Chapter 2. Sets
Foundation of Mathematics
Bo-Chih Huang
Foundation of Mathematics
Chapter 2. Sets
Chapter 2. Sets
§2.1 Introduction
§2.2 Operations on sets
§2.3 Indexed families
§2.4 An axiomatic approach to sets
Bo-Chih Huang
Foundation of Mathematics
Chapter 2. Sets
§2.1 Introduction
Definition
A set is a collection of distinct objects. If A is a set and x is an object that
belongs to A, we say that x is an element of A or that x is a member of A, or
that x belongs to A, and write x P A. If an object y is not belongs to A, we
will denote as y R A. To describe a set, we can either list all elements belongs
to the set or use the notation
tx P px qu,
where the open sentence P px q describes the property that defines the set.
Examples:
t1, 2, 3u.
N t1, 2, 3, 4, u.
Z t , 2, 1, 0, 1, 2, u.
3Z t3nn P Zu.
tnn P N, n is odd, and n 14u.
Bo-Chih Huang
Foundation of Mathematics
Chapter 2. Sets
§2.1 Introduction
Remark
The object in a set could be anything, even the set itself, for example, the
set of all abstract ideas certainly is an abstract idea. We also can easily
construct
a set that does not contain itself. For example, if
A tx x is a chairu, then A R A, because A is not a chair.
However, there is a logical difficulty arises from the idea of calling any
collection of objects a set. More precisely, not arbitrary open sentence
P px q can be used to define a set.
The Russell Paradox
Let A tx x is a set and x R x u. That is, A be the set of all sets that do not
contain themselves as a member.
Question: A P A? or A R A?
If A P A, then by definition of A, A R A. If A R A, then again by definition of A,
we have A P A.
Bo-Chih Huang
Foundation of Mathematics
Chapter 2. Sets
§2.1 Introduction
Remark
The resolution of Russell’s paradox involving making a distinction between
sets and arbitrary collections of objects by defining set from a system of
axioms. Such theory is called Axiomatic set theory. The theory of set
that uses a natural language to describe sets and operations on sets is
called Naive set theory.
Recall that the universe is the collection of objects that can replace the
variable in the open sentence. Some ambiguity may arise unless
the
universe is known. For example, elements in the set A tx x 2 6x 0u
depends on an agreed-upon universe. The set A is t0, 6u when the
universe is R, but A is t6u when the universe is N.
Bo-Chih Huang
Foundation of Mathematics
Chapter 2. Sets
§2.1 Introduction
Definition
If A and B are sets, B is said to be a subset of A, written B „ A, provided
every element of B is an element of A, that is, p@x qpx P B Ñ x P Aq. Sets A
and B are said to be equal, written A B, provided A „ B and B „ A. If B is
a subset of A and A B, written B ˆ A, then B said to be a proper subset of
A.
Axiom 1.
∅ tx x
x u is a set. ∅ is called empty set or Null set.
Note that since for every object x in every universe, x is equal (identical) to x,
there are no elements in the collection ∅. That is, the statement x P ∅ is false
for every object x.
Definition
The number of elements in a finite set A is called the cardinality of A and is
denoted by |A| (or #A, npAq).
Bo-Chih Huang
Foundation of Mathematics
Chapter 2. Sets
§2.1 Introduction
Proposition 2.1
Let A, B are two sets with |A| 0 |B |, that is, A and B are sets that has no
elements, then A B.
Proof:
Suppose the statement
pA B q is true. Since
pA B q ô rpA „ B q ^ pB „ Aqs ô r pA „ B qs _ r pB „ Aqs
This shows that one of the statements pA „ B q or pB „ Aq is true. If
pA „ B q is true. Then the following statements
pA „ B q ô rp@x qpx P A Ñ x P B qs ô pDx q px P A Ñ x P B q
ô pDx qpx P A ^ x R B q
is true. But the statement pDx qpx P Aq implies |A| 0, a contradiction. Hence
A „ B. According to similar argument, we can prove B „ A. Therefore,
A B.
