Tychonoff’s theorem implies AC

Tychonoff’s theorem implies AC

In this entry, we prove that

Tychonoff

’s

theorem

implies that

product of non-empty set of non-empty sets is non-empty

, which is

equivalent

to the

axiom of choice

(AC). This fact, together with the fact that AC implies Tychonoff’s theorem, shows that Tychonoff’s theorem is equivalent to AC (under ZF). The proof was first discovered by John Kelley in 1950, and is now an exercise in

axiomatic set theory

.

Proof.

Let

C

be a non-empty

collection

of non-empty sets. Let

Y

be the

generalized cartesian product

of all the

elements

in

C

. Our objective is show that

Y

is non-empty.

First, some notations: for each

A

∈

C

, set

X

A

:=

A

∪

{

A

}

,

D

:=

{

X

A

∣

A

∈

C

}

,

X

the generalized cartesian product of all the

X

A

’s, and

p

A

the projection from

X

onto

X

A

.

We break down the proof into several steps:

1.

Y

is

equipollent

to

Z

:=

⋂

{

p

A

-

1

⁢

(

A

)

∣

A

∈

C

}

.

An element of

X

is a

function

f

:

D

→

⋃

D

, such that

f

⁢

(

X

A

)

∈

X

A

for each

A

∈

C

. In other words, either

f

⁢

(

X

A

)

∈

A

, or

f

⁢

(

X

A

)

=

A

. An element of

Y

is a function

g

:

C

→

⋃

C

such that

g

⁢

(

A

)

∈

A

for each

A

∈

C

. Finally,

h

∈

p

A

-

1

⁢

(

A

)

iff

h

⁢

(

X

A

)

∈

A

.

Given

g

∈

Y

, define

g

*

∈

X

by

g

*

⁢

(

X

A

)

:=

g

⁢

(

A

)

∈

A

. Since

A

is arbitrary,

g

*

∈

Z

.

Conversely

, given

h

∈

Z

, define

h

′

∈

Y

by

h

′

⁢

(

A

)

:=

h

⁢

(

X

A

)

, which is

well-defined

, since

h

⁢

(

X

A

)

∈

A

. Now, it is easy to see that the function

ϕ

:

Y

→

Z

given by

ϕ

⁢

(

g

)

=

g

*

is a

bijection

, whose

inverse

ϕ

-

1

:

Z

→

Y

is given by

ϕ

-

1

⁢

(

h

)

=

h

′

. This shows that

Y

and

Z

are equipollent.

2.

Next, we topologize each

X

A

in such a way that

X

A

is

compact

.

Let

𝒯

A

be the coarsest

topology

containing the

cofinite topology

on

X

A

and the

singleton

{

A

}

. A typical

open set

of

X

A

is either the

empty set

, or has the form

S

∪

{

A

}

, where

S

is cofinite in

A

.

To show that

X

A

is compact under

𝒯

A

, let

𝒟

be an

open cover

for

X

A

. We want to show that there is a finite subset of

𝒟

covering

X

A

. If

X

A

∈

𝒟

, then we are done. Otherwise, pick a non-empty element

S

∪

{

A

}

in

𝒟

, so that

A

-

S

≠

∅

, and is finite. By

assumption

, each element in

A

-

S

belongs to some open set in

𝒟

. So to cover

A

-

S

, only a finite number of open sets in

𝒟

are needed. These open sets, together with

S

∪

{

A

}

, cover

X

A

. Hence

X

A

is compact.

3.

Finally, we prove that

Z

, and therefore

Y

, is non-empty.

Apply Tychonoff’s theorem,

X

is compact under the

product topology

. Furthermore,

π

A

is

continuous

for each

A

∈

C

. Since

{

A

}

is open in

X

A

, and

A

=

X

A

-

{

A

}

,

A

is closed in

X

A

, and thus so is

p

A

-

1

⁢

(

A

)

closed in

X

.

To show that

Z

is non-empty, we employ a

characterization

of

compact space

:

X

is compact iff every collection of

closed sets

in

X

having FIP has non-empty

intersection

(

http://planetmath.org/ASpaceIsCompactIfAndOnlyIfTheSpaceHasTheFiniteIntersectionProperty

). Let us look at the collection

𝒮

:=

{

p

A

-

1

⁢

(

A

)

∣

A

∈

C

}

. Given

A

1

,

…

,

A

n

∈

C

, pick an element

a

i

∈

A

i

, since

A

i

≠

∅

by assumption. Note that this is possible, since there are only a finite number of sets. Define

f

:

D

→

⋃

D

as follows:

f

⁢

(

X

A

)

:=

{

a

i

if

⁢

A

=

A

i

⁢

for some

⁢

i

=

1

,

…

,

n

,

A

otherwise.

Since

f

⁢

(

X

A

i

)

=

a

i

∈

A

i

,

f

∈

p

A

i

-

1

⁢

(

A

i

)

for each

i

=

1

,

…

,

n

. Therefore,

f

∈

p

A

1

-

1

⁢

(

A

1

)

∩

⋯

∩

p

A

n

-

1

⁢

(

A

n

)

.

Since

p

A

1

-

1

⁢

(

A

1

)

,

…

,

p

A

n

-

1

⁢

(

A

n

)

are arbitrarily picked from

𝒮

, the collection

𝒮

has

finite intersection property

, and since

X

is compact,

Z

=

⋂

S

must be non-empty.

This

completes

the proof.
∎

Remark

. In the proof, we see that the trick is to adjoin the set

{

A

}

to each set

A

∈

C

. Instead of

{

A

}

, we could have picked some arbitrary, but fixed singleton

{

B

}

, as long as

B

∉

A

for each

A

∈

C

, and the proof follows essentially the same way.

References

1

T. J. Jech,

The Axiom of Choice

. North-Holland Pub. Co., Amsterdam, 1973.

2

J. L. Kelley,

The Tychonoff’s

product

theorem implies the axiom of choice

. Fund. Math. 37, pp. 75-76, 1950.

Title

Tychonoff’s theorem implies AC

Canonical

name

TychonoffsTheoremImpliesAC

Date of creation

2013-03-22 18:45:38

Last modified on

2013-03-22 18:45:38

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

7

Author

CWoo (3771)

Entry type

Proof

Classification

msc 54D30

Classification

msc 03E25

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