aarjaneiro/planetmath huggingface dataset:
Boolean ring

Boolean ring

A

Boolean ring

is a ring

R

that has a

multiplicative identity

,
and in which every

element

is

idempotent

, that is,

x

2

=

x

⁢

for all

⁢

x

∈

R

.

Boolean rings are necessarily

commutative

(

http://planetmath.org/CommutativeRing

).
Also, if

R

is a Boolean ring, then

x

=

-

x

for each

x

∈

R

.

Boolean rings are

equivalent

to

Boolean algebras

(or

Boolean

lattices (

http://planetmath.org/BooleanLattice

)).
Given a Boolean ring

R

,
define

x

∧

y

=

x

⁢

y

and

x

∨

y

=

x

+

y

+

x

⁢

y

and

x

′

=

x

+

1

for all

x

,

y

∈

R

,
then

(

R

,

∧

,

∨

,

′

,

0

,

1

)

is a Boolean algebra.
Given a Boolean algebra

(

L

,

∧

,

∨

,

′

,

0

,

1

)

,
define

x

⋅

y

=

x

∧

y

and

x

+

y

=

(

x

′

∧

y

)

∨

(

x

∧

y

′

)

,
then

(

L

,

⋅

,

+

)

is a Boolean ring.
In particular, the category of Boolean rings is

isomorphic

to the category of Boolean lattices.

Examples

As mentioned above, every Boolean algebra can be considered as a Boolean ring. In particular, if

X

is any set, then the

power set

𝒫

⁢

(

X

)

forms a Boolean ring, with

intersection

as

multiplication

and

symmetric difference

as addition.

Let

R

be the ring

ℤ

2

×

ℤ

2

with the

operations

being coordinate-wise.
Then we can check:

(

1

,

1

)

×

(

1

,

1

)

=

(

1

,

1

)

(

1

,

0

)

×

(

1

,

0

)

=

(

1

,

0

)

(

0

,

1

)

×

(

0

,

1

)

=

(

0

,

1

)

(

0

,

0

)

×

(

0

,

0

)

=

(

0

,

0

)

the four elements that form the ring are idempotent. So

R

is Boolean.

Title

Boolean ring

Canonical name

BooleanRing

Date of creation

2013-03-22 12:27:28

Last modified on

2013-03-22 12:27:28

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

24

Author

yark (2760)

Entry type

Definition

Classification

msc 06E99

Classification

msc 03G05

Related topic

Idempotency

Related topic

BooleanLattice

Related topic

BooleanIdeal

Generated on Fri Feb 9 10:04:27 2018 by

LaTeXML
generalized Boolean algebra

generalized Boolean algebra

A

lattice

L

is called a

generalized Boolean algebra

if

•

L

is

distributive

,

•

L

is relatively

complemented

, and

•

L

has

0

as the bottom.

Clearly, a

Boolean algebra

is a generalized Boolean algebra.

Conversely

, a generalized Boolean algebra

L

with a top

1

is a Boolean algebra, since

L

=

[

0

,

1

]

is a

bounded

distributive

complemented lattice

, so each

element

a

∈

L

has a unique

complement

a

′

by distributivity. So

′

is a unary

operator

on

L

which makes

L

into a

de Morgan algebra

. A complemented de Morgan algebra is, as a result, a Boolean algebra.

As an example of a generalized Boolean algebra that is not

Boolean

, let

A

be an

infinite set

and let

B

be the set of all finite subsets of

A

. Then

B

is generalized Boolean: order

B

by

inclusion

, then

B

is a distributive as the

operation

is inherited from

P

⁢

(

A

)

, the

powerset

of

A

. It is also relatively complemented: if

C

∈

[

X

,

Y

]

where

C

,

X

,

Y

∈

B

, then

(

Y

-

C

)

∪

X

is the

relative complement

of

C

in

[

X

,

Y

]

. Finally,

∅

is, as usual, the bottom element in

B

.

