regular open algebra

regular open algebra

A

regular open algebra

is an

algebraic system

𝒜

whose

universe

is the set of all

regular open sets

in a

topological space

X

, and whose

operations

are given by

1.

a

constant

1

such that

1

:=

X

,

2.

a unary operation

′

such that for any

U

,

U

′

:=

U

⊥

, where

U

⊥

is the

complement

of the

closure

of

U

in

X

,

3.

a

binary operation

∧

such that for any

U

,

V

∈

𝒜

,

U

∧

V

:=

U

∩

V

, and

4.

a binary operation

∨

such that for any

U

,

V

∈

𝒜

,

U

∨

V

:=

(

U

∪

V

)

⊥

⊥

.

From the parent entry, all of the operations above are well-defined (that the result sets are regular open). Also, we have the following:

Theorem 1

.

𝒜

is a

Boolean algebra

Proof.

We break down the proof into steps:

1.

𝒜

is a

lattice

. This amounts to verifying various laws on the operations:

–

(

idempotency

of

∨

and

∧

): Clearly,

U

∧

U

=

U

. Also,

U

∨

U

=

(

U

∪

U

)

⊥

⊥

=

U

⊥

⊥

=

U

, since

U

is regular open.

–

Commutativity

of the binary operations are obvious.

–

The

associativity

of

∧

is also obvious. The associativity of

∨

goes as follows:

U

∨

(

V

∨

W

)

=

(

U

∪

(

V

∪

W

)

⊥

⊥

)

⊥

⊥

=

U

⊥

∩

(

V

∪

W

)

⊥

⁣

⊥

⊥

=

U

⊥

∩

(

V

∪

W

)

⊥

, since

V

∪

W

is open (which implies that

(

V

∪

W

)

⊥

is regular open). The last expression is equal to

U

⊥

∩

(

V

⊥

∩

W

⊥

)

. Interchanging the roles of

U

and

W

, we obtain the

equation

W

∨

(

V

∨

U

)

=

W

⊥

∩

(

V

⊥

∩

U

⊥

)

, which is just

U

⊥

∩

(

V

⊥

∩

W

⊥

)

, or

U

∨

(

V

∨

W

)

. The commutativity of

∨

completes

the proof of the associativity of

∨

.

–

Finally, we verify the

absorption laws

. First,

U

∧

(

U

∨

V

)

=

U

∩

(

U

∪

V

)

⊥

⊥

=

U

⊥

⊥

∩

(

U

∪

V

)

⊥

⊥

=

(

U

⊥

∪

(

U

∪

V

)

⊥

)

⊥

=

(

U

⊥

∪

(

U

⊥

∩

V

⊥

)

⊥

=

(

U

⊥

)

⊥

=

U

. Second,

U

∨

(

U

∧

V

)

=

(

U

∪

(

U

∨

W

)

⊥

⊥

=

U

⊥

⊥

=

U

.

2.

𝒜

is complemented. First, it is easy to see that

∅

and

X

are the bottom and top elements of

𝒜

. Furthermore, for any

U

∈

𝒜

,

U

∧

U

′

=

U

∩

U

⊥

=

U

∩

(

X

∖

U

¯

)

⊆

U

¯

∩

(

X

∖

U

¯

)

=

∅

. Finally,

U

∨

U

′

=

(

U

∪

U

⊥

)

⊥

⊥

=

(

U

⊥

∩

U

⊥

⊥

)

⊥

=

(

U

⊥

∩

U

)

⊥

=

∅

⊥

=

X

.

3.

𝒜

is

distributive

. This can be easily proved once we show the following: for any

open sets

U

,

V

:

(

*

)

U

⊥

⊥

∩

V

⊥

⊥

=

(

U

∩

V

)

⊥

⊥

.

To begin, note that since

U

∩

V

⊆

U

, and

⊥

is order reversing,

(

U

∩

V

)

⊥

⊥

⊆

U

⊥

⊥

by applying

⊥

twice. Do the same with

V

and take the

intersection

, we get one of the

inclusions

:

(

U

∩

V

)

⊥

⊥

⊆

U

⊥

⊥

∩

V

⊥

⊥

. For the other inclusion, we first observe that

U

∩

V

¯

⊆

U

∩

V

¯

.

