Issues with regularity: $Y$ is $T_1$ iff there is regular space $X$ s.t. all continuous function from $X$ to $Y$ is constant.

Question

I am interested in the following proposition:

$Y$ is $T_1$ iff there is regular space $X$ s.t. all continuous function from $X$ to $Y$ is constant.

The translation by Martin Sleziak of the paper Wann sind alle stetigen Abbildungen in Y konstant? written by Horst Herrlich contains the following construction:

Definition of a space $Q$. First we start with some given space $Y$.

• The spaces $R_i$ for $i=1,2$ and points $r_i\in R_i$ are constructed in such way that every continuous map from $R_i$ to $Y$ is constant on some neighborhood of $R_i$.
• A space $T=R_1\times R_2\setminus \{(r_1,r_2)\}$. This space has the property, that for every continuous map $f$ from $T$ to $Y$ there exist neighborhoods $U_i$ of $r_i$ such that $f$ is constant on $U_1\times U_2 - \{(r_1; r_2)\}$.
• We take countably many homeomorphic copies $T\times\{n\}$ of the space $T$ for each integer $n$. We add two new points $a$, $b$ with local neighborhood bases $\{\bigcup T_m; m\ge n\}\cup \{a\}\subseteq B$ and $\{\bigcup T_m; m\le n\}\cup \{b\}\subseteq B$, $\color{red}{\text{then we get a regular space } R}$.

I have issues with the red sentence. It seems to me that every neighbourhood of $a$ is sure to intersect $T_n$ for some $n\in \Bbb Z$. In particular, if the closed set $A$ has nonempty intersections with each $T_n$, then it is impossible to separate $A$ and $a$. Am I missing something obvious?

Answer 1

(Just to kick it out of the unanswered queue)

Thanks to @BrianM.Scott comment:

The problem does not arise: the definition of the topology ensures that a closed set that meets each $T_n$ already contains both $$ and $$, so of course it cannot be separated from them. A set that meets $T_n$for infinitely many positive integers $$ has $$ in its closure, and one that meets $T_n$ for infinitely many negative integers $n$ has $$ in its closure.