B

is not a Boolean algebra, because the union of all the

singletons

(all in

B

) is

A

, which is

infinite

, thus not in

B

.

One

property

of a generalized Boolean algebra

L

is the following: if

y

and

z

are complements of

x

∈

[

a

,

b

]

, then

y

=

z

; in other words, relative complements are uniquely determined. This is true because in any

distributive lattice

, complents are uniquely determined. As

L

is distributive, so is each

lattice interval

[

a

,

b

]

in

L

.

In fact, because of the

existence

of

0

, we can actually construct the relative complement. Let

b

-

x

denote the unique complement of

x

in

[

0

,

b

]

. Then

(

b

-

x

)

∨

a

is the unique complement of

x

∈

[

a

,

b

]

:

x

∧

(

(

b

-

x

)

∨

a

)

=

(

x

∧

(

b

-

x

)

)

⁢

v

⁢

e

⁢

e

⁢

(

x

∧

a

)

=

0

∨

a

=

a

and

x

∨

(

(

b

-

x

)

∨

a

)

=

(

x

∨

(

b

-

x

)

)

∨

a

=

b

∨

a

=

b

.

Conversely, if

L

is a distributive lattice with

0

such that any lattice interval

[

0

,

a

]

is complemented, then

L

is a generalized Boolean algebra. Again,

(

b

-

x

)

∨

a

provides the necessary complement of

x

in

[

a

,

b

]

.

Title

generalized Boolean algebra

Canonical

name

GeneralizedBooleanAlgebra

Date of creation

2013-03-22 17:08:37

Last modified on

2013-03-22 17:08:37

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

6

Author

CWoo (3771)

Entry type

Definition

Classification

msc 06E99

Classification

msc 06D99

Synonym

generalized Boolean lattice

Generated on Fri Feb 9 10:04:47 2018 by

LaTeXML
regular open set

regular open set

Let

X

be a

topological space

. A subset

A

of

X

is called a

regular open

set if

A

is equal to the

interior

of the

closure

of itself:

A

=

int

⁡

(

A

¯

)

.

Clearly, every regular open set is open, and every

clopen set

is regular open.

Examples

. Let

ℝ

be the

real line

with the

usual
topology

(generated by

open intervals

).

•

(

a

,

b

)

is regular open whenever

-

∞

<

a

≤

b

<

∞

.

•

(

a

,

b

)

∪

(

b

,

c

)

is not regular open for

-

∞

<

a

≤

b

≤

c

<

∞

and

a

≠

c

. The interior of the closure of

(

a

,

b

)

∪

(

b

,

c

)

is

(

a

,

c

)

.

If we examine the

structure

of

int

⁡

(

A

¯

)

a
little more closely, we see that if we define

A

⊥

:=

X

-

A

¯

,

then

A

⊥

⊥

=

int

⁡

(

A

¯

)

.

So an alternative definition of
a regular open set is an

open set

A

such that

A

⊥

⊥

=

A

.

Remarks

.

•

For any

A

⊆

X

,

A

⊥

is always open.

•

∅

⊥

=

X

and

X

⊥

=

∅

.

•

A

∩

A

⊥

=

∅

and

A

∪

A

⊥

is

dense in

X

.

•

A

⊥

∪

B

⊥

⊆

(

A

∩

B

)

⊥

and

A

⊥

∩

B

⊥

=

(

A

∪

B

)

⊥

.

•

It can be shown that if

A

is open, then

A

⊥

is
regular open. As a result, following from the first property,

int

⁡

(

A

¯

)

, being

A

⊥

⊥

, is regular open for any subset

A

of

X

.

•

In

addition

, if both

A

and

B

are regular open, then

A

∩

B

is regular open.

•

It is not true, however, that the union of two regular open
sets is regular open, as illustrated by the second example above.

