Notes on the Foundations
of Mathematics and Analysis
Eduardo Dueñez
Lucio Tavernini
March 13, 2014
Contents
0 Background
0.1
Introduction . . . . . . . . . . . . .
0.2
Proof and Paradox . . . . . . . . .
0.3
Essential Linguistic Concepts . . .
0.4
A Dog’s World . . . . . . . . . . . .
0.5
Peano’s Postulates, Induction . . .
0.6
Formal Grammars . . . . . . . . . .
0.7
Warning About the Notation Used
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
1
1
3
6
7
11
13
16
1 The Propositional Calculus
1.1
Introduction . . . . . . . . . . . . . . . . . .
1.2
The Language of the Propositional Calculus
1.3
Substitution . . . . . . . . . . . . . . . . . .
1.4
Interpretations . . . . . . . . . . . . . . . . .
1.5
Tautologies and Contradictions . . . . . . .
1.6
Examples . . . . . . . . . . . . . . . . . . .
1.7
Some Useful Tautologies . . . . . . . . . . .
1.8
Proofs . . . . . . . . . . . . . . . . . . . . .
1.9
Appendix A: More Examples . . . . . . . . .
1.10 Appendix B: The Greek Alphabet . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
1-1
1-1
1-2
1-5
1-6
1-8
1-11
1-16
1-18
1-21
1-24
.
.
.
.
2-1
2-1
2-2
2-4
2-8
.
.
.
.
3-1
3-1
3-3
3-8
3-9
2 The Predicate Calculus
2.1
Introduction . . . . . . . . . . . . .
2.2
Quantification of Predicates . . . .
2.3
Formulas of the Predicate Calculus
2.4
Interpretations Revisited . . . . . .
3 Sets
3.1
3.2
3.3
3.4
.
.
.
.
.
.
.
.
.
.
.
Introduction . . . . . . . . . . . . . .
Axioms of Extension and Separation
Intersections and Differences . . . . .
Unions . . . . . . . . . . . . . . . . .
ii
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
iii
CONTENTS
3.5
3.6
Powers, Products and Ordered Pairs . . . . . . . . . . . . . . . . . 3-12
Appendix A: The Axiom Schema of Separation . . . . . . . . . . . 3-16
4 The Natural Numbers
4.1
Introduction . . . . . . . . .
4.2
The Set of Natural Numbers
4.3
The Peano Postulates . . . .
4.4
Trichotomy . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
4-1
4-1
4-2
4-3
4-8
5 Functions and Relations
5.1
Functions, Forward and Inverse Images . . . .
5.2
Compositions and Identities . . . . . . . . . .
5.3
Injections, Surjections, Bijections and Inverses
5.4
Properties of Functions . . . . . . . . . . . . .
5.5
Indexed Sets . . . . . . . . . . . . . . . . . . .
5.6
Equinumerous Sets . . . . . . . . . . . . . . .
5.7
Relations . . . . . . . . . . . . . . . . . . . . .
5.8
The Recursion Theorem . . . . . . . . . . . .
5.9
Cantor and Schröder-Bernstein Theorems . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
5-1
5-1
5-5
5-7
5-11
5-15
5-17
5-21
5-26
5-30
6 Numbers and Arithmetic
6.1
The Arithmetic of the Natural
6.2
Order . . . . . . . . . . . . . .
6.3
The Integers . . . . . . . . . .
6.4
The Rationals . . . . . . . . .
6.5
Algebraic Structures . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
6-1
6-1
6-12
6-20
6-24
6-29
.
.
.
.
7-1
7-1
7-2
7-6
7-8
.
.
.
.
.
.
.
8-1
8-1
8-2
8-6
8-12
8-14
8-21
8-27
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Numbers
. . . . . .
. . . . . .
. . . . . .
. . . . . .
7 The Axiom of Choice
7.1
Introduction . . . . . . . . . . . . .
7.2
Products and the Axiom of Choice
7.3
One-Sided Inverses . . . . . . . . .
7.4
Countable and Uncountable Sets . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
8 The Real Numbers
8.1
Introduction . . . . . . . . . . . . . . . . . .
