Arens square

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Short description: Topological space mathematics

}} In mathematics, the Arens square is a topological space, named for Richard Friederich Arens. Its role is mainly to serve as a counterexample.


The Arens square is the topological space [math]\displaystyle{ (X,\tau), }[/math] where

[math]\displaystyle{ X=((0,1)^2\cap\mathbb{Q}^2)\cup\{(0,0)\}\cup\{(1,0)\}\cup\{(1/2,r\sqrt{2})|\ r\in\mathbb{Q},\ 0\lt r\sqrt{2}\lt 1\} }[/math]

The topology [math]\displaystyle{ \tau }[/math] is defined from the following basis. Every point of [math]\displaystyle{ (0,1)^2\cap\mathbb{Q}^2 }[/math] is given the local basis of relatively open sets inherited from the Euclidean topology on [math]\displaystyle{ (0,1)^2 }[/math]. The remaining points of [math]\displaystyle{ X }[/math] are given the local bases

  • [math]\displaystyle{ U_n(0,0)=\{(0,0)\}\cup\{(x,y)|\ 0\lt x\lt 1/4,\ 0\lt y\lt 1/n\} }[/math]
  • [math]\displaystyle{ U_n(1,0)=\{(1,0)\}\cup\{(x,y)|\ 3/4\lt x\lt 1,\ 0\lt y\lt 1/n\} }[/math]
  • [math]\displaystyle{ U_n(1/2,r\sqrt{2})=\{(x,y)|1/4\lt x\lt 3/4,\ |y-r\sqrt{2}|\lt 1/n\} }[/math]


The space [math]\displaystyle{ (X,\tau) }[/math] is:

  1. T, since neither points of [math]\displaystyle{ (0,1)^2\cap\mathbb{Q}^2 }[/math], nor [math]\displaystyle{ (0,0) }[/math], nor [math]\displaystyle{ (0,1) }[/math] can have the same second coordinate as a point of the form [math]\displaystyle{ (1/2,r\sqrt{2}) }[/math], for [math]\displaystyle{ r\in\mathbb{Q} }[/math].
  2. not T3 or T, since for [math]\displaystyle{ (0,0)\in U_n(0,0) }[/math] there is no open set [math]\displaystyle{ U }[/math] such that [math]\displaystyle{ (0,0)\in U\subset \overline{U}\subset U_n(0,0) }[/math] since [math]\displaystyle{ \overline{U} }[/math] must include a point whose first coordinate is [math]\displaystyle{ 1/4 }[/math], but no such point exists in [math]\displaystyle{ U_n(0,0) }[/math] for any [math]\displaystyle{ n\in\mathbb{N} }[/math].
  3. not Urysohn, since the existence of a continuous function [math]\displaystyle{ f:X\to [0,1] }[/math] such that [math]\displaystyle{ f(0,0)=0 }[/math] and [math]\displaystyle{ f(1,0)=1 }[/math] implies that the inverse images of the open sets [math]\displaystyle{ [0,1/4) }[/math] and [math]\displaystyle{ (3/4,1] }[/math] of [math]\displaystyle{ [0,1] }[/math] with the Euclidean topology, would have to be open. Hence, those inverse images would have to contain [math]\displaystyle{ U_n(0,0) }[/math] and [math]\displaystyle{ U_m(1,0) }[/math] for some [math]\displaystyle{ m,n\in\mathbb{N} }[/math]. Then if [math]\displaystyle{ r\sqrt{2}\lt \min\{1/n,1/m\} }[/math], it would occur that [math]\displaystyle{ f(1/2,r\sqrt{2}) }[/math] is not in [math]\displaystyle{ [0,1/4)\cap(3/4,1]=\emptyset }[/math]. Assuming that [math]\displaystyle{ f(1/2,r\sqrt{2})\notin[0,1/4) }[/math], then there exists an open interval [math]\displaystyle{ U\ni f(1/2,r\sqrt{2}) }[/math] such that [math]\displaystyle{ \overline{U}\cap[0,1/4)=\emptyset }[/math]. But then the inverse images of [math]\displaystyle{ \overline{U} }[/math] and [math]\displaystyle{ \overline{[0,1/4)} }[/math] under [math]\displaystyle{ f }[/math] would be disjoint closed sets containing open sets which contain [math]\displaystyle{ (1/2,r\sqrt{2}) }[/math] and [math]\displaystyle{ (0,0) }[/math], respectively. Since [math]\displaystyle{ r\sqrt{2}\lt \min\{1/n,1/m\} }[/math], these closed sets containing [math]\displaystyle{ U_n(0,0) }[/math] and [math]\displaystyle{ U_k(1/2,r\sqrt{2}) }[/math] for some [math]\displaystyle{ k\in\mathbb{N} }[/math] cannot be disjoint. Similar contradiction arises when assuming [math]\displaystyle{ f(1/2,r\sqrt{2})\notin(3/4,1] }[/math].
  4. semiregular, since the basis of neighbourhood that defined the topology consists of regular open sets.
  5. second countable, since [math]\displaystyle{ X }[/math] is countable and each point has a countable local basis. On the other hand [math]\displaystyle{ (X,\tau) }[/math] is neither weakly countably compact, nor locally compact.
  6. totally disconnected but not totally separated, since each of its connected components, and its quasi-components are all single points, except for the set [math]\displaystyle{ \{(0,0),(1,0)\} }[/math] which is a two-point quasi-component.
  7. not scattered (every nonempty subset [math]\displaystyle{ A }[/math] of [math]\displaystyle{ X }[/math] contains a point isolated in [math]\displaystyle{ A }[/math]), since each basis set is dense-in-itself.
  8. not zero-dimensional, since [math]\displaystyle{ (0,0) }[/math] doesn't have a local basis consisting of open and closed sets. This is because for [math]\displaystyle{ x\in[0,1] }[/math] small enough, the points [math]\displaystyle{ (x,1/4) }[/math] would be limit points but not interior points of each basis set.


  • Lynn Arthur Steen and J. Arthur Seebach, Jr., Counterexamples in Topology. Springer-Verlag, New York, 1978. Reprinted by Dover Publications, New York, 1995. ISBN:0-486-68735-X (Dover edition).