Lindenbaum’s lemma

Lindenbaum’s lemma

In this entry, we prove the following assertion known as Lindenbaum’s lemma: every consistent set is a subset of a

maximally consistent

set. From this, we automatically conclude the

existence

of a maximally consistent set based on the fact that the

empty set

is

consistent

(

vacuously

).

Proposition 1

.

(Lindenbaum’s lemma) Every consistent set can be extended to a maximally consistent set.

We give two proofs of this, one uses Zorn’s lemma, and the other does not, and is based on the countability of the set of wff’s in the logic. Both proofs require the following:

Lemma 1

.

The union of a chain of consistent sets, ordered by

⊆

, is consistent.

Proof.

Let

𝒞

:=

{

Γ

i

∣

i

∈

I

}

be a chain of consistent sets ordered by

⊆

and

indexed by

some set

I

. We want to show that

Γ

:=

⋃

𝒞

is also consistent. Suppose not. Then

Γ

⊢

⟂

. Let

A

1

,

…

,

A

n

be a

deduction

of

⟂

(which is

A

n

) from

Γ

. Then for each

j

=

1

,

…

,

n

-

1

, there is some

Γ

i

⁢

(

j

)

∈

𝒞

such that

Γ

i

⁢

(

j

)

⊢

A

i

. Since

𝒞

is a chain, take the largest of the

Γ

i

⁢

(

j

)

’s, say

Γ

k

, so that

Γ

k

⊢

A

i

for all

i

=

1

,

…

,

n

-

1

. This implies that

Γ

k

⊢

A

n

, or

Γ

k

⊢

⟂

, contradicting the assumption that

Γ

k

is consistent. As a result,

Γ

is consistent.
∎

First Proof

. Suppose

Δ

is consistent. Let

P

be the

partially ordered set

of all consistent

supersets

of

Δ

, ordered by

inclusion

⊆

. If

𝒞

is a chain of

elements

in

P

, then

⋃

𝒞

is consistent by the lemma above, so

⋃

𝒞

∈

P

as each element of

𝒞

is a superset of

Δ

. By Zorn’s lemma,

P

has a

maximal element

, call it

Γ

. To see that

Γ

is maximally consistent, suppose

Γ

is not maximal. Then there is a wff

A

such that

Γ

⊬

A

and

Γ

⊬

¬

⁢

A

, the first of which implies that

A

∉

Γ

, and the second of which implies that

Γ

∪

{

A

}

is consistent, and therefore in

P

. The two imply that

Γ

∪

{

A

}

is a consistent

proper superset

of

Γ

, contradicting the maximality of

Γ

in

P

. Therefore,

Γ

is maximally consistent.

□

Second Proof

. Let

A

1

,

…

,

A

n

,

…

be an enumeration of all wff’s of the logic in question (this can be achieved if the set of

propositional variables

can be enumerated). Let

Δ

be a consistent set of wff’s. Define sets

Γ

1

,

Γ

2

,

…

,

Γ

of wff’s inductively as follows:

Γ

1

:=

Δ

Γ

n

+

1

:=

{

Γ

n

∪

{

A

n

}

if

⁢

Γ

n

⊢

A

n

Γ

n

∪

{

¬

⁢

A

n

}

otherwise

Γ

:=

⋃

i

=

1

∞

Γ

i

.

First, notice that each

Γ

i

is consistent. This is done by

induction

on

i

. By assumption, the case is true when

i

=

1

. Now, suppose

Γ

n

is consistent. Then its

deductive closure

Ded

(

Γ

n

)

is also consistent. If

Γ

n

⊢

A

n

, then clearly

Γ

n

∪

{

A

n

}

is consistent since it is a subset of Ded

(

Γ

)

. Otherwise,

Γ

n

⊬

A

n

, or

Γ

n

⊬

¬

⁢

¬

⁢

A

n

by the

substitution theorem

, and therefore

Γ

n

∪

{

¬

⁢

A

n

}

by one of the properties of consistency (see here (

http://planetmath.org/PropertiesOfConsistency

)). In either case,

Γ

n

+

1

is consistent.

Next,

Γ

is maximally consistent.

Γ

is consistent because, by the lemma above, it is the union of a chain of consistent sets. To see that

Γ

is maximal, pick any wff

A

. Then

A

is some

A

n

in the enumerated list of all wff’s. Therefore, either

A

∈

Γ

n

+

1

or

¬

⁢

A

∈

Γ

n

+

1

. Since

Γ

n

+

1

⊆

Γ

, we have

A

∈

Γ

or

¬

⁢

A

∈

Γ

, which implies that

Γ

is maximal (see here (

http://planetmath.org/MaximallyConsistent

)).

□

.

Given a logic

L

, let

W

L

be the set of all maximally

L

-consistent sets. By Lindebaum’s lemma,

W

L

≠

∅

. We record two useful

corollaries

:

•

For any consistent set

Δ

, Ded

(

Δ

)

=

⋂

{

Γ

∈

W

L

∣

Δ

⊆

Γ

}

.

•

L

=

⋂

W

L

.

Proof.

The second statement is a corollary of the first, for

L

=

Ded

⁢

(

∅

)

. To see the first, let

𝒟

:=

{

Γ

∈

W

L

∣

Δ

⊆

Γ

}

. Then

𝒟

≠

∅

by Lindebaum’s lemma. Also, for any

Γ

∈

𝒟

, Ded

(

Δ

)

⊆

Γ

since

Γ

is

deductively closed

. On the other hand if

A

∉

Ded

⁢

(

Δ

)

, then

Δ

⊬

A

, so

Δ

∪

{

¬

⁢

A

}

is consistent, and therefore is contained in a maximally consistent set

Γ

′

∈

𝒟

by Lindenbaum’s lemma. Since

¬

⁢

A

∈

Γ

′

,

A

∉

Γ

′

, so that

A

∉

⋂

𝒟

.
∎

Title

Lindenbaum’s lemma

Canonical

name

LindenbaumsLemma

Date of creation

2013-03-22 19:35:16

Last modified on

2013-03-22 19:35:16

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

10

Author

CWoo (3771)

Entry type

Theorem

Classification

msc 03B45

Classification

msc 03B10

Classification

msc 03B05

Classification

msc 03B99

Related topic

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