10.3 Ordinal numbers

10.3 Ordinal numbers

Definition 10.3.1

.

Let

A

be a set and

(

–

<

–

)

:

A

→

A

→

𝖯𝗋𝗈𝗉

a

binary relation

on

A

.
We define by

induction

what it means for an

element

a

:

A

to be

accessible

by

<

:

•

If

b

is accessible for every

b

<

a

, then

a

is accessible.

We write

acc

⁢

(

a

)

to mean that

a

is accessible.

It may seem that such an inductive definition can never get off the ground, but of course if

a

has the property that there are

no

b

such that

b

<

a

, then

a

is vacuously accessible.

Note that this is an inductive definition of a family of types, like the type of vectors considered in

\autoref

sec:

generalizations

.
More precisely, it has one constructor, say

𝖺𝖼𝖼

<

, with type

𝖺𝖼𝖼

<

:

∏

a

:

A

(

∏

b

:

A

(

b

<

a

)

→

𝖺𝖼𝖼

(

b

)

)

→

𝖺𝖼𝖼

(

a

)

.

The induction principle for

𝖺𝖼𝖼

says that for any

P

:

∏

(

a

:

A

)

𝖺𝖼𝖼

⁢

(

a

)

→

𝒰

, if we have

f

:

∏

(

a

:

A

)

∏

(

h

:

∏

(

b

:

A

)

(

b

<

a

)

→

𝖺𝖼𝖼

(

b

)

)

(

∏

(

b

:

A

)

∏

(

l

:

b

<

a

)

P

⁢

(

b

,

h

⁢

(

b

,

l

)

)

)

→

P

⁢

(

a

,

𝖺𝖼𝖼

<

⁢

(

a

,

h

)

)

,

then we have

g

:

∏

(

a

:

A

)

∏

(

c

:

𝖺𝖼𝖼

(

a

)

)

P

⁢

(

a

,

c

)

defined by induction, with

g

(

a

,

𝖺𝖼𝖼

<

(

a

,

h

)

)

≡

f

(

a

,

h

,

λ

b

.

λ

l

.

g

(

b

,

h

(

b

,

l

)

)

)

.

This is a mouthful, but generally we apply it only in the simpler case where

P

:

A

→

𝒰

depends only on

A

.
In this case the second and third

arguments

of

f

may be combined, so that what we have to prove is

f

:

∏

a

:

A

(

∏

b

:

A

(

b

<

a

)

→

𝖺𝖼𝖼

(

b

)

×

P

(

b

)

)

→

P

(

a

)

.

That is, we assume every

b

<

a

is accessible and

g

⁢

(

b

)

:

P

⁢

(

b

)

is defined, and from these define

g

⁢

(

a

)

:

P

⁢

(

a

)

.

The omission of the second argument of

P

is justified by the following lemma, whose proof is the only place where we use the more general form of the induction principle.

Lemma 10.3.2

.

Accessibility is a mere property.

Proof.

We must show that for any

a

:

A

and

s

1

,

s

2

:

𝖺𝖼𝖼

⁢

(

a

)

we have

s

1

=

s

2

.
We prove this by induction on

s

1

, with

P

1

(

a

,

s

1

)

:

≡

∏

s

2

:

𝖺𝖼𝖼

⁢

(

a

)

(

s

1

=

s

2

)

.

Thus, we must show that for any

a

:

A

and

h

1

:

∏

(

b

:

A

)

(

b

<

a

)

→

𝖺𝖼𝖼

(

b

)

and

k

1

:

∏

(

b

:

A

)

∏

(

l

:

b

<

a

)

∏

(

t

:

𝖺𝖼𝖼

(

b

)

)

h

1

⁢

(

b

,

l

)

=

t

,

we have

𝖺𝖼𝖼

<

⁢

(

a

,

h

)

=

s

2

for any

s

2

:

𝖺𝖼𝖼

⁢

(

a

)

.
We regard this statement as

∏

(

a

:

A

)

∏

(

s

2

:

𝖺𝖼𝖼

(

a

)

)

P

2

⁢

(

a

,

s

2

)

, where

P

2

(

a

,

s

2

)

:

≡

∏

(

h

1

:

⋯

)

∏

(

k

1

:

⋯

)

(

𝖺𝖼𝖼

<

(

a

,

h

1

)

=

s

2

)

