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. == See also ==