Download 1. Sets and Set Builder Notation

CHAPTER 3 NOTES: SET THEORY
1. Sets and Set Builder Notation
The notion of set can be made very precise in mathematics(see below). We will deal mostly
naive set theory ).
with set theory in an intuitive manner (often called
This is the set
theory that almost all mathematicians use anyway.
On the one hand, you can specify a set simply by listing all of its elements:
A=f
A
p
is a set. The numbers
1
2 A, 0 2 A,
2
1; 0; 1; and
2 A and so forth.
Denition
for every
p
1; 0; 1;
Example:
A and B , we write A = B
g
2
2 are called the
If A and B are sets, we say
x, x 2 A implies that x 2 B .
For two sets
p
that
A
is a subset of
if and only if
AB
Answer the following questions about the sets
A
B
C
A B?
Is B A?
Is A C ?
Is C A?
Is A = C ?
=
=
=
elements
f1; 2; 3; 4; 5; 6; 7g
f2; 4; 6g
f3; 2; 7; 5; 1; 4; 6g
and
of the set. We write
B,
written
A
B,
if
B A.
A; B; C :
(1) Is
(2)
(3)
(4)
(5)
There is a set which contains no elements.
denoted as
It is called the empty set and is typically
;.
There is no way to dene a set of all sets. However, it is often the case that we only want
to talk about the set of integers or the set real numbers. In this event, all the sets that we
care about in a particular instance are subsets of some
generally denoted as
U
universal set.
The universal set is
U and a predicate P (x) quantied over U , we may form a set S
in set-builder notation as follows. We say that and element s 2 U lies in S whenever
P (s) is true. This is typically written as follows:
fx 2 U : P (x)g
Given a universal set
1
Example:
It is impossible to list all even integers.
However, they can be described
fx 2 Z : x is even g.
Example: Note that f1; 1g = f1g. In general, sets do not contain multiple occurrences of
the same element. How many elements does the following set have:
p
f 1; 0; 2; 1; 5; 5; ; 2g
in set-builder notation as
Example:
Write the following sets in set builder notation.
(1) The set of odd integers.
(2) The set of primes.
(3) The set of multiples of 5.
(4) The set of integers whose remainder when divided by 6 is 4.
Example:
Intervals of real numbers are dened using set builder notation.
be real numbers,
a < b.
(a; b)
=
(a; b]
=
[a; b)
=
[a; b]
=
1)
[a; 1)
( 1; b)
( 1; b]
( 1; 1)
(a;
Example:
(1)
(2)
(3)
(4)
p;
=
=
=
=
List some elements of [1 10].
integers in the interval [
Example:
p
=
2; ]? Is
2
;
List some elements of (0 1).
Write the following sets as intervals of real numbers.
Recall that:
jxj =
(
a, b
fx 2 R : a < x < bg
f x 2 R : a < x bg
fx 2 R : a x < bg
f x 2 R : a x bg
f x 2 R : a < xg
fx 2 R : a xg
fx 2 R : x < bg
f x 2 R : a < x bg
fx 2 R : 1 = 1 g
2 [1; 2]?
x 2 R such that jxj < 3
x 2 R such that jx 1j 4
x 2 R such that jx 2j > 2
x 2 R such that jx + 3j < 5
Let
x;
x;
2
if
if
x0
x<0
Are there any
Also recall that
ja
bj represents the distance between a; b on R.
2. Intersection, Union and Complement
Sets have many operations on them. The operations help us to dene new sets. First we
dene all of the operations.
Denition:
The set
The set
If
A and B
are subsets of some universal set
A [ B = fx 2 U : x 2 A or x 2 B g
A[B
is called the
A\B
is called the
union
of
A
0
intersection
is called the
A
Example:
(1)
(2)
(3)
(4)
(5)
(6)
B
Let
0
=
complement
of
fx 2 U : x 2= Ag
of
A in U .
is called the set dierence of
A and B
A and B .