8.2
The Reals . . . . . . . . . . . . . . . . . . .
8.3
Addition . . . . . . . . . . . . . . . . . . . .
8.4
Multiplication . . . . . . . . . . . . . . . . .
8.5
The Real Field, Density . . . . . . . . . . .
8.6
Monotonic Functions. No-Gaps Condition. .
8.7
Powers, Roots, Exponentials and Logarithms
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
iv
CONTENTS
8.8
Sequences and Numerals . . . . . . . . . . . . . . . . . . . . . . . 8-37
9 Important Subsets of the Reals
9.1
Preliminaries . . . . . . . . . . . . .
9.2
Countable and Uncountable Sets . .
9.3
Open and Closed Sets . . . . . . . .
9.4
Properties of Open and Closed Sets
9.5
Compactness . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
9-1
9-1
9-3
9-14
9-20
9-26
10 Sequences and Series
10.1 The Extended Reals . . . . . . . . . . .
10.2 The Topology of the Extended Reals .
10.3 Sequences and their Limits . . . . . . .
10.4 Subsequences and Subsequential Limits
10.5 Cauchy Sequences . . . . . . . . . . . .
10.6 Some Convergence Results . . . . . . .
10.7 Contractive Sequences . . . . . . . . .
10.8 The Binomial Theorem . . . . . . . . .
10.9 Monotonicity . . . . . . . . . . . . . .
10.10 Series . . . . . . . . . . . . . . . . . . .
10.11 Euler’s Number e . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
10-1
10-1
10-5
10-9
10-15
10-19
10-21
10-24
10-26
10-33
10-42
10-58
.
.
.
.
.
11 Continuity
11-1
11.1 Limits of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 11-1
11.2 Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . . . 11-10
12 Answers to the Exercises
12.1 Note 1 . . . . . . . . .
12.2 Note 2 . . . . . . . . .
12.3 Note 3 . . . . . . . . .
12.4 Note 4 . . . . . . . . .
12.5 Note 5 . . . . . . . . .
12.6 Note 6 . . . . . . . . .
12.7 Note 8 . . . . . . . . .
12.8 Note 9 . . . . . . . . .
12.9 Note 10 . . . . . . . .
12.10 Note 11 . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
12-1
12-1
12-10
12-12
12-24
12-28
12-41
12-58
12-69
12-76
12-87
Foundations Note 4
The Natural Numbers
Eduardo Dueñez and Lucio Tavernini
March 13, 2014
Contents
4.1
4.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-1
4.2
The Set of Natural Numbers . . . . . . . . . . . . . . . . . . 4-2
4.3
The Peano Postulates . . . . . . . . . . . . . . . . . . . . . . 4-3
4.4
Trichotomy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-8
Introduction
Assuming, for the time being, that a set exists and, recalling that
• if a set exists so does the empty set ∅ and
• if a is set so are {a} and a ∪ {a},
we can define the first four natural numbers as:
0
1 = 0 ∪ {0}
2 = 1 ∪ {1} =
{0} ∪ {{0}}
= {0, {0}}
3 = 2 ∪ {2} = {0, 1} ∪ {{0, 1}} = {0, 1, {0, 1}}
Expressed just in terms of the empty set, the above are:
0 = ∅,
1 = {∅},
2 = {∅, {∅}},
3 = {∅, {∅}, {∅, {∅}}}.
4-1
=
=
=
=
∅,
{0},
{0, 1},
{0, 1, 2}.
Note 4: The Natural Numbers – March 13, 2014
4-2
The numbers just defined are all distinct from each other. Other consequences
of the above definitions are 0 ∈ 1 ∈ 2 ∈ 3 and 0 ⊂ 1 ⊂ 2 ⊂ 3.
We have defined just the first four natural numbers. To define them all and
package them in a set that can replace our intuitive notion N we need a new axiom.
Below, we show how to accomplish this.
4.2
The Set of Natural Numbers
Let us abstract the properties we require to define the natural numbers as a set.
