What does the closure of an open neigborhood look like in this example of countable connected Hausdorff space?

Question

I am trying to understand the following nice exposition of Bing's example of countable connected Hausdorff space by Brian M. Scott:

The points of the space $X$ are the points $\langle p,q\rangle\in\Bbb Q^2$ such that $q\ge 0$. Let $Y=\Bbb Q\times\{0\}$. For each $\epsilon>0$ and $x=\langle q,0\rangle\in Y$ let $B(x,\epsilon)=\{\langle p,0\rangle\in Y:|p-q|<\epsilon\}$. For each $x=\langle p,q\rangle\in X\setminus Y$ let $T_x$ be the equilateral triangle in $\Bbb R^2$ with one vertex at $x$ and the other two on the $x$-axis; the two base vertices have the irrational $x$-coordinates $\ell(x)=p-\frac{q}{\sqrt3}$ and $r(x)=p+\frac{q}{\sqrt3}$, so they don’t actually lie in $X$. For $\epsilon>0$ let

$$B(x,\epsilon)=\{x\}\cup\{\langle s,0\rangle\in Y:|s-\ell(x)|<\epsilon\text{ or }|s-r(x)|<\epsilon\}\;.$$

For both kinds of point $\{B(x,\epsilon):\epsilon>0\}$ is a local base at $x$.

If you think of $\ell(x)$ and $r(x)$ as the left and right ‘feet’ of the point $x$, you can see where the name sticky foot comes from: open nbhds of $x$ are little clots of points on $Y$ ‘stuck to’ the feet of $x$.

Clearly $X$ is countable, and it’s easy to check that $X$ is Hausdorff. To see that $X$ is connected, observe that for any $x,y\in X$ and $\epsilon,\delta>0$ we have $\operatorname{cl}B(x,\epsilon)\cap\operatorname{cl}B(y,\delta)\ne\varnothing$. (A picture helps here: once you see what the closures of these open sets are, the result is very clear.) Thus, no two non-empty open sets have disjoint closures, and it follows easily that $X$ is connected.

In the last paragraph, I do not know how to visualize the closure $\operatorname{cl}B(x,\epsilon)$ in this "weird" topology. I guess it may be the union of four half infinite "rational" strips but I am not sure. Could anyone give a picture or description of the picture showing $\operatorname{cl}B(x,\epsilon)$?

Answer 1

As you correctly guessed the closure of every basic open set $B(x,\varepsilon)$ contains four stripes, since every point inside one of those stripes projects to a point within $\varepsilon$ of either $l(x)$ or $r(x)$ along a line of slope $1/\sqrt{3}$. In other words if $y$ is in one of the stripes every neighbourhood of $y$ intersects $B(x,\varepsilon)$, so $y\in\mathrm{cl}B(x,\varepsilon)$.

The Bing space is space n.75 in "Counterexamples in Topology" by Steen ad Seebach, where you can find the following illustration of the closure of a basic open set, together with many more properties of this space (there's also other connected countable Hausdorff spaces in the book, such as 61 and 125).

It should now be clear why every pair of basic open sets have intersecting closure and hence why the space is connected.