10.2 Cardinal numbers

10.2 Cardinal numbers

Definition 10.2.1

.

The

type of

cardinal numbers

is the 0-truncation of the type

\set

of sets:

𝖢𝖺𝗋𝖽

:

≡

∥

\set

∥

0

Thus, a

cardinal number

, or

cardinal

, is an inhabitant of

Card

≡

∥

\set

∥

0

.
Technically, of course, there is a separate type

Card

U

associated to each

universe

U

.

As usual for truncations, if

A

is a set, then

|

A

|

0

denotes its image under the

canonical projection

\set

→

∥

\set

∥

0

≡

𝖢𝖺𝗋𝖽

; we call

|

A

|

0

the

cardinality

of

A

.
By definition,

𝖢𝖺𝗋𝖽

is a set.
It also inherits the

structure

of a semiring from

\set

.

Definition 10.2.2

.

The

operation

of

cardinal addition

(

–

+

–

)

:

𝖢𝖺𝗋𝖽

→

𝖢𝖺𝗋𝖽

→

𝖢𝖺𝗋𝖽

is defined by

induction

on truncation:

|

A

|

0

+

|

B

|

0

:

≡

|

A

+

B

|

0

.

Proof.

Since

𝖢𝖺𝗋𝖽

→

𝖢𝖺𝗋𝖽

is a set, to define

(

α

+

–

)

:

𝖢𝖺𝗋𝖽

→

𝖢𝖺𝗋𝖽

for all

α

:

𝖢𝖺𝗋𝖽

, by induction it suffices to assume that

α

is

|

A

|

0

for some

A

:

\set

.
Now we want to define

(

|

A

|

0

+

–

)

:

𝖢𝖺𝗋𝖽

→

𝖢𝖺𝗋𝖽

, i.e. we want to define

|

A

|

0

+

β

:

𝖢𝖺𝗋𝖽

for all

β

:

𝖢𝖺𝗋𝖽

.
However, since

𝖢𝖺𝗋𝖽

is a set, by induction it suffices to assume that

β

is

|

B

|

0

for some

B

:

\set

.
But now we can define

|

A

|

0

+

|

B

|

0

to be

|

A

+

B

|

0

.
∎

Definition 10.2.3

.

Similarly, the operation of

cardinal multiplication

(

–

⋅

–

)

:

𝖢𝖺𝗋𝖽

→

𝖢𝖺𝗋𝖽

→

𝖢𝖺𝗋𝖽

is defined by induction on truncation:

|

A

|

0

⋅

|

B

|

0

:

≡

|

A

×

B

|

0

Lemma 10.2.4

.

𝖢𝖺𝗋𝖽

is a

commutative

semiring, i.e. for

α

,

β

,

γ

:

Card

we have the following.

(

α

+

β

)

+

γ

=

α

+

(

β

+

γ

)

α

+

0

=

α

α

+

β

=

β

+

α

(

α

⋅

β

)

⋅

γ

=

α

⋅

(

β

⋅

γ

)

α

⋅

1

=

α

α

⋅

β

=

β

⋅

α

α

⋅

(

β

+

γ

)

=

α

⋅

β

+

α

⋅

γ

where

0

:

≡

|

0

|

0

and

1

:

≡

|

1

|

0

.

Proof.

We prove the commutativity of multiplication,

α

⋅

β

=

β

⋅

α

; the others are exactly analogous.
Since

𝖢𝖺𝗋𝖽

is a set, the type

α

⋅

β

=

β

⋅

α

is a mere proposition, and in particular a set.
Thus, by induction it suffices to assume

α

and

β

are of the form

|

A

|

0

and

|

B

|

0

respectively, for some

A

,

B

:

\set

.
Now

|

A

|

0

⋅

|

B

|

0

≡

|

A

×

B

|

0

and

|

B

|

0

×

|

A

|

0

≡

|

B

×

A

|

0

, so it suffices to show

A

×

B

=

B

×

A

.
Finally, by univalence, it suffices to give an equivalence

A

×

B

≃

B

×

A

.
But this is easy: take

(

a

,

b

)

↦

(

b

,

a

)

and its obvious

inverse

.
∎

Definition 10.2.5

.

The operation of

cardinal exponentiation

is also defined by induction on truncation:

|

A

|

0

|

B

|

0

:

≡

|

B

→

A

|

0

.

Lemma 10.2.6

.

For

α

,

β

,

γ

:

Card

we have

α

0

=

1

1

α

=

1

α

1

=

α

α

β

+

γ

=

α

β

⋅

α

γ

α

β

⋅

γ

=

(

α

β

)

γ

(

α

⋅

β

)

γ

=

α

γ

⋅

β

γ

Proof.

Exactly like

\autoref

card:semiring.
∎

Definition 10.2.7

.

The

relation

of

cardinal inequality

(

–

≤

–

)

:

𝖢𝖺𝗋𝖽

→

𝖢𝖺𝗋𝖽

→

𝖯𝗋𝗈𝗉

is defined by induction on truncation:

|

A

|

0

≤

|

B

|

0

:

≡

∥

𝗂𝗇𝗃

(

A

,

B

)

∥

where

inj

⁢

(

A

,

B

)

is the type of

injections

from

A

to

B

.
In other words,

|

A

|

0

≤

|

B

|

0

means that there merely exists an injection from

A

to

B

.

Lemma 10.2.8

.

Cardinal inequality is a

preorder

, i.e. for

α

,

β

:

Card

we have

α

≤

α

(

α

≤

β

)

→

(

β

≤

γ

)

→

(

α

≤

γ

)

Proof.