4.2.1 Definition. Inductive Sets: A set A is called inductive if
(4.1)
(∅ ∈ A) ∧ (∀x)(x ∈ A → x ∪ {x} ∈ A).
Define the successor of any set x to be the set x+ = x ∪ {x}. Then, an inductive
set is a set that contains the empty set and the successor of every of its members.
4.2.2 Remark. At this point we need an axiom to ensure that an inductive set
exists.
4.2.3 Axiom of Infinity. There exists an inductive set, i.e.:
(∃A)[(∅ ∈ A) ∧ (∀x)(x ∈ A → x ∪ {x} ∈ A)].
4.2.4 Remarks. Note that the above axiom can also be written
(∃A)[{x ∈ A | x 6= x} ∈ A ∧ (∀x)(x ∈ A → x ∪ {x} ∈ A)].
The existence of ∅ = {x ∈ A | x 6= x} is part of the axiom.
The axiom of infinity asserts that an inductive set exists. This axiom does not
say how many inductive sets there are and does not say what they may look like.
We want the smallest (in an appropriate sense) inductive set. This is accomplished
via the construction shown below.
4.2.5 Definition. The Set of Natural Numbers: The set of natural numbers
ω is defined as the smallest (by inclusion) inductive set. Let φ(B) denote (4.1) with
A replaced by B. Let A denote the inductive set whose existence is guaranteed by
the axiom of infinity. Define
ω = {x ∈ A | (∀B)(φ(B) ↔ x ∈ B)}.
4.2.6 Proposition. The set ω of natural numbers is the smallest inductive set.
Note 4: The Natural Numbers – March 13, 2014
4-3
Proof. By the axiom of extension, the set ω is defined uniquely. We show that ω is
inductive. We have ∅ ∈ ω because ∅ belongs to every inductive set. If x ∈ ω, then
x must belong to every inductive set B. Therefore, x ∪ {x} also belongs to every
inductive set B, and, therefore, to ω. This shows that ω is inductive. We now show
the minimality of ω. Note that for every x, x belongs to ω if and only if x belongs
to every inductive set. Therefore, ω is a subset of every inductive set.
4.3
The Peano Postulates
In our setting the Peano postulates of Note 0 for the natural numbers ω can be
proved, i.e.: they are not taken as axioms. As a matter of fact, several of the
postulates were used in the definition of ω. The postulates are stated as (P 1)
through (P 5) below. The proof of (P 5) requires a simple technical result.
The set ω of natural numbers is inductive. That is, we have
(P 1)
0∈ω
and
(P 2)
(∀n) (n ∈ ω → n+ ∈ ω).
4.3.1 Theorem. The set ω of natural numbers has the properties
(P 3)
(∀A)[A ⊂ ω ∧ 0 ∈ A ∧ (∀n)(n ∈ A → n+ ∈ A) → A = ω],
i.e.: every inductive subset of ω must be ω; and
(P 4)
(∀n) (n ∈ ω → n+ 6= 0),
i.e.: 0 is not the successor of any natural number.
Proof. To prove (P 3), let A by any set such that
0 ∈ A ∧ (∀n)(n ∈ A → n+ ∈ A).
Then, A is an inductive set. Therefore, by the minimality of ω, we have ω ⊂ A. If
A ⊂ ω also, it follows that A = ω.
To prove (P 4), we note that n+ always contains n (since n+ = n ∪ {n}) and
that 0 is empty. Therefore, no n+ can be 0.
4.3.2 Remarks. Postulate (P 3) is called the induction principle. We show below
how to use it to construct an inductive proof, a proof where we show that if for
some set A we have (i) 0 ∈ A and (ii) n ∈ A → n+ ∈ A we can conclude that
A = ω.
Note 4: The Natural Numbers – March 13, 2014
4-4
The induction principle is sometimes formulated in a slightly generalized form
called the principle of strong induction:
The set ω of natural numbers has the property
(∀A)[A ⊂ ω ∧ (∀n)(n ∈ ω ∧ n ⊂ A → n ∈ A) → A = ω].