•

It can also be shown that the set of all regular open sets of
a topological space

X

forms a

Boolean algebra

under the following
set of

operations

:

(a)

1

=

X

and

0

=

∅

,

(b)

a

∧

b

=

a

∩

b

,

(c)

a

∨

b

=

(

a

∪

b

)

⊥

⊥

, and

(d)

a

′

=

a

⊥

.

This is an example of a Boolean algebra coming from a

collection

of
subsets of a set that is not formed by the standard set operations
union

∪

,

intersection

∩

, and

complementation

′

.

The definition of a regular open set can be dualized. A

closed set

A

in a topological space is called a

regular closed set

if

A

=

int

⁡

(

A

)

¯

.

References

1

P. Halmos (1970).

Lectures on Boolean Algebras

, Springer.

2

S. Willard (1970).

General Topology

, Addison-Wesley Publishing Company.

Title

regular open set

Canonical name

RegularOpenSet

Date of creation

2013-03-22 15:04:03

Last modified on

2013-03-22 15:04:03

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

9

Author

CWoo (3771)

Entry type

Definition

Classification

msc 06E99

Synonym

regularly open

Synonym

regularly closed

Synonym

regularly closed set

Defines

regular open

Defines

regular closed

Generated on Fri Feb 9 10:05:02 2018 by

LaTeXML
ordered topological vector space

ordered topological vector space

Let

k

be either

ℝ

or

ℂ

considered as a field. An

ordered topological vector space

L

, (

ordered t.v.s

for short) is

•

a

topological vector space

over

k

, and

•

an

ordered vector space

over

k

, such that

•

the

positive cone

L

+

of

L

is a closed subset of

L

.

The last statement can be interpreted as follows: if a

sequence

of non-negative

elements

x

i

of

L

converges

to an element

x

, then

x

is non-negative.

Remark

. Let

L

,

M

be two ordered t.v.s., and

f

:

L

→

M

a

linear transformation

that is

monotone

. Then if

0

≤

x

∈

L

,

0

≤

f

⁢

(

x

)

∈

M

also. Therefore

f

⁢

(

L

+

)

⊆

M

+

.

Conversely

, a linear map that is

invariant

under positive cones is monotone.

Title

ordered topological vector space

Canonical name

OrderedTopologicalVectorSpace

Date of creation

2013-03-22 17:03:23

Last modified on

2013-03-22 17:03:23

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

4

Author

CWoo (3771)

Entry type

Definition

Classification

msc 06F20

Classification

msc 46A40

Classification

msc 06F30

Synonym

ordered topological linear space

Generated on Fri Feb 9 10:05:51 2018 by

LaTeXML
rational rank of a group

rational rank of a group

In the following,

G

is an

abelian group

.

Definition 1

.

The group

Q

⊗

Z

G

is called the

divisible hull

of

G

.

It is a

ℚ

-

vector space

such that the scalar

ℤ

-

multiplication

of

G

is extended to

ℚ

.

Definition 2

.

The elements

g

1

,

g

2

,

…

,

g

r

∈

G

are called rationally independent if they are

linearly independent

over

Z

, i.e. for all

n

1

,

…

,

n

r

∈

Z

:

n

1

⁢

g

1

+

…

+

n

r

⁢

g

r

=

0

⇒

n

1

=

…

=

n

r

=

0

.

Definition 3

.

The

dimension

of

Q

⊗

Z

G

over

Q

is called the rational rank of

G

.

We denote the rational rank of

G

by

r

⁢

(

G

)

.

Example

:

r

⁢

(

ℤ

×

ℤ

)

=

2

because

ℚ

⊗

ℤ

(

ℤ

×

ℤ

)

=

(

ℚ

⊗

ℤ

ℤ

)

×

(

ℚ

⊗

ℤ

ℤ

)

=

ℚ

×

ℚ

.

Properties

:

•

If

H

is a

subgroup

of

G

then we have:

r

⁢

(

G

)

=

r

⁢

(

H

)

+

r

⁢

(

G

/

H

)

.

It results from the fact that

ℚ

ℤ

is a

flat module

.

