I give short and constructive proofs of Tarski’s fixed-point theorem, and of Zhou’s extension of Tarski’s fixed-point theorem to set-valued maps

Echenique, Federico

English

Federico Echenique

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A short and constructive proof of Tarski’s fixed-point theorem

Tarski’s fixed-point theorem, games of strategic complimentarities, supermodular games,

Research Papers in Economics

Caltech Authors - Main

Published in the International Journal of Game Theory
Volume 33(2), June 2005, pages 215–218.
A SHORT AND CONSTRUCTIVE PROOF OF
TARSKI’S FIXED-POINT THEOREM
FEDERICO ECHENIQUE
Abstract. I give short and constructive proofs of Tarski’s fixed-
point theorem, and of Zhou’s extension of Tarski’s fixed-point the-
orem to set-valued maps.
1. Introduction
I give short and constructive proofs of two related fixed-point the-
orems. The first is Tarski’s fixed-point theorem: If F is a monotone
function on a non-empty complete lattice, the set of fixed points of F
forms a non-empty complete lattice. The second is Zhou’s [9] extension
of Tarski’s fixed-point theorem to set-valued functions: If ϕ : X → 2X
is monotone—when 2X is endowed with the induced set order—the
set of fixed-points of ϕ forms a non-empty complete lattice. Zhou’s
extension is important in the theory of games with strategic comple-
mentarities (see, for example, [6] or [8]).
When F is continuous as well as monotone, my proof is very simple
(see Section 4). The proof when F is continuous is thus useful for
teaching game theory—if one wishes to prove a fixed-point theorem,
but finds Kakutani’s too involved, one can teach Tarski’s. 1
Tarski’s [5] original proof is beautiful and elegant, but non-constructive
and somewhat uninformative. Cousot and Cousot [1] give a construc-
tive proof of Tarski’s fixed-point theorem. But their proof is long and
quite involved. I present a simpler, and succinct, proof. On the other
hand, Cousot and Cousot obtain certain sub-products from their ap-
proach that I do not obtain; I shall only be concerned with Tarski’s
fixed-point theorem, and its extension to set-valued functions.
The extension to set-valued functions was developed by Smithson [4]
and Zhou [9]. Earlier, Vives [7] proved a stronger version of the exten-
sion, which applied to games with strict strategic complementarities.
Date: October 6, 2004.
I thank Charles Blair, William Thomson, an associated editor and a referee for
their helpful suggestions.
1I have taught Tarski’s Theorem with F continuous to Caltech undergraduates.
1
2 F. ECHENIQUE
I give a constructive proof of Zhou’s version of the result. Smithson
has a weaker monotonicity requirement than Zhou, but Smithson does
not obtain a lattice structure of the set of fixed-points. In addition,
Smithson needs a continuity assumption.
2. Definitions
An in-depth discussion of the following concepts can be found in [6].
A set X endowed with a partial order ≤ is denoted 〈X,≤〉; 〈X,≤〉 is a
complete lattice if, for all nonempty B ⊆ X, the greatest lower bound∧
X B and the least upper bound
∨
X B exist in X. If A ⊆ X, say that
A is a subcomplete sublattice of 〈X,≤〉 if, for all nonempty B ⊆ A,∧
X B ∈ A, and
∨
X B ∈ A.
Note that 〈A,≤〉 may be a complete lattice, even if A is not a sub-
lattice of 〈A,≤〉. So, for A′ ⊆ A, ∨AA′ may differ from ∨X A′.
Say that A ⊆ X is smaller than B ⊆ X in the induced set order
(denoted A v B) if
(x ∈ A, y ∈ B)⇒ (x ∧ y ∈ A, x ∨ y ∈ B).
