Sikorski’s extension theorem

Sikorski’s extension theorem

Theorem 1

(Sikorski’s ).

Let

A

be a

Boolean subalgebra

of a

Boolean algebra

B

, and

f

:

A

→

C

a

Boolean algebra homomorphism

from

A

to a

complete Boolean algebra

C

. Then

f

can be extended to a Boolean algebra homomorphism

g

:

B

→

C

.

Remark

. In the

category

of Boolean algebras and Boolean algebra homomorphisms, this

theorem

says that every complete Boolean algebra is an

injective object

.

Proof.

We prove this using Zorn’s lemma. Let

M

be the set of all pairs

(

h

,

D

)

such that

D

is a

subalgebra

of

B

containing

A

, and

h

:

D

→

C

is an algebra homomorphism extending

f

. Note that

M

is not empty because

(

f

,

A

)

∈

M

. Also, if we define

(

h

1

,

D

1

)

≤

(

h

2

,

D

2

)

by requiring that

D

1

⊆

D

2

and that

h

2

extending

h

1

, then

(

M

,

≤

)

becomes a poset. Notice that for every chain

𝒞

in

M

,

(

⋃

{

h

∣

(

h

,

D

)

∈

𝒞

}

,

⋃

{

D

∣

(

h

,

D

)

∈

𝒞

}

)

is an

upper bound

of

𝒞

(in fact, the least upper bound). So

M

has a

maximal element

, say

(

g

,

E

)

, by Zorn’s lemma. We want to show that

E

=

B

.

If

E

≠

B

, pick

a

∈

B

-

E

. Let

r

be the join of all elements of the form

g

⁢

(

x

)

where

x

∈

E

and

x

≤

a

, and

t

the meet of all elements of the form

g

⁢

(

y

)

where

y

∈

E

and

a

≤

y

.

r

and

t

exist because

C

is

complete

. Since

g

preserves order, it is evident that

r

≤

t

. Pick an

element

s

∈

C

such that

r

≤

s

≤

t

.

Let

F

=

⟨

E

,

a

⟩

. Every element in

F

has the form

(

e

1

∧

a

)

∨

(

e

2

∧

a

′

)

, with

e

1

,

e

2

∈

E

. Define

h

:

F

→

C

by setting

h

⁢

(

b

)

=

(

g

⁢

(

e

1

)

∧

s

)

∨

(

g

⁢

(

e

2

)

∧

s

′

)

, where

b

=

(

e

1

∧

a

)

∨

(

e

2

∧

a

′

)

. We now want to show that

h

is a Boolean algebra homomorphism extending

g

. There are three steps to showing this:

1.

h

is a

function

. Suppose

(

e

1

∧

a

)

∨

(

e

2

∧

a

′

)

=

(

e

3

∧

a

)

∨

(

e

4

∧

a

′

)

. Then, by the last remark of this entry (

http://planetmath.org/BooleanSubalgebra

),

e

2

⁢

Δ

⁢

e

4

≤

a

≤

e

1

↔

e

3

, so that

g

(

e

2

)

Δ

g

(

e

4

)

=

g

(

e

2

Δ

e

4

)

≤

s

≤

g

(

e

1

↔

e

3

)

=

g

(

e

1

)

↔

g

(

e

3

)

, which in turn implies that

(

g

⁢

(

e

1

)

∧

s

)

∨

(

g

⁢

(

e

2

)

∧

s

′

)

=

(

g

⁢

(

e

3

)

∧

s

)

∨

(

g

⁢

(

e

4

)

∧

s

′

)

. Hence

h

is well-defined.

2.

h

is a Boolean homomorphism. All we need to show is that

h

respects

∨

and

′

. Let

x

=

(

e

1

∧

a

)

∨

(

e

2

∧

a

′

)

and

y

=

(

e

3

∧

a

)

∨

(

e

4

∧

a

′

)

. Then

x

∨

y

=

(

e

5

∧

a

)

∨

(

e

6

∧

a

′

)

, where

e

5

=

e

1

∨

e

3

and

e

6

=

e

2

∨

e

4

. So

h

⁢

(

x

∨

y

)

=

(

g

⁢

(

e

5

)

∧

s

)

∨

(

g

⁢

(

e

6

)

∧

s

′

)

=

(

(

g

⁢

(

e

1

)

∨

g

⁢

(

e

3

)

)

∧

s

)

∨

(

(

g

⁢

(

e

2

)