(P 30 )
Any member m of a natural number n is called a predecessor of n (said differently,
each natural number is the set of all its predecessors). A proof by strong induction
concludes that a set A of natural numbers is equal to ω by showing that, whenever
every predecessor of a natural number n is in A (this is the statement n ⊂ A), n
itself is in A. Note that, since 0 is a natural number having no predecessors, the
∀n statement, with n = 0, simply states that 0 ∈ A unconditionally; therefore, just
like a regular inductive proof, a proof by strong induction should start by showing
that 0 ∈ A.
What can we say about a natural number m that is a member of some successor
n+ = n ∪ {n} of a natural number n? Of course, for any pair of sets m and n,
m ∈ n ∪ {n} ⇔ m ∈ n ∨ m ∈ {n}
⇔ m ∈ n ∨ m = n.
(4.2)
Can m ∈ n and m = n both be true? If so, then we have n ∈ n. If m and n are
natural numbers, this cannot be, as we show below.∗
4.3.3 Proposition. The natural numbers have the following properties.
(I) Every element of any natural number is a natural number:
(4.3)
(∀n)(n ∈ ω → n ⊂ ω).
(II) No natural number is a subset of any of its elements; i.e.:
(4.4)
(∀n ∈ ω)(∀m ∈ ω)(m ∈ n → n 6⊂ m),
or, equivalently, by contraposition,
(∀n ∈ ω)(∀m ∈ ω)(n ⊂ m → m ∈
/ n).
(III) If m and n are natural numbers such that m ∈ n+ , then either m ∈ n or
m = n (exclusive or).
We shall leave unanswered whether it is possible for a set to be a member of itself. This will
not be true for any set we encounter in these notes.
∗
Note 4: The Natural Numbers – March 13, 2014
4-5
Proof. First we prove (I) by induction. Consider the set A of all natural numbers
having the property (4.3):
A = {n ∈ ω | n ⊂ ω}.
To prove (I), it suffices to show that A = ω. We need to prove that A fulfills the
criteria of (P 3):
(i) We show that 0 ∈ A (the base step of the inductive proof). This is obvious,
since 0 = ∅ ⊂ ω.
(ii) We show that n ∈ A ⇒ n+ ∈ A (the inductive step of the proof). So we start
by assuming that n ∈ A (the inductive hypothesis), and seek to conclude that
n+ ∈ A. Thus, we need to show that m ∈ n+ ⇒ m ∈ ω. So let m ∈ n+ .
Since n+ = n ∪ {n}, we have m ∈ n or m = n. In case m ∈ n, the inductive
hypothesis n ∈ A means simply that n ⊂ ω, which implies that m ∈ ω. In
case m = n, the inductive hypothesis n ∈ A immediately gives n ∈ ω since
A ⊂ ω.
Now we prove (II): Let A be the set of all natural numbers defined by
(4.5)
A = {n ∈ ω | (∀m) (m ∈ n → n 6⊂ m)}
= {n ∈ ω | (∀m) (n ⊂ m → m ∈
/ n)}.
We need to show that A = ω. The proof is inductive.
(i) We prove that 0 ∈ A: Since 0 has no element, m ∈ 0 is always false. Hence,
the implication m ∈ 0 ⇒ 0 6⊂ m is always true. Therefore, 0 ∈ A.
(ii) We prove that n ∈ A ⇒ n+ ∈ A: Suppose that n ∈ A.
(a) Since n ⊂ n, we have n ∈
/ n by (4.5).
(b) Since n ∈ n ∪ {n} and n ∈
/ n, it follows that n+ = n ∪ {n} is not a subset
of n.
(c) If n+ ⊂ m, for some natural number m, then n ⊂ m also and, since
n ∈ A, we have that m ∈
/ n, by (4.5).
Thus, n+ is not a subset of n and n+ is not a subset of any element of n. In
other words,
(∀m)(m ∈ n+ → n+ 6⊂ m).
Therefore n+ ∈ A.
Note 4: The Natural Numbers – March 13, 2014
4-6
We have shown that 0 ∈ A and
(∀n)(n ∈ A → n+ ∈ A).
Therefore, thanks to (P 3), A = ω.
