Download Some important sets: ∅ or {}: the empty set Z: the set of integers R

Sets
Some important sets:
∅ or {}: the empty set
Z: the set of integers
R: the set of real numbers
Q: the set of rational numbers
C: the set of complex numbers
N: the set of natural numbers, meaning either {1, 2, 3, . . . } or
{0, 1, 2, 3, . . . }. Ambiguous notation.
Z≥1 , Z≥0 , Z≤5 , Q≥0 , Q>3 , etc.
(2, 7), [2, 7], [2, 7), (2, 7], (2, ∞), (−∞, 3], etc.
Sets
To define a set: see the book for the full description and lots of
examples.
Two examples: let S = {1, 3, 5, 7, 9}. Define a set E by
E= {
n∈Z
:
n is even
}.
Read this as “ the set of n ∈ Z such that n is even ” or “the
set of all integers n such that n is even” or just “the set of all even
integers.”
Sets
If A is a set, the notation x ∈ A means that x is an element of A.
For example, 3 ∈ Z and π 6∈ Z.
If A and B are sets, then A ⊆ B means that A is a subset of B.
This means that every element of A is an element of B. If E is the
set of even integers, then E ⊆ Z and E ⊆ R while Z 6⊆ E . Be
careful: E 6∈ Z: E is a subset of Z, not an element of Z. Note also
that for any set A, ∅ ⊆ A: the empty set is a subset of every set.
To prove that A ⊆ B, prove the implication “x ∈ A ⇒ x ∈ B”.
To prove that A = B, prove both A ⊆ B and A ⊇ B; that is, prove
“x ∈ A ⇒ x ∈ B and x ∈ B ⇒ x ∈ A.”
Sets
Let A and B be sets. Basic definitions:
A ∪ B: the union of A and B
A ∩ B: the intersection of A and B
A − B: the difference of A and B (sometimes written A \ B
instead)
Example
If A = {1, 2, 3, 4} and B = {1, 3, 5}, then
A ∪ B = {1, 2, 3, 4, 5},
A ∩ B = {1, 3},
A − B = {2, 4},
B − A = {5}.
B
3
4
1
2
5
A
Sets
Ac : the complement of A. This is only defined if we have
specified a “universal set” U, and then Ac = U − A.
Example
If Q is the universal set, then
c
1
1
,3 = Q −
,3
2
2
and
(Q>4 )c = Q≤4
P(A): the power set of A, the set of all subsets of A
Example
n
o
P({1, 2, 5}) = ∅, {1}, {2}, {5}, {1, 2}, {1, 5}, {2, 5}, {1, 2, 5} .
Working within {1, 2, 5}, we have {1}c = {2, 5} and {1, 5}c = {2}.
Sets
Proposition
Let A, B and C be sets.
1
(A ∪ B)c = Ac ∩ B c .
2
A ∪ (B ∩ C ) = (A ∪ B) ∩ (A ∪ C ).
3
A ∩ (B ∪ C ) = (A ∩ B) ∪ (A ∩ C ) (done in the book).
4
A ⊆ B if and only if A ∪ B = B.
Proof.
For 1, 2, and 3, we need to prove that two sets are equal, so we
have to prove two inclusions: (left side) ⊆ (right side) and
(left side) ⊇ (right side).
Number 4 will also have two parts, since it is an “if and only if”
statement: A ⊆ B ⇒ (A ∪ B ⊆ B and A ∪ B ⊇ B), and
(A ∪ B = B) ⇒ A ⊆ B.
Sets
Proof that (A ∪ B)c = Ac ∩ B c .
We need to show that (A ∪ B)c ⊆ Ac ∩ B c and that
(A ∪ B)c ⊇ Ac ∩ B c . To show that (A ∪ B)c ⊆ Ac ∩ B c , suppose
that x ∈ (A ∪ B)c . Then x 6∈ A ∪ B, so x is not in A or B;
equivalently, x is not in A and x is not in B. Thus x 6∈ A and
x 6∈ B: x ∈ Ac and x ∈ B c . So x ∈ Ac ∩ B c . This proves that
(A ∪ B)c ⊆ Ac ∩ B c .
To show that (A ∪ B)c ⊇ Ac ∩ B c , suppose that x ∈ Ac ∩ B c .
