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:


 * 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 $$f^{i+1}(\bot) \sqsubseteq k.$$