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