Boolean algebras
edit
Within the subject of algebra, there is a structure called algebra . In order to meet our needs, we need to strongly modify this concept to obtain Boolean algebras.
Definition 1.1 (Boolean algebras) :
A Boolean algebra is a set
A
{\displaystyle A}
together with two binary operations
∨
{\displaystyle \vee }
and
∧
{\displaystyle \wedge }
, an unary operation
¬
{\displaystyle \neg }
and
0
,
1
∈
A
{\displaystyle 0,1\in A}
such that the following axioms hold for all
a
,
b
,
c
∈
A
{\displaystyle a,b,c\in A}
:
Associativity of
∨
{\displaystyle \vee }
and
∧
{\displaystyle \wedge }
:
a
◻
(
b
◻
c
)
=
(
a
◻
b
)
◻
c
{\displaystyle a\Box (b\Box c)=(a\Box b)\Box c}
,
◻
∈
{
∨
,
∧
}
{\displaystyle \Box \in \{\vee ,\wedge \}}
Commutativity of
∨
{\displaystyle \vee }
and
∧
{\displaystyle \wedge }
:
a
◻
b
=
b
◻
a
{\displaystyle a\Box b=b\Box a}
,
◻
∈
{
∨
,
∧
}
{\displaystyle \Box \in \{\vee ,\wedge \}}
Absorbtion laws:
a
◻
(
a
▽
b
)
=
a
{\displaystyle a\Box (a\triangledown b)=a}
,
{
◻
,
▽
}
=
{
∨
,
∧
}
{\displaystyle \{\Box ,\triangledown \}=\{\vee ,\wedge \}}
Distributivity laws:
a
◻
(
b
▽
c
)
=
(
a
◻
b
)
▽
(
a
◻
c
)
{\displaystyle a\Box (b\triangledown c)=(a\Box b)\triangledown (a\Box c)}
,
{
◻
,
▽
}
=
{
∨
,
∧
}
{\displaystyle \{\Box ,\triangledown \}=\{\vee ,\wedge \}}
Neutral elements:
a
∨
0
=
a
{\displaystyle a\vee 0=a}
,
a
∧
1
=
a
{\displaystyle a\wedge 1=a}
Complementation laws:
a
∨
¬
a
=
1
{\displaystyle a\vee \neg a=1}
,
a
∧
¬
a
=
0
{\displaystyle a\wedge \neg a=0}
Fundamental example 1.2 (logic) :
If we take
A
=
{
⊥
,
⊤
}
{\displaystyle A=\{\bot ,\top \}}
and
∨
,
∧
,
¬
{\displaystyle \vee ,\wedge ,\neg }
to be the usual operations from logic, we obtain a Boolean algebra.
Fundamental example and theorem 1.3 :
Let
S
{\displaystyle S}
be an arbitrary set, and let
A
⊂
2
S
{\displaystyle {\mathcal {A}}\subset 2^{S}}
such that
S
,
∅
∈
A
{\displaystyle S,\emptyset \in {\mathcal {A}}}
A
,
B
∈
S
⇒
A
∪
B
∈
S
,
A
∩
B
∈
S
{\displaystyle A,B\in S\Rightarrow A\cup B\in S,A\cap B\in S}
A
∈
S
⇒
A
¯
∈
S
{\displaystyle A\in S\Rightarrow {\overline {A}}\in S}
, where
A
¯
{\displaystyle {\overline {A}}}
denotes the complement of
A
{\displaystyle A}
.
We set
1
:=
S
{\displaystyle 1:=S}
,
0
:=
∅
{\displaystyle 0:=\emptyset }
,
∨
:=
∪
{\displaystyle \vee :=\cup }
,
∧
:=
∩
{\displaystyle \wedge :=\cap }
, and
¬
A
:=
A
¯
{\displaystyle \neg A:={\overline {A}}}
for all
A
⊆
S
{\displaystyle A\subseteq S}
.
Then
(
A
,
0
,
1
,
∨
,
∧
,
¬
)
{\displaystyle ({\mathcal {A}},0,1,\vee ,\wedge ,\neg )}
is a Boolean algebra, called an algebra of subsets of
S
{\displaystyle S}
.
Proof : Closedness under the operations follows from 1. - 3. We have to verify 1. - 6. from definition 1.1.
1.
