# Is Thomas' Corkscrew completely regular? (from Counterexamples in Topology)

In Example 94 of Counterexamples in Topology. In the example itself, it is written that the space is completely regular. But in the appendix at the end of the book, it is written that the space is regular but not completely regular. Which one is correct?

Thank you!

94. Thomas' Corkscrew

Let $X = \bigcup_{i=0}^\infty L_i$ be the union of lines in the plane where $L_0 = \{ (x,0) \mid x \in (0,1) \}$, and for $i \geq 1$, $L_i = \{ ( x , \frac{1}{i} \mid x \in [0,1) \}$. If $i > 0$, each point of $L_i$ except for $(0,\frac{1}{i})$ is open; basis neighborhoods of $(0,\frac{1}{i})$ are subsets of $L_i$ with finite complements. Similarly, the sets $U_i(x,0) = \{ (x,0) \} \cup \{ (x,\frac{1}{n}) \mid n > i \}$ form a basis for the points in $L_0$.

1. Every basis neighborhood of $X$ is closed as well as open, so $X$ is zero dimensional and therefore regular since it is clearly T1.

2. $X$ is also completely regular since if $C$ is a closed set and $p \notin C$, ...

## 1 Answer

Just so that this question doesn't go unanswered:

Steen and Seebach define both Thomas' Plank (Example 93) and Thomas' Corkscrew (Example 94) at essentially the same time. The first part of this section (which is quoted in the question) defines Thomas' Plank, and it is this space that is completely regular (as mentioned in point 2). Also note that in the table Example 93 is indicated to be completely regular (i.e., T0 and T3 1/2).

Only later do they give some indication of the construction of Thomas' Corkscrew, which I will quote (p.114 of the first edition):

5 If we use copies of Thomas' plank to build a corkscrew (as in the Tychonoff corkscrew) we can obtain a regular space which is not Urysohn, since every continuous function $f$ will be constant except for a countable set on coordinate axes at each level of the corkscrew. Thus if $p^+$ and $p^-$ are the infinity points, $f(p^+) = f(p^-)$.

Note that this implies that Thomas' Corkscrew is not completely regular, since $\{ p^- \}$ would be a closed set not containing $p^+$ which cannot be separated from $p^+$ by a continuous real-valued function.