•

The rational rank of the group

G

can be defined as the least upper bound (finite or infinite) of the cardinals

r

such that there exist

r

rationally independent elements in

G

.

Title

rational rank of a group

Canonical

name

RationalRankOfAGroup

Date of creation

2013-03-22 16:53:25

Last modified on

2013-03-22 16:53:25

Owner

polarbear (3475)

Last modified by

polarbear (3475)

Numerical id

9

Author

polarbear (3475)

Entry type

Definition

Classification

msc 06F20

Classification

msc 20K20

Classification

msc 20K15

Classification

msc 20K99

Defines

rationally independent

Defines

rational rank

Defines

divisible hull

Generated on Fri Feb 9 10:05:53 2018 by

LaTeXML
vector lattice

vector lattice

An

ordered vector space

whose underlying poset is a

lattice

is called a

vector lattice

. A vector lattice is also called a

Riesz space

.

For example, given a

topological space

X

, its

ring of continuous functions

C

⁢

(

X

)

is a vector lattice. In particular, any finite dimensional

Euclidean space

ℝ

n

is a vector lattice.

A

vector sublattice

is a

subspace

of a vector lattice that is also a sublattice.

Below are some properties of the join (

∨

) and meet (

∧

)

operations

on a vector lattice

L

. Suppose

u

,

v

,

w

∈

L

, then

1.

(

u

+

w

)

∨

(

v

+

w

)

=

(

u

∨

v

)

+

w

2.

u

∧

v

=

(

u

+

v

)

-

(

u

∨

v

)

3.

If

λ

≥

0

, then

λ

⁢

u

∨

λ

⁢

v

=

λ

⁢

(

u

∨

v

)

4.

If

λ

≤

0

, then

λ

⁢

u

∨

λ

⁢

v

=

λ

⁢

(

u

∧

v

)

5.

If

u

≠

v

, then the

converse

holds for 3 and 4

6.

If

L

is an ordered vector space, and if for any

u

,

v

∈

L

, either

u

∨

v

or

u

∧

v

exists, then

L

is a vector lattice. This is basically the result of property 2 above.

7.

(

u

∧

v

)

+

w

=

(

u

+

w

)

∧

(

v

+

w

)

(dual of statement 1)

8.

u

∧

v

=

-

(

-

u

∨

-

v

)

(a direct

consequence

of statement 4, with

λ

=

-

1

)

9.

(

-

u

)

∧

u

≤

0

≤

(

-

u

)

∨

u

Proof.

(

-

u

)

∧

u

≤

u

and

(

-

u

)

∧

u

≤

-

u

imply that

2

⁢

(

(

-

u

)

∧

u

)

≤

u

+

(

-

u

)

=

0

, so

(

-

u

)

∧

u

≤

0

, which means

0

≤

-

(

(

-

u

)

∧

u

)

=

u

∨

(

-

u

)

.
∎

10.

(

a

∨

b

)

+

(

c

∨

d

)

=

(

a

+

c

)

∨

(

a

+

d

)

∨

(

b

+

c

)

∨

(

b

+

d

)

, by repeated application of 1 above.

Remark

. The first five properties are also satisfied by an ordered vector space, with the

assumptions

that the suprema exist for the appropriate pairs of

elements

(see the entry on ordered vector space for detail).

Title

vector lattice

Canonical name

VectorLattice

Date of creation

2013-03-22 17:03:13

Last modified on

2013-03-22 17:03:13

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

7

Author

CWoo (3771)

Entry type

Definition

Classification

msc 06F20

Classification

msc 46A40

Synonym

Riesz Space

Defines

vector sublattice

Generated on Fri Feb 9 10:06:38 2018 by

LaTeXML
basic facts about ordered rings

basic facts about ordered rings

Throughout this entry,

(

R

,

≤

)

is an

ordered ring

.

Lemma 1

.

If

a

,

b

,

c

∈

R

with

a

<

b

, then

a

+

c

<

b

+

c

.

Proof.

