While studying some examples of regular spaces which are not completely regular, I came across Steen's and Seebach's "Counterexamples in Topology". In this book, after searching for examples, I only found four examples of regular not completely regular spaces: Tychonoff Corkscrew, deleted Tychonoff Corkscrew, Hewitt's Condensed Corkscrew and Thomas' Corkscrew.

All this examples are variations of the same example: the Tychonoff Corkscrew. So my question is the following one: is there a regular not completely regular space which is not a corkscrew? Or in other way, is it possible to define the property of being a corkscrew is some sense in order to have a result of the form that every regular space is completely regular except if it is a corkscrew?

Alternatively, is it possible to construct a regular not completely regular space which is not a corkscrew in the sense of being radically different -in the sense that use totally different ideas and it is not homeomorphic to the cited ones- from the previous examples?

  • 1
    $\begingroup$ Pretty interesting question. Simply put, my interpretation of your question is to find a counterexample to the following conjecture: the Tychonoff corkscrew is the only regular not completely regular space. EDIT: I was searching for this on Google, and found this MO question: mathoverflow.net/questions/17371/… . Hope this helps! $\endgroup$ – user122283 Mar 4 '14 at 0:54

The following example (which I am fairly certain is not a corkscrew-like construction) is essentially taken from

A. Mysior, A regular space which is not completely regular, Proc. Amer. Math. Soc. 81 (1981), pp.652-653, MR601748, AMS link

Let $X = (\;\mathbb{R} \times [ 0 , 2 )\;) \cup \{ \langle 0 , -1 \rangle \}$, and topologise $X$ as follows:

  • all points in $\mathbb{R} \times ( 0 , 2 )$ are isolated;
  • for $\langle x , 0 \rangle$ the basic open neighbourhoods are of the form $V_x \setminus A$ where $V_x = \{ \langle x,y \rangle : 0 \leq y \leq 2 \} \cup \{ \langle x + y , y \rangle : 0 \leq y < 2 \}$, and $A \subseteq V_x \setminus \{ \langle x,0 \rangle \}$ is finite; and
  • the basic open neighbourhoods of $\langle 0 , -1 \rangle$ are of the form $U_n = \{ \langle 0 , -1 \rangle \} \cup \{ \langle x , y \rangle \in \mathbb{R} \times [0,2) : x > n \}$ for $n \in \mathbb{N}$.

It is fairly easy to check that $X$ is regular, the basic idea being that the basic open neighbourhoods of the points in $\mathbb{R} \times [ 0 , 2 )$ described above are actually clopen, and $\overline{U_{n+2}} \subseteq U_n$ for all $n \in \mathbb{N}$.

However there is no continuous $f : X \to [0,1]$ such that $f(0,-1) = 0$ and $F = \{ \langle x , 0 \rangle \in X : 0 \leq x \leq 1 \} \subseteq f^{-1} [ \{ 1 \} ]$. The basic idea here is to show that if $F \subseteq f^{-1} [ \{ 1 \} ]$, then $f( 0 , -1 ) = 1$. For this we first prove by induction that $B_n = \{ x \in ( n-1 , n ] : f( x , 0 ) = 1 \}$ is infinite for all $n \geq 1$.

For $n = 1$ the result is trivial. So suppose that $B_n$ is infinite for some $n \geq 1$. For each $x \in B_n$, one can show that there is a countable $C_x \subseteq [0,2)$ such that $f( x + y , y ) = 1$ for all $y \in [ 0 , 2 ) \setminus C_x$. Fixing a countably infinite $B^\prime \subseteq B_n$, it follows that $C = \bigcup_{x \in B^\prime} \{ x + y : y \in C_x \}$ is countable, and so $(n,n+1] \setminus C$ is (uncountably) infinite. For $z \in [n,n+1] \setminus C$, note that for each $x \in B^\prime$ we have that $z = x + ( z - x ) \notin C_x$, and so $f( z , z-x ) = 1$. Since there are infinitely many $\langle z,y \rangle \in V_z$ such that $f(z,y) = 0$, by continuity it follows that $f ( z, 0 ) = 1$. Therefore $B_{n+1} \supseteq (n,n+1] \setminus C$ is infinite.

Now note that if $f ( 0,-1 ) \neq 1$, then there would be an $n \geq 1$ such that $U_n \subseteq f^{-1} [\;[0,1)\;]$, however as $B_{n+1} \subseteq U_n$ this is impossible.

  • $\begingroup$ After thinking some days, I think the following reformulation make it similar to a corkscrew. Say me if you agree. We consider the "Mysior's plank" given by $X=[0,1]\times [0,2)$, where the points $(x,y)$ with $y>0$ are isolated and those with $y=0$ have as neighborhoods of the form $V_x\backslash A$, with $V_x=\{(x,t)\,|\,t\in[0,2)\}$ and $A$ finite. Then we paste this "Mysior's planks" by dividing them by the diagonal and identifying one trinagle with one of the next in the following way: -Put them in a sequence of planks $M_n$. $\endgroup$ – Josué Tonelli-Cueto Mar 6 '14 at 22:27
  • $\begingroup$ -If $n$ is even, consider the triangles define by the diagonal that goes from bottom-left to up-right and otherwise, if $n$ is odd, the diagonal that goes from up-left to bottom-right. -After this, identify the triangles of $M_n$ and $M_{n+1}$ that have the horizontal edge int he bottom part if $n$ is even by identifying horizontal edges and each diagonal with the vertical one and similarly the ones with the horizontal edges in the upper part if $n$ is odd. Finally, add the infinity point. $\endgroup$ – Josué Tonelli-Cueto Mar 6 '14 at 22:33
  • $\begingroup$ In this sense, I feel that it is the same idea since you are pasting planks in an appropriate way to arrive to an infinity point in the end. Do you agree? Comment please. $\endgroup$ – Josué Tonelli-Cueto Mar 6 '14 at 22:35
  • 1
    $\begingroup$ @SomeStrangeUser: Note that given $x \in B_n$ for each $n > 0$ there must be a basic open set of the form $V_x \setminus A_{x,n}$ such that $| f(u,v) - 1 | < \frac{1}{n}$ for all $\langle u,v \rangle \in V_x \setminus A_{x,n}$. Taking $C_x = \bigcup_n A_{x,n}$ will give the desired countable set. $\endgroup$ – user642796 Feb 25 '15 at 18:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.