;

thus we may prove it by induction on

s

2

.
Therefore, we assume

h

2

:

∏

(

b

:

A

)

(

b

<

a

)

→

𝖺𝖼𝖼

(

b

)

, and

k

2

with a monstrous but irrelevant type,
and must show that for any

h

1

and

k

1

with types as above,
we have

𝖺𝖼𝖼

<

⁢

(

a

,

h

1

)

=

𝖺𝖼𝖼

<

⁢

(

a

,

h

2

)

.
By function extensionality, it suffices to show

h

1

⁢

(

b

,

l

)

=

h

2

⁢

(

b

,

l

)

for all

b

:

A

and

l

:

b

<

a

.
This follows from

k

1

.
∎

Definition 10.3.3

.

A binary relation

<

on a set

A

is

well-founded

if every element of

A

is accessible.

The point of well-foundedness is that for

P

:

A

→

𝒰

, we can use the induction principle of

𝖺𝖼𝖼

to conclude

∏

(

a

:

A

)

𝖺𝖼𝖼

⁢

(

a

)

→

P

⁢

(

a

)

, and then apply well-foundedness to conclude

∏

(

a

:

A

)

P

⁢

(

a

)

.
In other words, if from

∀

(

b

:

A

)

.

(

b

<

a

)

→

P

(

b

)

we can prove

P

⁢

(

a

)

, then

∀

(

a

:

A

)

.

P

(

a

)

.
This is called

well-founded induction

.

Lemma 10.3.4

.

Well-foundedness is a mere property.

Proof.

Well-foundedness of

<

is the type

∏

(

a

:

A

)

𝖺𝖼𝖼

⁢

(

a

)

, which is a mere proposition since each

𝖺𝖼𝖼

⁢

(

a

)

is.
∎

Example 10.3.5

.

Perhaps the most familiar well-founded relation is the usual strict ordering on

N

.
To show that this is well-founded, we must show that

n

is accessible for each

n

:

N

.
This is just the usual proof of “strong induction” from ordinary induction on

N

.

Specifically, we prove by induction on

n

:

N

that

k

is accessible for all

k

≤

n

.
The

base case

is just that

0

is accessible, which is

vacuously true

since nothing is strictly less than

0

.
For the inductive step, we assume that

k

is accessible for all

k

≤

n

, which is to say for all

k

<

n

+

1

; hence by definition

n

+

1

is also accessible.

A different relation on

N

which is also well-founded is obtained by setting only

n

<

succ

⁢

(

n

)

for all

n

:

N

.
Well-foundedness of this

relation

is almost exactly the ordinary induction principle of

N

.

Example 10.3.6

.

Let

A

:

\set

and

B

:

A

→

\set

be a family of sets.
Recall from

\autoref

sec:w-types that the

W

-type

W

(

a

:

A

)

⁢

B

⁢

(

a

)

is inductively generated by the single constructor

•

𝗌𝗎𝗉

:

∏

(

a

:

A

)

(

B

(

a

)

→

𝖶

(

x

:

A

)

B

(

x

)

)

→

𝖶

(

x

:

A

)

B

(

x

)

We define the relation

<

on

W

(

x

:

A

)

⁢

B

⁢

(

x

)

by

recursion

on its second argument:

•

For any

a

:

A

and

f

:

B

⁢

(

a

)

→

𝖶

(

x

:

A

)

⁢

B

⁢

(

x

)

, we define

w

<

𝗌𝗎𝗉

⁢

(

a

,

f

)

to mean that there merely exists a

b

:

B

⁢

(

a

)

such that

w

=

f

⁢

(

b

)

.

Now we prove that every

w

:

W

(

x

:

A

)

⁢

B

⁢

(

x

)

is accessible for this relation, using the usual induction principle for

W

(

x

:

A

)

⁢

B

⁢

(

x

)

.
This means we assume given

a

:

A

and

f

:

B

⁢

(

a

)

→

W

(

x

:

A

)

⁢

B

⁢

(

x

)

, and also a lifting

f

′

:

∏

(

b

:

B

(

a

)

)

acc

⁢

(

f

⁢

(

b

)

)

.
But then by definition of

<

, we have

acc

⁢

(

w

)

for all

w

<

sup

⁢

(

a

,

f

)

; hence

sup

⁢

(

a

,

f

)

is accessible.

