Kleene fixed-point theorem

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

Let (L, ⊑) be a CPO (complete partial order), and let f : L → L be a Scott-continuous (and therefore monotone) function. Then f has a least fixed point, which is the supremum of the ascending Kleene chain of f.

The ascending Kleene chain of f is the chain

\bot \;\sqsubseteq \;f(\bot )\;\sqsubseteq \;f\left(f(\bot )\right)\;\sqsubseteq \;\dots \;\sqsubseteq \;f^{n}(\bot )\;\sqsubseteq \;\dots

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

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

where ${\textrm {lfp}}$ denotes the least fixed point.

This result is often attributed to Alfred Tarski, but 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).

Proof

We first have to show that the ascending Kleene chain of f exists in L. To show that, we prove the following lemma:

Lemma 1:If L is CPO, and f : L → L is a Scott-continuous function, then $f^{n}(\bot )\sqsubseteq f^{n+1}(\bot ),n\in \mathbb {N} _{0}$

Proof by induction:

Assume n = 0. Then $f^{0}(\bot )=\bot \sqsubseteq f^{1}(\bot )$ , since ⊥ is the least element.
Assume n > 0. Then we have to show that $f^{n}(\bot )\sqsubseteq f^{n+1}(\bot )$ . By rearranging we get $f(f^{n-1}(\bot ))\sqsubseteq f(f^{n}(\bot ))$ . By inductive assumption, we know that $f^{n-1}(\bot )\sqsubseteq f^{n}(\bot )$ holds, and because f is monotone (property of Scott-continuous functions), the result holds as well.

Immediate corollary of Lemma 1 is the existence of the chain.

Let $\mathbb {M}$ be the set of all elements of the chain: $\mathbb {M} =\{\bot ,f(\bot ),f(f(\bot )),\ldots \}$ . This set is clearly a directed/ω-chain, as a corollary of Lemma 1. From definition of CPO follows that this set has a supremum, we will call it $m$ . What remains now is to show that $m$ is the least fixed-point.

First, we show that $m$ is a fixed point, i.e. that $f(m)=m$ . Because $f$ is Scott-continuous, $f(\sup(\mathbb {M} ))=\sup(f(\mathbb {M} ))$ , that is $f(m)=\sup(f(\mathbb {M} ))$ . Also, since $f(\mathbb {M} )=\mathbb {M} \setminus \{\bot \}$ and because $\bot$ has no influence in determining $\sup$ , we have that $\sup(f(\mathbb {M} ))=\sup(\mathbb {M} )$ . It follows that $f(m)=m$ , making $m$ a fixed-point of $f$ .

The proof that $m$ is in fact the least fixed point can be done by showing that any Element in $\mathbb {M}$ is smaller than any fixed-point of $f$ (because by property of supremum, if all elements of a set $D\subseteq L$ are smaller than an element of $L$ then also $\sup(D)$ is smaller than that same element of $L$ ). This is done by induction: Assume $k$ is some fixed-point of $f$ . We now prove by induction over $i$ that $\forall i\in \mathbb {N} \colon f^{i}(\bot )\sqsubseteq k$ . For the induction start, we take $i=0$ : $f^{0}(\bot )=\bot \sqsubseteq k$ obviously holds, since $\bot$ is the smallest element of $L$ . As the induction hypothesis, we may assume that $f^{i}(\bot )\sqsubseteq k$ . We now do the induction step: From the induction hypothesis and the monotonicity of $f$ (again, implied by the Scott-continuity of $f$ ), we may conclude the following: $f^{i}(\bot )\sqsubseteq k~\implies ~f^{i+1}(\bot )\sqsubseteq f(k)$ . Now, by the assumption that $k$ is a fixed-point of $f$ , we know that $f(k)=k$ , and from that we get $f^{i+1}(\bot )\sqsubseteq k$ .

Proof

See also