canonical ordering on pairs of ordinals

canonical ordering on pairs of ordinals

The

lexicographic ordering

on

On

×

On

, the class of all pairs of

ordinals

, is a well-order in the broad sense, in that every

subclass

of

On

×

On

has a least

element

, as

proposition

2 of the parent entry readily shows. However, with this type of

ordering

, we get

initial segments

which are not sets. For example, the initial segment of

(

1

,

0

)

consists of all ordinal pairs of the form

(

0

,

α

)

, where

α

∈

On

, and is easily seen to be a

proper class

. So the questions is: is there a way to order

On

×

On

such that every initial segment of

On

×

On

is a set? The answer is yes. The construction of such a

well-ordering

in the following discussion is what is known as the

canonical well-ordering

of

On

×

On

.

To begin, let us consider a strictly

linearly ordered set

(

A

,

<

)

. We

construct

a

binary relation

≺

on

A

×

A

as follows:

(

a

1

,

a

2

)

≺

(

b

1

,

b

2

)

iff

{

max

⁡

{

a

1

,

a

2

}

<

max

⁡

{

b

1

,

b

2

}

,

or

max

⁡

{

a

1

,

a

2

}

=

max

⁡

{

b

1

,

b

2

}

,

and

⁢

a

1

<

b

1

,

or

max

⁡

{

a

1

,

a

2

}

=

max

⁡

{

b

1

,

b

2

}

,

and

⁢

a

1

=

b

1

,

and

⁢

a

2

<

b

2

.

For example, consider the usual ordering on

ℤ

. Given

(

p

,

q

)

∈

ℤ

×

ℤ

. Suppose

p

≤

q

. Then the set of all

(

m

,

n

)

∈

ℤ

×

ℤ

such that

(

m

,

n

)

≺

(

p

,

q

)

is the union of the three

pairwise disjoint

sets

{

(

m

,

n

)

∣

max

⁡

{

m

,

n

}

<

q

}

∪

{

(

m

,

q

)

∣

m

<

p

}

∪

{

(

p

,

n

)

∣

n

<

q

}

.

Proposition 1

.

.

≺

is a strict

linear ordering

on

A

×

A

.

Proof.

It is

irreflexive

because

(

a

1

,

a

2

)

is never

comparable

with itself. It is linear because, first of all, given

(

a

1

,

a

2

)

≠

(

b

1

,

b

2

)

, exactly one of the three conditions is true, and hence either

(

a

1

,

a

2

)

≺

(

b

1

,

b

2

)

, or

(

b

1

,

b

2

)

≺

(

a

1

,

a

2

)

. It remains to show that

≺

is

transitive

, suppose

(

a

1

,

a

2

)

≺

(

b

1

,

b

2

)

and

(

b

1

,

b

2

)

≺

(

c

1

,

c

2

)

.

The two cases

1.

max

⁡

{

a

1

,

a

2

}

<

max

⁡

{

b

1

,

b

2

}

and

max

⁡

{

b

1

,

b

2

}

≤

max

⁡

{

c

1

,

c

2

}

,

2.

max

⁡

{

a

1

,

a

2

}

≤

max

⁡

{

b

1

,

b

2

}

and

max

⁡

{

b

1

,

b

2

}

<

max

⁡

{

c

1

,

c

2

}

,

produce

max

⁡

{

a

1

,

a

2

}

<

max

⁡

{

c

1

,

c

2

}

. Now, assume

max

⁡

{

a

1

,

a

2

}

=

max

⁡

{

b

1

,

b

2

}

=

max

⁡

{

c

1

,

c

2

}

, which result in three more cases

1.

a

1

<

b

1

and

b

1

≤

c

1

,

2.

a

1

≤

b

1

and

b

1

<

c

1

,

3.

a

1

=

b

1

=

c

1

, and

a

2

<

b

2

and

b

2

<

c

2

,

the first two produce

a

1

<

c

1

, and the last

a

1

=

c

1

and

a

2

<

c

2

. In all cases, we get

(

a

1

,

a

2

)

≺

(

c

1

,

c

2

)

.
∎

Proposition 2

.

