10.4 Classical well-orderings

10.4 Classical well-orderings

We now show the

equivalence

of our

ordinals

with the more familiar classical

well-orderings

.

Lemma 10.4.1

.

Assuming excluded middle, every ordinal is

trichotomous

:

∀

(

a

,

b

:

A

)

.

(

a

<

b

)

∨

(

a

=

b

)

∨

(

b

<

a

)

.

Proof.

By

induction

on

a

, we may assume that for every

a

′

<

a

and every

b

′

:

A

, we have

(

a

′

<

b

′

)

∨

(

a

′

=

b

′

)

∨

(

b

′

<

a

′

)

.
Now by induction on

b

, we may assume that for every

b

′

<

b

, we have

(

a

<

b

′

)

∨

(

a

=

b

′

)

∨

(

b

′

<

a

)

.

By excluded middle, either there merely exists a

b

′

<

b

such that

a

<

b

′

, or there merely exists a

b

′

<

b

such that

a

=

b

′

, or for every

b

′

<

b

we have

b

′

<

a

.
In the first case, merely

a

<

b

by

transitivity

, hence

a

<

b

as it is a mere proposition.
Similarly, in the second case,

a

<

b

by transport.
Thus, suppose

∀

(

b

′

:

A

)

.

(

b

′

<

b

)

→

(

b

′

<

a

)

.

Now analogously, either there merely exists

a

′

<

a

such that

b

<

a

′

, or there merely exists

a

′

<

a

such that

a

′

=

b

, or for every

a

′

<

a

we have

a

′

<

b

.
In the first and second cases,

b

<

a

, so we may suppose

∀

(

a

′

:

A

)

.

(

a

′

<

a

)

→

(

a

′

<

b

)

.
However, by

extensionality

, our two suppositions now imply

a

=

b

.
∎

Lemma 10.4.2

.

A

well-founded relation

contains no cycles, i.e.

∀

(

n

:

ℕ

)

.

∀

(

a

:

ℕ

n

→

A

)

.

¬

(

(

a

0

<

a

1

)

∧

…

∧

(

a

n

-

1

<

a

n

)

∧

(

a

n

<

a

0

)

)

.

Proof.

We prove by induction on

a

:

A

that there is no cycle containing

a

.
Thus, suppose by induction that for all

a

′

<

a

, there is no cycle containing

a

′

.
But in any cycle containing

a

, there is some

element

less than

a

and

contained

in the same cycle.
∎

In particular, a well-founded relation must be

irreflexive

, i.e.

¬

⁡

(

a

<

a

)

for all

a

.

Theorem 10.4.3

.

Assuming excluded middle,

(

A

,

<

)

is an ordinal if and only if every nonempty subset

B

⊆

A

has a least element.

Proof.

If

A

is an ordinal, then by

\autoref

thm:wfmin every nonempty subset merely has a

minimal element

.
But trichotomy implies that any minimal element is a least element.
Moreover, least elements are unique when they exist, so merely having one is as good as having one.

Conversely

, if every nonempty subset has a least element, then by

\autoref

thm:wfmin,

A

is well-founded.
We also have trichotomy, since for any

a

,

b

the subset

\setof

a

,

b

:

≡

\setof

x

:

A

|

x

=

a

∨

x

=

b

merely has a least element, which must be either

a

or

b

.
This implies transitivity, since if

a

<

b

and

b

<

c

, then either

a

=

c

or

c

<

a

would produce a cycle.
Similarly, it implies extensionality, for if

∀

(

c

:

A

)

.

(

c

<

a

)

⇔

(

c

<

b

)

, then

a

<

b

implies (letting

c

be

a

) that

a

<

a

, which is a cycle, and similarly if

b

<

a

; hence

a

=

b

.
∎

In classical mathematics, the

characterization

of

\autoref

thm:wellorder is taken as the definition of a

well-ordering

, with the

ordinals

being a canonical set of representatives of isomorphism classes for well-orderings.
In our

context

, the

structure

identity

principle means that there is no need to look for such representatives: any well-ordering is as good as any other.

