distributivity in po-groups

distributivity in po-groups

Let

G

be a

po-group

and

A

be a set of elements of

G

. Denote the

supremum

of elements of

A

, if it exists, by

⋁

A

. Similarly, denote the

infimum

of elements of

A

, if it exists, by

⋀

A

. Furthermore, let

A

-

1

=

{

a

-

1

∣

a

∈

A

}

, and for any

g

∈

G

, let

g

⁢

A

=

{

g

⁢

a

∣

a

∈

A

}

and

A

⁢

g

=

{

a

⁢

g

∣

a

∈

A

}

.

1.

If

⋁

A

exists, so do

⋁

g

⁢

A

and

⋁

A

⁢

g

.

2.

If 1. is true, then

g

⁢

⋁

A

=

⋁

g

⁢

A

=

⋁

A

⁢

g

.

3.

⋁

A

exists iff

⋀

A

-

1

exists; when this is the case,

⋀

A

-

1

=

(

⋁

A

)

-

1

.

4.

If

⋀

A

exists, so do

⋀

g

⁢

A

, and

⋀

A

⁢

g

.

5.

If 4. is true, then

g

⁢

⋀

A

=

⋀

g

⁢

A

=

⋀

A

⁢

g

.

6.

If 1. is true and

A

=

{

a

,

b

}

, then

a

∧

b

exists and is equal to

a

⁢

(

a

∨

b

)

-

1

⁢

b

.

Proof.

Suppose

⋁

A

exists.

•

(1. and 2.) Clearly, for each

a

∈

A

,

a

≤

⋁

A

, so that

g

⁢

a

≤

g

⁢

⋁

A

, and therefore elements of

g

⁢

A

are

bounded from above

by

g

⁢

⋁

A

. To show that

g

⁢

⋁

A

is the least

upper bound

of elements of

g

⁢

A

, suppose

b

is the upper bound of elements of

g

⁢

A

, that is,

g

⁢

a

≤

b

for all

a

∈

A

, this means that

a

≤

g

-

1

⁢

b

for all

a

∈

A

. Since

⋁

A

is the least upper bound of the

a

’s,

⋁

A

≤

g

-

1

⁢

b

, so that

g

⁢

⋁

A

≤

b

. This shows that

g

⁢

⋁

A

is the supremum of elements of

g

⁢

A

; in other words,

g

⁢

⋁

A

=

⋁

g

⁢

A

. Similarly,

⋁

A

⁢

g

exists and

g

⁢

⋁

A

=

⋁

A

⁢

g

as well.

•

(3.) Write

c

=

⋁

A

. Then

a

≤

c

for each

a

∈

A

. This means

c

-

1

≤

a

-

1

. If

b

≤

a

-

1

for all

a

∈

A

, then

a

≤

b

-

1

for all

a

∈

A

, so that

c

≤

b

-

1

, or

b

≤

c

-

1

. This shows that

c

-

1

is the

greatest lower bound

of elements of

A

-

1

, or

(

⋁

A

)

-

1

=

⋀

A

-

1

. The

converse

is proved likewise.

•

(4. and 5.) This is just the dual of 1. and 2., so the proof is omitted.

•

(6.) If

A

=

{

a

,

b

}

, then

a

⁢

A

-

1

⁢

b

=

A

, and the

existence

of

⋀

A

is the same as the existence of

⋀

(

a

⁢

A

-

1

⁢

b

)

, which is the same as the existence of

a

⁢

(

⋀

A

-

1

)

⁢

b

by 4 and 5 above. Since

⋁

A

exists, so does

⋀

A

-

1

, and hence

a

⁢

(

⋀

A

-

1

)

⁢

b

, by 3 above. Also by 3, we have the

equality

a

⁢

(

⋀

A

-

1

)

⁢

b

=

a

⁢

(

⋁

A

)

-

1

⁢

b

. Putting everything together, we have the result:

a

∧

b

=

a

⁢

(

a

∨

b

)

-

1

⁢

b

.

This

completes

the proof.
∎

Remark

. From the above result, we see that group

multiplication

distributes over

arbitrary joins and meets, if these joins and meets exist.

One can use this result to prove the following: every

Dedekind complete

po-group is an Archimedean po-group.

Proof.

Suppose

a

n

≤

b

for all integers

n

. Let

A

=

{

a

n

∣

n

∈

ℤ

}

. Then

A

is bounded from above by

b

so has least upper bound

⋁

A

. Then

a

⁢

⋁

A

=

⋁

a

⁢

A

=

⋁

A

, since

a

⁢

A

=

A

. As a result, multiplying both sides by

(

⋁

A

)

-

1

, we get

a

=

e

.
∎

Remark

. The above is a

generalization

of a famous property of the

real numbers

:

ℝ

has the Archimedean property.

Title

distributivity in po-groups

Canonical name

DistributivityInPogroups

Date of creation

2013-03-22 17:05:12

Last modified on

2013-03-22 17:05:12

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

6

Author

CWoo (3771)

Entry type

Definition

Classification

msc 06F05

Classification

msc 06F20

Classification

msc 06F15

Classification

msc 20F60

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