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Bo-Chih Huang
Foundation of Mathematics
Chapter 2. Sets
§2.1 Introduction
Proposition 2.2
Let A be any set. Then ∅ „ A and A „ A.
Proof:
It is clear that the statement x P A Ñ x P A is a tautology, so we have A „ A.
To show that ∅ „ A. Suppose the statement p∅ „ Aq is true. Then the
statements
rp@x qpx P ∅ Ñ x P Aqs ô pDx q px P ∅ Ñ x P Aq
ô pDx qrpx P ∅q ^ px R Aqs
is true. By definition of ∅, we have x x, a contradiction. Therefore ∅ „ A.l
Corollary 2.3
Every nonempty set has at least two subsets.
Bo-Chih Huang
Foundation of Mathematics
Chapter 2. Sets
§2.1 Introduction
Proposition 2.4
Let A, B and C be sets. If A „ B and B
„ C , then A „ C .
Proof:
If A ∅, then A „ C by Proposition 2.2. If A ∅, let x
x P B. Since B „ C , so x P C . Therefore, A „ C .
P A. Since A „ B, so
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Remark
It is easy to confuse two notions: “P” (an element of) and “„” (a subset of).
Consider the set t17u which is the set whose only element is 17. Clearly
17 P t17u, but 17 is NOT a subset of t17u. It only has two subsets,
namely ∅ and t17u itself.
From Proposition 2.4, for any sets A, B and C , if A „ B and B „ C , then
A „ C . However, if A t3u, B tt3u, 5u, and C tB, 17u, we have
A P B and B P C , but A R C .
Bo-Chih Huang
Foundation of Mathematics
Chapter 2. Sets
§2.1 Introduction
Proposition 2.5
For any sets A and B, if A „ B and A ∅, then B
∅.
Proof:
If A „ B and A ∅. Since A is nonempty, there is an element x
A „ B, and x P A, so x P B. Therefore B ∅.
P A. Since
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Definition
If A is a set, then the set of all subsets of A is called the power set of A and is
denoted by P pAq or 2A .
Examples:
Let A t2, 3, 4u, Then
P pAq t∅, t2u, t3u, t4u, t2, 3u, t3, 4u, t2, 4u, Au.
Let A tt1, 2, 3u, t4, 5u, 6u. Then
P pAq t∅, tt1, 2, 3uu, tt4, 5uu, t6u, tt1, 2, 3u, t4, 5uu, tt1, 2, 3u, 6u, tt4, 5u, 6u, Au
Bo-Chih Huang
Foundation of Mathematics
Chapter 2. Sets
§2.1 Introduction
Remark
For every set A, we have ∅ „ A, ∅ P P pAq, t∅u „ P pAq, and
∅ „ P pAq.
For the previous case A tt1, 2, 3u, t4, 5u, 6u, we have
tt4, 5uu P P pAq and ttt4, 5uuu „ P pAq (7 t4, 5u P A).
t4, 5u R P pAq (7 t4, 5u † A)
t∅u † A, t∅u R P pAq, and tt∅uu † P pAq (7 ∅ R A)
Proposition 2.6
If A is a set and |A| n, then |P pAq| 2n .
Proof:
If n 0, then A ∅ by Proposition 2.1. then P p∅q t∅u and
|P pAq| 20 1. Suppose |A| n, n ¥ 1. Write A tx1 , x2 , , xn u. If
B „ A, then for each xi , either xi P B or xi R B. Thus there are exactly
2 2 2 2n distinct ways to construct a subset B „ A. Therefore,
looomooon
|P pAq| 2n .
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n factors
Bo-Chih Huang
Foundation of Mathematics
Chapter 2. Sets
§2.1 Introduction
Proposition 2.7
Let A, B are sets. Then A „ B if and only if P pAq „ P pB q.
Proof:
“Ñ”: Suppose X P P pAq, then X „ A. Since A „ B, then X „ B by
Proposition 2.4. Hence X P P pB q and therefore P pAq „ P pB q.
“Д: By Proposition 2.2, we have that A „ A, so A P P pAq. Since
P pAq „ P pB q, so A P P pB q. Therefore A „ B.