We now move on to consider

consequences

of the

axiom of choice

.
For any set

X

, let

𝒫

+

⁢

(

X

)

denote the type of merely inhabited subsets of

X

:

𝒫

+

(

X

)

:

≡

\setof

Y

:

𝒫

(

X

)

|

∃

(

x

:

X

)

.

x

∈

Y

.

Assuming excluded middle, this is equivalently the type of

nonempty

subsets of

X

, and we have

𝒫

⁢

(

X

)

≃

(

𝒫

+

⁢

(

X

)

)

+

𝟏

.

Theorem 10.4.4

.

Assuming excluded middle, the following are

equivalent

.

1.

For every set

X

, there merely exists a

function

f

:

𝒫

+

⁢

(

X

)

→

X

such that

f

⁢

(

Y

)

∈

Y

for all

Y

:

𝒫

⁢

(

X

)

.

2.

Every set merely admits the structure of an ordinal.

Of course,

1

is a standard classical version of the axiom of choice; see

\autoref

ex:choice-function.

Proof.

One direction is easy: suppose

2

.
Since we aim to prove the mere proposition

1

, we may assume

A

is an ordinal.
But then we can define

f

⁢

(

B

)

to be the least element of

B

.

Now suppose

1

.
As before, since

2

is a mere proposition, we may assume given such an

f

.
We extend

f

to a function

f

¯

:

𝒫

⁢

(

X

)

≃

(

𝒫

+

⁢

(

X

)

)

+

𝟏

⟶

X

+

𝟏

in the obvious way.
Now for any ordinal

A

, we can define

g

A

:

A

→

X

+

𝟏

by

well-founded recursion

:

g

A

(

a

)

:

≡

f

¯

(

X

∖

\setof

g

A

(

b

)

|

(

b

<

a

)

∧

(

g

A

(

b

)

∈

X

)

)

(regarding

X

as a subset of

X

+

𝟏

in the obvious way).

Let

A

′

:

≡

\setof

a

:

A

|

g

A

(

a

)

∈

X

be the

preimage

of

X

⊆

X

+

𝟏

; then we claim the

restriction

g

A

′

:

A

′

→

X

is

injective

.
For if

a

,

a

′

:

A

with

a

≠

a

′

, then by trichotomy and without loss of generality, we may assume

a

′

<

a

.
Thus

g

A

(

a

′

)

∈

\setof

g

A

(

b

)

|

b

<

a

, so since

f

⁢

(

Y

)

∈

Y

for all

Y

we have

g

A

⁢

(

a

)

≠

g

A

⁢

(

a

′

)

.

Moreover,

A

′

is an

initial segment

of

A

.
For

g

A

⁢

(

a

)

lies in

𝟏

if and only if

\setof

g

A

(

b

)

|

b

<

a

=

X

, and if this holds then it also holds for any

a

′

>

a

.
Thus,

A

′

is itself an ordinal.

Finally, since

𝖮𝗋𝖽

is an ordinal, we can take

A

:

≡

𝖮𝗋𝖽

.
Let

X

′

be the image of

g

𝖮𝗋𝖽

′

:

𝖮𝗋𝖽

′

→

X

; then the

inverse

of

g

𝖮𝗋𝖽

′

yields an injection

H

:

X

′

→

𝖮𝗋𝖽

.
By

\autoref

thm:ordunion, there is an ordinal

C

such that

H

⁢

x

≤

C

for all

x

:

X

′

.
Then by

\autoref

thm:ordsucc, there is a further ordinal

D

such that

C

<

D

, hence

H

⁢

x

<

D

for all

x

:

X

′

.
Now we have

g

𝖮𝗋𝖽

⁢

(

D

)

=

f

¯

(

X

∖

\setof

g

𝖮𝗋𝖽

(

B

)

|

B

<

D

∧

(

g

𝖮𝗋𝖽

(

B

)

∈

X

)

)

=

f

¯

(

X

∖

\setof

g

𝖮𝗋𝖽

(

B

)

|

g

𝖮𝗋𝖽

(

B

)