Finally, we prove (III): If m and n are natural numbers such that m ∈ n+ ,
then, by (4.2), m ∈ n or m = n. If m ∈ n and m = n, then n ∈ n. This contradicts
(4.4) which implies
n ∈ n → n 6⊂ n.
Therefore, m ∈ n ∧ m = n is impossible. This leaves us with
m ∈ n ∪ {n} ⇔ (m ∈ n ∨ m = n) ∧ ¬(m ∈ n ∧ m = n)
⇔ ¬(m ∈ n ⇔ m = n),
i.e.: if m ∈ n+ , then either m ∈ n or m = n (exclusive or).
4.3.4 Definition. Transitive Sets: A set A is called transitive if
(∀a)(a ∈ A → a ⊂ A),
i.e.: a set is transitive if it includes everything it contains.
4.3.5 Examples.
1. The set A = {∅, {∅}} is transitive. Its elements are ∅ and {∅}. Its subsets
are ∅, {∅}, {{∅}} and {∅, {∅}}. Every element of A is also a subset of A.
2. The set A = {∅, {{∅}}} is not transitive. Its elements are ∅ and {{∅}}.
Its subsets are ∅, {∅}, {{{∅}}} and {∅, {{∅}}}. We have {{∅}} ∈ A, but
{{∅}} 6⊂ A.
4.3.6 Proposition. Every natural number is transitive, i.e.:
(∀n ∈ ω)(∀m ∈ ω)(m ∈ n → m ⊂ n).
Proof. Define A to be the set of all transitive natural numbers, i.e.:
A = {n ∈ ω | (∀m) (m ∈ n → m ⊂ n)}.
We will show that A = ω. The proof is by induction:
(i) We prove that 0 ∈ A: Since 0 has no element, m ∈ 0 is false for all m.
Therefore, the implication m ∈ 0 ⇒ m ⊂ 0 is true for all m. Hence, 0 ∈ A.
(ii) We prove that n ∈ A ⇒ n+ ∈ A: Suppose that n ∈ A, i.e.: suppose that
n is transitive. For every m ∈ n+ we have either m ∈ n or m = n (by
Proposition 4.3.3). We have two cases to consider, (a) m ∈ n and (b) m = n:
Note 4: The Natural Numbers – March 13, 2014
4-7
(a) If m ∈ n then m ⊂ n (because n is transitive), therefore m ∈ n+
(because n+ = n ∪ {n}).
(b) If m = n, then m ⊂ n+ (because n+ = n ∪ {n}.
It follows that every element of n+ is a subset of n+ ; that is, n+ ∈ A.
We have shown that 0 ∈ A and (∀n)(n ∈ A ⇒ n+ ∈ A). Therefore, thanks to (P 3),
A = ω.
4.3.7 Theorem. If two natural numbers are distinct then so are their successors,
i.e.: the set ω of natural numbers has the property
(∀n ∈ ω)(∀m ∈ ω)(n 6= m → n+ 6= m+ ),
or, by contraposition,
(P 5)
(∀n ∈ ω)(∀m ∈ ω)(n+ = m+ → n = m).
Proof. Suppose that m and n are in ω and that n+ = m+ .
Since n ∈ n+ and n+ = m+ , we have that n ∈ m+ . Hence, since m+ = m∪{m},
either n ∈ m or n = m. We have shown that
(4.6)
¬(n ∈ m ↔ n = m).
Since m ∈ m+ and m+ = n+ , we have that m ∈ n+ . Hence, since n+ = n ∪ {n},
either m ∈ n or m = n.
(4.7)
¬(m ∈ n ↔ m = n).
If m = n, we are finished. If m 6= n, since (4.6) and (4.7) are both true, we have
m ∈ n and n ∈ m. Therefore, since n and m are transitive by Proposition 4.3.6,
we have m ⊂ n and n ⊂ m. That is: m = n.
4.3.8 Exercise. Prove that every natural number other than zero is the successor
of a natural number.