Bo-Chih Huang
Foundation of Mathematics
l
Chapter 2. Sets
§2.2 Operations on sets
Definition
We have seen in the previous section that the set of all sets in NOT a set.
Nonetheless, in naive set theory, we still use the word universal set to denote
the collection that contains all the sets we are considering, and assume that all
sets under consideration are subsets of the universal set U.
The above assumption allows us to draw diagrams
that enable us to picture the behavior of sets.
This diagram is called Venn diagrams which is a
diagram that shows all possible logical relations
between a finite collection of different sets.
The diagram on the left picture the relation
A „ B „ C.
Bo-Chih Huang
Foundation of Mathematics
Chapter 2. Sets
§2.2 Operations on sets
Definition
If A and B are sets, then the union of A and B is the set of all objects that
belongs to A or to B, which is denoted by A Y B and read “A union B.” More
precisely,
A Y B tx P U x P A _ x P B u.
The intersection of A and B is the set of all objects that belong to both A and
B, which is denoted by A X B and read “A intersect B.”. More precisely,
AXB
tx P U px P Aq ^ px P B qu.
Bo-Chih Huang
Foundation of Mathematics
Chapter 2. Sets
§2.2 Operations on sets
Examples:
Let U
Then
t1, 2, 3, 4, 5, 6, 7, 8, 9, 10u, A t2, 3, 4, 6u, and B t3, 4, 7, 9u.
AYB
t2, 3, 4, 6, 7, 9u,
AXB
t3, 4u.
Consider A,B „ R with A r1, 3q, B p2, 4s, that is,
A tx P R1 ¤ x 3u, and B tx P R2 x ¤ 4u. Then
AYB
r1, 4s,
AXB
p2, 3q.
Consider A, B „ Z, where A is the set of all even integers, and B is the set
of all odd integers. Then
AYB
Z,
Bo-Chih Huang
AXB
∅.
Foundation of Mathematics
Chapter 2. Sets
§2.2 Operations on sets
Definition
If A and B are sets and A X B ∅, we say that A and B are disjoint sets.
Note that for any set A (even A ∅), then A and ∅ are disjoint.
Bo-Chih Huang
Foundation of Mathematics
Chapter 2. Sets
§2.2 Operations on sets
Proposition 2.8
Let A, B are C be sets.
(a) ∅ X A ∅ and ∅ Y A A.
(b) A X B
„ A.
(d) A Y B
B Y A and A X B B X A. (Commutative Law)
X C q pA X B q X C .
(c) A „ A Y B.
(e) A Y pB Y C q pA Y B q Y C and A X pB
(Associative Law)
(f) A Y A A A X A.
(g) If A „ B, then A Y C
„ B Y C and A X C „ B X C .
Bo-Chih Huang
Foundation of Mathematics
Chapter 2. Sets
§2.2 Operations on sets
Proof of Proposition 2.8:
(f) By (b) and (c), we have that A X A „ A and A „ A Y A. It is enough to
show that A Y A „ A X A. Let x P A Y A, then x P A or x P A, so x P A.
It follows that x P A X A. Hence A Y A „ A X A.
(g) Let x P A Y C , then x P A or x P C . If x P C . Since C „ B Y C by (c), so
x P B Y C . If x P A. Since A „ B, so x P B „ B Y C by (c) again. Hence
A Y C „ B Y C . On the other hand, Let y P A X C , then y P A and
y P C . Since A „ B, so y P B. Also we have y P C , so y P B X C . Hence
A X C „ B X C.
The remain statements of Proposition 2.8 are left as an exercise.
Bo-Chih Huang
Foundation of Mathematics
l
Chapter 2. Sets
§2.2 Operations on sets
Definition
Let A and
B be sets. Then the complement of
A relative to B is the set
tx P B x R Au, which is denoted by B A and read “B minus A.”
If U is the universal set and A „ U, then U A is written A1 and is called
the complement of A.
Examples:
Consider U Z and let A be the set of all odd integers. Then A1 is the
set of all even integers.
