# Arens square

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 $(X,\tau ),$ where
    $X=((0,1)^{2}\cap \mathbb {Q} ^{2})\cup \{(0,0)\}\cup \{(1,0)\}\cup \{(1/2,r{\sqrt {2}})|\ r\in \mathbb {Q} ,\ 0<r{\sqrt {2}}<1\}$
The topology $\tau$ is defined from the following basis. Every point of $(0,1)^{2}\cap \mathbb {Q} ^{2}$ is given the local basis of relatively open sets inherited from the Euclidean topology on $(0,1)^{2}$. The remaining points of $X$ are given the local bases
* $U_{n}(0,0)=\{(0,0)\}\cup \{(x,y)|\ 0<x<1/4,\ 0<y<1/n\}$
* $U_{n}(1,0)=\{(1,0)\}\cup \{(x,y)|\ 3/4<x<1,\ 0<y<1/n\}$
* $U_{n}(1/2,r{\sqrt {2}})=\{(x,y)|1/4<x<3/4,\ |y-r{\sqrt {2}}|<1/n\}$

## Properties

The space $(X,\tau )$ is:
* T$_{2½}$, since neither points of $(0,1)^{2}\cap \mathbb {Q} ^{2}$, nor $(0,0)$, nor $(0,1)$ can have the same second coordinate as a point of the form $(1/2,r{\sqrt {2}})$, for $r\in \mathbb {Q}$.
* not T$_3$ or T$_{3½}$, since for $(0,0)\in U_{n}(0,0)$ there is no open set $U$ such that $(0,0)\in U\subset {\overline {U}}\subset U_{n}(0,0)$ since ${\overline {U}}$ must include a point whose first coordinate is $1/4$, but no such point exists in $U_{n}(0,0)$ for any $n\in \mathbb {N}$.
* not Urysohn, since the existence of a continuous function $f:X\to [0,1]$ such that $f(0,0)=0$ and $f(1,0)=1$ implies that the inverse images of the open sets $[0,1/4)$ and $(3/4,1]$ of $[0,1]$ with the Euclidean topology, would have to be open. Hence, those inverse images would have to contain $U_{n}(0,0)$ and $U_{m}(1,0)$ for some $m,n\in \mathbb {N}$. Then if $r{\sqrt {2}}<\min\{1/n,1/m\}$, it would occur that $f(1/2,r{\sqrt {2}})$ is not in $[0,1/4)\cap (3/4,1]=\emptyset$. Assuming that $f(1/2,r{\sqrt {2}})\notin [0,1/4)$, then there exists an open interval $U\ni f(1/2,r{\sqrt {2}})$ such that ${\overline {U}}\cap [0,1/4)=\emptyset$. But then the inverse images of ${\overline {U}}$ and ${\overline {[0,1/4)}}$ under $f$ would be disjoint closed sets containing open sets which contain $(1/2,r{\sqrt {2}})$ and $(0,0)$, respectively. Since $r{\sqrt {2}}<\min\{1/n,1/m\}$, these closed sets containing $U_{n}(0,0)$ and $U_{k}(1/2,r{\sqrt {2}})$ for some $k\in \mathbb {N}$ cannot be disjoint. Similar contradiction arises when assuming $f(1/2,r{\sqrt {2}})\notin (3/4,1]$.
* semiregular, since the basis of neighbourhood that defined the topology consists of regular open sets.
* second countable, since $X$ is countable and each point has a countable local basis. On the other hand $(X,\tau )$ is neither weakly countably compact, nor locally compact.
* totally disconnected but not totally separated, since each of its connected components, and its quasi-components are all single points, except for the set $\{(0,0),(1,0)\}$ which is a two-point quasi-component.
* not scattered (every nonempty subset $A$ of $X$ contains a point isolated in $A$), since each basis set is dense-in-itself.
* not zero-dimensional, since $(0,0)$ doesn't have a local basis consisting of open and closed sets. This is because for $x\in [0,1]$ small enough, the points $(x,1/4)$ would be limit points but not interior points of each basis set.