DISCRETE AND CONTINUOUS
DYNAMICAL SYSTEMS
SUPPLEMENT 2007
Website: www.AIMSciences.org
pp. 568–572
FIXED POINTS AND COMPLETE LATTICES
Paula Kemp
Department of Mathematics
Missouri State University
Springfield, MO 65897
Abstract. Tarski proved in 1955 that every complete lattice has the fixed
point property. Later, Davis proved the converse that every lattice with the
fixed point property is complete. For a chain complete ordered set, there is the
well known Abian-Brown fixed point result. As a consequence of the AbianBrown result, every chain complete ordered set with a smallest element has
the fixed point property. In this paper, a new characterization of a complete
lattice is given. Also, fixed point theorems are given for decreasing functions
where the partially ordered set need not be dense as is the usual case for fixed
point results for decreasing functions.
1. Introduction. Throughout the paper, a partially ordered set will be called a
poset. First we give some definitions.
Definition 1.1. A function from a poset (P, ≤) into itself is decreasing if
1. whenever x < y, then f (y) ≤ f (x)
A mapping from a poset (P, ≤) into itself has a fixed point iff
there exists an x ∈ P such that f (x) = x. The poset (P, ≤) has the fixed point
property if every increasing mapping of (P, ≤) into itself has a fixed point.
In [8], Tarski proved that every complete lattice has the fixed point property. In
[2], Davis proved the converse, ie., ”Every lattice with the fixed point property is
complete”.
Without the poset (P, ≤) being a lattice, we have the well known Abian/Brown
[1] result which says: Let (P, ≤) be a chain complete poset. Let f be an increasing
mapping from (P, ≤) into itself such that for some a ∈ P , a ≤ f (a). Then the
mapping f has a fixed point.
In this paper, the author proves fixed point theorems for decreasing functions
where the partially ordered sets need not be dense as is the usual case for fixed
point results for decreasing functions.
Before proving some theorems, we give some more definitions.
2. The mapping f from a poset P into itself has a fixed apex u if there exists a
v in P such that f (v) = u and f (u) = v.
2. Fixed points and complete lattices. In the following, the set P will denote
a poset where
3. P 0 = P × {1, −1}
We define an ordering on P 0 as follows:
2000 Mathematics Subject Classification. Primary 03E25.
Key words and phrases. Complete Lattices, Decreasing, Increasing, and Fixed points.
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4. If a ≤ b, define (a, −1) ≤ (b, −1) and (b, 1) ≤ (a, 1)
5. For all c, d ∈ P , define (c, −1) ≤ (d, 1)
In view of 4. and 5., the set P 0 is clearly a poset. We will denote an element of
the form (x, i) by x where i ∈ {1, −1}. Also, we denote π1 as the projection from
P × {1, −1} onto P .
Before giving some characterizations of Complete lattices, we give some additional definitions below: Let P be a poset and let f be a mapping from the poset
P into itself. Define
6. f (P ) = f (P ) ∪ {a ∈ P : a = supC for some chain C ∈ f (P )}
7. A mapping f of a poset P into itself is called relatively isotone if for x ≤ y, x ≤
f (y), f (y) ≤ y, implies f (x) ≤ f (y), for every x, y ∈ P .
In [9], the following theorem is given:
Theorem 2.1. Let (P, ≤) be a lattice. If (P, ≤) has a minimum, then the following
statements are equivalent:
(a) (P, ≤) is a complete lattice
(b) (Tarski [8], Davis [2]) P has the fixed point property
(c) Every mapping of P into itself with x ≤ f 2 (x), for all x ∈ f (P ) has a fixed
apex
(d) Every mapping of P into itself with x ≤ f (x) for all x ∈ f (P ) has a fixed
point
(e) (Markowski [5] Every inf-preserving map P into itself has a least fixed point
(f) (Klimes [3]). Every relatively isotone mapping of (P, ≤) into itself which is
comparable has a fixed point.
In the above theorem, complete lattices are characterized in terms of increasing
functions and functions which satisfy x ≤ f (x) or x ≤ f 2 (x) for all x ∈ f (P ).
For other results see [3], [6], and [7].
In the following definition, we say a non empty subset D of (P 0 , ≤) has the
bounded property if the set D is bounded and if every non empty subset of π1 (D)
has a supremum and an infimum in (P, ≤).