Well-foundedness allows us to define

functions

by recursion and prove statements by induction, such as for

instance

the following.
Recall from

\autoref

subsec:prop-subsets that

𝒫

⁢

(

B

)

denotes the

power set

𝒫

(

B

)

:

≡

(

B

→

𝖯𝗋𝗈𝗉

)

.

Lemma 10.3.7

.

Suppose

B

is a set and we have a function

g

:

𝒫

⁢

(

B

)

→

B

Then if

<

is a well-founded relation on

A

, there is a function

f

:

A

→

B

such that for all

a

:

A

we have

f

(

a

)

=

g

(

\setof

f

(

a

′

)

|

a

′

<

a

)

.

(We are using the notation for images of subsets from

\autoref

sec:image.)

Proof.

We first define, for every

a

:

A

and

s

:

𝖺𝖼𝖼

⁢

(

a

)

, an element

f

¯

⁢

(

a

,

s

)

:

B

.
By induction, it suffices to assume that

s

is a function assigning to each

a

′

<

a

a witness

s

⁢

(

a

′

)

:

𝖺𝖼𝖼

⁢

(

a

′

)

, and that moreover for each such

a

′

we have an element

f

¯

⁢

(

a

′

,

s

⁢

(

a

′

)

)

:

B

.
In this case, we define

f

¯

(

a

,

s

)

:

≡

g

(

\setof

f

¯

(

a

′

,

s

(

a

′

)

)

|

a

′

<

a

)

.

Now since

<

is well-founded, we have a function

w

:

∏

(

a

:

A

)

𝖺𝖼𝖼

⁢

(

a

)

.
Thus, we can define

f

(

a

)

:

≡

f

¯

(

a

,

w

(

a

)

)

.
∎

In

classical logic

, well-foundedness has a more well-known reformulation.
In the following, we say that a subset

B

:

𝒫

⁢

(

A

)

is

nonempty

if it is unequal to the empty subset

(

λ

x

.

⊥

)

:

𝒫

(

X

)

.
We leave it to the reader to verify that assuming excluded middle, this is

equivalent

to mere inhabitation, i.e. to the condition

∃

(

x

:

A

)

.

x

∈

B

.

Lemma 10.3.8

.

Assuming excluded middle,

<

is well-founded if and only if every nonempty subset

B

:

P

⁢

(

A

)

merely has a

minimal element

.

Proof.

Suppose first

<

is well-founded, and suppose

B

⊆

A

is a subset with no minimal element.
That is, for any

a

:

A

with

a

∈

B

, there merely exists a

b

:

A

with

b

<

a

and

b

∈

B

.

We claim that for any

a

:

A

and

s

:

𝖺𝖼𝖼

⁢

(

a

)

, we have

a

∉

B

.
By induction, we may assume

s

is a function assigning to each

a

′

<

a

a proof

s

⁢

(

a

′

)

:

𝖺𝖼𝖼

⁢

(

a

)

, and that moreover for each such

a

′

we have

a

′

∉

B

.
If

a

∈

B

, then by

assumption

, there would merely exist a

b

<

a

with

b

∈

B

, which contradicts this assumption.
Thus,

a

∉

B

; this

completes

the induction.
Since

<

is well-founded, we have

a

∉

B

for all

a

:

A

, i.e.

B

is empty.

Now suppose each nonempty subset merely has a minimal element.
Let

B

=

\setof

a

:

A

|

¬

𝖺𝖼𝖼

(

a

)

.
Then if

B

is nonempty, it merely has a minimal element.
Thus there merely exists an

a

:

A

with

a

∈

B

such that for all

b

<

a

, we have

𝖺𝖼𝖼

⁢

(

b

)

.
But then by definition (and induction on

truncation

),

a

is merely accessible, and hence accessible, contradicting

a

∈

B

.
Thus,

B

is empty, so

<

is well-founded.
∎

Definition 10.3.9

.

A well-founded relation

<

on a set

A

is

extensional

if for any

a

,

b

:

A

, we have

(

∀

(

c

:

A

)

.

(

c

<

a

)

⇔

(

c

<

b

)

)

→

(

a

=

b

)

.

Note that since

A

is a set,

extensionality

is a mere proposition.
This notion of “extensionality” is unrelated to function extensionality, and also unrelated to the extensionality of identity types.
Rather, it is a “local” counterpart of the

axiom of extensionality

in classical

set theory

.

Theorem 10.3.10

.

