Is Arens Square a Urysohn space?

Example 80 in Steen-Seebach1 is called Arens Square. It is defined in the book as follows:

Let $S$ be the set of rational lattice points in the interior of the unit square except those whose $x$-coordinate is $\frac12$.
Define $X$ to be $S\cup\{(0,0)\}\cup\{(1,0)\} \cup \{(\frac12,\sqrt2r); r\in\mathbb Q, 0<r\sqrt2<1\}$.
We define a basis for a topology on $X$ by granting to each point of $S$ the local basis of relatively open sets which $S$ inherits from the Euclidean topology on the unit square, and to the other points of $X$ the following local bases:
\begin{align*}
U_n(0,0) &= \{(0,0)\} \cup \{(x,y); 0<x<\frac14, 0<y<\frac1n\},\\
U_n(1,0) &= \{(0,0)\} \cup \{(x,y); \frac34<x<1, 0<y<\frac1n\},\\
U_n(\frac12,r\sqrt2) &= \{(x,y); \frac14<x<\frac34, |y-r\sqrt2|<\frac1n\}
\end{align*}

The authors claim that:

With this topology $X$ is $T_{2½}$.
This may be seen by direct consideration of cases noting that neither any point of $S$ nor $(0,0)$ nor $(0,1)$ may have the same $y$ coordinate as a point of the form $(\frac12,r\sqrt2)$.

A Urysohn space or $T_{2½}$-space is a space in which any two distinct points can be separated by closed neighborhoods. (I.e., they have disjoint closed neighborhoods. In Steen-Seebach such spaces are call such spaces completely Hausdorff. As far as I can say, the name Urysohn space is more frequent.)

The problem I see with the above claim is the following: Let us take two points $a=(\frac38,y)$ and $b=(\frac58,y)$, $y\in\mathbb Q$, which have the same $y$-coordinate and belong to the “middle part”. If we take any neighborhoods $O_a\ni a$ and $O_b\ni b$ then they contain neighborhoods of the form $B(a,\frac1n)\cap S$, $B(b,\frac1n)\cap S$ for some $n$. Now let $c$ be any point of the form $(\frac12,r\sqrt2)$ with $y-1/n<\sqrt2r<y+1/n$. Then every basic neighborhood of $c$ intersects these two balls and, consequently, it intersects $O_a$ and $O_b$. Which means that $c\in \overline{O_a} \cap \overline{O_b}$. So the points $a$ and $b$ do not have neighborhoods with disjoint closures.

What am I missing here? Or is it a mistake in this book? (In fact, the above problem was found by a colleague of mine. But since I was not able to answer his doubts, I decided to ask here.)

The authors also say that this is a modification of an example given in a paper by Hewitt. I had a look at that paper, but I did not find there exactly the same example as this one.

  1. Hewitt [51] credits Arens with constructing an example of this type; we present a modified version, and then a simplified version.
    Arens square is both semiregular and completely Hausdorff but not regular.

[51] Hewitt, E. On two problems of Urysohn. Annals of Math. 47 (1946) 503-509. DOI: 0.2307/1969089

It is perhaps also worth mentioning that π-Base list this space as $T_{2½}$-space: http://topology.jdabbs.com/traits/237

1 Lynn Arthur Steen and J. Arthur Seebach, Jr., Counterexamples in Topology. Springer-Verlag, New York, 1978. Reprinted by Dover Publications, New York, 1995.

Solutions Collecting From Web of "Is Arens Square a Urysohn space?"

You appear to have found a genuine mistake, and not just in the book: the same mistake appears in connection with the slightly more complicated example in Hewitt’s paper. The example can be repaired as follows.

Let $Q=(0,1)\cap\Bbb Q$, and let $\{Q_q:q\in Q\}$ be a partition of $Q$ into countably many dense subsets. Let

$$S=\bigcup_{q\in Q}\big(\{q\}\times Q_q\big)\;,$$

and let $X=\{\langle 0,0\rangle,\langle 1,0\rangle\}\cup S$. Let $M=\left\{\frac12\right\}\times Q_{1/2}$. Points of $S\setminus M$ have the open nbhds that they inherit from the Euclidean topology. For each $q\in Q_{1/2}$ we let

$$U_n\left(\frac12,q\right)=\left(\left(\frac14,\frac34\right)\times\left(q-\frac1n,q+\frac1n\right)\right)\cap S$$

and take $\left\{U_n\left(\frac12,q\right):n\in\Bbb Z^+\right\}$ as a local base at $\left\langle\frac12,q\right\rangle$. Local bases at $\langle 0,0\rangle$ and $\langle 1,0\rangle$ are defined as in Steen & Seebach.

Now no two points of $S$ have the same $y$-coordinate, so the problem that you pointed out no longer arises, but the basic structure of the example is intact.