Knaster–Tarski theorem

Since complete lattices cannot be empty (they must contain supremum of empty set), the theorem in particular guarantees the existence of at least one fixed point of f, and even the existence of a least (or greatest) fixed point. In many practical cases, this is the most important implication of the theorem.

The least fixpoint of f is the least element x such that f(x) = x, or, equivalently, such that f(x) ≤ x; the dual holds for the greatest fixpoint, the greatest element x such that f(x) = x.

If f(lim x_n)=lim f(x_n) for all ascending sequences x_n, then the least fixpoint of f is lim fⁿ(0) where 0 is the least element of L, thus giving a more "constructive" version of the theorem. (See: Kleene fixed-point theorem.) More generally, if f is monotonic, then the least fixpoint of f is the stationary limit of f^α(0), taking α over the ordinals, where f^α is defined by transfinite induction: f^α+1 = f ( f^α) and f^γ for a limit ordinal γ is the least upper bound of the f^β for all β ordinals less than γ. The dual theorem holds for the greatest fixpoint.

For example, in theoretical computer science, least fixed points of monotone functions are used to define program semantics. Often a more specialized version of the theorem is used, where L is assumed to be the lattice of all subsets of a certain set ordered by subset inclusion. This reflects the fact that in many applications only such lattices are considered. One then usually is looking for the smallest set that has the property of being a fixed point of the function f. Abstract interpretation makes ample use of the Knaster–Tarski theorem and the formulas giving the least and greatest fixpoints.

Knaster–Tarski theorem can be used for a simple proof of the Cantor–Bernstein–Schroeder theorem.[4][5]

Weaker versions of the Knaster–Tarski theorem can be formulated for ordered sets, but involve more complicated assumptions. For example:

Let L be a partially ordered set with the smallest element (bottom) and let f : L → L be an order-preserving function. Further, suppose there exists u in L such that f(u) ≤ u and that any chain in the subset {x in L : x ≤ f(x), x ≤ u} has supremum. Then f admits a least fixed point.

This can be applied to obtain various theorems on invariant sets, e.g. the Ok's theorem:

For the monotone map F : P(X) → P(X) on the family of (closed) nonempty subsets of X the following are equivalent: (o) F admits A in P(X) s.t. ${\displaystyle A\subseteq F(A)}$, (i) F admits invariant set A in P(X) i.e. ${\displaystyle A=F(A)}$, (ii) F admits maximal invariant set A, (iii) F admits the greatest invariant set A.

In particular, using the Knaster-Tarski principle one can develop the theory of global attractors for noncontractive discontinuous (multivalued) iterated function systems. For weakly contractive iterated function systems Kantorovitch fixpoint theorem (known also as Tarski-Kantorovitch fixpoint principle) suffices.

Other applications of fixed-point principles for ordered sets come from the theory of differential, integral and operator equations.

Let's restate the theorem.

For a complete lattice ${\displaystyle \langle L,\leq \rangle }$ and a monotone function ${\displaystyle f\colon L\rightarrow L}$ on L, the set of all fixpoints of f is also a complete lattice ${\displaystyle \langle P,\leq \rangle }$, with:

• ${\displaystyle \bigvee P=\bigvee \{x\in L\mid x\leq f(x)\}}$ as the greatest fixpoint of f
• ${\displaystyle \bigwedge P=\bigwedge \{x\in L\mid x\geq f(x)\}}$ as the least fixpoint of f.

Proof. We begin by showing that P has both a least element and a greatest element. Let D = { x | x ≤ f(x) } and x ∈ D (we know that at least 0_L belongs to D). Then because f is monotone we have f(x) ≤ f(f(x)), that is f(x) ∈ D.

Now let ${\displaystyle u=\bigvee D}$ (u exists because D ⊆ L and L is a complete lattice). Then for all x ∈ D it is true that x ≤ u and f(x) ≤ f(u), so x ≤ f(x) ≤ f(u). Therefore, f(u) is an upper bound of D, but u is the least upper bound, so u ≤ f(u), i.e. u ∈ D. Then f(u) ∈ D (because f(u) ≤ f(f(u))) and so f(u) ≤ u from which follows f(u) = u. Because every fixpoint is in D we have that u is the greatest fixpoint of f.

The function f is monotone on the dual (complete) lattice ${\displaystyle \langle L^{op},\geq \rangle }$. As we have just proved, its greatest fixpoint exists. It is the least fixpoint of L, so P has least and greatest elements, that is more generally, every monotone function on a complete lattice has a least fixpoint and a greatest fixpoint.

If a ∈ L and b ∈ L, we'll write [a, b] for the closed interval with bounds a and b: { x ∈ L | a ≤ x ≤ b }. If a ≤ b, then ${\displaystyle \langle }$[a, b], ${\displaystyle \leq \rangle }$ is a complete lattice.

It remains to be proven that P is a complete lattice. Let ${\displaystyle 1_{L}=\bigvee L}$, W ⊆ P and ${\displaystyle w=\bigvee W}$. We′ll show that f([w, 1_L]) ⊆ [w, 1_L]. Indeed, for every x ∈ W we have x = f(x) and since w is the least upper bound of W x ≤ f(w). In particular w ≤ f(w). Then from y ∈ [w, 1_L] follows that w ≤ f(w) ≤ f(y), giving f(y) ∈ [w, 1_L] or simply f([w, 1_L]) ⊆ [w, 1_L]. This allows us to look at f as a function on the complete lattice [w, 1_L]. Then it has a least fixpoint there, giving us the least upper bound of W. We′ve shown that an arbitrary subset of P has a supremum, that is, P is a complete lattice.

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• An application to an elementary combinatorics problem: Given a book with 100 pages and 100 lemmas, prove that there is some lemma written on the same page as its index