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Kleene fixed-point theorem
Theorem in order theory and lattice theory
Theorem in order theory and lattice theory
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, the Kleene fixed-point theorem can be extended to monotone functions using transfinite iterations.
Proof
Source:
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.
References
References
- Alfred Tarski. (1955). "A lattice-theoretical fixpoint theorem and its applications". [[Pacific Journal of Mathematics]].
- Patrick Cousot and Radhia Cousot. (1979). "Constructive versions of Tarski's fixed point theorems". Pacific Journal of Mathematics.
- (1994). "Mathematical Theory of Domains by V. Stoltenberg-Hansen". Cambridge University Press.
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