properties of ordinals

properties of ordinals

Recall that an

ordinal

is a

transitive set

well-ordered

by the membership

relation

∈

. Below we list and prove some basic

properties of ordinals

.

1.

∅

is an ordinal. This is

vacuously true

.

2.

If

α

is an ordinal, then

α

∉

α

, for otherwise

α

∈

α

∈

α

, contradicting the fact that

∈

is

irreflexive

on

α

. Notice that by the

axiom of foundation

, this statement is true even if

α

is not an ordinal. But in this case, there is no need to resort to the axiom.

3.

If

α

is an ordinal, and

β

∈

α

, then

β

is an ordinal.

Proof.

Since

α

is

transitive

,

β

∈

α

implies that

β

⊆

α

, and hence

β

is well-ordered by

∈

(inherited from

α

). Now, suppose

δ

∈

γ

∈

β

. Then

γ

∈

α

and

δ

∈

α

. Since

∈

is a

linear ordering

in

α

,

δ

∈

β

, as desired.
∎

4.

If

α

is an ordinal, so is

α

+

:=

α

∪

{

α

}

(

α

+

is the

successor

of

α

).

Proof.

If

γ

∈

α

+

, then either

γ

=

α

or

γ

∈

α

, and in the latter case,

γ

⊆

α

, since

α

is transitive. In either case,

γ

⊆

α

+

so that

α

+

is transitive. Now, suppose

δ

⊆

α

+

. Then

δ

-

{

α

}

⊆

α

, so has a least

element

η

. Since

η

∈

α

,

η

is also the least element of

δ

. This shows that

α

+

is well-ordered by

∈

, and therefore

α

+

is an ordinal.
∎

5.

If

α

is an ordinal and

β

⊂

α

is transitive, then

β

∈

α

. As a result, if

α

≠

β

are ordinals such that

β

⊂

α

, then

β

∈

α

.

Proof.

β

is well-ordered since

α

is, and since

β

is transitive by

assumption

,

β

is an ordinal. Let

γ

be the least element of

α

-

β

, and suppose

δ

∈

β

. Since

∈

is a linear ordering on

α

, either

δ

∈

γ

or

γ

∈

δ

. If the latter is true, then

γ

∈

δ

∈

β

, and therefore

γ

∈

β

since

β

is transitive, contradicting

γ

∈

α

-

β

. So

δ

∈

γ

, and therefore

β

⊆

γ

. Moreover, if

δ

∈

γ

, then

δ

∉

α

-

β

since

γ

is the least element in

α

-

β

. Therefore

δ

∈

β

, so

γ

⊆

β

, and hence

β

=

γ

∈

α

.
∎

6.

(

Law of Trichotomy

) For any ordinals

α

,

β

, exactly one of the following is true:

α

=

β

,

α

∈

β

,

and

β

∈

α

.

Proof.

If

θ

∈

δ

∈

α

∩

β

, then

θ

∈

α

∩

β

since both

α

and

β

are transitive. So

α

∩

β

is transitive, and hence an ordinal. If

α

∩

β

⊂

α

and

α

∩

β

⊂

β

, then

α

∩

β

∈

α

∩

β

by 5, which is impossible by 2. This means that either

α

∩

β

=

α

or

α

∩

β

=

β

. If

α

≠

β

, then

α

⊂

β

, and hence by 5,

α

∈

β

in the former case, and

β

∈

α

in the latter.
∎

7.

If

A

is a non-empty class of ordinals, then

⋂

A

is the least element of

A

. Furthermore,

A

is well-ordered.

Proof.

Set

α

:=

⋂

A

. There are three steps:

–

α

is an ordinal: First, notice that

α

is well-ordered, as the

ordering

is inherited from members of

A

. Next, if

β

∈

α

, then

β

∈

γ

for every

γ

∈

A

. Since each

γ

is transitive,

β

⊆

γ

, and

β

⊆

α

as a result, showing that

α

is transitive. Finally,

α

is a set, since it is a

subclass

of each element (which is a set) of

A

. This shows that

α

is an ordinal.

–

α

∈

A

: By

definition

,

α

⊆

γ

for each

γ

∈

A

. Suppose in

addition

that

α

⊂

γ

. Since

α

is an ordinal, then

α

∈

γ

by 5. This shows that

α

∈

⋂

{

γ

∣

γ

∈

A

}

=

α

, which is impossible by 2. Therefore

α

=

γ

for some

γ

∈

A

, as desired.

–

α

is the least element in

A

: If there is some

δ

∈

A

such that

δ

∈

α

. But

α

=

⋂

{

γ

∣

γ

∈

A

}

⊆

δ

, implying that

δ

∈

δ

, a

contradiction

.