Let U t1, 2, 3u, A t1, 3u, B t2u, and C ∅. Then
U1
C,
A1
B,
Bo-Chih Huang
B1
A,
and C 1
Foundation of Mathematics
U.
Chapter 2. Sets
§2.2 Operations on sets
Theorem 2.9
Let A, B be subsets of some universal set U. Then
(a) pA1 q1 A.
(b) pA Y B q1 A1 X B 1 .
(c) pA X B q1 A1 Y B 1 .
+
(d) A B A X B 1 .
(e) A „ B if and only if B 1
pDe Morgan’s Lawq
„ A1 .
Proof:
(e) “Ñ” Let x P B 1 , then x R B. Since A „ B, so x R A and thus x P A1 .
Hence B 1 „ A1 .
“Д Suppose B 1 „ A1 , then according to above argument, we have
pA1 q1 „ pB 1 q1 . By (a), we have that pA1 q1 A and pB 1 q1 B. so we get
A „ B.
The proof of the remaining parts of Theorem 2.9 as an exercise.
Bo-Chih Huang
Foundation of Mathematics
l
Chapter 2. Sets
§2.2 Operations on sets
Remark
The Venn diagrams often motivate or guide to prove that two sets are equal.
For example, to show Theorem 2.9(b), we consider the following Venn
diagrams:
pA Y B q1
AYB
B1
A1
Bo-Chih Huang
A1 X B 1
Foundation of Mathematics
Chapter 2. Sets
§2.2 Operations on sets
Theorem 2.10 (Distribution Law)
Let A, B and C are sets. Then
A X pB
Y C q pA X B q Y pA X C q and A Y pB X C q pA Y B q X pA Y C q.
Proof:
“„” Let x P A X pB Y C q, then x P A and x P B Y C . Thus x P B or x P C .
Suppose that x R A X B. Then x P A and x R B. It follows that x P A X C .
Thus x P A X B or x P A X C , so x P pA X B q Y pA X C q
“ ” Let x P pA X B q Y pA X C q. Then x P A X B or x P A X C and, in either
case, x P A. If x P A X B, then x P B „ B Y C , and if x P A X C ,
x P C „ B Y C . In either case, x P A and x P B Y C , and we have
x P A X p B Y C q.
The case A Y pB X C q pA Y B q X pA Y C q is left as exercise.
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Bo-Chih Huang
Foundation of Mathematics
Chapter 2. Sets
§2.2 Operations on sets
We can use Venn diagram to illustrate what we have just proved–namely, that
intersection distributes over union.
A X pB
Y Cq
AXC
Bo-Chih Huang
AXB
pA X B q Y pA X C q
Foundation of Mathematics
Chapter 2. Sets
§2.2 Operations on sets
Theorem 2.11
Let A, B are C be sets.
(a) A pB
Y C q pA B q X pA C q.
(b) P pAq X P pB q P pA X B q.
(c) P pAq Y P pB q „ P pA Y B q.
A X B 1.
1
(e) A X B pA1 Y B q1
(d) A B
Example for (c):
Let A t1u, B t2u. Then P pAq Y P pB q t∅, t1u, t2uu, and
P pA Y B q t∅, t1u, t2u, t1, 2u.
The proof is left as exercise.
Bo-Chih Huang
Foundation of Mathematics
Chapter 2. Sets
§2.2 Operations on sets
Further remark
Recall if A tt1, 2, 3u, t4, 5u, 6u. Then
P pAq ,
$
& ∅, tt1, 2, 3uu, tt4, 5uu, t6u, .
%
tt1, 2, 3u, t4, 5uu,
tt1, 2, 3u, 6u, tt4, 5u, 6u, A
-
Note that P pAq is the set of all subsets in A. However, we see that
t2u „ t1, 2, 3u and t3, 4u „ t1, 2, 3u Y t4, 5u.
Why both t2u and t3, 4u are not in P pAq?
Although t2u „ t1, 2, 3u, but t1, 2, 3u is an “element” in A, that is,
t1, 2, 3u P A, NOT a subset of A. So t2u can not be a subset of A.
