Boolean subalgebra

Boolean subalgebra

Boolean Subalgebras

Let

A

be a

Boolean algebra

and

B

a non-empty subset of

A

. Consider the following conditions:

1.

if

a

∈

B

, then

a

′

∈

B

,

2.

if

a

,

b

∈

B

, then

a

∨

b

∈

B

,

3.

if

a

,

b

∈

B

, then

a

∧

b

∈

B

,

It is easy to see that, by de Morgan’s laws, conditions 1 and 2 imply conditions 1 and 3, and vice versa. Also, If

B

satisfies

1 and 2, then

1

=

a

∨

a

′

∈

B

by picking an

element

a

∈

B

. As a result,

0

=

1

′

∈

B

as well.

A non-empty subset

B

of a Boolean algebra

A

satisfying conditions 1 and 2 (or equivalently 1 and 3) is called a

Boolean subalgebra

of

A

.

Examples

.

•

Every Boolean algebra contains a unique two-element Boolean algebra consisting of just

0

and

1

.

•

If

A

contains a

non-trivial element

a

(

≠

0

), then

0

,

1

,

a

,

a

′

form a four-element Boolean subalgebra of

A

.

•

Here is a concrete example. Let

A

be the field of all sets (

http://planetmath.org/RingOfSets

) of a set

X

(the

power set

of

X

). Let

B

be the set of all finite and all cofinite subsets of

A

. Then

B

is a

subalgebra

of

A

.

•

Continuing from the example above, if

X

is equipped with a topology

𝒯

, then the set of all clopen sets in

X

is a Boolean subalgebra of the

algebra

of all

regular open sets

of

X

.

Remarks

. Let

A

be a Boolean algebra.

•

Arbitrary

intersections

of Boolean subalgebras of

A

is a Boolean subalgebra.

•

An arbitrary union of an

increasing

chain of Boolean subalgebras of

A

is a Boolean subalgebra.

•

A subalgebra

B

of

A

is said to be

dense

in

A

if it is dense as a subset of the underlying poset

A

. In other words,

B

is

dense in

A

iff for any

a

∈

A

, there is a

b

∈

B

such that

b

≤

a

. It can be easily shown that

B

is dense in

A

is

equivalent

to any one of the following:

(a)

for any

a

∈

A

, there is

b

∈

B

such that

a

≤

b

;

(b)

for any

x

,

y

∈

A

with

x

≤

y

, there is

z

∈

B

such that

x

≤

z

≤

y

.

To see the last equivalence, notice first that by picking

x

=

0

we see that

B

is dense, and

conversely

, if

x

≤

y

, then there is

r

∈

B

such that

r

≤

y

, so that

x

≤

x

∨

r

≤

y

.

Subalgebras Generated by a Set

Let

A

be a Boolean algebra and

X

a subset of

A

. The intersection of all Boolean subalgebras (of

A

) containing

X

is called the Boolean subalgebra generated by

X

, and is denoted by

⟨

X

⟩

. As indicated by the remark above,

⟨

X

⟩

is a Boolean subalgebra of

A

, the smallest subalgebra containing

X

. If

⟨

X

⟩

=

A

, then we say that the Boolean algebra

A

itself is generated by

X

.

Proposition 1

.

Every element of

⟨

X

⟩

⊆

A

has a

disjunctive normal form

(DNF). In other words, if

a

∈

⟨

X

⟩

, then

a

=

⋁

i

=

1

n

⋀

j

=

1

ϕ

⁢

(

i

)

a

i

⁢

j

=

(

a

11

∧

⋯

∧

a

1

⁢

ϕ

⁢

(

1

)

)

∨

(

a

21

∧

⋯

∧

a

2

⁢

ϕ

⁢

(

2

)

)

∨

⋯

∨

(

a

n

⁢

1

∧

⋯

∧

a

n

⁢

ϕ

⁢

(

n

)

)

,

where either

a

i

⁢

j

or

a

i

⁢

j

′

belongs to

X

.

Proof.

