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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?

Josué Tonelli-Cueto
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    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: http://mathoverflow.net/questions/17371/regular-spaces-that-are-not-completely-regular . Hope this helps! –  Mar 04 '14 at 00:54

1 Answers1

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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.

user642796
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  • 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 $M_n$. – Josué Tonelli-Cueto Mar 06 '14 at 22:27
  • -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. – Josué Tonelli-Cueto Mar 06 '14 at 22:33
  • 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. – Josué Tonelli-Cueto Mar 06 '14 at 22:35
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    @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. – user642796 Feb 25 '15 at 18:47
  • @user642796: Something is wrong with your proffered proof: You have at the end $B_{n+1} \subset U_n$. But $B_{n+1}$ is a set of $x$-values whereas $U_n$ is a subset of the plane (together with the adjointed point $a$). Do you mean instead to define $B_n = {(x, 0) : x \in (n-1, n], f(x, 0) = 1}$? – murray May 24 '22 at 21:40
  • @user642796: Also, it seems that all that is really needed to get the contradiction is that $B_n$ is nonempty for each $n$. Indeed, then for $m$ with $U_m \subset f^{-1}([0, 1))$ we have some $x \in B_{m+1}$ with $(x, 0) \in U_m$ whence $f(x,0) = 1$, yet $f(x,0) < 1$. – murray May 25 '22 at 01:09
  • @user642796: Your comment about how to get $C_x$ seems to run afoul of the logic: You have a fixed $n$ within the induction and are trying to deduce that $B_{n+1}$ is infinite from the assumption tht $B_n$ is infinite. Yet your construction of $C_x$ has $n$ running through all positive integers. – murray May 25 '22 at 17:22
  • @user642796: Also, since your $A_{x, n}$ is finite, it could be empty, in which case it could be that your $C_x$ is empty as well, despite taking a union over denumerably many indices $n$. – murray May 25 '22 at 17:36
  • @user642796: Moreover, the set $C_x$ you produce in your comment is a subset of the upper half-plane but not a subset of the horizontal axis. And this conflicts with your claim inside the induction that $C_x \subset [0, 2)$. – murray May 26 '22 at 01:04
  • @user642796: I think in your definition you actually want $B_n = {x \in [n-1, n] : f(x, 0) = 1}$, i.e., interval closed at both ends, in order to prove that $B_1$ is infinite, since to do that you want to use that the projection $E = [0, 1]$ of the set $F$ onto the horizontal axis is contained in $B_1$. – murray May 26 '22 at 15:43
  • This discussion is continued in https://math.stackexchange.com/q/4458777/32337. – murray May 28 '22 at 14:02
  • A detailed complete and correct proof appears in https://math.stackexchange.com/a/3925760/32337. – murray May 29 '22 at 16:49