Similarly, t1, 2, 3u and t4, 5u are “elements” in A, that is,
t1, 2, 3u, t4, 5u P A, NOT subsets of A. Thus, although t1, 2, 3u Y t4, 5u is
a set, but it is not a subset of A, it is even not an element in A. Hence
t3, 4u „ t1, 2, 3u Y t4, 5u can not imply t3, 4u is a subset of A.
Bo-Chih Huang
Foundation of Mathematics
Chapter 2. Sets
§2.2 Operations on sets
Proposition 2.12
Let U be the universal set and let A, B, C , D
and D „ C . Show that B 1 „ A1 .
„ U. If A X C ∅, B Y D U,
Proof:
Let x P B 1 then x R B. Since B Y D U and x R B, so we have x P D. Since
D „ C , so x P C . Since A X C ∅ and x P C , we have that x R A and hence
x P A1 . Therefore, B 1 „ A1 .
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Bo-Chih Huang
Foundation of Mathematics
Chapter 2. Sets
§2.3 Indexed families
Definition
A set of sets is called a family or a collection of sets.
Examples:
For any set A, the power set P pAq of A is the family of all subsets of A.
A tt1, 2, 3u, t3, 4, 5u, t3, 6u, t2, 3, 6, 7, 9, 10uu is a finite family
consisting of four sets.
tpx, x qx P R and x ¡ 0u is an infinite family of open intervals.
? ?
Note that the sets p1, 1q, p 2, 2q, and p5, 5q are elements of C .
C
Bo-Chih Huang
Foundation of Mathematics
Chapter 2. Sets
§2.3 Indexed families
Example:
For the family A tt1, 2, 3u, t3, 4, 5u, t3, 6u, t2, 3, 6, 7, 9, 10uu, there are four
elements (which are sets) in A , namely t1, 2, 3u, t3, 4, 5u, t3, 6u, and
t2, 3, 6, 7, 9, 10u. We may say that
the set t1, 2, 3u is the 1st element in A and denote it by A1 .
the set t3, 4, 5u is the 2nd element in A and denote it by A2 .
the set t3, 6u is the 3rd element in A and denote it by A3 .
the set t2, 3, 6, 7, 9, 10u is the 4th element in A and denote it by A4 .
Thus the family A can be represented as
A
tA1 , A2 , A3 , A4 u or A tAi i 1, 2, 3, 4u or A tAi i P t1, 2, 3, 4uu
We can say that the family A is indexed by 1,2,3,4 (or indexed by the set
is an indexed family, and the set t1, 2, 3, 4u is the
indexing set of the family A .
In this case, we see that A is a finite family with |A | 4, which can be
indexed by a finite set t1, 2, 3, 4u.
t1, 2, 3, 4u). We say A
Bo-Chih Huang
Foundation of Mathematics
Chapter 2. Sets
§2.3 Indexed families
Example for infinite families:
For each n P N, let An t2n 1, 2nu (e.g. A1 t1, 2u, A2 t3, 4u, ).
Then A tAn n P Nu is an indexed family with indexing set N.
For each n P Z, let An be the closed interval rn 1, ns, that is,
An tx P Rn 1 ¤ x ¤ nu. Then A tAn n P Zu is an indexed family
with indexing set Z.
For the family C tpx, x qx P R and x ¡ 0u. We see that for each
positive real number x, we find an element (which is an open interval),
namely px, xq in C . Hence C is an indexed family with indexing set
R : tx P Rx ¡ 0u.
Note that the above families have been indexed in the way that Ar
and only if r s.
As if
P N, let
An tp P Np is a prime number that divides nu.
Then A tAn n P Nu is an indexed family with indexing set N. Note that
A12 A24 t2, 3u and that if k P N and k ¡ 1, there are infinitely many
n P N such that Ak An . Finally, we note that A1 ∅.