∨

g

⁢

(

e

4

)

)

∧

s

′

)

=

(

g

⁢

(

e

1

)

∧

s

)

∨

(

g

⁢

(

e

2

)

∧

s

′

)

∨

(

g

⁢

(

e

3

)

∧

s

)

∨

(

g

⁢

(

e

4

)

∧

s

′

)

=

h

⁢

(

x

)

∨

h

⁢

(

y

)

,

so

h

respects

∨

. In

addition

,

h

respects

′

, as

x

′

=

(

e

2

′

∧

a

)

∨

(

e

1

∧

a

′

)

, so that

h

⁢

(

x

′

)

=

h

⁢

(

(

e

2

′

∧

a

)

∨

(

e

1

∧

a

′

)

)

=

(

g

⁢

(

e

2

′

)

∧

s

)

∨

(

g

⁢

(

e

1

)

∧

s

′

)

=

(

g

⁢

(

e

2

)

′

∧

s

)

∨

(

g

⁢

(

e

1

)

∧

s

′

)

=

(

(

g

⁢

(

e

1

)

∧

s

)

∨

(

g

⁢

(

e

2

)

∧

s

′

)

)

′

=

h

⁢

(

x

)

′

.

3.

h

extends

g

. If

x

∈

E

, write

x

=

(

x

∧

a

)

∨

(

x

∧

a

′

)

. Then

h

⁢

(

x

)

=

(

g

⁢

(

x

)

∧

s

)

∨

(

g

⁢

(

x

)

∧

s

′

)

=

g

⁢

(

x

)

.

This implies that

(

g

,

E

)

<

(

h

,

F

)

, and with this, we have a

contradiction

that

(

g

,

E

)

is

maximal

. This completes the proof.
∎

One of the

consequences

of this theorem is the following variant of the

Boolean prime ideal theorem

:

Corollary 1

.

Every

Boolean ideal

of a Boolean algebra is

contained

in a

maximal ideal

.

Proof.

Let

I

be an ideal of a Boolean algebra

A

. Let

B

=

⟨

I

⟩

, the Boolean subalgebra generated by

I

. The function

f

:

B

→

{

0

,

1

}

given by

f

⁢

(

a

)

=

0

iff

a

∈

I

is a Boolean homomorphism. First, notice that

f

⁢

(

a

)

=

0

iff

a

∈

I

iff

a

′

∉

I

iff

f

⁢

(

a

′

)

=

1

. Next, if at least one of

a

,

b

is in

I

,

a

∧

b

∈

I

, so that

f

⁢

(

a

∧

b

)

=

0

=

f

⁢

(

a

)

∧

f

⁢

(

b

)

. If neither are in

I

, then

a

′

,

b

′

∈

I

, so

(

a

∧

b

)

′

=

a

′

∨

b

′

∈

I

, or

a

∧

b

∉

I

. This means that

f

⁢

(

a

∧

b

)

=

1

=

f

⁢

(

a

)

∧

f

⁢

(

b

)

.

Now, by Sikorski’s

extension theorem

,

f

can be extended to a

homomorphism

g

:

A

→

{

0

,

1

}

. The kernel of

g

clearly contains

I

, and is in addition maximal (either

a

or

a

′

is in the kernel of

g

).
∎

Remarks

.

•

As the proof of the theorem shows, ZF+AC (the

axiom of choice

) implies Sikorski’s extension theorem (SET). It is still an open question whether the ZF+SET implies AC.

•

Next, comparing with the Boolean prime ideal theorem (BPI), the proof of the corollary above shows that ZF+SET implies BPI. However, it was proven by John Bell in 1983 that SET is

independent

from ZF+BPI: there is a model satisfying all axioms of ZF, as well as BPI (considered as an axiom, not as a consequence of AC), such that SET fails.

References

1

R. Sikorski,

Boolean Algebras

, 2nd Edition, Springer-Verlag, New York (1964).

2

J. L. Bell,

http://plato.stanford.edu/entries/axiom-choice/

The Axiom of Choice

, Stanford Encyclopedia of Philosophy (2008).

Title

Sikorski’s extension theorem

Canonical name

SikorskisExtensionTheorem

Date of creation

2013-03-22 18:01:31

Last modified on

2013-03-22 18:01:31

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

21

Author

CWoo (3771)

Entry type

Theorem

Classification

msc 06E10

Synonym

Sikorski extension theorem

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