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:
 * 1) T2½, 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 $$r\in\mathbb{Q}$$.
 * 2) not T3 or T3½, 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 $$n\in\mathbb{N}$$.
 * 3) 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 $$f(1/2,r\sqrt{2})\notin(3/4,1]$$.
 * 4) semiregular, since the basis of neighbourhood that defined the topology consists of regular open sets.
 * 5) 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.
 * 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 $$\{(0,0),(1,0)\}$$ which is a two-point quasi-component.
 * 7) not scattered (every nonempty subset $$A$$ of $$X$$ contains a point isolated in $$A$$), since each basis set is dense-in-itself.
 * 8) 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.