For each n
Bo-Chih Huang
Foundation of Mathematics
Chapter 2. Sets
§2.3 Indexed families
Definition
Let A
tAα α P Λu be an indexed family of sets with indexing set Λ. Then
The union of A or the union over A , denoted by
”
Aα ), is the set
αPΛ
tx x P Aα for some Aα P A u tx x
Therefore,
x
P
¤
A if and only if
”
A (or
”
Aα or
Aα PA
P Aα for some α P Λu.
pDAα P A qpx P Aα q
The
“ intersection
“ of A or the intersection over A , denoted by
Aα or
Aα ), is the set
P
Aα A
P
tx x P Aα for all Aα P A u tx x
“
α Λ
Therefore,
x
P
£
A if and only if
Bo-Chih Huang
P Aα for all α P Λu.
p@Aα P A qpx P Aα q
Foundation of Mathematics
A (or
Chapter 2. Sets
§2.3 Indexed families
Examples:
Let A tA1 , A2 , A3 u where A1
A3 t12, 13, 14, 15, 16u. Then
¤
£
Let A
A
A
t12, 13, 14u, A2 t13, 14, 15u and
A1 Y A2 Y A3 t12, 13, 14, 15, 16u.
A1 X A2 X A3 t13, 14u.
tAn n P Nu, where An n, n1
¤
£
A
¤
P
£
P
. Then
An
! x n
x
for some n
P N p8, 1s.
An
! x n
¤ n1
x
¤ n1
for each n
P N p1, 0s.
n N
A
n N
Bo-Chih Huang
)
)
Foundation of Mathematics
Chapter 2. Sets
§2.3 Indexed families
Definition
Let A be a family of sets. We say that A is pairwise disjoint provided that if
A, B P A such that A B, then A X B ∅.
“
Clearly, if A is a pairwise disjoint family, then A ∅. However, the
converse is NOT true.
“
Consider A tAn n P Nu where An pn, 8q. Then suppose x P
An , then
P
x P An pn, 8q for all n P N. But this“
implies that x ¡ n for all n P N. Such
real number x does not exists. Hence
An ∅. However, choose m, n P N
n N
with m n. It is clear that An
assume that m ¡ n, then
A m X An
P
Am . Without loss of generality, we may
n N
pm, 8q X pn, 8q pm, 8q Am ∅.
Therefore, A is not pairwise disjoint.
Bo-Chih Huang
Foundation of Mathematics
Chapter 2. Sets
§2.3 Indexed families
Proposition 2.13
tAα α P Λu be an indexed family of sets and let β P Λ. Then
”
Aβ „
Aα .
“ αPΛ
Aα „ Aβ .
Let A
(a)
(b)
P
α Λ
Proof:
P Aβ . Since β P Λ, then Aβ P A and thus x P ” A ” Aα .
αPΛ
”
Therefore Aβ „
Aα .
αPΛ
“
Let x P
Aα . Then x P Aα , @α P Λ. In particular, x P Aβ since β P Λ.
αPΛ“
Therefore
Aα „ Aβ .
l
(a) Let x
(b)
P
α Λ
Bo-Chih Huang
Foundation of Mathematics
Chapter 2. Sets
§2.3 Indexed families
Proposition 2.14
Let A tAα α P Λu be an indexed family of sets and let B be a set. Then the
following statements
”
”hold:
,
(a) B X
Aα αPΛ pB X Aα q. .
αPΛ
“
“
pDistribution Lawq
(b) B Y
Aα αPΛ pB Y Aα q. -
“ αPΛ 1 ”
Aα A1α .
αPΛ
αPΛ
”
1 “ 1
Aα .
(d)
Aα αPΛ
αPΛ
(c)
,
/
/
.
/
/
-
pDe Morgan’s Lawq
Proof of (d):
x
P
¤
P
α Λ
Aα
1
Øx R
¤
P
Aα
Ø
x
α Λ
P
¤
P
Aα is true
α Λ
Ø rpDα P Λqpx P Aα qs is true Ø p@α P Λq px P Aα q is true
£
Ø p@α P Λqpx R Aα q Ø p@α P Λqpx P A1α q Ø x P A1α l
P
α Λ
Bo-Chih Huang
Foundation of Mathematics
Chapter 2. Sets
§2.4 An axiomatic approach to sets
We have known that not all collections of sets are set. Strictly speaking, we
have to restrict our notion of set by demanding that sets obey certain axioms.
