Open Access

The Alexandroff-Urysohn Square and the Fixed Point Property

Fixed Point Theory and Applications20092009:310832

DOI: 10.1155/2009/310832

Accepted: 17 September 2009

Published: 18 October 2009

Abstract

Every continuous function of the Alexandroff-Urysohn Square into itself has a fixed point. This follows from G. S. Young's general theorem (1946) that establishes the fixed-point property for every arcwise connected Hausdorff space in which each monotone increasing sequence of arcs is contained in an arc. Here we give a short proof based on the structure of the Alexandroff-Urysohn Square.

Alexandroff and Urysohn [1] in Mémoire sur les espaces topologiques compacts defined a variety of important examples in general topology. The final manuscript for this classical paper was prepared in 1923 by Alexandroff shortly after the death of Urysohn. On [1, page 15], Alexandroff denoted a certain space by . While Steen and Seebach in Counterexamples in Topology [2, Example  101] refer to this space as the Alexandroff Square, we concur with Cameron [3, pages 791-792], who attributes it to Urysohn. Hence we refer to as the Alexandroff-Urysohn Square and for convenience denote it by . The following definition of is given by Steen and Seebach [2, Example  101, pages 120-121]. Define to be the closed unit square with the topology defined by taking as a neighborhood basis of each point ( ) off the diagonal the intersection of with open vertical line segments centered at ( ) (e.g., ). Neighborhoods of each point are the intersection with of open horizontal strips less a finite number of vertical lines (e.g., and ). Note is not first countable, and therefore not metrizable. However, is a compact arcwise-connected Hausdorff space [2].

In Young's paper [4] of 1946, local connectivity is introduced on a space by a change of topology with consequent implications on generalized dendrites. A non-specialist may not notice that the fixed-point property for the Alexandroff-Urysohn Square follows from a result in Young's paper. We offer the following short proof based on the structure of the Alexandroff-Urysohn Square. The proof is direct and uses a dog-chases-rabbit argument [5, page 123–125]; first having the dog run up the diagonal, and then up (or down) a vertical fiber. The Alexandroff-Urysohn Square is a Hausdorff dendroid. For a dog-chases-rabbit argument that metric dendroids have the fixed point property, see [6], and also see [7].

Definition 1.

A set in is an ordered segment if is a connected vertical linear neighborhood or is a component of the intersection of and a horizontal strip neighborhood.

Note the relative topology induced on each ordered segment by is the Euclidean topology. Each point of is contained in arbitrarily small ordered segments.

Let be the function defined by . Since each neighborhood in of a point of is projected by onto the complement of a finite set in , the function is discontinuous at each point of .

Let be the function defined by . Note is continuous.

Lemma 2.

Let be a continuous function. Let be a point of . If , then there is an ordered segment containing such that is in one component of .

Proof.

Suppose . We consider two cases.

Case 1.

Assume . Let be a vertical ordered segment containing .

Since p and f is continuous, there is a horizontal strip neighborhood H in of p such that and . Let be the -component of . Note is an ordered segment containing and . The point is contained in one component of .

Case 2.

Assume . Let be a horizontal strip neighborhood in of such that and is connected. Let be the -component of . Note is a square set with diagonal .

Let be a horizontal strip neighborhood in of such that and . Let be the ordered segment that is the -component of . Note is a connected subset of and . Hence is in one component of . This completes the proof of our lemma.

Theorem 3.

The Alexandroff-Urysohn Square has the fixed-point property.

Proof.

Let be a continuous function. We will show there exists a point of that is not moved by .

Let . Note . Let be the least upper bound of .

Note . To see this assume . Then, by the lemma, there is an ordered segment in containing such that is in one component of . However since is the least upper bound of , there exist points and in such that and , a contradiction. Hence, .

If , then as desired.

If , then either or . Assume without loss of generality that .

Let denote the interval .

Let be the function defined by if and if .

Note is an open and closed subset of . It follows that is continuous. Thus, is a retraction of to .

Let be the restriction of to . Since is a continuous function of the interval into itself, there is a point such that .

Since every point of that is sent into by is moved by , it follows that . Hence .

Authors’ Affiliations

(1)
Alcatel-Lucent
(2)
Department of Mathematics, California State University

References

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