# Kleene fixed-point theorem 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 $(L,\sqsubseteq )$ is a directed-complete partial order (dcpo) with a least element, and let $f:L\to L$ 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

$\bot \sqsubseteq f(\bot )\sqsubseteq f(f(\bot ))\sqsubseteq \cdots \sqsubseteq f^{n}(\bot )\sqsubseteq \cdots$ obtained by iterating f on the least element ⊥ of L. Expressed in a formula, the theorem states that

${\textrm {lfp}}(f)=\sup \left(\left\{f^{n}(\bot )\mid n\in \mathbb {N} \right\}\right)$ where ${\textrm {lfp}}$ 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, Kleene Fixed-Point Theorem can be extended to monotone functions using transfinite iterations.

## Proof

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

Lemma. If $L$ is a dcpo with a least element, and $f:L\to L$ is Scott-continuous, then $f^{n}(\bot )\sqsubseteq f^{n+1}(\bot ),n\in \mathbb {N} _{0}$ Proof. We use induction:
• Assume n = 0. Then $f^{0}(\bot )=\bot \sqsubseteq f^{1}(\bot ),$ since $\bot$ 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.

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

$\mathbb {M} =\{\bot ,f(\bot ),f(f(\bot )),\ldots \}.$ From the definition of a dcpo it follows that $\mathbb {M}$ has a supremum, 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 $\mathbb {M} =f(\mathbb {M} )\cup \{\bot \}$ and because $\bot$ has no influence in determining the supremum we have: $\sup(f(\mathbb {M} ))=\sup(\mathbb {M} )$ . 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 $\forall i\in \mathbb {N} :f^{i}(\bot )\sqsubseteq k$ . The base of the induction $(i=0)$ obviously holds: $f^{0}(\bot )=\bot \sqsubseteq k,$ since $\bot$ is the least 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 $f^{i+1}(\bot )\sqsubseteq k.$ 