Tarski-Knaster theorem

Tarski-Knaster theorem

The aim of this article is to prove

Theorem 1

(LATTICE-THEORETICAL FIXPOINT THEOREM).

Let

(i)

𝔄

=

(

A

,

≤

)

be a

complete lattice

,

(ii)

f

be an increasing function on

A

to

A

,

(iii)

P

the set of all fixpoints of

f

.

Then the set

P

is not empty and the system

(

P

,

≤

)

is a complete
lattice; in particular we have

∪

P

=

∪

E

x

[

f

(

x

)

≥

x

]

∈

P

and

∩

P

=

∩

E

x

[

f

(

x

)

≤

x

]

∈

P

.

This article follows, with clarification, Tarski’s original
1955 article

[

6

]

, which may not be available to some readers.

1

Definitions

The symbol

∅

denotes the

empty set

. The symbol

∀

is
shorthand for ”for all”.

Definition 1

.

A

poset

(

partially ordered set

)

A

=

(

A

,

≤

)

consists of
a

nonempty set

A

and a

relation

≤

:

A

×

A

⟶

{

𝐹𝑎𝑙𝑠𝑒

,

𝑇𝑟𝑢𝑒

}

that is

reflexive

(

a

≤

a

∀

a

∈

A

), antisymmetric (

∀

a

,

b

∈

A

,

if

a

≤

b

and

b

≤

a

,

then

a

=

b

). and

transitive

(

∀

a

,

b

,

c

∈

A

,

if

a

≤

b

and

b

≤

c

,

then

a

≤

c

).

Definition 2

.

If

A

=

(

A

,

≤

)

is a poset and

B

is a nonempty subset of

A

,

then

sup

⁡

(

B

)

is an element of

A

,

if it exists, with these

properties

:

1.

∀

b

∈

B

,

b

≤

sup

⁡

(

B

)

, (in shorthand,

B

≤

sup

⁡

(

B

)

).

2.

If

a

∈

A

and

B

≤

a

,

then

sup

⁡

(

B

)

≤

a

.

Similarly

inf

⁡

(

B

)

is an element of

A

,

if it exists, with these properties:

(3)

inf

⁡

(

B

)

≤

B

,

(4)

If

a

∈

A

and

a

≤

B

,

then

a

≤

inf

⁡

(

B

)

.

Theorem 2

.

If

sup

⁡

(

B

)

exists, then it is unique; if

inf

⁡

(

B

)

exists, it is unique.

Definition 3

.

Let

A

=

(

A

,

≤

)

be a poset. Then

A

is a

lattice

if

∀

a

,

b

∈

A

,

the set

{

a

,

b

}

has a

sup

and an

inf

.

We write

a

∪

b

=

sup

⁡

(

{

a

,

b

}

)

𝑎𝑛𝑑

a

∩

b

=

inf

⁡

(

{

a

,

b

}

)

.

As an exercise, prove the following:

Theorem 3

.

Let

A

=

(

A

,

≤

)

be a lattice. Then

1.

a

∪

b

=

b

∪

a

,

a

∩

b

=

b

∩

a

2.

(

a

∪

b

)

∪

c

=

a

∪

(

b

∪

c

)

,

(

a

∩

b

)

∩

c

=

a

∩

(

b

∩

c

)

3.

a

∪

(

b

∩

c

)

≤

(

a

∪

b

)

∩

(

a

∪

c

)

.

(

a

∩

b

)

∪

(

a

∩

c

)

≤

a

∩

(

b

∪

c

)

.

Definition 4

.

A poset

A

=

(

A

,

≤

)

is a

complete lattice

if for each
nonempty

B

⊆

A

,

both

sup

⁡

(

B

)

and

inf

⁡

(

B

)

exist.

In particular, this is the case when

B

=

{

a

,

b

}

,

so a complete lattice is a lattice.

2

The Tarski-Knaster Theorem

Statement of the

theorem

, as it is in

[

6

]

:

Theorem 4

(LATTICE-THEORETICAL FIXPOINT THEOREM).

Let

(i)

𝔄

=

(

A

,

≤

)

be a complete lattice,

(ii)

f

be an increasing function on

A

to

A

,

(iii)

P

the set of all fixpoints of

f

.

Then the set

P

is not empty and the system

(

P

,

≤

)

is a complete
lattice; in particular we have

∪

P

=

∪

E

x

[

f

(

x

)

≥

x

]

∈

P

and

∩

P

=

∩

E

x

[

f

(

x

)

≤

x

]

∈

P

.

The symbols

∪

P

and

∩

P

are what I am calling

sup

⁡

(

P

)

and

inf

⁡

(

C

)

.

Now let me restate the theorem:

Theorem 5

.

Let

A

=

(

A

,

≤

)

be a complete lattice. Let

f

be an increasing
function on

A

to

A

,

that is, whenever

a

≤

b

,

then

f

⁢

(

a

)

≤

f

⁢

(

b

)

.

Let

P

be the set of all

fixed points

of

f

:

P

=

{

a

∈

A

|

f

⁢

(

a

)

=

a

}

.

Then

1.

P

≠

∅

.

2.

The system

𝔓

=

(

P

,

≤

)

is a complete lattice.

3.

Set

B

=

{

b

∈

A

|

b

≤

f

⁢

(

b

)

}

. Then

B

≠

∅

and

sup

⁡

(

B

)

=

sup

⁡

(

P

)

∈

P

.

4.

Set

C

=

{

c

∈

A

|

f

⁢

(

c

)

≤

c

}

. Then

C

≠

∅

and

inf

⁡

(

C

)

=

inf

⁡

(

P

)

∈

P

.

Lemma 1

.

Let

A

=

(

A

,

≤

)

be a complete lattice and

∅

≠

A

0

⊆

A

.