4.3.9 Example. We end this section with an informal example of a proof by strong
induction. We have not yet introduced either the arithmetic operations or the order
of ω (a task we will postpone until Note 6), so we are forced to provide an informal
example that relies on our physical intuition and some (yet unproven) properties
of natural numbers.
Suppose that n is a natural number and we are in possession of a rectangular
chocolate bar of size n + 1 by 1. (We take the length of the bar to be n + 1 so that
even when n = 0 we have a 1-by-1 bar rather than no chocolate at all.) Suppose
Note 4: The Natural Numbers – March 13, 2014
4-8
this bar is meant to be cut into 1-by-1 pieces (n+1 of them). It is very easy to show,
using ordinary induction, that n cuts will be necessary if, at each successive step,
we cut off (say) the rightmost 1-by-1 square, repeating n times over. It should also
be intuitively clear that, regardless of the position of the successive cuts exactly
n cuts will be necessary. The latter statement is most easily proved by strong
induction.
Claim. It takes n cuts to break an (n + 1)-by-1 chocolate bar into 1-by-1 pieces,
regardless of the order in which the cuts are performed.
Proof. Let A be the set of natural numbers n for which the statement above is
true. By strong induction on n, we will show that A = ω.
The truth of the statement for n = 0 is obvious, since 0 + 1 = 1 and a 1-by-1
chocolate bar is already cut into 1-by-1 pieces, so zero cuts are needed.
Assume now that n is a nonzero natural number such that n ⊂ A; in other
words, every predecessor m of n is an member of A (this is the strong inductive
hypothesis). Perform a first cut on the bar in any allowable place; this splits the bar
into two smaller bars of sizes a-by-1 and b-by-1. Neither a nor b is zero; therefore
(by Exercise 4.3.8), we have a = k + = k + 1, b = m+ = m + 1 for some natural
numbers k, m. Moreover, the fact that no chocolate was lost nor added in the
process of cutting means that a + b = n + 1.
Since neither a, b are zero, it follows from a + b = n+ that both a, b are predecessors of n+ (though we haven’t proved this). Hence, k, m are predecessors of n
(this is proved in Proposition 4.4.7 below).
By the strong inductive hypothesis, both k, m are members of A; therefore,
the a-by-1 bar requires k cuts to be split into 1-by-1 pieces and the b-by-1 bar
requires m.
We conclude that k + m cuts will always be required to cut the two smaller
bars into 1-by-1 pieces. Adding the first cut that split the (n + 1)-by-1 bar into
two, we see that the altogether number of cuts needed is equal to 1 + k + m =
a + b − 1 = n; hence the statement is true for n, so n ∈ A, and the strong induction
is complete.
4.4
Trichotomy
If m and n are natural numbers, m ∈ n does not formalize the intuitive notion
that “m is less than n” unless exactly one of m ∈ n, m = n and n ∈ m is true. We
show that this is indeed the case.
4.4.1 Proposition. For any two natural numbers m and n at most one of the
statements m ∈ n, m = n and n ∈ m can be true.
Note 4: The Natural Numbers – March 13, 2014
4-9
Proof. Recall that
(4.8)
no natural number is a subset of any of its elements.
If m = n and n ∈ m then m ∈ m and m ⊂ m, contradicting (4.8). If m = n and
m ∈ n then, again, m ∈ m and m ⊂ m, contradicting (4.8).
Recall that every natural number is a transitive set (it include everything it
contains). Therefore, m ∈ n implies m ⊂ n and n ∈ m implies n ⊂ m. Hence,
m ∈ n and n ∈ m together imply m = n. Then, m = n together with m ∈ n (or
n ∈ m), once again, contradicting (4.8).
4.4.2 Definition. Two natural numbers m and n are called comparable if m ∈ n,
or m = n, or n ∈ m. For each natural number n define
Cn = {m ∈ ω | m ∈ n ∨ m = n ∨ n ∈ m}
(the set of all natural numbers comparable with n) and define
N = {n ∈ ω | Cn = ω}
(the set of all natural numbers comparable with every natural number).
4.4.3 Remarks. Note that n ∈ N is equivalent to Cn = ω. From the definition of
Cn it is clear that
(4.9)
n ∈ Cn
and
(4.10)
m ∈ n → m ∈ Cn .