If

x

∈

LHS, then

x

∈

U

and for any open set

W

with

x

∈

W

, we have that

W

∩

V

≠

∅

. In particular,

U

∩

W

is such an open set (for

x

∈

U

∩

W

), so that

(

U

∩

W

)

∩

V

≠

∅

, or

W

∩

(

U

∩

V

)

≠

∅

. Since

W

is arbitrary,

x

∈

RHS. Now, apply the set complement, we have

(

U

∩

V

)

⊥

⊆

U

∁

∪

V

⊥

. Applying

⊥

next we get

(

U

∩

V

)

⊥

⊥

for the LHS, and

(

U

∁

∪

V

⊥

)

⊥

=

U

∁

-

∁

∩

V

⊥

⊥

=

U

∁

⁢

∁

∩

V

⊥

⊥

=

U

∩

V

⊥

⊥

for RHS, since

U

∁

is closed. As

⊥

reverses order, the new inclusion is

(

*

*

)

U

∩

V

⊥

⊥

⊆

(

U

∩

V

)

⊥

⊥

.

From this, a direct calculation shows

U

⊥

⊥

∩

V

⊥

⊥

⊆

(

U

⊥

⊥

∩

V

)

⊥

⊥

⊆

(

U

∩

V

)

⊥

⁣

⊥

⁣

⊥

⊥

=

(

U

∩

V

)

⊥

⊥

, noticing that the first and second inclusions use

(

*

*

)

above (and the fact that

⊥

⊥

preserves order), and the last equation uses the fact that for any open set

W

,

W

⊥

is regular open. This proves the

(

*

)

.

Finally, to finish the proof, we only need to show one of two distributive laws, say,

U

∧

(

V

∨

W

)

=

(

U

∧

V

)

∨

(

U

∧

W

)

, for the other one follows from the use of the

distributive inequalities

. This we do be direct computation:

U

∧

(

V

∨

W

)

=

U

∩

(

V

∪

W

)

⊥

⊥

=

U

⊥

⊥

∩

(

V

∪

W

)

⊥

⊥

=

(

U

∩

(

V

∪

W

)

)

⊥

⊥

=

(

(

U

∩

V

)

∪

(

U

∩

W

)

)

⊥

⊥

=

(

(

U

∧

V

)

∪

(

U

∧

W

)

)

⊥

⊥

=

(

U

∧

V

)

∨

(

U

∧

W

)

.

Since a complemented

distributive lattice

is

Boolean

, the proof is complete.
∎

Theorem 2

.

The subset

B

of all

clopen sets

in

X

forms a

Boolean subalgebra

of

A

.

Proof.

Clearly, every clopen set is regular open. In

addition

,

1

∈

ℬ

. If

U

is clopen, so is the complement of its closure, and hence

U

′

∈

ℬ

. If

U

,

V

are clopen, so is their intersection

U

∧

V

. Similarly,

U

∪

V

is clopen, so that

U

∨

V

=

U

∪

V

is clopen also. ∎

Theorem 3

.

In fact,

A

is a

complete Boolean algebra

. For an arbitrary subset

K

of

A

, the meet and join of

K

are

(

⋂

{

U

∣

U

∈

K

}

)

⊥

⊥

and

(

⋃

{

U

∣

U

∈

K

}

)

⊥

⊥

respectively.

Proof.

Let

V

=

(

⋃

{

U

∣

U

∈

𝒦

}

)

⊥

⊥

. For any

U

∈

𝒦

,

U

⊆

⋃

{

U

∣

U

∈

𝒦

}

so that

U

=

U

⊥

⊥

=

(

⋃

{

U

∣

U

∈

𝒦

}

)

⊥

⊥

=

V

. This shows that

V

is an

upper bound

of elements of

𝒦

. If

W

is another such upper bound, then

U

⊆

W

, so that

⋃

{

U

∣

U

∈

𝒦

}

⊆

W

, whence

V

=

(

⋃

{

U

∣

U

∈

𝒦

}

)

⊥

⊥

⊆

W

⊥

⊥

=

W

. The

infimum

is proved similarly.
∎

Theorem 4

.

𝒜

is the smallest complete Boolean subalgebra of

P

⁢

(

X

)

extending

B

.

More to come…

Title

regular open algebra

Canonical name

RegularOpenAlgebra

Date of creation

2013-03-22 17:56:21

Last modified on

2013-03-22 17:56:21

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

16

Author

CWoo (3771)

Entry type

Definition

Classification

msc 06E99

Generated on Fri Feb 9 10:05:00 2018 by

LaTeXML