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abian-themostfixed.html
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<!doctype html public "-//w3c//dtd html 4.0 transitional//en">
<html>
<head>
<BASE HREF="http://www.math.ucdavis.edu/~suh/abian/abian-themostfixed.html">
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
<meta name="GENERATOR" content="Mozilla/4.77 [en] (WinNT; U) [Netscape]">
<title> A MOST FUNDAMENTAL FIXED POINT THEOREM </title>
</head>
<body bgcolor="#FFFFFF">
<I><FONT COLOR=green>(Note: This is a copy of a page from
Alexander Abian's web site circa 1999. - N. Megill 11-Jun-2009)</FONT></I>
<center>
<h2>
A MOST FUNDAMENTAL FIXED POINT THEOREM <br>
<br>
<b>Alexander Abian</b></h2></center>
<p><br><DOUBLE> The following a most Fundamental Fixed Point Theorem
is proved in ZF set theory
<p><b>without the Axiom of Choice.</b> It is "<b> a most fundamental</b>"
in the sense that it gives a
<p>necessary and sufficient condition for the existence of a fixed point
of a mapping of a set S
<p>into itself where absolutely no algebraic or analytic or order theoretic
or any other special
<p>structures are imposed on S.
<p> <b>THEOREM</b> (<b>Abian</b>). Let f
be a mapping from a set S into itself. Then f has
a fixed point if
<p>and only if:
<br> There exists an element a of
P such that f<sup>k</sup>(a) is an
element of P for every ordinal k,
and
<br>for every ordinal v
<p>(1) if f<sup>v</sup>(a) is not a
fixed point of f then the f<sup>u</sup>(a)
's are all distinct for every ordinal u < v.
<p> <b>PROOF.</b> First we show that (1) implies that
f has a fixed point. Assume on the contrary that
f
<p> has no fixed point and let p and
q be any two distinct ordinal numbers. Clearly, there
always
<p>exists an ordinal v such that
p < v and q < v. But then
since by our assumption f<sup>v</sup>(a)
<p>cannot be a fixed point of f hence (1) implies
that f<sup>p</sup>(a) and f<sup>q</sup>(a)
are two distinct elements
<p>of S. Thus, our assumption implies that for every
two distinct ordinals p and q
there
<p>correspond two distinct elements f<sup>p</sup>(a)
and f<sup>q</sup>(a) of the set S.
Consequently, every ordinal can
<p>be assigned in a one-to-one way to every element of a subset of
S. But then the Axiom Scheme
<p>of Replacement of ZF would imply that the set
of all ordinals exists, which is a contradiction.
<p>Thus, our assumption is false and f has a fixed
point.
<p> Next, assume that f has a fixed point
a. Then (1 ) is obviously satisfied by setting
<p>a = f(a) = f<sup>k</sup>(a) for every ordinal
k.
<p> Thus, the Theorem is proved.
<p> <b>REMARK.</b> Note that f<sup>k</sup>(a)
in the above does not necessarily indicate the k-th or any other
<p>iterates of f. Thus, it is not even required that
f(f<sup>k</sup>(a)) be equal to f<sup>k+1</sup>(a).
Obviously, by
<p>" f<sup>k</sup>(a) is a fixed point of f
" is meant that f(f<sup>k</sup>(a))
= f<sup>k</sup>(a), and as mentioned above,
<p>without necessitating that f(f<sup>k</sup>(a))
be equal to f<sup>k+1</sup>(a).
<p> The fundamental significance of the Theorem lies in the fact
that a great many fixed point
<p>theorems can be reduced to the special cases of the Theorem, in as much
as, no special structures
<p>are required by the set S to have, and, no iterative
rules are imposed on f.
<p><b><a href="http://www.math.iastate.edu/abian/">Alexander Abian</a></b>
<br>Dept .of Math. Iowa state Univ.
<br>Ames, IA, 50011, USA.
<br><b>e-mail:</b> <a href="mailto:[email protected]"><b>abian</b>@iastate.edu</a>
<br>
</body>
</html>