In this paper https://www.ams.org/journals/proc/1981-081-04/S0002-9939-1981-0601748-4/S0002-9939-1981-0601748-4.pdf, Mysior gives a example of a regular space which is not completely regular, but I didn't understand the construction of space (X). Is there any easier way to visualize, such as a picture?

# A regular space which is not completely regular: Mysior’s example.

general-topology

#### Related Solutions

Yes, $X$ is completely Hausdorff. There are two nontrivial cases: separating two points of $\mathbb{R}\times\{0\}$ and separating a point of $\mathbb{R}\times\{0\}$ and $a$. For both of these cases, observe that for any $x\in\mathbb{R}$ the set $U_x=\{(x,0)\}\cup I_x\cup I_x'$ is clopen in $X$. Thus the characteristic function of $U_x$ is continuous and separates $(x,0)$ from every other point of $\mathbb{R}\times\{0\}$ and from $a$.

An interesting example is the subspace $X=A\cup B$ of the Niemitzki plane, where $A=\Bbb R\times \{0\}$ and $B=\{(a/n, 1/n):a\in\Bbb Z\land n\in\Bbb N\}.$ The subspace $A$ is closed in $X$ & discrete, with $|A|=2^{\aleph_0}.$ ... $B$ is a countable union of open singletons of $X$, so $B$ is also discrete. $B$ is dense in $X,$ so $X$ is not normal by the Jones Lemma. Any nbhd in $X$ of any $(r,0)\in A$ has an open-and-closed sub-nbhd $V=\{(r,0)\}\cup W$ where $W\subset B$ and $\overline W=V,$ so $V$ is homeomorphic to the real subspace $\{0\}\cup \{2^{-j}:j\in\Bbb N\}.$

In General Topology by R. Engelking, an example non-normal $Y=A\cup \Bbb N \subset \beta \Bbb N$ is given, where $A$ is a closed discrete subspace of $Y,$ with $A\subset \beta \Bbb N\setminus \Bbb N$ and $|A|=2^{\aleph_0}.$ I noticed that $Y$ is homeomorphic to $X$ of my 1st paragraph.

It seems the hypotheses of the Q are redundant. Wouldn't a locally metrizable space be $T_1$ and $T_{3\frac 1 2}$?

It would be interesting to see an example (if there is one) that does not involve the Jones Lemma.

BTW, in the Niemietzki plane, a basic open set $\{(r,0)\}\cup D,$ where $D$ is an open disc in $\Bbb R^+\times \Bbb R,$ of radius $s>0$ and centered at $(r,s),$ can also be seen to be metrizable by another strong metrization theorem: A $T_1,T_3$ space is metrizable iff it has a $\sigma$-locally finite base (basis). From this theorem, if $D$ is a metrizable open subspace of $E=D\cup \{p\}$ where $E$ is $T_1, T_3$ and $p$ has countable character in $E$, then $E$ is metrizable.

## Best Answer

The description is quite simple (it's two paragraphs); the set of points is just $\{(x,y)\in \Bbb R^2\mid y \ge 0\} \cup \{a\}$, where $a$ is some point outside the plane (sort of infinity, one could say).

All points in the plane part with $y>0$ are isolated (so lots of scattered dust there) and the "glue" is provided by points of the form $(x,0)$. A basic neighbourhood of $(x,0)$ is of the form

$$U((x,0),F) = \{(x,0)\} \cup \left( I_x:=\{(x,y)\mid 0 \le y < 2\} \cup I'_x:=\{(x,x+y)\mid 0 \le y < 2\right) \setminus F$$

where $F \subseteq I_x\cup I'_x$ is finite. $I_x$ is a fixed length vertical open-ended line segment emanating from $(x,0)$ and $I'_x$ is an angle-45-degrees open-ended line segment emanating from $(x,0)$ to the right side. The substracted finite set is just to ensure that the space will be Hausdorff as otherwise we could not separate $(x,0)$ from $(x,1)$ by open sets and now we can, by using $F=\{(x,1)\}\subseteq I_x$ e.g.

A basic neighbourhood of $a$ looks like $U_n(a)= \{a\} \cup \{(x,y)\mid y \ge 0, x > n\}$ where $n$ varies over $\Bbb N$ (so $a$ lies "to the right" of the half plane.

All this is nice and consistent (e.g. all $U_n(a)$ consist of interior points only when we consider the points in it, and likewise for the sets $U((x,0),F)$) so in total this defines all open sets in the usual way.

So visually it's not that hard: scattered upper plane points that can be in two-pronged open sets for the axis points and one infinity point on the right.

Hope this helps a little bit.