Let

B

be the set of all elements written in DNF using elements

X

. Clearly

B

⊆

⟨

X

⟩

. What we want to show is that

⟨

X

⟩

⊆

B

. First, notice that the join of two elements in

B

is in

B

. Second, the

complement

of an element in

B

is also in

B

. We prove this in three steps:

1.

If

a

∈

B

and either

b

or

b

′

is in

X

, then

b

∧

a

∈

B

.

If

a

=

(

a

11

∧

⋯

∧

a

1

⁢

ϕ

⁢

(

1

)

)

∨

(

a

21

∧

⋯

∧

a

2

⁢

ϕ

⁢

(

2

)

)

∨

⋯

∨

(

a

n

⁢

1

∧

⋯

∧

a

n

⁢

ϕ

⁢

(

n

)

)

, then

b

∧

a

=

b

∧

(

(

a

11

∧

⋯

∧

a

1

⁢

ϕ

⁢

(

1

)

)

∨

(

a

21

∧

⋯

∧

a

2

⁢

ϕ

⁢

(

2

)

)

∨

⋯

∨

(

a

n

⁢

1

∧

⋯

∧

a

n

⁢

ϕ

⁢

(

n

)

)

)

=

(

b

∧

a

11

∧

⋯

∧

a

1

⁢

ϕ

⁢

(

1

)

)

∨

(

b

∧

a

21

∧

⋯

∧

a

2

⁢

ϕ

⁢

(

2

)

)

∨

⋯

∨

(

b

∧

a

n

⁢

1

∧

⋯

∧

a

n

⁢

ϕ

⁢

(

n

)

)

which is in DNF using elements of

X

.

2.

If

a

,

b

∈

B

, then

a

∧

b

∈

B

.

In the last expression of the previous step, notice that each term

b

∧

a

i

⁢

1

∧

⋯

∧

a

i

⁢

ϕ

⁢

(

i

)

can be written in DNF by iteratively using the result of the previous step. Hence, the join of all these terms is again in DNF (using elements of

X

).

3.

If

a

∈

B

, then

a

′

∈

B

.

If

a

=

(

a

11

∧

⋯

∧

a

1

⁢

ϕ

⁢

(

1

)

)

∨

(

a

21

∧

⋯

∧

a

2

⁢

ϕ

⁢

(

2

)

)

∨

⋯

∨

(

a

n

⁢

1

∧

⋯

∧

a

n

⁢

ϕ

⁢

(

n

)

)

, then

a

′

=

(

a

11

′

∨

⋯

∨

a

1

⁢

ϕ

⁢

(

1

)

′

)

∧

(

a

21

′

∨

⋯

∨

a

2

⁢

ϕ

⁢

(

2

)

′

)

∧

⋯

∧

(

a

n

⁢

1

′

∨

⋯

∨

a

n

⁢

ϕ

⁢

(

n

)

′

)

,

which is the meet of

n

elements in

B

. Consequently, by step 2,

a

′

∈

B

.

Since

B

is

closed under

join and

complementation

,

B

is a Boolean subalgebra of

A

. Since

B

contains

X

,

⟨

X

⟩

⊆

B

.
∎

Similarly, one can show that

⟨

X

⟩

is the set of all elements in

A

that can be written in

conjunctive normal form

(CNF) using elements of

X

.

Corollary 1

.

Let

⟨

B

,

x

⟩

be the Boolean subalgebra (of

A

) generated by a Boolean subalgebra

B

and an element

x

∈

A

. Then every element of

⟨

B

,

x

⟩

has the form

(

b

1

∧

x

)

∨

(

b

2

∧

x

′

)

for some

b

1

,

b

2

∈

B

.

Proof.

