The Alexandroff-Urysohn Square and the Fixed Point Property
© T. H. Foregger et al. 2009
Received: 9 June 2009
Accepted: 17 September 2009
Published: 18 October 2009
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  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 .
In Young's paper  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 , and also see .
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 .
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.
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, .
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