If m is comparable with n then n is comparable with m and vice versa. Therefore
(4.11)
m ∈ Cn ↔ n ∈ Cm .
4.4.4 Proposition. 0 ∈ N, i.e.: C0 = ω.
Proof. The proof is inductive. 0 ∈ C0 trivially. We prove that m ∈ C0 ⇒ m+ ∈ C0
for all natural numbers m to conclude that C0 = ω. Suppose that m ∈ C0. There
are three cases to consider: m ∈ 0, m = 0 and 0 ∈ m
(a) The case m ∈ 0 is impossible for any m.
(b) If m = 0 then 0 ∈ m+ , i.e.: 0 ∈ 0+ . Thus, 0 ∈ Cm+ . Hence, m+ ∈ C0 .
4-10
Note 4: The Natural Numbers – March 13, 2014
(c) If 0 ∈ m then 0 ∈ m+ , since m+ contains all of its predecessors. Thus,
0 ∈ Cm+ . Hence, m+ ∈ C0 .
In all possible cases m+ ∈ C0 . We have shown that C0 = ω.
4.4.5 Proposition. Suppose that Cn = ω for some natural number n. Then, for
all natural numbers m
m ∈ n+ → m+ ∈ Cn+ .
Proof. If m ∈ n+ , since n+ = n ∪ {n}, we have two cases to consider: m = n and
m ∈ n.
(a) If m = n then m+ = n+ . Hence, m+ ∈ Cn+ .
(b) Consider the case m ∈ n. Since Cn = ω, we have m+ ∈ Cn . Therefore, we
have three possible cases: m+ ∈ n, m+ = n and n ∈ m+ .
(i) If m+ ∈ n then m+ ∈ n+ , since m ∈ n (hypothesis). Hence, m+ ∈ Cn+ .
(ii) If m+ = n then m+ ∈ n+ , since n ∈ n+ . Hence, m+ ∈ Cn+ .
(iii) If n ∈ m+ then either n ∈ m or n = m, since m+ = m∪{m}. Since m is a
transitive set (it includes everything it contains), n ∈ m implies m ⊂ m.
Hence, n ⊂ m in either case. Because we also have m ∈ n (hypothesis),
n is a subset of one of its elements, a contradiction. Therefore, n ∈ m+
is impossible.
In all possible cases m+ ∈ Cn+ .
4.4.6 Proposition. Trichotomy Law. Any two natural numbers are comparable,
i.e.: N = ω. In other words, given any two natural numbers m and n, exactly one
of the following holds:
m ∈ n,
m = n,
n ∈ m.
Proof. The proof is inductive. Thanks to Proposition 4.4.4, 0 ∈ N. We prove that
n ∈ N ⇒ n+ ∈ N, i.e.: that
(∀n)(Cn = ω → Cn+ = ω),
to conclude that N = ω.
Suppose that n ∈ N for some natural number n, i.e.: suppose that Cn = ω.
The proof that Cn+ = ω is inductive. We prove that 0 ∈ Cn+ . Then we prove that
m ∈ Cn+ → m+ ∈ Cn+
for all natural numbers m to conclude that Cn+ = ω.
Note 4: The Natural Numbers – March 13, 2014
4-11
(a) We prove that 0 ∈ Cn+ . Since C0 = ω we have n+ ∈ C0. Hence, 0 ∈ Cn+ .
(b) We prove that m ∈ Cn+ → m+ ∈ Cn+ . Suppose that m ∈ Cn+ . There are
three cases to consider: m ∈ n+ , m = n+ and n+ ∈ m.
(i) If m ∈ n+ then m+ ∈ Cn+ , thanks to Proposition 4.4.5.
(ii) If n+ = m then n+ ∈ m+ , since m+ = m ∪ {m}. Thus, n+ ∈ Cm+ .
Hence, m+ ∈ Cn+ .
(iii) If n+ ∈ m then n+ ∈ m+ , since m+ = m ∪ {m}. Thus, n+ ∈ Cm+ .
