Embedding Thomas's plank à la Steen & Seebach into a space that is not completely regular

Question

In https://math.stackexchange.com/a/386811/32337, it is shown how to embed John Thomas's original "Thomas plank" into a regular space that is not completely regular. This is done by adjoining two new points to Thomas's plank and prescribing a suitable neighborhood base at each of these.

Can one make a similar construction — but using the modified version of Thomas's plank as presented in Example 93 of Steen and Seebach's Counterexamples in Topology?

The analog of https://math.stackexchange.com/a/386811/32337 for the Steen & Seebach version would be, naively, to:

• start with Steen & Seebach's modified Thomas plank $P = \bigcup_{n=0}^{\infty} L_{n}$, where $L_{n} = [0, 1) \times \{1/n\}$ for $n = 1, 2, \dots$, and $L_{0} = (0, 1) \times \{0\}$, with neighborhood bases as Steen and Seebach describe them (see also https://math.stackexchange.com/questions/4245107/what-is-the-correct-definition-of-the-topology-of-thomass-plank);
• form $X = P \cup \{p^{-}, p^{+}\}$ where $p^{-}$ and $p^{+}$ are distinct objects not already in $P$;
• take as a neighborhood base at $p^{-}$ sets of the form $U_{\alpha} = \{(x, y) \in P : x < \alpha\}$ for $0 < \alpha < 1$; and
• take as a neighborhood base at $p^{+}$ sets of the form $V_{\beta} = \{(x, y) \in P : x > \beta\}$ for $0 < \beta < 1$.

Unfortunately, this specification does not satisfy the requirements for neighborhood bases. Specifically, any such $U_{\alpha}$ contains the point $(0, 1)$, yet no basic open neighborhood of $(0, 1)$ can be contained in $U_{\alpha}$ (since such a basic open neighborhood has a finite complement in $L_{1}$).

Is it possible to suitably define neighborhood systems at the two new points so as to obtain a topology on $X$ that is still regular but not completely regular?

(Steen and Seebach do provide a different way of embedding their $P$ into a regular, non-completely regular space, not by adding two new points, but instead forming an analog of the "Tychonoff corkscrew.")

Answer 1

The construction you refer to does not use one copy of $P$ but infinitely many of them, indexed by $\mathbb{Z}$. You start with $P\times\mathbb{Z}$ and add two points, $p^+$ en $p^-$, to that product and let the basic neighbourhoods of $p^-$ be the sets $\{p^-\}\cup(P\times(-\infty,n])$, and those of $p^+$ the sets $\{p^+\}\cup(P\times[n,\infty))$. This space is still completely regular. Let $A$ be the set $\{(0,1/n):n\ge1\}$ and $B=\{(x,0):0<x\le1\}$. Now make a quotient of this space by identifying $(a,2k)$ and $(a,2k+1)$ whenever $a\in A$ and $k\in\mathbb{Z}$, and $(b,2k+1)$ and $(b,2k+2)$ whenever $b\in B$ and $k\in\mathbb{Z}$. See E. Hewitt, On two problems of Urysohn, Ann. Math. 47 (1946), 503-509

Note: this is the simpler example in the answer you quoted. The set $Y$ is the corkscrew over Thomas' plank. The points $(x,y)$ of $Y$ with $2n\le x\le2n+1$ provide a copy of $P$: the points $p_{2n+1,k}$ correspond to the set $A$ above and $\{2n\}\times[0,\frac12)$ corresponds to $B$ (and mirrorwise over the interval $[2n+1,2n]$.

What you want cannot be done: assume that we add two points $p$ and $q$ to $P$ and create a regular space. Then to $p$ and $q$ correspond families of open subsets of $P$, say $\mathcal{U}_p$ and $\mathcal{U}_q$, such that $\{U\cup\{x\}:U\in\mathcal{U}_x\}$ is a local base at $x$ (for $x=p,q$). Assume $p\in\bar{A}$, it then follows that $\bar{U}\cap\bar{B}$ is co-countable in $B$ for all $U\in\mathcal{U}_p$ (follow the proof that $P$ is not normal). And so $q\notin\bar A$ and there is $V\in\mathcal{U}_q$ such that $\bar V\cap A=\emptyset$. This imlies that $V$ is clopen in $P$ ($V$ meets every $L_n$ in a finite set) and so there is a continuous function that separates $p$ from $q$. If $p$ and $q$ are both not in $\bar A$ then both have neighbourhoods like $V$ above and, again, there is a continuous function that separates them.