# Arens square

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.

## Definition

The Arens square is the topological space $\displaystyle{ (X,\tau), }$ where

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

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

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

## Properties

The space $\displaystyle{ (X,\tau) }$ is:

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