Let

D

=

{

d

∈

A

|

d

≤

A

0

}

,

E

=

{

e

∈

A

|

A

0

≤

e

}

.

(

d

≤

A

o

means

d

≤

a

0

∀

a

0

∈

A

0

.

) Then both

B

and

C

are

complete

lattices, with

≤

inherited from that of

A

.

Proof (of Lemma).

We’ll do the case of

D

;

that for

E

is

similar

(dual).
First,

D

≠

∅

because

inf

⁡

(

A

)

∈

D

.

Suppose

∅

≠

G

⊆

D

.

Obviously

inf

⁡

(

G

)

≤

g

≤

A

0

for any

g

∈

G

,

so

inf

⁡

(

G

)

∈

D

.

As

g

≤

A

0

for all

g

∈

G

,

we
have

sup

⁡

(

G

)

≤

A

0

by (2) in the

definition

of

sup

.

So

sup

⁡

(

G

)

∈

D

.

As

G

was any nonempty subset of

D

,

these statements say that

D

is a
complete lattice.
∎

Proof (of Theorem).

B

≠

∅

because

inf

⁡

(

A

)

∈

B

.

Hence

b

1

=

sup

⁡

(

B

)

exists. If

b

∈

B

,

then

b

≤

b

1

,

which implies

f

⁢

(

b

)

≤

f

⁢

(

b

1

)

.

because

f

is increasing., so

b

≤

f

⁢

(

b

)

≤

f

⁢

(

b

1

)

.

This is true

∀

b

∈

B

.

Hence

b

1

=

sup

⁡

(

B

)

≤

f

⁢

(

b

1

)

.

Therefore

b

1

∈

B

.

Because

f

is increasing, this implies

f

⁢

(

b

1

)

≤

f

⁢

(

f

⁢

(

b

1

)

)

,

from which

f

⁢

(

b

1

)

∈

B

.

Hence

f

⁢

(

b

1

)

≤

sup

⁡

(

B

)

=

b

1

.

From

b

1

≤

f

⁢

(

b

1

)

≤

b

1

we have

f

⁢

(

b

1

)

=

b

1

,

that is,

b

1

∈

P

.

This settles (1):

P

≠

∅

,

and incidentally proves

sup

⁡

(

B

)

∈

P

.

From this,

sup

⁡

(

B

)

≤

sup

⁡

(

P

)

.

But obviously

P

⊆

B

,

so

sup

⁡

(

P

)

≤

sup

⁡

(

B

)

.

Therefore

sup

⁡

(

P

)

=

sup

⁡

(

B

)

,

completing the proof of
(3). Statement (4) is proved similarly (or dually).

Next we prove (2). Let

∅

≠

P

0

⊆

P

.

Set

Q

=

{

q

∈

A

|

q

≤

P

0

}

.

By the Lemma,

Q

is a complete lattice. If

q

∈

Q

,

then

∀

p

0

∈

P

0

,

we have

q

≤

p

0

,

hence

f

⁢

(

q

)

≤

f

⁢

(

p

0

)

=

p

0

,

that is,

f

⁢

(

q

)

∈

Q

.

In other words,

f

:

Q

⟶

Q

.

By what we have already proved for

A

,

applied to

Q

,

we have

sup

⁡

(

Q

)

∈

Q

,

and

sup

⁡

(

Q

)

is a fixed point of

f

.

That is,

sup

⁡

(

Q

)

∈

P

.

But
obviously

sup

⁡

(

Q

)

=

inf

⁡

(

P

0

)

,

so

inf

⁡

(

P

0

)

∈

P

Similarly

sup

⁡

(

P

0

)

∈

P

,

completing the proof of (2):

P

is a complete lattice.
∎

This theorem was proved by A. Tarski

[

6

]

. A special case of
this theorem (for lattices of sets) appeared in a paper of B. Knaster

[

3

]

. Kind of

converse

of this theorem was proved by Anne C.
Davis

[

1

]

: If every order-preserving function

f

:

L

→

L

has a fixed point, then

L

is a complete lattice.

References

1

Anne C. Davis,

A

characterization

of complete lattices.

, Pac. J. Math.

5

(1955), 311–319.

2

Thomas Forster,

Logic,

induction

and sets

, Cambridge University Press, Cambridge, 2003.

3

B. Knaster,

Un théorème sur les fonctions d’ensembles.

, Annales Soc. Polonaise

6

(1928), 133–134.

4

M. Kolibiar, A. Legéň, T. Šalát, and Š. Znám,

Algebra

a príbuzné disciplíny

, Alfa, Bratislava, 1992 (Slovak).

5

J. B. Nation:

http://bigcheese.math.sc.edu/ mcnulty/alglatvar/

Notes on

lattice theory

6

Alfred Tarski,

A lattice-theoretical fixpoint theorem and its applications.

, Pac. J. Math.

5

(1955), 285–309.

7

Wikipedia’s entry on

http://en.wikipedia.org/wiki/Knaster-Tarski_theorem

Knaster-Tarski theorem

Title

Tarski-Knaster theorem

Canonical

name

TarskiKnasterTheorem

Date of creation

2013-03-22 15:30:19

Last modified on

2013-03-22 15:30:19

Owner

kompik (10588)

Last modified by

kompik (10588)

Numerical id

18

Author

kompik (10588)

Entry type

Theorem

Classification

msc 06B99

Synonym

Tarski theorem

Synonym

Knaster-Tarski theorem

Related topic

lattice

Related topic

completelattice

Related topic

fixedpoint

Related topic

ProofOfSchroederBernsteinTheoremUsingTarskiKnasterTheorem

Related topic

Lattice

Related topic

CompleteLattice

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