10.1.5 The axiom of choice implies excluded middle

10.1.5 The axiom of choice implies excluded middle

We begin with the following lemma.

Lemma 10.1.1

.

If

A

is a mere proposition then its suspension

Σ

⁢

(

A

)

is a set,
and

A

is

equivalent

to

N

=

Σ

⁢

(

A

)

S

.

Proof.

To show that

Σ

⁢

(

A

)

is a set, we define a
family

P

:

Σ

⁢

(

A

)

→

Σ

⁢

(

A

)

→

𝒰

with the

property

that

P

⁢

(

x

,

y

)

is a mere proposition for each

x

,

y

:

Σ

⁢

(

A

)

,
and which is equivalent to its identity type

𝖨𝖽

Σ

⁢

(

A

)

.
We make the following

definitions

:

P

⁢

(

𝖭

,

𝖭

)

:

≡

𝟏

P

⁢

(

𝖲

,

𝖭

)

:

≡

A

P

⁢

(

𝖭

,

𝖲

)

:

≡

A

P

⁢

(

𝖲

,

𝖲

)

:

≡

𝟏

.

We have to check that the definition

preserves

paths.
Given any

a

:

A

, there is a

meridian

𝗆𝖾𝗋𝗂𝖽

⁢

(

a

)

:

𝖭

=

𝖲

,
so we should also have

P

⁢

(

𝖭

,

𝖭

)

=

P

⁢

(

𝖭

,

𝖲

)

=

P

⁢

(

𝖲

,

𝖭

)

=

P

⁢

(

𝖲

,

𝖲

)

.

But since

A

is inhabited by

a

, it is equivalent to

𝟏

, so we have

P

⁢

(

𝖭

,

𝖭

)

≃

P

⁢

(

𝖭

,

𝖲

)

≃

P

⁢

(

𝖲

,

𝖭

)

≃

P

⁢

(

𝖲

,

𝖲

)

.

The univalence axiom turns these into the desired equalities. Also,

P

⁢

(

x

,

y

)

is a mere
proposition for all

x

,

y

:

Σ

⁢

(

A

)

, which is proved by

induction

on

x

and

y

, and
using the fact that being a mere proposition is a mere proposition.

Note that

P

is a

reflexive relation

.
Therefore we may apply

\autoref

thm:h-set-refrel-in-paths-sets, so it suffices to

construct

τ

:

∏

(

x

,

y

:

Σ

(

A

)

)

P

(

x

,

y

)

→

(

x

=

y

)

. We do this by a double induction.
When

x

is

𝖭

, we define

τ

⁢

(

𝖭

)

by

τ

(

𝖭

,

𝖭

,

u

)

:

≡

𝗋𝖾𝖿𝗅

𝖭

and

τ

(

𝖭

,

𝖲

,

a

)

:

≡

𝗆𝖾𝗋𝗂𝖽

(

a

)

.

If

A

is inhabited by

a

then

𝗆𝖾𝗋𝗂𝖽

⁢

(

a

)

:

𝖭

=

𝖲

so we also need

𝗆𝖾𝗋𝗂𝖽

(

a

)

*

(

τ

(

𝖭

,

𝖭

)

)

=

τ

(

𝖭

,

𝖲

)

.

This we get by function extensionality using the fact that, for all

x

:

A

,

𝗆𝖾𝗋𝗂𝖽

(

a

)

*

(

τ

(

𝖭

,

𝖭

,

x

)

)

=

τ

(

𝖭

,

𝖭

,

x

)

\centerdot

𝗆𝖾𝗋𝗂𝖽

⁢

(

a

)

-

1

≡

𝗋𝖾𝖿𝗅

𝖭

⁢

\centerdot

⁢

𝗆𝖾𝗋𝗂𝖽

⁢

(

a

)

=

𝗆𝖾𝗋𝗂𝖽

⁢

(

a

)

=

𝗆𝖾𝗋𝗂𝖽

⁢

(

x

)

≡

τ

⁢

(

𝖭

,

𝖲

,

x

)

.