As before, by induction on truncation.
For

instance

, since

(

α

≤

β

)

→

(

β

≤

γ

)

→

(

α

≤

γ

)

is a mere proposition, by induction on 0-truncation we may assume

α

,

β

, and

γ

are

|

A

|

0

,

|

B

|

0

, and

|

C

|

0

respectively.
Now since

|

A

|

0

≤

|

C

|

0

is a mere proposition, by induction on

(

-

1

)

-truncation we may assume given injections

f

:

A

→

B

and

g

:

B

→

C

.
But then

g

∘

f

is an injection from

A

to

C

, so

|

A

|

0

≤

|

C

|

0

holds.

Reflexivity

is even easier.
∎

We may likewise show that cardinal inequality is

compatible

with the semiring operations.

Lemma 10.2.9

.

Consider the following statements:

1.

There is an injection

A

→

B

.

2.

There is a

surjection

B

→

A

.

Then, assuming excluded middle:

•

Given

a

0

:

A

, we have

1

→

2

.

•

Therefore, if

A

is merely inhabited, we have

1

→

merely

2

.

•

Assuming the

axiom of choice

, we have

2

→

merely

1

.

Proof.

If

f

:

A

→

B

is an injection, define

g

:

B

→

A

at

b

:

B

as follows.
Since

f

is injective, the fiber of

f

at

b

is a mere proposition.
Therefore, by excluded middle, either there is an

a

:

A

with

f

⁢

(

a

)

=

b

, or not.
In the first case, define

g

(

b

)

:

≡

a

; otherwise set

g

(

b

)

:

≡

a

0

.
Then for any

a

:

A

, we have

a

=

g

⁢

(

f

⁢

(

a

)

)

, so

g

is surjective.

The second statement follows from this by induction on truncation.
For the third, if

g

:

B

→

A

is surjective, then by the axiom of choice, there merely exists a

function

f

:

A

→

B

with

g

⁢

(

f

⁢

(

a

)

)

=

a

for all

a

.
But then

f

must be injective.
∎

Theorem 10.2.10

(Schroeder–Bernstein).

Assuming excluded middle, for sets

A

and

B

we have

𝗂𝗇𝗃

(

A

,

B

)

→

𝗂𝗇𝗃

(

B

,

A

)

→

(

A

≅

B

)

Proof.

The usual “back-and-forth”

argument

applies without significant changes.
Note that it actually constructs an

isomorphism

A

≅

B

(assuming excluded middle so that we can decide whether a given

element

belongs to a cycle, an

infinite chain

, a chain beginning in

A

, or a chain beginning in

B

).
∎

Corollary 10.2.11

.

Assuming excluded middle, cardinal inequality is a

partial order

, i.e. for

α

,

β

:

Card

we have

(

α

≤

β

)

→

(

β

≤

α

)

→

(

α

=

β

)

.

Proof.

Since

α

=

β

is a mere proposition, by induction on truncation we may assume

α

and

β

are

|

A

|

0

and

|

B

|

0

, respectively, and that we have injections

f

:

A

→

B

and

g

:

B

→

A

.
But then the Schroeder–Bernstein theorem gives an isomorphism

A

≃

B

, hence an

equality

|

A

|

0

=

|

B

|

0

.
∎

Finally, we can reproduce Cantor’s

theorem

, showing that for every cardinal there is a greater one.

Theorem 10.2.12

(Cantor).

For

A

:

\set

, there is no surjection

A

→

(

A

→

2

)

.

Proof.

Suppose

f

:

A

→

(

A

→

𝟐

)

is any function, and define

g

:

A

→

𝟐

by

g

(

a

)

:

≡

¬

f

(

a

)

(

a

)

.
If

g

=

f

⁢

(

a

0

)

, then

g

⁢

(

a

0

)

=

f

⁢

(

a

0

)

⁢

(

a

0

)

but

g

⁢

(

a

0

)

=

¬

⁢

f

⁢

(

a

0

)

⁢

(

a

0

)

, a

contradiction

.
Thus,

f

is not surjective.
∎

Corollary 10.2.13

.

Assuming excluded middle, for any

α

:

Card

, there is a cardinal

β

such that

α

≤

β

and

α

≠

β

.

Proof.

Let

β

=

2

α

.
Now we want to show a mere proposition, so by induction we may assume

α

is

|

A

|

0

, so that

β

≡

|

A

→

𝟐

|

0

.
Using excluded middle, we have a function

f

:

A

→

(

A

→

𝟐

)

defined by

f

(

a

)

(

a

′

)

:

≡

{

1

𝟐

a

=

a

′

0

𝟐

a

≠

a

′

.

And if

f

⁢

(

a

)

=

f

⁢

(

a

′

)

, then

f

⁢

(

a

′

)

⁢

(

a

)

=

f

⁢

(

a

)

⁢

(

a

)

=

1

𝟐

, so

a

=

a

′

; hence

f

is injective.
Thus,

α

≡

|

A

|

0

≤

|

A

→

𝟐

|

0

≡

2

α

.

On the other hand, if

2

α

≤

α

, then we would have an injection

(

A

→

𝟐

)

→

A

.
By

\autoref

thm:injsurj, since we have

(

λ

⁢

x

⁢

. 0

𝟐

)

:

A

→

𝟐

and excluded middle, there would then be a surjection

A

→

(

A

→

𝟐

)

, contradicting Cantor’s theorem.
∎

Title

10.2 Cardinal numbers

\metatable

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