A
∪
(
B
∪
C
)
=
{
x
∈
S
:
x
∈
A
∨
(
x
∈
B
∪
C
)
}
=
{
x
∈
S
:
x
∈
A
∨
(
x
∈
B
∨
x
∈
C
)
}
=
{
x
∈
S
:
(
x
∈
A
∨
x
∈
B
)
∨
x
∈
C
)
}
=
{
x
∈
S
:
(
x
∈
A
∪
B
)
∨
x
∈
C
}
=
(
A
∪
B
)
∪
C
{\displaystyle {\begin{aligned}A\cup (B\cup C)&=\{x\in S:x\in A\vee (x\in B\cup C)\}\\&=\{x\in S:x\in A\vee (x\in B\vee x\in C)\}\\&=\{x\in S:(x\in A\vee x\in B)\vee x\in C)\}\\&=\{x\in S:(x\in A\cup B)\vee x\in C\}\\&=(A\cup B)\cup C\end{aligned}}}
A
∩
(
B
∩
C
)
=
{
x
∈
S
:
x
∈
A
∧
(
x
∈
B
∩
C
)
}
=
{
x
∈
S
:
x
∈
A
∧
(
x
∈
B
∧
x
∈
C
)
}
=
{
x
∈
S
:
(
x
∈
A
∧
x
∈
B
)
∧
x
∈
C
)
}
=
{
x
∈
S
:
(
x
∈
A
∩
B
)
∧
x
∈
C
}
=
(
A
∩
B
)
∩
C
{\displaystyle {\begin{aligned}A\cap (B\cap C)&=\{x\in S:x\in A\wedge (x\in B\cap C)\}\\&=\{x\in S:x\in A\wedge (x\in B\wedge x\in C)\}\\&=\{x\in S:(x\in A\wedge x\in B)\wedge x\in C)\}\\&=\{x\in S:(x\in A\cap B)\wedge x\in C\}\\&=(A\cap B)\cap C\end{aligned}}}
2.
A
∪
B
=
{
x
∈
S
:
x
∈
A
∨
x
∈
B
}
=
{
x
∈
S
:
x
∈
B
∨
x
∈
A
}
=
B
∪
A
{\displaystyle A\cup B=\{x\in S:x\in A\vee x\in B\}=\{x\in S:x\in B\vee x\in A\}=B\cup A}
A
∩
B
=
{
x
∈
S
:
x
∈
A
∧
x
∈
B
}
=
{
x
∈
S
:
x
∈
B
∧
x
∈
A
}
=
B
∩
A
{\displaystyle A\cap B=\{x\in S:x\in A\wedge x\in B\}=\{x\in S:x\in B\wedge x\in A\}=B\cap A}
3.
A
∪
(
A
∩
B
)
=
{
x
∈
S
:
x
∈
A
∨
x
∈
(
A
∩
B
)
}
=
{
x
∈
S
:
x
∈
A
∨
(
x
∈
A
∧
x
∈
B
)
}
=
{
x
∈
S
:
(
x
∈
A
∨
x
∈
A
)
∧
(
x
∈
A
∨
x
∈
B
)
}
=
{
x
∈
S
:
x
∈
A
∧
(
x
∈
A
∨
x
∈
B
)
}
=
{
x
∈
S
:
x
∈
A
}
=
A
{\displaystyle {\begin{aligned}A\cup (A\cap B)&=\{x\in S:x\in A\vee x\in (A\cap B)\}\\&=\{x\in S:x\in A\vee (x\in A\wedge x\in B)\}\\&=\{x\in S:(x\in A\vee x\in A)\wedge (x\in A\vee x\in B)\}\\&=\{x\in S:x\in A\wedge (x\in A\vee x\in B)\}\\&=\{x\in S:x\in A\}=A\end{aligned}}}
A
∩
(
A
∪
B
)
=
{
x
∈
S
:
x
∈
A
∧
x
∈
(
A
∪
B
)
}
=
{
x
∈
S
:
x
∈
A
∧
(
x
∈
A
∨
x
∈
B
)
}
=
{
x
∈
S
:
(
x
∈
A
∨
x
∈
A
)
∧
(
x
∈
A
∨
x
∈
B
)
}
=
{
x
∈
S
:
x
∈
A
∨
(
x
∈
A
∧
x
∈
B
)
}
=
{
x
∈
S
:
x
∈
A
}
=
A
{\displaystyle {\begin{aligned}A\cap (A\cup B)&=\{x\in S:x\in A\wedge x\in (A\cup B)\}\\&=\{x\in S:x\in A\wedge (x\in A\vee x\in B)\}\\&=\{x\in S:(x\in A\vee x\in A)\wedge (x\in A\vee x\in B)\}\\&=\{x\in S:x\in A\vee (x\in A\wedge x\in B)\}\\&=\{x\in S:x\in A\}=A\end{aligned}}}
4.
