dual space of a Boolean algebra

dual space of a Boolean algebra

Let

B

be a

Boolean algebra

, and

B

*

the set of all

maximal ideals

of

B

. In this entry, we will equip

B

*

with a

topology

so it is a

Boolean space

.

Definition

. For any

a

∈

B

, define

M

⁢

(

a

)

:=

{

M

∈

B

*

∣

a

∉

M

}

, and

ℬ

:=

{

M

⁢

(

a

)

∣

a

∈

B

}

.

It is know that in a Boolean algebra, maximal ideals and

prime ideals

coincide. From this entry (

http://planetmath.org/RepresentingABooleanLatticeByFieldOfSets

), we have the three following properties concerning

M

⁢

(

a

)

:

M

⁢

(

a

)

∩

M

⁢

(

b

)

=

M

⁢

(

a

∧

b

)

,

M

⁢

(

a

)

∪

M

⁢

(

b

)

=

M

⁢

(

a

∨

b

)

,

B

*

-

M

⁢

(

a

)

=

M

⁢

(

a

′

)

.

Furthermore, if

M

⁢

(

a

)

=

M

⁢

(

b

)

, then

a

=

b

.

From these properties, we see that

M

⁢

(

0

)

=

∅

and

M

⁢

(

1

)

=

B

*

. As a result, we see that

Proposition 1

.

B

*

is a

topological space

, whose topology

T

is generated by the basis

B

.

Proof.

∅

and

B

*

are both open, as they are

M

⁢

(

0

)

and

M

⁢

(

1

)

respectively. Also, the

intersection

of

open sets

M

⁢

(

a

)

and

M

⁢

(

b

)

is again open, since it is

M

⁢

(

a

∧

b

)

. ∎

We may in fact treat

ℬ

as a

subbasis

for

𝒯

, since finite intersections of elements of

ℬ

remain in

ℬ

.

Proposition 2

.

Each member of

B

is closed, hence

T

is generated by a basis of

clopen sets

. In other words,

B

*

is

zero-dimensional

.

Proof.

Each

M

⁢

(

a

)

is open, by definition, and closed, since it is the

complement

of the open set

M

⁢

(

a

′

)

. ∎

Proposition 3

.

B

*

is

Hausdorff

.

Proof.

If

M

,

N

∈

B

*

such that

M

≠

N

, then there is some

a

∈

B

such that

a

∈

M

and

a

∉

N

. This means that

N

∈

M

⁢

(

a

)

and

M

∉

M

⁢

(

a

)

, which means that

M

∈

B

*

-

M

⁢

(

a

)

=

M

⁢

(

a

′

)

. Since

M

⁢

(

a

)

and

M

⁢

(

a

′

)

are open and

disjoint

, with

N

∈

M

⁢

(

a

)

and

M

∈

M

⁢

(

a

′

)

, we see that

B

*

is Hausdorff. ∎

Now, based on a topological fact, every zero-dimensional Hausdorff space is

totally disconnected

. Hence

B

*

is totally disconnected.

Proposition 4

.

B

*

is

compact

.

Proof.

Suppose

{

U

i

∣

i

∈

I

}

is a

collection

of open sets whose union is

B

*

. Since each

U

i

is a union of elements of

ℬ

, we might as well assume that

B

*

is covered by elements of

ℬ

. In other words, we may assume that each

U

i

is some

M

⁢

(

a

i

)

∈

ℬ

.

Let

J

be the

ideal generated by

the set

{

a

i

∣

i

∈

I

}

. If

J

≠

B

, then

J

can be extended to a maximal ideal

M

. Since each

a

i

∈

M

, we see that

M

∉

M

⁢

(

a

i

)

, so that

M

∉

⋃

{

M

⁢

(

a

i

)

∣

i

∈

I

}

=

B

*

, which is a

contradiction

. Therefore,

J

=

B

. In particular,

1

∈

J

, which means that

1

can be expressed as the join of a finite number of the

a

i

’s:

1

=

⋁

{

a

i

∣

i

∈

K

}

,

where

K

is a finite subset of

J

. As a result, we have

⋃

{

M

⁢

(

a

i

)

∣

i

∈

K

}

=

M

⁢

(

⋁

{

a

i

∣

i

∈

K

}

)

=

M

⁢

(

1

)

=

B

*

.

So

B

*

has a finite

subcover

, and hence is compact.
∎

Collecting the last three results, we see that

B

*

is a Boolean space.

Remark

. It can be shown that

B

is

isomorphic

to the Boolean algebra of clopen sets in

B

*

. This is the famous

Stone representation theorem

.

Title

dual space

of a Boolean algebra

Canonical name

DualSpaceOfABooleanAlgebra

Date of creation

2013-03-22 19:08:35

Last modified on

2013-03-22 19:08:35

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

6

Author

CWoo (3771)

Entry type

Definition

Classification

msc 06E05

Classification

msc 03G05

Classification

msc 06B20

Classification

msc 03G10

Classification

msc 06E20

Related topic

StoneRepresentationTheorem

Related topic

MHStonesRepresentationTheorem

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