# 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 $$S\cup\{(0,0)\}\cup\{(1,0)\} \cup \{(\frac12,\sqrt2r); r\in\mathbb Q, 0. We define a basis for a topology on $$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

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 $$y-1/n<\sqrt2r. Then every basic neighborhood of $$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: https://topology.pi-base.org/spaces/S000072/properties

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

• FWIW, I see nothing wrong with your argument. Commented Mar 27, 2016 at 10:32

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.

• Hi @BrianM.Scott FYI this example has now been added to pi-base: topology.pi-base.org/spaces/S000080 Commented Jun 30, 2023 at 22:51
• @PatrickR: Thanks for letting me know. Commented Jun 30, 2023 at 23:22

I'm working on a fix for this in pi-Base. But we need to recover that the Arens square is still $$T_2$$ since the proof for $$T_{2\frac{1}{2}}$$ was invalid.

Each $$U_n(0,0),U_n(1,0),U_n(\frac{1}{2},r\sqrt 2)$$ is pairwise disjoint. Given $$(\frac{1}{2},r\sqrt 2)$$ and $$(\frac{1}{2},q\sqrt 2)$$, pick $$\frac{1}{n}<\frac{1}{2}|r\sqrt 2-q\sqrt 2|$$ and then $$U_n(\frac{1}{2},r\sqrt 2)\cap U_n(\frac{1}{2},q\sqrt 2)=\emptyset$$.

Finally given some $$(x,y)\in S$$, let $$\frac{1}{n}<\frac{y}{2}$$. Then $$B_{y/2}(x,y)\cap U_n(0,0)=\emptyset$$ and $$B_{y/2}(x,y)\cap U_n(1,0)=\emptyset$$. Finally consider $$(\frac{1}{2},r\sqrt 2)$$, and pick $$\frac{1}{n}<\frac{1}{2}|y-r\sqrt 2|$$; then, $$B_{1/n}(x,y)\cap U_n(\frac{1}{2},r\sqrt 2)=\emptyset$$.