Let

X

be a generating set for

B

(pick

X

=

B

if necessary). By the

proposition

above, every element in

⟨

B

,

x

⟩

is the join of elements of the form

a

1

∧

a

2

∧

⋯

∧

a

n

, where each

a

i

or

a

i

′

is in

X

or is

x

. By the

absorption laws

, the form is

reduced

to one of the three forms

a

,

a

∧

x

, or

a

∧

x

′

, where

a

∈

B

. Joins of elements in the form of the first kind is an element of

B

, since

B

is a subalgebra. Joins of elements in the form of the second kind is again an element in the form of the second kind (for

(

a

1

∧

x

)

∨

(

a

2

∧

x

)

=

(

a

1

∨

a

2

)

∧

x

), and similarly for the last case. Therefore, any element of

⟨

B

,

x

⟩

has the form

a

∨

(

a

1

∧

x

)

∨

(

a

2

∧

x

′

)

. Since

a

=

(

a

∧

x

)

∨

(

a

∧

x

′

)

, by setting

b

1

=

a

1

∨

a

and

b

2

=

a

2

∨

a

, we have the desired form.
∎

Remarks

.

•

If

a

=

(

b

1

∧

x

)

∨

(

b

2

∧

x

′

)

, then

a

′

=

(

b

2

′

∧

x

)

∨

(

b

1

′

∧

x

′

)

.

•

Representing

a

in terms of

(

b

1

∧

x

)

∨

(

b

2

∧

x

′

)

is not unique. Actually, we have the following fact:

(

b

1

∧

x

)

∨

(

b

2

∧

x

′

)

=

(

c

1

∧

x

)

∨

(

c

2

∧

x

′

)

iff

b

2

⁢

Δ

⁢

c

2

≤

x

≤

b

1

↔

c

1

, where

Δ

and

↔

are the

symmetric difference

and

biconditional

operators

.

Proof.

(

⇒

)

. The LHS can be rewritten as

(

b

1

∨

b

2

)

∧

(

b

2

∨

x

)

∧

(

b

1

∨

x

′

)

. As a result,

c

2

∧

x

′

≤

b

2

∨

x

so that

c

2

∨

x

=

(

c

2

∧

x

′

)

∨

x

≤

(

b

2

∨

x

)

∨

x

=

b

2

∨

x

. Therefore,

c

2

-

b

2

=

c

2

∧

b

2

′

≤

(

c

2

∨

x

)

∧

b

2

′

≤

(

b

2

∨

x

)

∧

b

2

′

=

b

2

′

∧

x

≤

x

. Similarly,

b

2

-

c

2

≤

x

. Taking the join and we get

b

2

⁢

Δ

⁢

c

2

≤

x

. Dually,

b

1

⁢

Δ

⁢

c

1

≤

x

′

, and complementing this to get

x

≤

(

b

1

⁢

Δ

⁢

c

1

)

′

=

b

1

↔

c

1

.

(

⇐

)

. If

b

2

-

c

2

≤

x

, then

b

2

∨

c

2

=

(

b

2

-

c

2

)

∨

c

2

≤

x

∨

c

2

, so that

(

b

2

∨

c

2

)

-

x

≤

c

2

-

x

, or

(

b

2

-

x

)

∨

(

c

2

-

x

)

≤

c

2

-

x

, which implies that

b

2

-

x

≤

c

2

-

x

. Similarly,

c

2

-

x

≤

b

2

-

x

. Putting the two

inequalities

together and we have

b

2

-

x

=

c

2

-

x

. Since

x

≤

b

1

↔

c

1

is equivalent to

b

1

⁢

Δ

⁢

c

1

≤

x

′

(dual statements), we also have

b

1

-

x

′

=

c

1

-

x

′

or

b

1

∧

x

=

c

1

∧

x

. Combining (via meet) this

equality

with the last one, we get

(

b

1

∧

x

)

∧

(

b

2

∧

x

′

)

=

(

c

1

∧

x

)

∧

(

c

2

∧

x

′

)

.
∎

Title

Boolean subalgebra

Canonical name

BooleanSubalgebra

Date of creation

2013-03-22 17:57:55

Last modified on

2013-03-22 17:57:55

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

9

Author

CWoo (3771)

Entry type

Definition

Classification

msc 06E05

Classification

msc 03G05

Classification

msc 06B20

Classification

msc 03G10

Synonym

dense subalgebra

Defines

dense Boolean subalgebra

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