B = fx 2 U : x 2 A and x 2= B g
A
The set
A and B .
A \ B = fx 2 U : x 2 A and x 2 B g
A
The set
U:
A and B
or the complement of
A
B
=
(3)
(4)
(5)
(6)
A.
=
f1; 2; 3; 4; 5; 6; 7; 8; 9; 10g
f0; 1; 2; 4; 6; 8; 10; 12; 14g
0
0
Let
A
and
B
be given as below.
Find each of the following.
draw a number line in order to answer your question.
(2)
in
be given as below. Find each of the following:
A[B
A\B
A
B
A B
B A
Example:
(1)
B
A[B
A\B
A
B
A B
B A
A
B
=
=
fx 2 R : jx 1j < 1g
fx 2 R : jxj > 1g
0
0
3
Make sure to
3. Venn Diagrams
These are useful tools for visualizing sets.
Suppose you have several sets
A; B; C .
You
draw a region in the plane corresponding to each set. The intersection of the two sets is
the overlapping region. The union of two sets is the region inside
both
regions that you
drew. The complement of a set is the stu on the outside of your set.
Example:
Draw a Venn Diagram representing the following sets:
U
A
B
C
=
=
=
=
f1; 2; 3; 4; 5; 6; 7; 8; 9; 10g
f2; 4; 5; 6; 8g
f1; 3; 5; 6; 7g
f1; 2; 3; 5; 6; 9g
Use the Venn Diagram to compute each of the following:
(1)
(2)
(3)
(4)
A \ B \ C.
A \ (B [ C ).
A B.
(A [ B )
0
4. Cartesian Product
Let
A and B
be two sets. For
a 2 A and b 2 B , the ordered pair (a; b) is dene as:
(a; b) =
ffag; fa; bgg
Note that there is a dierence between the ordered pair (
a; b), and the interval (a; b).
Theorem 1. (a; b) = (c; d) if and only if a = c and b = d
Proof.
If
a = c and b = d, then
(a; b) =
ffag; fa; bgg = ffcg; fc; dgg = (c; d)
For the reverse direction, note rst that
example, if 1
and one from
a
and
b
are considered as in dierent sets. For
2 A and 1 2 B , the set f1A; 1B g contains two dierent elements: one from A
B.
Suppose then that:
ffag; fa; bgg = ffcg; fc; dgg
Then we must have that fag = fcg, since these sets both have one element. This means
that a = c. Also, we must have fa; bg = fc; dg, since these are the only sets having two
elements. Since
a = c, we are left with b = d.
This denition of ordered pair is due to the Polish mathematician Kuratowski.
Denition:
The Cartesian product of
A
and
pairs.
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B,
denoted
AB
is the set of ordered
Example:
A,B , and C .
f1; 2; 3; 4g
f7; 8; 9g
f1; 3g
Answer the following questions about
A B?
What is B A?
Is A B = B A?
What is (A A)
(C C )?
A
B
C
=
=
=
(1) What is
(2)
(3)
(4)
Example:
Answer the following questions about
the Cartesian plane
RR
A
B
A and B .
fx 2 R : jx
fx 2 R : jx
=
=
A B?
What is B A?
Is A B = B A?
What is (A A)
(B B )?
Draw a picture of the sets in
j 2g
3j < 1g
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(1) What is
(2)
(3)
(4)
Example:
Consider the subset
A R and B R?
C; D
of
R given below.
R
Why or why not?
C
D
=
=
Is
D
=
AB
for some
f(x; y) 2 R R : x2 + y2 1g
f(x; y) 2 R R : jxj + jyj 1g
The rst set is a circle and the second set is a diamond.
5. Finite Union, Intersections, and Products
We can also dene union of more than one set. Let
n
[
i=1
n
\
i=1
A1 ; : : : ; An
Ai = fx 2 U
:(
9i 2 f1; 2; : : : ; ng)(x 2 Ai)g
Ai = fx 2 U
:(
8i 2 f1; 2; : : : ; ng)(x 2 Ai)g
Example:
Let
Ai = [i; i + 1] for i 2 N, i 100.