If

B

is any non-empty subclass of

A

, then

⋂

B

is the least element of

B

by the result above. Hence

A

is well-ordered.
∎

8.

On

, the class of all ordinals is transitive, well-ordered (by

∈

), and proper (not a set).

Proof.

If

α

∈

β

∈

On

, then

α

∈

On

by 3, so

On

is transitive. It is well-ordered by 7. Finally, if

On

were a set, then

On

is an ordinal since it is transitive and well-ordered. This means that

On

∈

On

, contradicts 2 above.
∎

9.

If

A

is a non-empty set of ordinals, then

⋃

A

is an ordinal, and

⋃

A

=

sup

⁡

A

.

Proof.

Set

α

:=

⋃

A

. First,

α

is a set because

A

is. If

γ

∈

β

∈

α

, then

γ

∈

β

∈

δ

for some

δ

∈

A

, and therefore

γ

∈

δ

⊆

α

, showing that

α

is transitive. If

x

⊆

α

, then every element of

x

is an element of some

δ

∈

A

. Hence

x

is a set of ordinals, and therefore well-ordered by 7. This shows that

α

is an ordinal.

Now, pick any

δ

∈

A

. Then

δ

⊆

⋃

A

=

α

, so either

δ

=

α

or

δ

⊂

α

. In the latter case, we get

δ

∈

α

by 5. This shows that

α

is an

upper bound

of

A

. If

θ

is an ordinal such that

δ

∈

θ

for every

δ

∈

A

, then

δ

⊂

θ

, which implies

α

⊂

θ

, so that

α

∈

θ

by 5, showing that

α

is the least upper bound of

A

.
∎

10.

A transitive set of ordinals is an ordinal, true since the set is well-ordered by 7. This is a partial

converse

of 3. In particular, this provides an alternative proof of 4. Another

corollary

is that an

initial segment

of an ordinal is an ordinal.

11.

Two ordinals are

isomorphic

iff they are the same.

Proof.

One direction is

obvious

. Suppose now that

ϕ

:

α

→

β

is an order isomorphism between ordinals

α

and

β

. Suppose

α

≠

β

. Then either

α

∈

β

or

β

∈

α

, which means that

α

⊂

β

or

β

⊂

α

. In either case, we get a contradiction since

ϕ

is a

bijection

.
∎

12.

Every well-ordered set is

order isomorphic

to exactly one ordinal.

Proof.

Uniqueness is provided by 12. Now, let us prove the

existence

of such an ordinal. First, notice that if there is an order preserving

injection

W

into an ordinal

α

, then the image of

W

is an initial segment of

α

, which is an ordinal, and we are done. Now, assume the contrary. Since ordinals are well-ordered, the assumption implies that every

α

∈

On

has a strict injection into

W

(this is an important

property

on well-ordered sets, see a proof here (

http://planetmath.org/PropertiesOfWellOrderedSets

)). We shall derive a contradiction by showing that

On

is a set. First, for each ordinal, there is a unique

order-preserving

injection into

W

. In addition, if two ordinals are such that their corresponding injections into

W

agree, then they are isomorphic and hence identical. Thus, elements of

On

are completely determined by subsets (in fact, initial segments) of

W

. Since

W

is a set, so is

On

, a contradiction!
∎

13.

An ordinal has exactly one of the following forms:

∅

,

⋃

α

, or

β

+

for some ordinals

α

,

β

.

Proof.

Suppose an ordinal

α

≠

∅

. Being a transitive set of ordinals (by 3),

β

:=

⋃

α

is also an ordinal by 9. So

β

⊆

α

. Then either

β

=

α

or

β

⊂

α

. In the former case,

α

=

⋃

α

. In the latter case,

β

∈

α

. On the one hand, since

β

=

sup

⁡

α

, any

γ

∈

α

, either

γ

=

β

or

γ

∈

β

, so that

γ

∈

β

+

in any case. This means that

α

⊆

β

+

. On the other hand,

β

∈

α

combined with

β

⊂

α

give us the other

inclusion

β

+

⊆

α

. As a result,

α

=

β

+

.
∎

14.

A set is an ordinal iff it is a

von Neumann ordinal

. Every von Neumann ordinal is clearly an ordinal by 1, 4, and 9. The converse is true by 13.

Remark

. Property 8 above also resolves one of the

set theory

paradoxes

called the

Burali-Forti paradox

: the

collection

of all ordinals is an ordinal, which is not true because the collection is not a set.

Title

properties of ordinals

Canonical

name

PropertiesOfOrdinals

Date of creation

2013-03-22 18:48:58

Last modified on

2013-03-22 18:48:58

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

15

Author

CWoo (3771)

Entry type

Result

Classification

msc 03E10

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