The induced set order is a partial order on the set of sublattices of X.
Denote by ¹ the usual linear order on ordinal numbers.
Let 〈X,≤X〉 be a lattice and 〈Y,≤Y 〉 be a partially ordered set. A
function F : X → Y is monotone if x ≤X y implies F (x) ≤Y F (y).
Say that a set-valued map ϕ : X → 2X is monotone if it is monotone
when ϕ(X) is ordered by the induced set order.
3. Results
Let 〈X,≤〉 be a complete lattice and F : X → X be monotone. The
set of fixed points of F is E(F ) = {x ∈ X : x = F (x)}.
Lemma 1. 〈E(F ),≤〉 has a smallest element.
Proof. Let η be an ordinal number with cardinality greater than X, let
ξ = η + 1. Define f : ξ → X by transfinite recursion as f(0) = ∧X X,
and
f(β) =
∨
X
{F (f(α)) : α < β}
for β > 0.
That (β ≺ α) ⇒ (f(β) ≤ f(α)) is immediate from the definition
of f . Then, for all α ∈ η, it follows that f(α + 1) = F (f(α)), as
f(β) ≤ f(α), for all β < α, and F is monotone. Since η has cardinality
greater than X, there is γ ∈ η such that f(γ) = f(γ+1). Let γ be the
smallest such γ; γ is well-defined because any set of ordinal numbers
TARSKI’S THEOREM 3
has a smallest element (see [3] for an informal introduction to ordinal
numbers). Let e = f(γ). Then e = F (e). So e ∈ E(F ).
I shall prove that e is the smallest element in 〈E(F ),≤〉. Let e ∈
E(F ), and consider the proposition Pα : f(α) ≤ e. Proposition Pα
implies that f(α+1) = F (f(α)) ≤ F (e) = e. By transfinite induction,
then, e ≤ e. 
A version of Lemma 1 is also crucial in [1]’s proof of Tarski’s fixed-
point theorem. It was apparently first proved in [2]; my proof is more
direct than the one in [2].
Theorem 2. 〈E(F ),≤〉 is a non-empty complete lattice.
Proof. By Lemma 1, E(F ) is nonempty. Let E ⊆ E(F ) be nonempty.
I shall find
∨
E(F )E.
Let x =
∨
X E, and let Y = {z ∈ X : x ≤ z} be the set of upper
bounds on E. If z ∈ Y , then, for all e ∈ E, e ≤ F (z), as e = F (e) ≤
F (z). Thus F (Y ) ⊆ Y . Let G = F |Y . Then G maps Y into Y , and G
is monotone.
By Lemma 1, 〈E(G),≤〉 has a smallest element. By definition of G,
this smallest element is
∨
E(F )E.
The construction of
∧
E(F )E is symmetric. 
My proof is constructive in the sense that it gives a procedure for
finding a fixed point—and if E is a collection of fixed points, for finding∨
E(F )E and
∧
E(F )E. The proof in [1] is constructive in this sense
as well. As both proofs use ordinal numbers, there are notions of
constructiveness that neither my proof or [1]’s would satisfy.
Let ϕ : X → 2X be a set-valued map such that, for all x ∈ X,
ϕ(x) is a non-empty subcomplete sublattice of X. Suppose that ϕ is
monotone. The set of fixed points of ϕ is E(ϕ) = {x ∈ X : x ∈ ϕ(x)}.
Lemma 3. 〈E(ϕ),≤〉 has a smallest element.
Proof. Let F (x) =
∧
X ϕ(x). Note that, for all x, F (x) ∈ ϕ(x), and
that F is monotone. By Lemma 1, there is a smallest element, say e
of 〈E(F ),≤〉. Note that e = F (e) ∈ ϕ(e), so e ∈ E(ϕ).
I shall prove that e is the smallest element in 〈E(ϕ),≤〉. Let e ∈ E(ϕ).
Let f be as in the proof of Lemma 1. Consider the proposition Pα :
f(α) ≤ e. Proposition Pα implies that f(α+1) = F (f(α)) ≤ F (e) ≤ e,
as F (e) =
∧
X ϕ(e), and e ∈ ϕ(e). By transfinite induction, then,
e ≤ e. 
4 F. ECHENIQUE
Theorem 4. 〈E(ϕ),≤〉 is a non-empty complete lattice.
Proof. Lemma 3 implies that E(ϕ) is nonempty. I shall prove that it is
a complete lattice.
Let E ⊆ E(ϕ) be nonempty. I shall prove that ∨E(ϕ)E exists.
Let x =
∨
X E, and let Y = {z ∈ X : x ≤ z}. Define ψ : Y → 2Y by
ψ(z) = Y ∩ ϕ(z).