If

<

is a well-order on

A

, then so is

≺

on

A

×

A

.

Proof.

Let

R

⊆

A

×

A

be non-empty. Let

B

:=

{

max

⁡

{

b

1

,

b

2

}

∣

(

b

1

,

b

2

)

∈

R

}

.

Then

∅

≠

B

⊆

A

, and therefore has a least element

b

, since

<

is a well-order on

A

. Next, let

C

:=

{

c

1

∣

max

⁡

{

c

1

,

c

2

}

=

b

,

where

⁢

(

c

1

,

c

2

)

∈

R

}

.

Then

C

≠

∅

, and has a least element

c

. Finally, let

D

:=

{

d

2

∣

max

⁡

{

c

,

d

2

}

=

b

,

where

⁢

(

c

,

d

2

)

∈

R

}

.

Again,

D

≠

∅

, so has a least element

d

. So

(

c

,

d

)

∈

R

. We want to show that

(

c

,

d

)

is the least element of

R

.

Pick any

(

x

,

y

)

∈

R

distinct from

(

c

,

d

)

. Then

max

⁡

{

x

,

y

}

∈

B

is at least

b

=

max

⁡

{

c

,

d

}

. If

b

<

max

⁡

{

x

,

y

}

, then

(

c

,

d

)

≺

(

x

,

y

)

. Otherwise,

b

=

max

⁡

{

x

,

y

}

, so that

x

∈

C

is at least

c

. If

c

<

x

, then again we have

(

c

,

d

)

≺

(

x

,

y

)

. But if

c

=

x

, then

y

∈

D

, so that

d

≤

y

. Since

(

x

,

y

)

≠

(

c

,

d

)

, and

x

=

c

,

y

≠

d

. Therefore

d

<

y

, and

(

c

,

d

)

≺

(

x

,

y

)

as a result.
∎

The

ordering relation

above can be generalized to arbitrary classes. Since

On

is well-ordered by

∈

, the canonical ordering on

On

×

On

is a well-ordering by proposition 2. Moreover,

Proposition 3

.

Given the canonical ordering

≺

on

On

×

On

, every initial segment is a set.

Proof.

Given ordinals

α

,

β

∈

On

, suppose

λ

=

max

⁡

{

α

,

β

}

. The initial segment of

(

α

,

β

)

is the union of the following

collections

1.

{

(

γ

,

δ

)

∣

max

⁡

{

γ

,

δ

}

<

λ

}

, which is a subcollection of

λ

×

λ

,

2.

{

(

γ

,

δ

)

∣

max

⁡

{

γ

,

δ

}

=

λ

,

and

⁢

γ

<

α

}

, which again is a subcollection

λ

×

λ

, and

3.

{

(

α

,

δ

)

∣

max

⁡

{

α

,

δ

}

=

λ

,

and

⁢

δ

<

β

}

, which is a subcollection of

{

α

}

×

β

.

Since

λ

×

λ

and

{

α

}

×

β

are both sets, so is the initial segment of

(

α

,

β

)

.
∎

Remark

. The canonical well-ordering on

On

×

On

can be used to prove a well-known property of alephs:

ℵ

α

⋅

ℵ

α

=

ℵ

α

, for any ordinal

α

.

Title

canonical ordering on pairs of ordinals

Canonical name

CanonicalOrderingOnPairsOfOrdinals

Date of creation

2013-03-22 18:50:02

Last modified on

2013-03-22 18:50:02

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

8

Author

CWoo (3771)

Entry type

Definition

Classification

msc 03E10

Classification

msc 06A05

Synonym

canonical well-ordering

Related topic

IdempotencyOfInfiniteCardinals

Defines

canonical ordering

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