Axioms
1. The empty set is a set as is every member of a set.
2. If X is a set and, for each x in X , P px q is a proposition, then
tx P X P px qu is a set.
3. If X is a set and Y is a set, so is tX , Y u.
4. ”
If X is a set, tz z
X.
P x for some x P X u is a set. This set is denoted by
5. If X is a nonempty
set, tz z
“
denoted by X .
6. If X is a set, tz z
power set of X .
P x for each x P X u is a set. This set is
„ X u is a set. This set, denoted by P pX q, is called the
7. The set N of all natural numbers is a set.
8. No set is a member of itself.
9. If X is a set and Y is a set, so is X
Y.
10. The Axiom of Choice: If X is a set and ∅ R X , there is a function f with
domain X such that f px q P x for each x P X .
Bo-Chih Huang
Foundation of Mathematics
Chapter 2. Sets
§2.4 An axiomatic approach to sets
Constructing natural numbers from Axiom 16.
By Axiom 1, ∅ is a set, then by Axiom 3, t∅, ∅u t∅u is a set
Since t∅u has one member, we give this set a name, say “1”.
Since t∅u is a set, then by Axiom 3, t∅, t∅uu is a set.
Since t∅, t∅uu has two members, we give this set a name, say “2”.
Since t∅, t∅uu is a set and tt∅, t∅uuu is a set. By Axiom 3, the
collection A tt∅, t∅uu, tt∅, t∅uuuu is a set. By Axiom 4,
¤
A
t∅, t∅u, t∅, t∅uuu is a set
Since t∅, t∅u, t∅, t∅uuu has three members, we give this set a name, say
“3”.
..
.
Bo-Chih Huang
Foundation of Mathematics
Chapter 2. Sets
§2.4 An axiomatic approach to sets
Proposition 2.15
The collection B
tt1, 2u, t∅, 2u, t1, 2, 3uu is a set.
Proof:
In previous slide we have construct sets “1”,“2”, and “3”. Using Axiom 1 and
Axiom 3 to build sets t∅, 2u, t1, 2u and t3u. Using
” Axiom 3 again to build set
A1 tt1, 2u, t3uu. Using Axiom 4 to build set A1 t1, 2, 3u.
Using Axiom 3 to build sets tt1, 2u, t∅, 2uu and tt1, 2, 3uu. Using Axiom 3
again to build set A2 ttt1, 2u, t∅, 2uu, tt1, 2, 3uuu. Finally, using Axiom 4 to
build the set
¤
A2 tt1, 2u, t∅, 2u, t1, 2, 3uu B.
l
Bo-Chih Huang
Foundation of Mathematics
Chapter 2. Sets
§2.4 An axiomatic approach to sets
Remark
In axiomatic set theory, there is no distinction between
1 and t∅u;
2 and t∅, t∅uu t∅, 1u;
3 and t∅, t∅u, t∅, t∅uuu t∅, 1, 2u; Thus in axiomatic set theory, it would say that 3 is t∅, 1, 2u, 1 t∅u P 3.
In our class, we take N as given and hence do not view 1 as either an
element or a subset of 2.
Given
sets X and Y , we can use Axiom 3 and Axiom 4 to form the set
”
tX , Y u, and denote this set by X Y Y . Note that
X YY
¤
tX , Y u taa P A for some A P tX , Y uu taa P X or a P Y u.
“
Similarly, we can use Axiom 3 and Axiom 5 to form the set tX , Y u, and
denote this set by X X Y . Note that
£
X X Y tX , Y u taa P A for all A P tX , Y uu taa P X and a P Y u.
This means the operator of sets can also be deduce from axioms.
Bo-Chih Huang
Foundation of Mathematics
Chapter 2. Sets
Suggested exercises and homework
Red ones are homework that you have to turn in.
§2.1: 5,6,710,11,12,13,1820.
§2.2: 2125,29,30,31,32,33,34,35,37,3943,44,45,46,47,49,52.
§2.3: 5358,59,6062,65,67,70,71.
§2.4: 73,75,7779.
Bo-Chih Huang
Foundation of Mathematics