Computation of the least fixpoint of f(x) = 1/10x²+atan(x)+1 using Kleene's theorem in the real interval [0,7] with the usual order

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 ${\displaystyle (L,\sqsubseteq )}$ is a directed-complete partial order (dcpo) with a least element, and let ${\displaystyle f:L\to L}$ be a Scott-continuous (and therefore monotone) function. Then ${\displaystyle f}$ has a least fixed point, which is the supremum of the ascending Kleene chain of ${\displaystyle f.}$

The ascending Kleene chain of f is the chain

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

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

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

where ${\displaystyle {\textrm {lfp}}}$ 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 ^[1] Moreover, Kleene Fixed-Point Theorem can be extended to monotone functions using transfinite iterations.^[2]

Proof^[3]

We first have to show that the ascending Kleene chain of ${\displaystyle f}$ exists in ${\displaystyle L}$. To show that, we prove the following:

Lemma. If ${\displaystyle L}$ is a dcpo with a least element, and ${\displaystyle f:L\to L}$ is Scott-continuous, then ${\displaystyle f^{n}(\bot )\sqsubseteq f^{n+1}(\bot ),n\in \mathbb {N} _{0}}$

Proof. We use induction:

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

As a corollary of the Lemma we have the following directed ω-chain:

${\displaystyle \mathbb {M} =\{\bot ,f(\bot ),f(f(\bot )),\ldots \}.}$

From the definition of a dcpo it follows that ${\displaystyle \mathbb {M} }$ has a supremum, call it ${\displaystyle m.}$ What remains now is to show that ${\displaystyle m}$ is the least fixed-point.

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

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

See also

• Other fixed-point theorems

References