Hence, m+ ∈ Cn+ .
In all cases in (b) n+ ∈ Cm . Hence, m ∈ Cn+ .
We have shown that Cn+ = ω.
4.4.7 Proposition.
(4.12)
(∀m ∈ ω)(∀n ∈ ω)(m ∈ n ↔ m+ ∈ n+ ).
Proof. Recall that a natural number is a transitive set (it includes everything it
contains). In other words, for n and ` in ω we have
(4.13)
n ∈ ` ⇒ n ⊂ `.
Therefore for m, n and ` in ω we have, thanks to Exercise 4.4.9
(4.14)
m ∈ n ∧ n ∈ ` ⇒ m ∈ `.
We prove that
(4.15)
m+ ∈ n+ ⇒ m ∈ n.
Suppose that m+ ∈ n+ . Since n+ = n ∪ {n}, m+ ∈ n+ means m+ ∈ n ∪ {n} which
implies m+ ∈ n or m+ = n. Since m ∈ m+ always, we have m ∈ m+ ∧ m+ ∈ n or
m ∈ m+ ∧ m+ = n. Hence, m ∈ n in either case, thanks to (3) in the first of the
two cases.
Now, we prove that
(4.16)
m ∈ n → m+ ∈ n+ .
We proceed by induction on n. The case where n = 0 is vacuous. So, (4.16) holds
trivially for n = 0. Suppose (4.16) holds. Suppose that m ∈ n+ . Then m ∈ n or
m = n.
Note 4: The Natural Numbers – March 13, 2014
4-12
• If m ∈ n then, by the inductive hypothesis,
m+ ∈ n+ ⊂ (n+ )+ .
Hence, m+ ∈ (n+ )+ .
• If m = n then, by the inductive hypothesis,
m+ = n+ ∈ (n+ )+ .
Hence, m+ ∈ (n+ )+ .
In either case we obtain
m ∈ n+ ⇒ m+ ∈ (n+ )+ ,
which is (4.16) with n replaced by n+ . We have established (4.16).
Now, (4.15) and (4.16) establish (4.12).
4.4.8 Remarks. The last two propositions establish two crucial properties of the
natural numers. In Note 6 we formalize the intuitive notion m < n by the property
m ∈ n. Then, trichotomy states that for any two natural numbers m and n exactly
one of the following holds: m < n, m = n, n < m.
The sum m + n of two natural number shall be defined by repeated applications
of the successor operation. We begin by defining m + 0 = m. Then, for 0 < n, the
sum m + n will be defined by formalizing the intuitive idea
m + n = (. . . ((m + 1) + 1) . . .) + 1 .
|
{z
}
n times
4.4.9 Exercise. Prove that if A is a transitive set, then
(4.17)
(∀a)(∀x)(x ∈ a ∧ a ∈ A → x ∈ A).
4.4.10 Exercise. Prove that if A is a set such that (4.17) holds, then A is transitive.
4.4.11 Exercise. Prove that if A is a transitive set, then ∪A ⊂ A.
4.4.12 Exercise. Prove that if a is a transitive set, then ∪(a+ ) = a.
4.4.13 Exercise. Prove that A is a set such that ∪A ⊂ A, then A is transitive.
4.4.14 Exercise. Prove that if a is a transitive set, then a+ is a transitive set.
4.4.15 Exercise. Prove that ω is transitive.
4.4.16 Exercise. Prove that if A is a transitive set, then P(A) is a transitive set.
4.4.17 Exercise. Prove that if A is a transitive set, then ∪A is a transitive set.
Note 4: The Natural Numbers – March 13, 2014
4-13
4.4.18 Exercise. Prove that if every member of a set A is a transitive set, then
∩A is a transitive set.
4.4.19 Exercise. Prove that if every member of a set A is a transitive set, then
∪A is a transitive set.
4.4.20 Exercise. Prove Postulate (P 30 ) (the principle of strong induction).
Suggestion: Show that if A satisfies the premises of (P 30 ), then the set
B = {n ∈ ω | n ⊂ A}
is inductive. Conclude that A = ω.