The

contrapositive

will be proven.

Let

a

,

b

,

c

∈

R

with

a

+

c

≥

b

+

c

. Note that

-

c

∈

R

. Thus,

b

=

b

+

0

=

b

+

c

+

(

-

c

)

≤

a

+

c

+

(

-

c

)

=

a

+

0

=

a

.

∎

Lemma 2

.

If

|

R

|

≠

1

and

R

has a

characteristic

, then it must be

0

.

Proof.

Suppose not. Let

n

be a positive integer such that

char

⁡

R

=

n

. Since

|

R

|

≠

1

, it must be the case that

n

>

1

.

Let

r

∈

R

with

r

>

0

. By the previous lemma,

0

<

r

≤

…

≤

∑

j

=

1

n

-

1

r

≤

∑

j

=

1

n

r

=

0

, a

contradiction

.
∎

Lemma 3

.

If

a

,

b

∈

R

with

a

≤

b

and

c

∈

R

with

c

<

0

, then

a

⁢

c

≥

b

⁢

c

.

Proof.

Note that

-

c

∈

R

and

0

=

c

+

(

-

c

)

<

0

+

(

-

c

)

=

-

c

. Since

a

≤

b

,

-

(

a

⁢

c

)

=

a

⁢

(

-

c

)

≤

b

⁢

(

-

c

)

=

-

(

b

⁢

c

)

. Thus,

b

⁢

c

=

b

⁢

c

+

0

=

b

⁢

c

+

(

a

⁢

c

+

(

-

(

a

⁢

c

)

)

)

=

(

b

⁢

c

+

a

⁢

c

)

+

(

-

(

a

⁢

c

)

)

≤

(

b

⁢

c

+

a

⁢

c

)

+

(

-

(

b

⁢

c

)

)

=

-

(

b

⁢

c

)

+

(

b

⁢

c

+

a

⁢

c

)

=

(

-

(

b

⁢

c

)

+

b

⁢

c

)

+

a

⁢

c

=

0

+

a

⁢

c

=

a

⁢

c

.

∎

Lemma 4

.

Suppose further that

R

is a ring with

multiplicative identity

1

≠

0

. Then

0

<

1

.

Proof.

Suppose that

0

≮

1

. Since

R

is an ordered ring, it must be the case that

1

<

0

. By the previous lemma,

1

⋅

1

≥

0

⋅

1

. Thus,

1

≥

0

, a contradiction.
∎

Title

basic facts about ordered rings

Canonical name

BasicFactsAboutOrderedRings

Date of creation

2013-03-22 16:17:21

Last modified on

2013-03-22 16:17:21

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

11

Author

Wkbj79 (1863)

Entry type

Result

Classification

msc 06F25

Classification

msc 12J15

Classification

msc 13J25

Related topic

MathbbCIsNotAnOrderedField

Generated on Fri Feb 9 10:06:41 2018 by

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ordered integral domain with well-ordered positive elements

ordered integral domain with well-ordered positive elements

Theorem

.

If

(

R

,

≤

)

is an ordered (

http://planetmath.org/OrderedRing

)

integral domain

and if the set

R

+

=

{

r

∈

R

:

0

<

r

}

of its

positive elements

(

http://planetmath.org/PositivityInOrderedRing

) is

well-ordered

, then

R

and

R

+

can be expressed as sets of

multiples

of the unity as follows:

•

R

=

{

m

⋅

1

:

m

∈

ℤ

}

,

•

R

+

=

{

n

⋅

1

:

n

∈

ℤ

+

}

.

The theorem may be interpreted so that such an integral domain is

isomorphic

with the

ordered ring

ℤ

of rational integers.