The type of extensional well-founded relations is a set.

Proof.

By the univalence axiom, it suffices to show that if

(

A

,

<

)

is extensional and well-founded and

f

:

(

A

,

<

)

≅

(

A

,

<

)

, then

f

=

𝗂𝖽

A

.
We prove by induction on

<

that

f

⁢

(

a

)

=

a

for all

a

:

A

.
The inductive hypothesis is that for all

a

′

<

a

, we have

f

⁢

(

a

′

)

=

a

′

.

Now since

A

is extensional, to conclude

f

⁢

(

a

)

=

a

it is

sufficient

to show

∀

(

c

:

A

)

.

(

c

<

f

(

a

)

)

⇔

(

c

<

a

)

.

However, since

f

is an

automorphism

, we have

(

c

<

a

)

⇔

(

f

(

c

)

<

f

(

a

)

)

.
But

c

<

a

implies

f

⁢

(

c

)

=

c

by the inductive hypothesis, so

(

c

<

a

)

→

(

c

<

f

(

a

)

)

.
On the other hand, if

c

<

f

⁢

(

a

)

, then

f

-

1

⁢

(

c

)

<

a

, and so

c

=

f

⁢

(

f

-

1

⁢

(

c

)

)

=

f

-

1

⁢

(

c

)

by the inductive hypothesis again; thus

c

<

a

.
Therefore, we have

(

c

<

a

)

⇔

(

c

<

f

(

a

)

)

for any

c

:

A

, so

f

⁢

(

a

)

=

a

.
∎

Definition 10.3.11

.

If

(

A

,

<

)

and

(

B

,

<

)

are extensional and well-founded, a

simulation

is a function

f

:

A

→

B

such that

1.

if

a

<

a

′

, then

f

⁢

(

a

)

<

f

⁢

(

a

′

)

, and

2.

for all

a

:

A

and

b

:

B

, if

b

<

f

⁢

(

a

)

, then there merely exists an

a

′

<

a

with

f

⁢

(

a

′

)

=

b

.

Lemma 10.3.12

.

Any simulation is

injective

.

Proof.

We prove by double well-founded induction that for any

a

,

b

:

A

, if

f

⁢

(

a

)

=

f

⁢

(

b

)

then

a

=

b

.
The inductive hypothesis for

a

:

A

says that for any

a

′

<

a

, and any

b

:

B

, if

f

⁢

(

a

′

)

=

f

⁢

(

b

)

then

a

=

b

.
The inner inductive hypothesis for

b

:

A

says that for any

b

′

<

b

, if

f

⁢

(

a

)

=

f

⁢

(

b

′

)

then

a

=

b

′

.

Suppose

f

⁢

(

a

)

=

f

⁢

(

b

)

; we must show

a

=

b

.
By extensionality, it suffices to show that for any

c

:

A

we have

(

c

<

a

)

⇔

(

c

<

b

)

.
If

c

<

a

, then

f

⁢

(

c

)

<

f

⁢

(

a

)

by

\autoref

def:simulation

1

.
Hence

f

⁢

(

c

)

<

f

⁢

(

b

)

, so by

\autoref

def:simulation

2

there merely exists

c

′

:

A

with

c

′

<

b

and

f

⁢

(

c

)

=

f

⁢

(

c

′

)

.
By the inductive hypothesis for

a

, we have

c

=

c

′

, hence

c

<

b

.
The dual argument is symmetrical.
∎

In particular, this implies that in

\autoref

def:simulation

2

the word “merely” could be omitted without change of sense.

Corollary 10.3.13

.

If

f

:

A

→

B

is a simulation, then for all

a

:

A

and

b

:

B

, if

b

<

f

⁢

(

a

)

, there

purely

exists an

a

′

<

a

with

f

⁢

(

a

′

)

=

b

.

Proof.

Since

f

is injective,

∑

(

a

:

A

)

(

f

(

a

)

=

b

)

is a mere proposition.
∎

We say that a subset

C

:

𝒫

⁢

(

B

)

is an

initial segment

if

c

∈

C

and

b

<

c

imply

b

∈

C

.
The image of a simulation must be an initial segment, while the

inclusion

of any initial segment is a simulation.
Thus, by univalence, every simulation

A

→

B

is

equal

to the inclusion of some initial segment of

B

.

Theorem 10.3.14

.