In a

symmetric

fashion we may define

τ

⁢

(

𝖲

)

by

τ

(

𝖲

,

𝖭

,

a

)

:

≡

𝗆𝖾𝗋𝗂𝖽

⁢

(

a

)

-

1

and

τ

(

𝖲

,

𝖲

,

u

)

:

≡

𝗋𝖾𝖿𝗅

𝖲

.

To

complete

the construction of

τ

, we need to check

𝗆𝖾𝗋𝗂𝖽

(

a

)

*

(

τ

(

𝖭

)

)

=

τ

(

𝖲

)

,
given any

a

:

A

. The verification proceeds much along the same lines by induction on the
second

argument

of

τ

.

Thus, by

\autoref

thm:h-set-refrel-in-paths-sets we have that

Σ

⁢

(

A

)

is a set and that

P

(

x

,

y

)

≃

(

x

=

y

)

for all

x

,

y

:

Σ

⁢

(

A

)

.
Taking

x

:

≡

𝖭

and

y

:

≡

𝖲

yields

A

≃

(

𝖭

=

Σ

⁢

(

A

)

𝖲

)

as desired.
∎

Theorem 10.1.2

(Diaconescu).

The

axiom of choice

implies the

law of excluded middle

.

Proof.

We use the equivalent form of choice given in

\autoref

thm:ac-epis-split.
Consider a mere proposition

A

.
The

function

f

:

𝟐

→

Σ

⁢

(

A

)

defined by

f

(

0

𝟐

)

:

≡

𝖭

and

f

(

1

𝟐

)

:

≡

𝖲

is

surjective

.
Indeed, we have

(

0

𝟐

,

𝗋𝖾𝖿𝗅

𝖭

)

:

𝖿𝗂𝖻

f

(

𝖭

)

and

(

1

𝟐

,

𝗋𝖾𝖿𝗅

𝖲

)

:

𝖿𝗂𝖻

f

(

𝖲

)

.
Since

∥

𝖿𝗂𝖻

f

⁢

(

x

)

∥

is a mere proposition, by induction the claimed surjectivity follows.

By

\autoref

prop:trunc_of_prop_is_set the suspension

Σ

⁢

(

A

)

is a set, so by the axiom of choice there merely exists a

section

g

:

Σ

⁢

(

A

)

→

𝟐

of

f

.
As equality on

𝟐

is

decidable

we get

(

g

(

f

(

0

𝟐

)

)

=

g

(

f

(

1

𝟐

)

)

)

+

¬

(

g

(

f

(

0

𝟐

)

)

=

g

(

f

(

1

𝟐

)

)

)

,

and, since

g

is a section of

f

, hence

injective

,

(

f

(

0

𝟐

)

=

f

(

1

𝟐

)

)

+

¬

(

f

(

0

𝟐

)

=

f

(

1

𝟐

)

)

.

Finally, since

(

f

(

0

𝟐

)

=

f

(

1

𝟐

)

)

=

(

𝖭

=

𝖲

)

=

A

by

\autoref

prop:trunc_of_prop_is_set, we have

A

+

¬

⁢

A

.
∎

Theorem 10.1.3

.

If the axiom of choice holds then the

category

S

⁢

e

⁢

t

is a

well-pointed

boolean

elementary topos

with choice.

Proof.

Since

𝖠𝖢

implies

𝖫𝖤𝖬

, we have a boolean elementary topos with choice by

\autoref

thm:settopos and the remark following it. We leave the proof of well-pointedness as
an exercise for the reader (

\autoref

ex:well-pointed).
∎

Remark 10.1.4

.

The conditions on a category mentioned in the

theorem

are known as Lawvere’s
axioms for the

Elementary Theory of the

Category of Sets

[

lawvere:etcs-long

]

.

Title

10.1.5 The axiom of choice implies excluded middle

\metatable

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