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 ${\displaystyle (X,\tau ),}$ where

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

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

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

Properties

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

1. T_2½, 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_3½, 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 ${\displaystyle r{\sqrt {2}}<\min\{1/n,1/m\}}$, 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 ${\displaystyle r{\sqrt {2}}<\min\{1/n,1/m\}}$, 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.

References

• 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).