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Kleene fixed-point theorem

Theorem in order theory

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Record originEnglish Wikipedia
Text licenseCC BY-SA 4.0
Source revisionMay 10, 2025
Entity authorityQ3527263 ↗
Source-derived summary

In the mathematical areas of order and lattice theory, the Kleene fixed-point theorem, named after American mathematician Stephen Cole Kleene, states the following:

Kleene Fixed-Point Theorem. Suppose

(

L

,

⊑

)

{\displaystyle (L,\sqsubseteq )}

is a directed-complete partial order (dcpo) with a least element, and let

f

:

L

→

L

{\displaystyle f:L\to L}

be a Scott-continuous (and therefore monotone) function. Then

f

{\displaystyle f}

has a least fixed point, which is the supremum of the ascending Kleene chain of

f

.

{\displaystyle f.}

The ascending Kleene chain of f is the chain

⊥

⊑

f

(

⊥

)

⊑

f

(

f

(

⊥

)

)

⊑

⋯

⊑

f

n

(

⊥

)

⊑

⋯

{\displaystyle \bot \sqsubseteq f(\bot )\sqsubseteq f(f(\bot ))\sqsubseteq \cdots \sqsubseteq f^{n}(\bot )\sqsubseteq \cdots }

obtained by iterating f on the least element ⊥ of L. Expressed in a formula, the theorem states that

lfp

(

f

)

=

sup

(

{

f

n

(

⊥

)

∣

n

∈

N

}

)

{\displaystyle {\textrm {lfp}}(f)=\sup \left(\left\{f^{n}(\bot )\mid n\in \mathbb {N} \right\}\right)}

where

lfp

{\displaystyle {\textrm {lfp}}}

denotes the least fixed point.

Although Tarski's fixed point theorem

does not consider how fixed points can be computed by iterating f from some seed (also, it pertains to monotone functions on complete lattices), this result is often attributed to Alfred Tarski who proves it for additive functions. Moreover, the Kleene fixed-point theorem can be extended to monotone functions using transfinite iterations.

Proof

Source:

We first have to show that the ascending Kleene chain of

f

{\displaystyle f}

exists in

L

{\displaystyle L}

. To show that, we prove the following:

Lemma. If

L

{\displaystyle L}

is a dcpo with a least element, and

f

:

L

→

L

{\displaystyle f:L\to L}

is Scott-continuous, then

f

n

(

⊥

)

⊑

f

n

+

1

(

⊥

)

,

n

∈

N

0

{\displaystyle f^{n}(\bot )\sqsubseteq f^{n+1}(\bot ),n\in \mathbb {N} _{0}}

Proof. We use induction:

Assume n = 0.

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