Theorem 2.2. The lattice (P, ≤) is complete is equivalent to the following statement
(I) every decreasing function from the partially ordered set (P 0 , ≤) into itself has
a fixed point where the function f and the poset (P 0 , ≤) have the following
properties:
there exists a nonempty linearly ordered subset M of the lattice (P 0 , ≤) such that
f (M ) ⊆ M and
(i) if f (b) ≤ a < b ≤ f (a) holds in M , then there exists an element x ∈ M such
that a < f (x) < b and a < x < b
(ii) if D is a nonempty bounded subset of M with the bounded property, then
sup(D) exists in M .
Proof. Assume that the lattice (P, ≤) is complete. Let f be a decreasing mapping
of P 0 into itself. Let M be a nonempty linearly ordered subset of P 0 such that
f (M ) ⊆ M . Since the set M is not empty, let a ∈ M . Without loss of generality,
we assume that a ≤ f (a). But then by 1. we have
8. f (f (a)) = f 2 (a) ≤ f (a).
Let the sets A and B be defined as follows
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9. A = {x ∈ M : x ≤ f (x)} and B = {x ∈ M : f (x) ≤ x}
In view of our assumption a ≤ f (a), we see that a ∈ A and from 9. it follows
that f (a) ∈ B. Thus, A and B are nonempty.
We observe by 1. and by 9. the following can be readily verified
10. if x ∈ A, then f (x) ∈ B, f 2 (x) ∈ A and if y ∈ B, then f (y) ∈ A and
f 2 (y) ∈ B
Thus, f (A) ⊆ B and f (B) ⊆ A.
Since M is a linearly ordered set, we have
11. M = A ∪ B
We show that
12. x ≤ y for x ∈ A and every y ∈ B
For if y < x, by 9. we have f (y) ≤ y < x ≤ f (x). However, since f is a decreasing
function we must also have f (x) ≤ f (y). Thus, we arrive at a contradiction and
hence x ≤ y.
Since the poset (P, ≤) is complete for all nonempty subsets D of A, we have
13. sup(π1 (D)) and inf (π1 (D)) exist in the lattice (P, ≤)
Thus in view of (ii) and since A is a bounded subset of P 0 , we have that
14. e = supA exists in the subset M of P 0
Since M is a linearly ordered subset of (P 0 , ≤), we have that either e ≤ f (e) or
f (e) < e. If e < f (e), we have by 10. that f (e) ∈ B, and f 2 (e) ∈ A. Since e is the
least upper bound of A, we have that f 2 (e) ≤ e. Hence, f 2 (e) ≤ e < f (e) ≤ f (e).
By hypothesis of the Theorem, there exists x ∈ M such that:
15. e < x < f (e) and e < f (x) < f (e).
From 11. it follows that x ∈ A or x ∈ B. If x ∈ A, then x ≤ e by 14. which
contradicts 15. If x ∈ B, then f (x) ∈ A and thus in view of 14., f (x) ≤ e which also
contradicts 15. Therefore, e is not < f (e). The second case f (e) < e is similar to the
previous case and will be omitted. Thus, the two cases cannot happen. Therefore,
e must be a fixed point as desired.
Conversely, assume that Principle I holds and that the lattice (P, ≤) is not complete. Thus, there exists a maximal linearly ordered subset M1 of P which is not
complete. Thus, there exists a subset L of M1 such that supL or inf L does not
exist in the poset (P, ≤). Without loss of generality, assume that supL does not
exist in (P, ≤) . Clearly, the the set L does not have a maximum. Thus, there exists
an infinite subset L1 of L such that
16. L1 = {x0 < x1 < ... < xi < ...}
where i ∈ β and β is a limit ordinal where L1 is cofinal in L. Let L0 = {x ∈ M1 :
x ≤ xα for some α < β}. Then consider the poset P 0 and let M = L0 × {1, −1}.
Define a mapping f from P 0 into itself (where k = 1 or -1) by f (x, k) = (x, −k).