Then x ∈ Ac and x ∈ B c , so x is not in A and x is not in B. Thus
x is not in A ∪ B, so x ∈ (A ∪ B)c . This proves that
(A ∪ B)c ⊇ Ac ∩ B c , and thus the proof.
Sets
Proof that A ∪ (B ∩ C ) = (A ∪ B) ∩ (A ∪ C ).
We need to show that A ∪ (B ∩ C ) ⊆ (A ∪ B) ∩ (A ∪ C ) and that
A ∪ (B ∩ C ) ⊇ (A ∪ B) ∩ (A ∪ C ). To show that
A ∪ (B ∩ C ) ⊆ (A ∪ B) ∩ (A ∪ C ), suppose that x ∈ A ∪ (B ∩ C ).
Then x ∈ A or x ∈ B ∩ C , so there are two cases. If x ∈ A, then
x ∈ A ∪ B and x ∈ A ∪ C , so x ∈ (A ∪ B) ∩ (A ∪ C ). If x ∈ B ∩ C ,
then x ∈ B and x ∈ C , so x ∈ A ∪ B and x ∈ A ∪ C , so just as in
the first case, x ∈ (A ∪ B) ∩ (A ∪ C ). This proves that
A ∪ (B ∩ C ) ⊆ (A ∪ B) ∩ (A ∪ C ).
To show that A ∪ (B ∩ C ) ⊇ (A ∪ B) ∩ (A ∪ C ), suppose that
x ∈ (A ∪ B) ∩ (A ∪ C ). Then x ∈ A ∪ B and x ∈ A ∪ C . Therefore
either x ∈ A or x ∈ B, and x ∈ A or x ∈ C . If x ∈ A, then
x ∈ A ∪ (B ∩ C ). If x 6∈ A, then x ∈ B and x ∈ C , so x ∈ B ∩ C ,
so, again, x ∈ A ∪ (B ∩ C ). This proves that
A ∪ (B ∩ C ) ⊇ (A ∪ B) ∩ (A ∪ C ), which finishes the proof.
Sets
Proof that A ∩ (B ∪ C ) = (A ∩ B) ∪ (A ∩ C ).
(Done in book.)
We need to show that A ∩ (B ∪ C ) ⊆ (A ∩ B) ∪ (A ∩ C ) and that
A ∩ (B ∪ C ) ⊇ (A ∩ B) ∪ (A ∩ C ). To show that
A ∩ (B ∪ C ) ⊆ (A ∩ B) ∪ (A ∩ C ), suppose that x ∈ A ∩ (B ∪ C ).
Then x ∈ A and x ∈ B ∪ C . If x ∈ B, then x ∈ A ∩ B. If x ∈ C ,
then x ∈ A ∩ C . In either case, x ∈ (A ∩ B) ∪ (A ∩ C ). This proves
that A ∩ (B ∪ C ) ⊆ (A ∩ B) ∪ (A ∩ C ).
To show that A ∩ (B ∪ C ) ⊆ (A ∩ B) ∪ (A ∩ C ), suppose that
x ∈ (A ∩ B) ∪ (A ∩ C ). Then x ∈ A ∩ B or x ∈ A ∩ C . Therefore
either x ∈ A and x ∈ B, or x ∈ A and x ∈ C . In either case,
x ∈ A. Either x ∈ B or x ∈ C , so x ∈ B ∪ C . Thus
x ∈ A ∩ (B ∪ C ). This proves that
A ∩ (B ∪ C ) ⊇ (A ∩ B) ∪ (A ∩ C ), which finishes the proof.
Sets
Proof that A ⊆ B if and only if A ∪ B = B.
Suppose that A ⊆ B. We want to show that A ∪ B = B. If
x ∈ A ∪ B, then x ∈ A or x ∈ B. If x ∈ A, then x ∈ B since
A ⊆ B. So in either case, x ∈ B. Thus A ∪ B ⊆ B. It is always
true that A ∪ B ⊇ B. Combining these gives A ∪ B = B.
Now suppose that A ∪ B = B. We want to show that A ⊆ B, so
suppose that x ∈ A. Then x ∈ A ⊆ A ∪ B = B, so x ∈ B. Thus
every element of A is in B, so A ⊆ B.
This completes the proof.