First, I show ψ(z) 6= ∅. Note that x ≤ z implies that, for all e ∈ E,
there is zˆe ∈ ϕ(z) with e ≤ zˆe, as e ∈ ϕ(e) and ϕ is monotone. But
ϕ(z) is subcomplete, and thus
x ≤
∨
e∈E
zˆe ∈ ϕ(z),
so ψ(z) 6= ∅.
Second, I show that ψ is monotone. Let z ≤ z′, and fix y ∈ ψ(z)
and y′ ∈ ψ(z′). The mapping ϕ is monotone, so y ∧ y′ ∈ ϕ(z) and
y ∨ y′ ∈ ϕ(z′). For e ∈ E, e ∈ ϕ(e) implies e ∨ (y ∧ y′) ∈ ϕ(z). But
y∧ y′ ∈ Y , so ϕ(z) 3 e∨ (y ∧ y′) = y∧ y′. Similarly, e∨ (y ∨ y′) ∈ ϕ(z′)
and y ∨ y′ ∈ Y implies y ∨ y′ ∈ ψ(z′). Thus y ∧ y′ ∈ ψ(z) and y ∨ y′ ∈
ψ(z′).
Third, ψ(z) is a subcomplete sublattice because ψ(z) = Y ∩ ϕ(z)
and ϕ(z) is a subcomplete sublattice.
Thus, ψ satisfies the hypothesis of Lemma 3. Let e∗ ∈ E(ψ) be the
smallest ψ-fixed point. If e ∈ E(ϕ) is an upper bound on E, then e ∈ Y
and thus e ∈ E(ψ). Then e∗ ≤ e. But e∗ ∈ E(ϕ), so e∗ = ∨E(ϕ)E.
The proof that
∧
E(ϕ)E exists is symmetric. 
4. Continuous F
The proof of Tarski’s Theorem is elementary when F is order-continuous,
in addition to monotone.
First, the proof of Lemma 1 goes as follows: Let x =
∧
X X be the
smallest point in X, and let {xn} be the sequence of F -iterates from x;
so xn = F (xn−1) and x0 = x. Since F is monotone, {xn} is a monotone
sequence, and thus converges to a point e. The continuity of F implies
that e is a fixed point, as x2n+1 = F (x2n), and both {x2n} and {x2n+1}
converge to e. Further, if e is any other fixed point of F , x ≤ e, and
xn ≤ e implies xn+1 = F (xn) ≤ F (e) = e. By induction, e is the
smallest fixed point.
Second, Lemma 1 is used to prove Tarski’s Theorem as in the proof
of Theorem 2 above.
TARSKI’S THEOREM 5
References
[1] Cousot, P., and R. Cousot (1979): “Constructive Versions of
Tarski’s Fixed Point Theorems,” Pacific J. of Math., 82(1), 43–57.
[2] Devide´, V. (1964): “On monotonous mappings of complete lat-
tices,” Fundamenta Mathematicae, 53(2), 147–154.
[3] Halmos, P. R. (1974): Naive Set Theory. Springer Verlag.
[4] Smithson, R. E. (1971): “Fixed Points of Order Preserving Mul-
tifunctions,” Proc. Amer. Math. Soc., 28(1), 304–310.
[5] Tarski, A. (1955): “A lattice theoretical fixpoint theorem and its
applications,” Pacific J. of Math, 5, 285–309.
[6] Topkis, D. M. (1998): Supermodularity and Complementarity.
Princeton University Press, Princeton, New Jersey.
[7] Vives, X. (1990): “Nash Equilibrium with Strategic Complemen-
tarities,” Journal of Mathematical Economics, 19(3), 305–321.
[8] (1999): Oligopoly Pricing. MIT Press, Cambridge, Mas-
sachusetts.
[9] Zhou, L. (1994): “The Set of Nash Equilibria of a Supermodular
Game Is a Complete Lattice,” Games and Economic Behavior, 7(2),
295–300.
HSS 228-77, California Institute of Technology, Pasadena CA 91125.
E-mail address: fede@hss.caltech.edu
URL: http://www.hss.caltech.edu/~fede/


A lattice theoretical ﬁxpoint theorem and its applications,”

Constructive Versions of Tarski’s Fixed Point Theorems,”

Fixed Points of Order Preserving Multifunctions,”

Naive Set Theory.

Nash Equilibrium with Strategic Complementarities,”

Oligopoly Pricing.

On monotonous mappings of complete lattices,”

Supermodularity and Complementarity.

The Set of Nash Equilibria of a Supermodular Game Is a Complete Lattice,”

http://hdl.handle.net/10.1007/s001820400192

A short and constructive proof of Tarski’s fixed-point theorem

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Research Papers in Economics

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