On functions without fixed points

A N N A L E S S O C I E T A T I S M A T H E M A T I C A E P O L O N A E l e r ie s I : C O M M E N T A T I O N E S M A T H E M A T I C A E X V I I ( 1 9 7 3 ) C Z N I K I P O L S K I E G O T O W A R Z Y S T W A M A T E M A T Y C Z N E G O

S é r i a I : P R A C E M A T E M A T Y C Z N E X V I I ( 1 9 7 3 )

, Wi s n i e w s k i (Gdansk)

On functions without fixed points

The following fixed point theorem has been proved in [1] using e well-ordering principle:

(T) Let E be a set and f : JE -> E. Then f has a fixed point if and only E is not a union of three mutually disjoint sets Е г, Е г, E z such that Ei n f ( E i)

0 for г = 1 , 2 , 3 .

The purpose of this note is to show that (T) is not equivalent to the

;iom of choice. We shall show that (T) is a consequence of some weak rm of the axiom of choice, namely

FAC. F.or every family of finite sets there is a choice function.

On the other hand we showed (x), that (T) is independent of {[те] : n > 1}, lere [те] denotes the statement:

For every family of те-element sets there is a choice function.

Th e o r e m. FAC implies ^{(T ).}

Proof. Obviously, it is sufficient to show that FAC implies:

(I) If /: E E has no fixed point, then there are mutually disjoint ts E X, E2, E Z such that E = Ex\jEz\jEz and f ( E i) n E i —0 for i

1 , 2 , 3 .

We assume that f : E - + E has no fixed point. For every a and b э т E we set a ~ b if and only if there are non-negative integers h and l ch that f k(a) = f ( b ) (fk denotes here the &-th iteration of the function

in particular, /° is the identity).

The relation ~ is an equivalence relation in E. For each element X

3

о т E j ~ we are going to sets A ^X ) , A2(X), A3{X) such that X and А {(Х) r\f[Ai(X)] = 0 for г = 1 , 2 , 3 . г=1

It is obvious that for every X from E { ~ there is at most one finite t У с X such that f \ Y is a cycle.

О A. M ^ k o w s k i and K. W i s n i e w s k i , Generalisation o f A b ia n ’s fix e d p o in t югет, Prace Mat. 13 (1969), p. 63-65.

228 К . W i s n i e w s k i

Let F x be a set of these X e E / ~ that there is a finite subset Yx of X such that f\ Yx is a cycle. We set F2 = E j ^ — F x.

Using FAC we choose from every Y x such that X e F x exactly one element ax . For any non-negative integer Jc let Bk(X) be the set of ele­

ments xe X such that f k(x) — ax and for every non-negative integer l < Jc, f ( x ) Ф ax .

If the cardinality of Yx is odd we put A x(X) — [J B2n+x(X), A2(X) = [ J B2n+2(X) and A3(X) = B0(X).

П€ ft)

If the cardinality of Y x is even we put A x(X) = U B2n+x(X), A Z(X) = U B2n(X), A3(X) = 0 .

7 l € ft)

Let X e F 2. We put for every a and Ъ from X а ъ>х Ъ if and only if there is a non-negative integer Jc such that f k(a) = f k(b). This relation is an equivalence relation. Now we define a function f x on X [phx as follows : if Y belongs to Xj^ax , then f x {Y) is such an element Z of Х / ^ х that

!(?)<= Z.

We denote by F2 the set of such elements X of F2 that f x are per­

mutations of Xjpax . It is clear that these permutations are infinite cycles.

To each X from F2 we assign the set C(X) of orbits of the permutation f x . Because [2] is a consequence of FAC we can choose exactly one element D(X) of O(X). We define A x(X) = (J B ( X ) , A2(X) = (J

~ { B { X ) } ] , A3{X) = 0.

For every X from F2 — F2 there is exactly one element Vx of X / such that for any Ze XJ phx there is a non-negative integer Jc such that f x( Vx) = Z- Now we define A ^ X ) = \ J { f £ { V x ): Ice a}, A, {X) =

= U tfx*' (Vx): <«} and A3(X) = 0.

It is clear that the sets Et = и { А ф Х ) : X e E / ~ } (i = 1 , 2 , 3 ) satisfy the thesis of implication (I).

IN S T IT U T E O F M A T H E M A T IC S, U N IV E R S IT Y O F G D A N SK