Example:
Let
Ai = (0; 1=i) for i 2 N, i 10.
Compute
Compute
S100
T10
i=1 Ai .
i=1 Ai .
You can take cross products of more than two sets as well.
we can dene
be a nite list of sets.
n-tuples (a1 ; : : : ; an )
from these sets is denoted by:
where
ai
2 Ai for all i.
A1 A2 A3 : : : An
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Given sets
The set of all
A1 ; A2 ; : : : ; An ,
n-tuples taken
Example:
Let
Example:
Let
A = [0; 1].
An
have?
=
Describe the set
f1; : : : ; ng.
A A A as a subset of R R R.
How many elements does the set
A1 A2 : : : An
It is also possible to dene innite (or even arbitrary) unions, intersections, and products.
This can lead to some strange things.
A1 ; A2 ; : : :
Let
be an innite list of sets
indexed by the natural numbers which are subsets of some universal set
[
1
i=1
\
1
i=1
Example:
Ai = fx 2 U
:(
9i 2 N)(x 2 Ai)g
Ai = fx 2 U
:(
8i 2 N)(x 2 Ai)g
U.
Compute the following:
\
1
i=1
[
\ 1
; [ n; n] ;
0;
n n i=1
n
i=1
1
1 1
;
1
6. Power Sets and Cardinality
Let
A be a set.
Example:
Example:
Let
We dene the
power set
A = f0; 1; 2g.
Compute
The
cardinality
of a set
S
of
A, denoted P(A), as the set of all subsets of A.
P(A).
having nitely many elements is the number of
elements in the set. The cardinality of a nite
A = f0; 1; 2g.
Example:
Let
Example:
P(;) = f;g.
Hence,
Example: A = f1; 2; 3g.
Example:
Compute
S
is commonly denoted as
jS j or n(S ).
jP(A)j.
jP(A)j = 1.
Compute
jA A A Aj.
;
The interval [0 1] is not nite as it contains all of the numbers:
1 1 1 1
; ; ; ;:::
2 3 4 5
Hence, its cardinality is not dened using our terminology.
jP(A)j = 2 A
Proof. The proof is by induction on n = jAj.
Theorem 2. If A is a nite set, then
j
6
j
7. Proofs involving Sets And Set Identities
As sets are dened using logical connectives, it is not surprising that there are parallels
between set theory and logic. First, here are some general guidelines for proving things
about sets.
SUBSETS: To show
A B , take an x 2 A and show that x 2 B .
SET EQUALITY: To show
A = B , show A B and also show B A
x 2 A [ B , you have a proof by cases. CASE I: x 2 A.
x2B
INTERSECTION: If x 2 A \ B , you have two true statements: x 2 A
and x 2 B
COMPLEMENTS IN SET BUILDER: If x 2 A , then the negation of
x 2 A is true. Hence :P (x) is true.
UNION: If
CASE II:
0
Now with these principles in mind, we can set about proving theorems.
We will only
look at a few. For each of the below, rst give a Venn diagram argument which illustrates
the theorem.
Theorem 3. For any set A, ; A and A A.
Proof. For the rst, we must show that the predicate If x 2 ;, then x 2 A
is true. The
x 2 ; is false for all x in the universe, so the proposition For all x 2 U , If x 2 ;,
then x 2 A, is true. Hence ; A.
predicate
The predicate If
x 2 A, then x 2 A
x in the universe.
is true for all
Hence
A A.
Theorem 4. For any subsets A and B of a universal set U , A \ B A.
Proof. We must show that if x 2 A \ B then x 2 A. If x 2 A \ B , then x 2 A and x 2 B .
Thus,
x 2 A.
A \ B A.
This implies that
Theorem 5. For any subsets A; B of a universal set U :
[ B)
(A \ B )
(A
0
=
0
=
A
A
Proof. We will prove the second statement only.
and A [ B (A \ B ) .
