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| {{Unreferenced|date=December 2011}}
| | I'm Mari (22) from Reykjavik, Iceland. <br>I'm learning Hindi literature at a local high school and I'm just about to graduate.<br>I have a part time job in a the office.<br><br>Feel free to surf to my web site :: roofing contractor; [http://www.zooschule-heidelberg.de/search/node/http%3A/%252FZ23.co/oofinacramento98043 just click the next post], |
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| In the [[mathematics|mathematical]] areas of [[order theory|order]] and [[lattice theory]], the '''Kleene fixed-point theorem''', named after American mathematician [[Stephen Cole Kleene]], states the following:
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| :''Let L be a [[complete partial order]], and let f : L → L be a [[Scott continuity|Scott-continuous]] (and therefore [[monotone function|monotone]]) [[function (mathematics)|function]]. Then f has a [[least fixed point]], which is the [[supremum]] of the ascending Kleene chain of f.
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| The '''ascending Kleene chain''' of ''f'' is the [[chain (order theory)|chain]]
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| :<math>\bot \; \le \; f(\bot) \; \le \; f\left(f(\bot)\right) \; \le \; \dots \; \le \; f^n(\bot) \; \le \; \dots</math>
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| obtained by [[iterated function|iterating]] ''f'' on the [[least element]] ⊥ of ''L''. Expressed in a formula, the theorem states that
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| :<math>\textrm{lfp}(f) = \sup \left(\left\{f^n(\bot) \mid n\in\mathbb{N}\right\}\right)</math>
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| where <math>\textrm{lfp}</math> denotes the least fixed point.
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| This result is often attributed to [[Alfred Tarski]], but [[Tarski's fixed point theorem]] pertains to [[monotone function]]s on [[complete lattices]].
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| == Proof ==
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| We first have to show that the ascending Kleene chain of ''f'' exists in L. To show that, we prove the following lemma:
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| :Lemma 1:''If L is CPO, and f : L → L is a Scott-continuous, then <math>f^n(\bot) \le f^{n+1}(\bot), n \in \mathbb{N}_0</math> | |
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| Proof by induction:
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| * Assume n = 0. Then <math>f^0(\bot) = \bot \leq f^1(\bot)</math>, since ⊥ is the least element.
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| * Assume n > 0. Then we have to show that <math>f^n(\bot) \leq f^{n+1}(\bot)</math>. By rearranging we get <math>f(f^{n-1}(\bot)) \leq f(f^n(\bot))</math>. By inductive assumption, we know that <math>f^{n-1}(\bot) \leq f^n(\bot)</math> holds, and because f is monotone (property of Scott-continuous functions), the result holds as well.
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| Immediate corollary of Lemma 1 is the existence of the chain.
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| Let <math>\mathbb{M}</math> be the set of all elements of the chain: <math>\mathbb{M} = \{ \bot, f(\bot), f(f(\bot)), \ldots\}</math>. This set is clearly directed due to Lemma 1. From definition of CPO follows that this set has a supremum, we will call it <math>m</math>. What remains now is to show that <math>m</math> is the least fixed-point.
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| First, we show that <math>m</math> is a fixed point. That is, we have to show that <math>f(m) = m</math>. Because <math>f</math> is [[Scott continuity|Scott-continuous]], <math>f(\sup(\mathbb{M})) = \sup(f(\mathbb{M}))</math>, that is <math>f(m) = \sup(f(\mathbb{M}))</math>. Also, <math>\sup(f(\mathbb{M})) = \sup(\mathbb{M})</math> (from the property of the chain) and from that <math>f(m) = m</math>, making <math>m</math> a fixed-point of <math>f</math>.
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| The proof that <math>m</math> is in fact the ''least'' fixed point can be done by showing that any Element in <math>\mathbb{M}</math> is smaller than any fixed-point of <math>f</math>. This is done by induction: Assume <math>k</math> is some fixed-point of <math>f</math>. We now proof by induction over <math>i</math> that <math>\forall i \in \mathbb{N}\colon f^i(\bot) \sqsubseteq k</math>. For the induction start, we take <math>i = 0</math>: <math>f^0(\bot) = \bot \sqsubseteq k</math> obviously holds, since <math>\bot</math> is the smallest element of <math>L</math>. As the induction hypothesis, we may assume that <math>f^i(\bot) \sqsubseteq k</math>. We now do the induction step: From the induction hypothesis and the monotonicity of <math>f</math> (again, implied by the Scott-continuity of <math>f</math>), we may conclude the following: <math>f^i(\bot) \sqsubseteq k ~\implies~ f^{i+1}(\bot) \sqsubseteq f(k)</math>. Now, by the assumption that <math>k</math> is a fixed-point of <math>f</math>, we know that <math>f(k) = k</math>, and from that we get <math>f^{i+1}(\bot) \sqsubseteq k</math>.
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| == See also ==
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| * [[Knaster–Tarski theorem]]
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| * Other [[fixed-point theorem]]s
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| [[Category:Order theory]]
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| [[Category:Fixed-point theorems]]
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I'm Mari (22) from Reykjavik, Iceland.
I'm learning Hindi literature at a local high school and I'm just about to graduate.
I have a part time job in a the office.
Feel free to surf to my web site :: roofing contractor; just click the next post,