ∈ T and so m+ ∈ n+ ⊆ n++ .
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We are left with proving that for all n ∈ ω, n ∈
/ n. This is again proved by induction. We let
T = {n ∈ ω : n ∈
/ n}
and show that T is inductive. 0 ∈ T as ∅ ∈
/ ∅. For the inductive step we assume that n ∈ T
and prove that n+ ∈ T . This follows from the above as n ∈
/ n and so n+ ∈
/ n+ .
We can now prove the trichotomy law for ω.
Trichotomy law for ω. Let m and n be natural numbers. Exactly one of the following three
conditions holds
m ∈ n, n = m, n ∈ m.
Proof. At most of of the above may hold. m = n cannot hold simultaneously with any of
the other two properties by the special case of the Regularity Axiom in Lemma 4L above.
m ∈ n ∈ m cannot hold either as the transitivity of m gives m ∈ m.
To prove that at least one of these conditions holds we do induction (on n). We set
T = {n ∈ ω : ∀m ∈ ω (m ∈ n ∨ n = m ∨ n ∈ m )}
and show that T is transitive. To prove 0 ∈ T we prove by induction on m that as either m = 0
or 0 ∈ m. The statement is true for 0 and if it true for m, then it is true for m+ as 0 ∈ m ⊆ m+ .
We are left with proving n ∈ T =⇒ n+ ∈ T . As n ∈ T we have that any m ∈ ω satisfies
either (m ∈ n ∨ m = n) or n ∈ m. In the former case we are done as m ∈ n+ . In the latter
case by Lemma 4L we have n+ ∈ m+ . That is n+ = m or n+ ∈ m.
The usual inequality rules may now be proved (we will not do this). We first state them in the
customary notation and then in the language of sets.
Let m, n and p be natural numbers. Then
1. m < n =⇒ m + p < n + p. That is m ∈ω n =⇒ m + p ∈ω n + p or equivalently
m ∈ n =⇒ m + p ∈ n + p.
2. For non-zero p, m < n =⇒ m · p < n · p. That is m ∈ω n =⇒ m · p ∈ω n · p or equivalently
m ∈ n =⇒ m · p ∈ n · p.
The usual cancelation laws follow from the above statements and trichotomy.
1. m + p = n + p =⇒ m = n.
2. For non-zero p, m · p = n · p =⇒ m = n.
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6
Well ordering of ω
The final property we prove for ω is that every non-empty subset has a least element (with
respect to ∈ω ). The jargon used is that ∈ω is a ‘well-ordering’ on ω.
Well ordering of ω. Let A be a non-empty subset of ω. There exists a unique m ∈ A such
that for all n ∈ A either m = n or m ∈ n.
Proof. Let A be a subset that has no minimal element. We prove that A = ∅.
Let T be the following set
T = {n ∈ ω : p ∈ω n =⇒ p ∈
/ A}.
We will prove that T is inductive and hence must be the whole of ω. We will then be done as
for all p ∈ ω we have p ∈ω p+ and so p ∈
/ A (as p+ ∈ T ).
- 0 ∈ T : as no p ∈ ω satisfies p ∈ω 0.
- Next we assume n ∈ T and prove n+ ∈ T : given p ∈ω n+ we must prove that p ∈
/ A. We
either have p ∈ n or p = n. In the former case we are done by the induction hypothesis. In
the latter case we have n ∈
/ A, as otherwise n would be a least element in A (the induction
hypothesis ensures that all p ∈ω n do not lie in A).
Thus T is an inductive set and we are done.
An example of how strong this principle is the following (which is hard to prove with no reference
to the well ordering principle).
Corollary 4Q. There is no function G : ω 7→ ω such that F (n+ ) ∈ F (n) for all n ∈ ω.
Proof. Suppose that such F exists. Then ran(F ) is a non-empty subset of ω with no minimal
element.
An elegant way to complete proofs by induction by assuming that the statement we are trying
to prove is false for some n, consider the least element of the set of numbers on which the
statement is false and show that this set cannot have a least element.
In fact we have the following stronger principle of induction.
Strong Induction Principle. Let A be a subset of ω. Suppose that for all n ∈ ω we have
that if m ∈ A for all m less than n, then n must also lie in A. Then A = ω.
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In other words
(m ∈ω n =⇒ m ∈ A) =⇒ n ∈ A ,
gives A = ω.
Proof. We argue by contradiction. Suppose that A is a set that satisfies the condition yet is
not the whole of ω. Then ω \ A 6= ∅. So ω \ A has a least element m. By the minimality of m,
all numbers smaller than m cannot lie in ω \ A and hence must lie in A. The hypothesis on A
ensures that m ∈ A as well. This contradicts m ∈ ω \ A and we are done.
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