Title

ordered integral domain with well-ordered positive elements

Canonical name

OrderedIntegralDomainWithWellorderedPositiveElements

Date of creation

2013-03-22 14:46:43

Last modified on

2013-03-22 14:46:43

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

11

Author

Wkbj79 (1863)

Entry type

Theorem

Classification

msc 06F25

Classification

msc 12J15

Classification

msc 13J25

Related topic

TotalOrder

Related topic

OrderedRing

Defines

positive element

Generated on Fri Feb 9 10:06:43 2018 by

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ordered ring

ordered ring

An

ordered ring

is a

commutative ring

R

with a

total ordering

≤

such that, for every

a

,

b

,

c

∈

R

:

1.

If

a

≤

b

, then

a

+

c

≤

b

+

c

2.

If

a

≤

b

and

0

≤

c

, then

c

⋅

a

≤

c

⋅

b

An

ordered field

is an ordered ring

(

R

,

≤

)

where

R

is also a field.

Examples of ordered rings include:

•

The integers

ℤ

, under the standard

ordering

≤

.

•

The

real numbers

ℝ

under the standard ordering.

•

The

polynomial ring

ℝ

⁢

[

x

]

in one variable over

ℝ

, under the relation

f

≤

g

if and only if

g

-

f

has nonnegative

leading coefficient

.

Examples of rings

which do not admit any

ordering relation

making them into an ordered ring include:

•

The

complex numbers

ℂ

.

•

The

finite field

ℤ

/

p

⁢

ℤ

, where

p

is any prime.

Title

ordered ring

Canonical name

OrderedRing

Date of creation

2013-03-22 11:52:06

Last modified on

2013-03-22 11:52:06

Owner

djao (24)

Last modified by

djao (24)

Numerical id

13

Author

djao (24)

Entry type

Definition

Classification

msc 06F25

Classification

msc 12J15

Classification

msc 13J25

Classification

msc 11D41

Related topic

TotalOrder

Related topic

OrderingRelation

Defines

ordered field

Generated on Fri Feb 9 10:06:45 2018 by

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positive

positive

The

positive

is usually explained to that the number under consideration is greater than zero.  Without the

relation

“

>

”, the positivity of () numbers may be defined specifying which numbers of a given number kind are positive, e.g. as follows.

•

In the set

ℤ

of the

integers

, all numbers obtained from 1 via

addition

are positive.

•

In the set

ℚ

of the

rationals

, all numbers obtained from 1 via addition and

division

are positive.

•

In the set

ℝ

of the real numbers, the numbers defined by the

equivalence classes

of non-zero decimal

sequences

are positive; these sequences (

decimal expansions

) consist of

natural numbers

from 0 to 9 as digits and a single

decimal point

(where two decimal sequences are

equivalent

if they are identical, or if one has an

infinite

tail of 9’s, the other has an infinite tail of 0’s, and the leading portion of the first sequence is one lower than the leading portion of the second).

For example,

1

+

1

+

1

is a

positive integer

,

1

+

1

1

+

1

+

1

+

1

+

1

is a positive rational and

5.15115111511115

⁢

…

is a positive real number.

If

a

is positive and

a

+

b

=

0

, then the opposite number

b

is

negative

.

The sets of positive integers, positive rationals, positive (real)

algebraic numbers

and positive reals are

closed under

addition and

multiplication

, so also the set of positive

even numbers

.

Title

positive

Canonical

name

Positive

Date of creation

2013-03-22 14:35:05

Last modified on

2013-03-22 14:35:05

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

19

Author

pahio (2872)

Entry type

Definition

Classification

msc 06F25

Classification

msc 11B99

Classification

msc 00A05

Synonym

greater than zero

Defines

negative

Generated on Fri Feb 9 10:06:47 2018 by

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positivity in ordered ring

positivity in ordered ring

Theorem

.

If

(

R

,

≤

)

is an

ordered ring

, then it contains a subset

R

+

having the following :

•

R

+

is under

ring addition

and, supposing that the ring contains no

zero divisors

, also under ring multiplication.

•

Every

element

r

of

R

satisfies

exactly one of the conditions

(

1

)

⁢

r

=

0

,

(

2

)

⁢

r

∈

R

+

,

(

3

)

-

r

∈

R

+

.

