without the Axiom of Choice.
 

     THEOREM (Abian). Let    f    be a mapping from a partially ordered set  (P, )   into itself.

Then  f  has a fixed point if and only if:

     There exists an element  a  of  P  such that     f^k(a)     is an element of  P  for every ordinal  k  and

(1)           f^u(a)   f^v(a)      for every ordinal numbers   u   v     with     f^u+1(a)  =  f(f^u(a))
 

     PROOF.   Clearly, for a limit ordinal  w  the hypothesis (1) implies that    f^w(a)  exists (i.e., is given)

and  f^u(a)  f^w(a)  for every ordinal number  u  w.

    Now, assume on the contrary that  f  has no fixed point. Thus, in view of our assumption and (1)

       f⁰(a)  <  f¹(a)  <  f²(a)  <  ...  <  f^u(a)  <  ...    for every ordinal   u

But then to every term   f^u(a)   of the above sequence the unique ordinal   u   can be assigned (and

since the terms of the above sequence form a set) which, by virtue of the Axiom Scheme of

Replacement of   ZF, would imply the existence of the set of all ordinal numbers, yielding a contradiction.

     Conversely, if   a   is a fixed point of   f   then  (1)  is trivially satisfied.
 
 

NOTICE that it is not assumed that   (P, )   is a complete partially ordered set. Also, it is not assumed that every nonempty well ordered subset of P has an upper bound or a least upper bound. Notice also that   f   need not be an isotone or a noncontractive map. No special conditions on   (P, )   or on   f   are imposed other than that   P   has an element   a   which satisfies just   (1). In fact at a limit ordinal   w   the w-iterate   f^w(a)   need not be the supremum of the set of the previous iterates; since no such supremum may even exist. So, in this context, the above Fixed Point Theorem is a fundamental Fixed Point Theorem. Many other fixed point theorems of mappings in partially ordered sets are direct consequences of the above Fixed Point Theorem.
 

Alexander Abian
Dept .of Math. Iowa state Univ.
Ames, IA, 50011, USA.
e-mail: abian@iastate.edu