A
∪
(
B
∩
C
)
=
{
x
∈
S
|
x
∈
A
∨
x
∈
(
B
∩
C
)
}
=
{
x
∈
S
|
x
∈
A
∨
(
x
∈
B
∧
x
∈
C
)
}
=
{
x
∈
S
|
(
x
∈
A
∨
x
∈
B
)
∧
(
x
∈
A
∨
x
∈
C
)
}
=
{
x
∈
S
|
(
x
∈
A
∩
B
)
∧
(
x
∈
A
∩
C
)
}
=
(
A
∩
B
)
∪
(
A
∩
C
)
{\displaystyle {\begin{aligned}A\cup (B\cap C)&=\{x\in S|x\in A\vee x\in (B\cap C)\}\\&=\{x\in S|x\in A\vee (x\in B\wedge x\in C)\}\\&=\{x\in S|(x\in A\vee x\in B)\wedge (x\in A\vee x\in C)\}\\&=\{x\in S|(x\in A\cap B)\wedge (x\in A\cap C)\}\\&=(A\cap B)\cup (A\cap C)\end{aligned}}}
A
∩
(
B
∪
C
)
=
{
x
∈
S
|
x
∈
A
∧
x
∈
(
B
∪
C
)
}
=
{
x
∈
S
|
x
∈
A
∧
(
x
∈
B
∨
x
∈
C
)
}
=
{
x
∈
S
|
(
x
∈
A
∧
x
∈
B
)
∨
(
x
∈
A
∧
x
∈
C
)
}
=
{
x
∈
S
|
(
x
∈
A
∪
B
)
∨
(
x
∈
A
∪
C
)
}
=
(
A
∪
B
)
∩
(
A
∪
C
)
{\displaystyle {\begin{aligned}A\cap (B\cup C)&=\{x\in S|x\in A\wedge x\in (B\cup C)\}\\&=\{x\in S|x\in A\wedge (x\in B\vee x\in C)\}\\&=\{x\in S|(x\in A\wedge x\in B)\vee (x\in A\wedge x\in C)\}\\&=\{x\in S|(x\in A\cup B)\vee (x\in A\cup C)\}\\&=(A\cup B)\cap (A\cup C)\end{aligned}}}
5.
A
∩
S
=
{
x
∈
S
|
x
∈
A
∧
x
∈
S
}
=
{
x
∈
S
|
x
∈
A
∧
⊤
}
=
{
x
∈
S
|
x
∈
A
}
=
A
{\displaystyle A\cap S=\{x\in S|x\in A\wedge x\in S\}=\{x\in S|x\in A\wedge \top \}=\{x\in S|x\in A\}=A}
A
∪
∅
=
{
x
∈
S
|
x
∈
A
∨
x
∈
∅
}
=
{
x
∈
S
|
x
∈
A
∨
⊥
}
=
{
x
∈
S
|
x
∈
A
}
=
A
{\displaystyle A\cup \emptyset =\{x\in S|x\in A\vee x\in \emptyset \}=\{x\in S|x\in A\vee \bot \}=\{x\in S|x\in A\}=A}
6.
A
∪
A
¯
=
{
x
∈
S
|
x
∈
A
∨
(
x
∈
A
¯
)
}
=
{
x
∈
S
|
x
∈
A
∨
(
x
∉
A
)
}
=
{
x
∈
S
|
⊤
}
=
S
{\displaystyle {\begin{aligned}A\cup {\overline {A}}&=\{x\in S|x\in A\vee (x\in {\overline {A}})\}\\&=\{x\in S|x\in A\vee (x\notin A)\}\\&=\{x\in S|\top \}\\&=S\end{aligned}}}
A
∩
A
¯
=
{
x
∈
S
|
x
∈
A
∧
(
x
∈
A
¯
)
}
=
{
x
∈
S
|
x
∈
A
∧
(
x
∉
A
)
}
=
{
x
∈
S
|
⊥
}
=
S
{\displaystyle {\begin{aligned}A\cap {\overline {A}}&=\{x\in S|x\in A\wedge (x\in {\overline {A}})\}\\&=\{x\in S|x\in A\wedge (x\notin A)\}\\&=\{x\in S|\bot \}\\&=S\end{aligned}}}
We thus see that the laws of a Boolean algebra are "elevated" from the Boolean algebra of logic to the Boolean algebra of sets.
Exercises
edit
Exercise 1.1.1 : Let
A
{\displaystyle A}
be a Boolean algebra and
a
∈
A
{\displaystyle a\in A}
. Prove that
a
∧
a
=
a
{\displaystyle a\wedge a=a}
and
a
∨
a
=
a
{\displaystyle a\vee a=a}
.
Inclusion
edit
Infinite numbers of subsets
edit
Notation
edit
During the remainder of the book, we shall adhere to the following notation conventions (due to Felix Hausdorff ).
If the sets
A
1
,
…
,
A
n
{\displaystyle A_{1},\ldots ,A_{n}}
are pairwise disjoint, we shall write
∑
j
=
1
n
A
j
{\displaystyle \sum _{j=1}^{n}A_{j}}
for
⋃
j
=
1
n
A
j
{\displaystyle \bigcup _{j=1}^{n}A_{j}}
; with this notation we already indicate that the
A
j
{\displaystyle A_{j}}
are pairwise disjoint. That is, if we encounter an expression such as
∑
j
=
1
n
A
j
{\displaystyle \sum _{j=1}^{n}A_{j}}
and the
A
j
{\displaystyle A_{j}}
are sets, the
A
j
{\displaystyle A_{j}}
are assumed to be pairwise disjoint.
If
A
,
B
{\displaystyle A,B}
are sets and
A
⊆
B
{\displaystyle A\subseteq B}
, we replace
B
∖
A
{\displaystyle B\setminus A}
by
B
−
A
{\displaystyle B-A}
. This means: In any occasion where you find the notation
B
−
A
{\displaystyle B-A}
within this book, it means
B
∖
A
{\displaystyle B\setminus A}
and
A
⊆
B
{\displaystyle A\subseteq B}
(note that in this way a set obtains a unique "additive inverse").