Clearly, the mapping f is decreasing on P 0 and f (M ) ⊆ M . In order to show (i)
is true, assume that
17. f (b) ≤ a < b ≤ f (a) holds in M
In order for 17. to hold, we must have a = (a, −1), b = (b, 1) and a = b. In view
of 16., there exists xα ∈ L1 such that (xα , −1)and f (xα , −1) = (xα , 1) are between
(a, −1) and (b, 1). Thus part (i) is true.
In order to show part (ii), let D be a non empty bounded subset of M having
the bounded property.
FIXED POINTS AND COMPLETE LATTICES
571
We show that supD exists in M . If there exists an element of the form (x, 1) ∈ D,
then e = inf (π1 ({(x, 1)|(x, 1)) ∈ D}) exists in (P, ≤). Since M1 is a maximal
linearly ordered subset of (P, ≤), we have that e ∈ M1 . Thus, e ∈ L0 . Then
(e, 1) = supD ∈ M . If no element of the form (x, 1) is in D, then since e = supπ(D)
exists in the lattice (P, ≤) and M1 is a maximal linearly ordered subset of P we
have that e ∈ M1 . Thus e ∈ L0 . Thus, clearly the supremum of D in M is (e, −1).
In either case we have that supD exists in M .
Thus, the mapping f satisfies the hypothesis of I but the mapping f does not
have a fixed point.
From the above proof, we have the corollary given below. The following corollary
gives a sufficient condition for a decreasing function to have a fixed point. The poset
need not be dense as is the usual case and there needs to be no topology defined on
the poset P.
Corollary 2.3. Let (P, ≤) be a nonempty linearly ordered poset in which every
nonempty bounded above subset of P has a least upper bound and let f be a decreasing
function from (P, ≤) into itself. Assume also that for every elements
18. a, b ∈ P such that f (b) ≤ a < b ≤ f (a) then a < x < b and a < f (x) < b
for some x ∈ P .
Then the mapping f has a fixed point.
Below we give some remarks concerning the previous corollary.
Example 2.4. Any continuous function on the set of real numbers R satisfies 18.
of the previous Corollary. Thus, since the the set R has the the property that
every nonempty set which is bounded above has a least upper bound, then every
decreasing continuous function from R into R has a fixed point.
Example 2.5. The Corollary 2.3 is not true if 18. is deleted. For example let
P = {a, b} where a ≤ b. Let f : P −→ P be defined by f (a) = b and f (b) = a. The
mapping f is decreasing on P and the poset {a, b} is complete, but f does not have
a fixed point.
Example 2.6. To see in the above Corollary that 18. is not sufficient let f :
[0, 2] −→ [0, 2] be defined by f (x) = 2 if 0 ≤ x ≤ 1; f (x) = 0.5 if x ∈ [1, 2]−{1+1/n :
n ∈ N } and f (1 + 1/n) = 1 + 1/(2n), 1 ≤ n. Clearly, the mapping f satisfies 18.
but does not have a fixed point.
Example 2.7. In 18. of the previous corollary, a < f (x) < b can not be replaced
by
19. a < f (x) ≤ b
For let f map the closed interval [0, 1] into itself be defined by f (1) = 0 and
f (x) = 1, if 0 ≤ x < 1. The mapping satisfies 19. and the other hypotheses of the
theorem, but the mapping f does not have a fixed point.
Theorem 2.8. Let (P, ≤) be a complete poset. Then every decreasing function
from the poset (P 0 , ≤) into itself such that for some a in P 0 , a ≤ f 2 (a) has a fixed
apex or fixed point.
Proof. Assume the poset (P, ≤) is complete. Let f be a mapping of the lattice
(P 0 , ≤) into itself. Since P is a complete lattice, the poset (P 0 , ≤) is a complete
lattice. We consider the mapping F = f 2 . Then the mapping F is an increasing
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mapping. By Tarski’s Fixed Point Result [8], the mapping F has a fixed point and
thus the mapping f has a fixed point or a fixed apex.
From the Abian-Brown Fixed Point [1] result, we have the following corollary:
Corollary 2.9. Let (P, ≤) be a nonempty linearly ordered poset in which every
nonempty bounded subset has a least upper bound and let f be a decreasing mapping
from P into itself. Assume that there exists an element a ∈ M such that a ≤
f 2 (a) ≤ f (a). Then the mapping f has a fixed apex or a fixed point.
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Received September 2006; revised April 2007.
E-mail address: [email protected]