example of monadic algebra

example of monadic algebra

The canonical example of a

monadic algebra

is what is known as a

functional monadic algebra

, which is explained in this entry.

Let

A

be a

Boolean algebra

and

X

be a non-empty set. Then

A

X

, the set of all

functions

from

X

into

A

, has a natural

Boolean

algebraic structure

defined as follows:

(

f

∧

g

)

⁢

(

x

)

:=

f

⁢

(

x

)

∧

g

⁢

(

x

)

,

(

f

′

)

⁢

(

x

)

:=

f

⁢

(

x

)

′

,

1

⁢

(

x

)

=

1

where

f

,

g

:

X

→

A

are functions, and

1

:

X

→

A

is just the

constant function

mapping

everything to

1

∈

A

(the abuse of notation here is harmless).

For each

f

:

X

→

A

, let

f

⁢

(

X

)

⊆

A

be the range of

f

. Let

B

be the subset of

A

X

consisting of all functions

f

such that

⋁

f

⁢

(

X

)

and

⋀

f

⁢

(

X

)

exist, where

⋁

and

⋀

are the

infinite join

and infinite meet

operations

on

A

. In other words,

B

:=

{

f

∈

A

X

∣

⋁

f

⁢

(

X

)

∈

A

⁢

and

⁢

⋀

f

⁢

(

X

)

∈

A

}

.

Proposition 1

.

B

defined above is a

Boolean subalgebra

of

A

X

.

Proof.

We need to show that, (1):

1

∈

B

, (2): for any

f

∈

B

,

f

′

∈

B

, and (3): for any

f

,

g

∈

B

,

f

∧

g

∈

B

.

1.

⋁

1

⁢

(

X

)

=

⋁

{

1

}

=

1

and

⋀

1

⁢

(

X

)

=

⋀

{

1

}

=

1

so

1

∈

B

2.

Suppose

f

∈

B

. Then

⋁

f

′

⁢

(

X

)

=

⋁

{

f

′

⁢

(

x

)

∣

x

∈

X

}

=

⋁

{

f

⁢

(

x

)

′

∣

x

∈

X

}

. By de Morgan’s law on infinite joins, the last expression is

(

⋀

{

f

⁢

(

x

)

∣

x

∈

X

}

)

′

, which exists. Dually,

⋀

f

′

⁢

(

X

)

exists by de Morgan’s law on infinite meets. Therefore,

f

′

∈

B

.

3.

Suppose

f

,

g

∈

B

. Then

⋀

(

f

∧

g

)

⁢

(

X

)

=

⋀

{

f

⁢

(

x

)

∧

g

⁢

(

x

)

∣

x

∈

X

}

=

⋀

{

f

⁢

(

x

)

∣

x

∈

X

}

∧

⋀

{

g

⁢

(

x

)

∣

x

∈

X

}

=

⋀

f

⁢

(

X

)

∧

⋀

g

⁢

(

X

)

,

which exists because both

⋀

f

⁢

(

X

)

and

⋀

g

⁢

(

X

)

do. In

addition

,

⋁

(

f

∧

g

)

⁢

(

X

)

=

⋁

{

f

⁢

(

x

)

∧

g

⁢

(

x

)

∣

x

∈

X

}

=

⋁

f

⁢

(

X

)

∧

⋁

g

⁢

(

X

)

.

The last

equality

stems from the

distributive law

of infinite meets over finite joins. Since the last expression exists,

f

∧

g

∈

B

.

The three conditions are verified and the proof is

complete

.
∎

Remark

. Every constant function belongs to

B

.

For each

f

∈

B

, write

f

∨

:=

⋁

f

⁢

(

X

)

and

f

∧

:=

⋀

f

⁢

(

X

)

. Define two functions

f

∃

,

f

∀

∈

A

X

by

f

∃

⁢

(

x

)

:=

f

∨

and

f

∀

⁢

(

x

)

:=

f

∧

.

Since these are constant functions, they belong to

B

.

Now, we define

operators

∃

,

∀

on

B

by setting

∃

(

f

)

:=

f

∃

and

∀

(

f

)

:=

f

∀

.

By the remark above,

∃

and

∀

are well-defined functions on

B

(

f

∃

,

f

∀

∈

B

).

Proposition 2

.

∃

is an existential quantifier operator on

B

and

∀

is its dual.

Proof.

The following three conditions need to be verified:

•

∃

(

0

)

=

0

:

∃

(

0

)

⁢

(

x

)

=

0

∃

⁢

(

x

)

=

0

∨

=

⋁

0

⁢

(

X

)

=

⋁

0

=

0

.

•

f

≤

∃

(

f

)

:

f

⁢

(

x

)

≤

⋁

f

⁢

(

X

)

=

f

∨

=

f

∃

⁢

(

x

)

=

∃

(

f

)

⁢

(

x

)

.

•

∃

(

f

∧

∃

(

g

)

)

=

∃

(

f

)

∧

∃

(

g

)

:

∃

(

f

∧

∃

(

g

)

)

⁢

(

x

)

=

⋁

(

f

∧

∃

(

g

)

)

⁢

(

X

)

=

⋁

(

f

⁢

(

X

)

∧

∃

(

g

)

⁢

(

X

)

)

=

⋁

(

f

⁢

(

X

)

∧

∃

(

g

)

⁢

(

x

)

)

=

⋁

(

f

⁢

(

X

)

∧

⋁

g

⁢

(

X

)

)

=

⋁

f

⁢

(

X

)

∧

⋁

g

⁢

(

X

)

=

∃

(

f

)

⁢

(

x

)

∧

∃

(

g

)

⁢

(

x

)

=

(

∃

(

f

)

∧

∃

(

g

)

)

⁢

(

x

)

.

Finally, to see that

∀

is the dual of

∃

, we do the following

computations

:

∀

(

f

)

⁢

(

x

)

=

⋀

{

f

⁢

(

x

)

∣

x

∈

X

}

=

⋀

{

f

⁢

(

x

)

′′

∣

x

∈

X

}

=

(

⋁

{

f

⁢

(

x

)

′

∣

x

∈

X

}

)

′

=

(

⋁

{

f

′

⁢

(

x

)

∣

x

∈

X

}

)

′

=

(

∃

(

f

′

)

)

′

⁢

(

x

)

,

completing the proof.
∎

Based on

Propositions

1 and 2,

(

B

,

∃

)

is a monadic algebra, and is called the

functional monadic algebra

for the pair

(

A

,

X

)

.

Title

example of monadic algebra

Canonical name

ExampleOfMonadicAlgebra

Date of creation

2013-03-22 17:51:55

Last modified on

2013-03-22 17:51:55

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

10

Author

CWoo (3771)

Entry type

Example

Classification

msc 03G15

Defines

functional monadic algebra

Generated on Thu Feb 8 20:27:42 2018 by

LaTeXML