Proof.

We take

R

+

=

{

r

∈

R

:

0

<

r

}

=

{

r

∈

R

:

0

≤

r

∧

0

≠

r

}

.  Let

a

,

b

∈

R

+

.  Then

0

<

a

,

0

<

b

, and therefore we have

0

<

a

+

0

<

a

+

b

,  i.e.

a

+

b

∈

R

+

.  If

R

has no zero-divisors, then also

a

⁢

b

≠

0

and

0

=

a

⁢

0

<

a

⁢

b

,  i.e.

a

⁢

b

∈

R

+

.  Let

r

be an arbitrary non-zero element of

R

.  Then we must have either

0

<

r

or

r

<

0

(not both) because

R

is

totally ordered

.  The latter alternative gives that

0

=

-

r

+

r

<

-

r

+

0

=

-

r

.  The both cases that either

r

∈

R

+

or

-

r

∈

R

+

.

Title

positivity in ordered ring

Canonical name

PositivityInOrderedRing

Date of creation

2013-03-22 14:46:40

Last modified on

2013-03-22 14:46:40

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

12

Author

pahio (2872)

Entry type

Theorem

Classification

msc 06F25

Classification

msc 12J15

Classification

msc 13J25

Related topic

PositiveCone

Related topic

TopicEntryOnRealNumbers

Generated on Fri Feb 9 10:06:48 2018 by

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well-ordering principle for natural numbers

well-ordering principle for natural numbers

Every

nonempty set

S

of

natural numbers

contains a least

element

; that is, there is some number

a

in

S

such that

a

≤

b

for all

b

belonging to

S

.

Beware that there is another statement (which is

equivalent

to the

axiom of choice

) called the

well-ordering principle

. It asserts that every set can be

well-ordered

.

Note that the

well-ordering principle for natural numbers

is equivalent to the

principle of mathematical induction

(or, the

principle of finite induction

).

Title

well-ordering principle for natural numbers

Canonical

name

WellorderingPrincipleForNaturalNumbers

Date of creation

2013-03-22 11:46:38

Last modified on

2013-03-22 11:46:38

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

18

Author

CWoo (3771)

Entry type

Axiom

Classification

msc 06F25

Classification

msc 65A05

Classification

msc 11Y70

Related topic

MaximalityPrinciple

Related topic

WellOrderedSet

Related topic

ExistenceAndUniquenessOfTheGcdOfTwoIntegers

Generated on Fri Feb 9 10:06:50 2018 by

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ℂ is not an ordered field

ℂ

is not an ordered field

Theorem 1

.

ℂ

is not an

ordered field

.

First, the following

theorem

will be proven:

Theorem 2

.

ℤ

⁢

[

i

]

is not an ordered ring.

Proof.

Many facts that are used here are proven in the entry regarding

basic facts about ordered rings

.

Suppose that

ℤ

⁢

[

i

]

is an ordered ring under some

total ordering

≤

. Note that

0

<

1

and

-

1

=

-

1

+

0

<

-

1

+

1

=

0

.

Note also that

i

≠

0

. Thus, either

i

>

0

or

i

<

0

. In either case,

-

1

=

i

⋅

i

≥

0

⋅

i

=

0

, a

contradiction

.

It follows that

ℤ

⁢

[

i

]

is not an ordered ring.
∎

Because of theorem 2, no ring containing

ℤ

⁢

[

i

]

can be an ordered ring. It follows that

ℂ

is not an ordered field.

Title

ℂ

is not an ordered field

Canonical name

mathbbCIsNotAnOrderedField

Date of creation

2013-03-22 16:17:25

Last modified on

2013-03-22 16:17:25

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

10

Author

Wkbj79 (1863)

Entry type

Theorem

Classification

msc 06F25

Classification

msc 13J25

Classification

msc 12J15

Related topic

Complex

Related topic

BasicFactsAboutOrderedRings

Generated on Fri Feb 9 10:06:52 2018 by

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