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^2)\cup\\cup\\cup\$

The topology $\tau$ is defined from the following basis. Every point of $(0,1)^2\cap\mathbb^2$ 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
*$U_n(0,0)=\\cup\$
*$U_n(1,0)=\\cup\$
*$U_n(1/2,r\sqrt)=\$

Properties

The space $(X,\tau)$ is:
# T_2½, since neither points of $(0,1)^2\cap\mathbb^2$, nor $(0,0)$, nor $(0,1)$ can have the same second coordinate as a point of the form $(1/2,r\sqrt)$, for $r\in\mathbb$.
# not T₃ or T_3½, since for $(0,0)\in U_n(0,0)$ there is no open set $U$ such that $(0,0)\in U\subset \overline\subset U_n(0,0)$ since $\overline$ must include a point whose first coordinate is $1/4$, but no such point exists in $U_n(0,0)$ for any $n\in\mathbb$.
# not Urysohn, since the existence of a continuous function f:X\to ,1/math> such that $f(0,0)=0$ and $f(1,0)=1$ implies that the inverse images of the open sets ,1/4) and (3/4,1/math> of ,1/math> 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 $m,n\in\mathbb$. Then if $r\sqrt<\min\$, it would occur that $f(1/2,r\sqrt)$ is not in ,1/4)\cap(3/4,1\emptyset. Assuming that $f(1/2,r\sqrt)\notin[0,1/4)$, then there exists an open interval $U\ni f(1/2,r\sqrt)$ such that $\overline\cap[0,1/4)=\emptyset$. But then the inverse images of $\overline$ and $\overline$ under $f$ would be disjoint closed sets containing open sets which contain $(1/2,r\sqrt)$ and $(0,0)$, respectively. Since $r\sqrt<\min\$, these closed sets containing $U_n(0,0)$ and $U_k(1/2,r\sqrt)$ for some $k\in\mathbb$ cannot be disjoint. Similar contradiction arises when assuming $f(1/2,r\sqrt)\notin(3/4,1]$.
# Semiregular space, semiregular, since the basis of neighbourhood that defined the topology consists of regular open sets.
# Second-countable space, 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 $\$ 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 ,1/math> small enough, the points $(x,1/4)$ 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).

Topological spaces