For a set

A

, let

P

⁢

(

A

)

be the type of extensional well-founded relations on

A

.
If

<

A

:

P

⁢

(

A

)

and

<

B

:

P

⁢

(

B

)

and

f

:

A

→

B

, let

H

<

A

⁢

<

B

⁢

(

f

)

be the mere proposition that

f

is a simulation.
Then

(

P

,

H

)

is a standard notion of

structure

over

S

⁢

e

⁢

t

in the sense of

\autoref

sec:sip.

Proof.

We leave it to the reader to verify that

identities

are simulations, and that composites of simulations are simulations.
Thus, we have a notion of structure.
For standardness, we must show that if

<

and

≺

are two extensional well-founded relations on

A

, and

𝗂𝖽

A

is a simulation in both directions, then

<

and

≺

are equal.
Since extensionality and well-foundedness are mere propositions, for this it suffices to have

∀

(

a

,

b

:

A

)

.

(

a

<

b

)

⇔

(

a

≺

b

)

.
But this follows from

\autoref

def:simulation

1

for

𝗂𝖽

A

.
∎

Corollary 10.3.15

.

There is a category whose objects are sets equipped with extensional well-founded relations, and whose morphisms are simulations.

In fact, this category is a poset.

Lemma 10.3.16

.

For extensional and well-founded

(

A

,

<

)

and

(

B

,

<

)

, there is at most one simulation

f

:

A

→

B

.

Proof.

Suppose

f

,

g

:

A

→

B

are simulations.
Since being a simulation is a mere property, it suffices to show

∀

(

a

:

A

)

.

(

f

(

a

)

=

g

(

a

)

)

.
By induction on

<

, we may suppose

f

⁢

(

a

′

)

=

g

⁢

(

a

′

)

for all

a

′

<

a

.
And by extensionality of

B

, to have

f

⁢

(

a

)

=

g

⁢

(

a

)

it suffices to have

∀

(

b

:

B

)

.

(

b

<

f

(

a

)

)

⇔

(

b

<

g

(

a

)

)

.

But since

f

is a simulation, if

b

<

f

⁢

(

a

)

, then we have

a

′

<

a

with

f

⁢

(

a

′

)

=

b

.
By the inductive hypothesis, we have also

g

⁢

(

a

′

)

=

b

, hence

b

<

g

⁢

(

a

)

.
The dual argument is symmetrical.
∎

Thus, if

A

and

B

are equipped with extensional and well-founded relations, we may write

A

≤

B

to mean there exists a simulation

f

:

A

→

B

.

\autoref

thm:wfcat implies that if

A

≤

B

and

B

≤

A

, then

A

=

B

.

Definition 10.3.17

.

An

ordinal

is a set

A

with an extensional well-founded relation which is

transitive

, i.e.

satisfies

∀

(

a

,

b

,

c

:

A

)

.

(

a

<

b

)

→

(

b

<

c

)

→

(

a

<

c

)

.

Example 10.3.18

.

Of course, the usual

strict order

on

N

is transitive.
It is easily seen to be extensional as well; thus it is an ordinal.
As usual, we denote this ordinal by

ω

.

Let

𝖮𝗋𝖽

denote the type of ordinals.
By the previous results,

𝖮𝗋𝖽

is a set and has a natural

partial order

.
We now show that

𝖮𝗋𝖽

also admits a well-founded relation.

If

A

is an ordinal and

a

:

A

, let

A

/

a

:

≡

\setof

b

:

A

|

b

<

a

denote the initial segment.
Note that if

A

/

a

=

A

/

b

as ordinals, then that

isomorphism

must respect their inclusions into

A

(since simulations form a poset), and hence they are equal as subsets of

A

.
Therefore, since

A

is extensional,

a

=

b

.
Thus the function

a

↦

A

/

a

is an injection

A

→

𝖮𝗋𝖽

.

Definition 10.3.19

.

For ordinals

A

and

B

, a simulation

f

:

A

→

B

is said to be

bounded

if there exists

b

:

B

such that

A

=

B

/

b

.

The remarks above imply that such a

b

is unique when it exists, so that boundedness is a mere property.

We write

A

<

B

if there exists a bounded simulation from

A

to

B

.
Since simulations are unique,

A

<

B

is also a mere proposition.

Theorem 10.3.20

.

(

𝖮𝗋𝖽

,

<

)

is an ordinal.