0
0
0
0
0
0
We must show that (
0
\ B) A [ B
x 2= A or x 2= B .
(A
\B
[B
0
0
0
Let
x
2
(A
\ B) .
0
Then
x 2= A \ B .
Case 1: If
x 2= A, then x 2 A .
Since
x2A, x2A
Case 2: If
x 2= B , then x 2 B .
Since
x2B,x2A
0
0
0
0
0
7
0
A [B
0
0
Then it must be true that
[B.
or x 2 B . Hence x 2 A [ B .
or
0
A \ B)
x2B.
0
0
Hence
x2A
0
0
0
0
A
0
[ B (A \ B )
0
0
x2A
0
Suppose
[ B . Then x 2 A
0
0
x2B.
0
or
Case 1: If
x 2 A , then x 2= A.
Since
x 2= A, x 2= A \ B .
Hence,
x 2 (A \ B ) .
Case 2: If
x 2 B , then x 2= B .
Since
x 2= B , x 2= A \ B .
Hence,
x 2 (A \ B ) .
Hence, (
0
0
A \ B)
0
=
A
0
0
0
[B.
0
Theorem 6. For any subsets A; B; C of a universal set U , we have:
A \ (B [ C )
A [ (B \ C )
=
=
\ B ) [ (A \ C )
( A [ B ) \ (A [ C )
(A
Proof. We prove only the rst statement. The second will be an exercise. we must show
that A \ (B [ C ) (A \ B ) [ (A \ C ) and (A \ B ) [ (A \ C ) A \ (B [ C ).
A \ (B [ C ) (A \ B ) [ (A \ C ): Let x 2 A \ (B [ C ). Then x 2 A and x 2 B [ C .
x 2 B [ C . Case I: If x 2 B , then x 2 A and x 2 B . Hence x 2 A \ B [ A \ C . Case
II. If x 2 C , then x 2 A and x 2 C . Hence x 2 A \ B [ A \ C .
Then
\ B ) [ (A \ C ) A \ (B [ C ): Let x 2 (A \ B ) [ (A \ C ). Then x 2 A \ B or x 2 A \ C .
x 2 A \ B , then x 2 A and x 2 B . Then x 2 B or x 2 C . Hence x 2 A \ (B [ C ).
Case II: If x 2 A \ C , then x 2 A and x 2 C . Then x 2 B or x 2 C . Hence x 2 A \ (B [ C ).
(A
Case I: If
It follows that
A \ (B [ C ) = ( A \ B ) [ (A \ C )
Theorem 7. For any sets A; B; C ,
A (B [ C ) = ( A B ) [ (A C )
Of course, once you have enough of these properties, you can prove things simply by
manipulating the known expressions.
A
(B
\ C)
=
=
=
=
A \ (B \ C )
A \ (B [ C )
(A \ B ) [ (A \ C )
(A
B ) [ (A C )
0
0
0
0
0
Theorem 8. For all subsets A; B; C of some universal set U ,
A
(B
\ C ) = (A
B ) [ (A
C)
8. Axiomatic Set Theory
Set builder notation can lead to paradoxes. For example, there is the famous paradox of
Bertrand Russell:
Here we are thinking of
x as any set.
S = fx : x 2= xg
This leads to the paradox:
naive set theory is not good enough for all purposes.
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S 2 S and S 2= S .
Clearly,
However, it is possible to dene set theories which contain no paradoxes.
called
axiomatic set theories.
These are
For example, the Zermelo-Frankel system with the Axiom
of Choice is a suciently powerful set theory to give all of the theorems of calculus so that
there are no paradoxes. However, it is also true that there are innitely many statements
which cannot be proved from the ZFC axioms but which are not inconsistent with the
axioms either! This is the content of the famous Gödel-Cohen Incompleteness Theorem,
a triumph of twentieth century mathematics.
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