∈

X

)

since if

B

:

𝖮𝗋𝖽

and

(

g

𝖮𝗋𝖽

⁢

(

B

)

∈

X

)

, then

B

=

H

⁢

x

for some

x

:

X

′

, hence

B

<

D

.
Now if

\setof

g

𝖮𝗋𝖽

(

B

)

|

g

𝖮𝗋𝖽

(

B

)

∈

X

is not all of

X

, then

g

𝖮𝗋𝖽

⁢

(

D

)

would lie in

X

but not in this subset, which would be a

contradiction

since

D

is itself a potential value for

B

.
So this set must be all of

X

, and hence

g

𝖮𝗋𝖽

′

is

surjective

as well as injective.
Thus, we can transport the ordinal structure on

𝖮𝗋𝖽

′

to

X

.
∎

Remark 10.4.5

.

If we had given the wrong proof of

\autoref

thm:ordsucc or

\autoref

thm:ordunion, then the resulting proof of

\autoref

thm:wop would be

invalid

: there would be no way to consistently assign

universe

levels.
As it is, we require propositional resizing (which follows from

LEM

) to ensure that

X

′

lives in the same universe as

X

(up to equivalence).

Corollary 10.4.6

.

Assuming the axiom of choice, the function

Ord

→

\set

(which forgets the order structure) is a surjection.

Note that

𝖮𝗋𝖽

is a set, while

\set

is a 1-type.
In general, there is no reason for a 1-type to admit any surjective function from a set.
Even the axiom of choice does not appear to imply that

every

1-type does so (although see

\autoref

ex:acnm-surjset), but it readily implies that this is so for 1-types constructed out of

\set

, such as the types of objects of categories of structures as in

\autoref

sec:sip.
The following corollary also applies to such categories.

Corollary 10.4.7

.

Assuming

AC

,

S

⁢

e

⁢

t

admits a

weak equivalence

functor from a strict category.

Proof.

Let

X

0

:

≡

𝖮𝗋𝖽

, and for

A

,

B

:

X

0

let

hom

X

(

A

,

B

)

:

≡

(

A

→

B

)

.
Then

X

is a strict category, since

𝖮𝗋𝖽

is a set, and the above surjection

X

0

→

\set

extends to a weak equivalence functor

X

→

𝒮

⁢

e

⁢

t

.
∎

Now recall from

\autoref

sec:

cardinals

that we have a further surjection

|

¯

|

0

:

\set

→

𝖢𝖺𝗋𝖽

, and hence a composite surjection

𝖮𝗋𝖽

→

𝖢𝖺𝗋𝖽

which sends each ordinal to its

cardinality

.

Theorem 10.4.8

.

Assuming

AC

, the surjection

Ord

→

Card

has a

section

.

Proof.

There is an easy and wrong proof of this: since

𝖮𝗋𝖽

and

𝖢𝖺𝗋𝖽

are both sets,

𝖠𝖢

implies that any surjection between them

merely

has a section.
However, we actually have a canonical

specified

section: because

𝖮𝗋𝖽

is an ordinal, every nonempty subset of it has a uniquely specified least element.
Thus, we can map each cardinal to the least element in the corresponding fiber.
∎

It is traditional in

set theory

to identify cardinals with their image in

𝖮𝗋𝖽

: the least ordinal having that cardinality.

It follows that

𝖢𝖺𝗋𝖽

also canonically admits the structure of an ordinal: in fact, one

isomorphic

to

𝖮𝗋𝖽

.
Specifically, we define by well-founded recursion a function

ℵ

:

𝖮𝗋𝖽

→

𝖮𝗋𝖽

, such that

ℵ

⁢

(

A

)

is the least ordinal having cardinality greater than

ℵ

⁢

(

A

/

a

)

for all

a

:

A

.
Then (assuming

𝖠𝖢

) the image of

ℵ

is exactly the image of

𝖢𝖺𝗋𝖽

.

Title

10.4 Classical well-orderings

\metatable

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