More precisely, this

theorem

says that the type

𝖮𝗋𝖽

𝒰

i

of ordinals in one

universe

is itself an ordinal in the next higher universe, i.e.

(

𝖮𝗋𝖽

𝒰

i

,

<

)

:

𝖮𝗋𝖽

𝒰

i

+

1

.

Proof.

Let

A

be an ordinal; we first show that

A

/

a

is accessible (in

𝖮𝗋𝖽

) for all

a

:

A

.
By well-founded induction on

A

, suppose

A

/

b

is accessible for all

b

<

a

.
By definition of accessibility, we must show that

B

is accessible in

𝖮𝗋𝖽

for all

B

<

A

/

a

.
However, if

B

<

A

/

a

then there is some

b

<

a

such that

B

=

(

A

/

a

)

/

b

=

A

/

b

, which is accessible by the inductive hypothesis.
Thus,

A

/

a

is accessible for all

a

:

A

.

Now to show that

A

is accessible in

𝖮𝗋𝖽

, by definition we must show

B

is accessible for all

B

<

A

.
But as before,

B

<

A

means

B

=

A

/

a

for some

a

:

A

, which is accessible as we just proved.
Thus,

𝖮𝗋𝖽

is well-founded.

For extensionality, suppose

A

and

B

are ordinals such that

∏

(

C

:

𝖮𝗋𝖽

)

(

C

<

A

)

⇔

(

C

<

B

)

.

Then for every

a

:

A

, since

A

/

a

<

A

, we have

A

/

a

<

B

, hence there is

b

:

B

with

A

/

a

=

B

/

b

.
Define

f

:

A

→

B

to take each

a

to the corresponding

b

; it is straightforward to verify that

f

is an isomorphism.
Thus

A

≅

B

, hence

A

=

B

by univalence.

Finally, it is easy to see that

<

is transitive.
∎

Treating

𝖮𝗋𝖽

as an ordinal is often very convenient, but it has its pitfalls as well.
For instance, consider the following lemma, where we pay attention to how universes are used.

Lemma 10.3.21

.

Let

U

be a universe.
For any

A

:

Ord

U

, there is a

B

:

Ord

U

such that

A

<

B

.

Proof.

Let

B

=

A

+

𝟏

, with the element

⋆

:

𝟏

being greater than all elements of

A

.
Then

B

is an ordinal and it is easy to see that

A

≅

B

/

⋆

.
∎

This lemma illustrates a potential pitfall of the “typically ambiguous” style of using

𝒰

to denote an arbitrary, unspecified universe.
Consider the following alternative proof of it.

Another putative proof of

\autoref

thm:ordsucc.

Note that

C

<

A

if and only if

C

=

A

/

a

for some

a

:

A

.
This gives an isomorphism

A

≅

𝖮𝗋𝖽

/

A

, so that

A

<

𝖮𝗋𝖽

.
Thus we may take

B

:

≡

𝖮𝗋𝖽

.
∎

The second proof would be valid if we had stated

\autoref

thm:ordsucc in a typically ambiguous style.
But the resulting lemma would be less useful, because the second proof would constrain the second “

𝖮𝗋𝖽

” in the lemma statement to refer to a higher universe level than the first one.
The first proof allows both universes to be the same.

Similar

remarks apply to the next lemma, which could be proved in a less useful way by observing that

A

≤

𝖮𝗋𝖽

for any

A

:

𝖮𝗋𝖽

.

Lemma 10.3.22

.

Let

U

be a universe.
For any

X

:

U

and

F

:

X

→

Ord

U

, there exists

B

:

Ord

U

such that

F

⁢

x

≤

B

for all

x

:

X

.

Proof.

Let

B

be the

quotient

of the

equivalence relation

∼

on

∑

(

x

:

X

)

F

⁢

x

defined as follows:

(

x

,

y

)

∼

(

x

′

,

y

′

)

:

≡

(

(

F

x

)

/

y

≅

(

F

x

′

)

/

y

′

)

.

Define

(

x

,

y

)

<

(

x

′

,

y

′

)

if

(

F

⁢

x

)

/

y

<

(

F

⁢

x

′

)

/

y

′

.
This clearly descends to the quotient, and can be seen to make

B

into an ordinal.
Moreover, for each

x

:

X

the induced map

F

⁢

x

→

B

is a simulation.
∎

Title

10.3

Ordinal numbers

\metatable

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