Making Tychonoff Corkscrew in Counterexamples in Topology rigorous?

Question

I'm reading pages 109 and 110 of Seebach and Steen's Counterexamples in Topology (p. 61 here) and I don't understand one of their steps. In particular, at the bottom of page 109 they say, "by slitting each $P^\ast$ immediately below the positive $A_\Omega$ axis and then joining the fourth quadrant of the plane to the first quadrant of the one immediately below it."

Now I have two questions. First, how do you "join" the fourth quadrant of one plane to the first quadrant of the one below it?

Second, if you look at the picture it seems like they're actually joining the fourth quadrant of the one plane to the fourth quadrant of the plane beneath it. Is this a typo or is the picture very misleading? In this case it wouldn't be a vertical corkscrew but a slanted one. (EDIT: we're both folding the fourth quadrant of the top plane down and the first quadrant of the bottom plane up, connecting them at the north-south boundary, so I was wrong)

By the way my ultimate goal is to get a $T_3$-space which is not $T_{3\frac{1}{2}}$.

Answer 1

Index the copies of $P^*$ by the integers, so that you have $P^*\times\Bbb Z$, and let $P^*_k=P^*\times\{k\}$, the $k$-th level of the space, for each $k\in\Bbb Z$. Denote by $\langle\pm\alpha,\pm n,k\rangle$ the copy of $\langle\pm\alpha,\pm n\rangle$ in $P^*_k$.

So far you just have a stack of unrelated copies of $P^*$. The point $\langle\alpha,\omega,k\rangle$ on level $k$ is the limit from the ‘north’ of points $\langle\alpha,n,k\rangle$ and from the ‘south’ of points $\langle\alpha,-n,k\rangle$, all in level $k$. Now you angle the fourth quadrant of $P^*_k$ down so that the points $\langle\alpha,-n,k\rangle$ converge to $\langle\alpha,\omega,k-1\rangle$ instead of to $\langle\alpha,\omega,k\rangle$. You leave the first quadrant alone, however, so that the points $\langle\alpha,n,k\rangle$ still converge to $\langle\alpha,\omega,k\rangle$. The picture is correct: if you start in the first quadrant of the upper level in the picture, you can walk west into the second quadrant, then south into the third, then east into the fourth, and if you now walk north, when you cross that lower dashed line you’ll find yourself in the first quadrant of the lower level.

I suspect that Steen and Seebach chose this version of the construction because it looks the most like a real corkscrew, but it does make it a bit more difficult to give a rigorous description. Here’s one way to do it. Let $P$ now be the deleted Tikhonov plank. Let $Y=P\times\Bbb Z$, and let $P_k=P\times\{k\}$ for $k\in\Bbb Z$. The sets $P_{4k}$ for $k\in\Bbb Z$ will be the first quadrants of the S&S version; the sets $P_{4k+1}$ will be the second quadrants, the sets $P_{4k+2}$ the third quadrants, and the sets $P_{4k+3}$ the fourth quadrants. Define an equivalence relation $\sim$ on $Y$ as follows: $y\sim y$ for each $y\in Y$, of course, and in addition

• $\langle\Omega,n,4k\rangle\sim\langle\Omega,n,4k+1\rangle$ for $k\in\Bbb Z$ and $0\le n<\omega$;
• $\langle\alpha,\omega,4k+1\rangle\sim\langle\alpha,\omega,4k+2\rangle$ for $k\in\Bbb Z$ and $0\le\alpha<\Omega$;
• $\langle\Omega,n,4k+2\rangle\sim\langle\Omega,n,4k+3\rangle$ for $k\in\Bbb Z$ and $0\le n<\omega$; and
• $\langle\alpha,\omega,4k+3\rangle\sim\langle\alpha,\omega,4k-4$ for $k\in\Bbb Z$ and $0\le\alpha<\Omega$.

Now let $X=Y/\sim$, the quotient space.

The first line of the definition of the non-trivial part of $\sim$ sews $P_{4k}$ to $P_{4k+1}$ along the line that looks like the positive $y$-axis in the S&S pictures. The second sews the ‘second quadrant’ $P_{4k+1}$ to the ‘third quadrant’ $P_{4k+2}$. The third sews the ‘third quadrant’ $P_{4k+2}$ to the ‘fourth quadrant’ $P_{4k+3}$. And the last sews the ‘fourth quadrant’ $P_{4k+3}$ not to the ‘first quadrant’ $P_{4k}$, where we started, but to $P_{4k-4}$, the ‘first quadrant’ one place lower in the stack. Thus, the corkscrew (without the ideal points called $a^+$ and $a^-$ in S&S) is just a quotient space of the product space $P\times\Bbb Z$.

A simpler example of a $T_3$-space that’s not Tikhonov was constructed by John Thomas; apart from two ideal points, it lives in $\Bbb R^2$, and I recommend making a sketch.

For each even $n\in\Bbb Z$ let $L_n=\{n\}\times\left[0,\frac12\right)$. For each odd integer $n$ and each integer $k\ge 2$ let $p_{n,k}=\left\langle n,1-\frac1k\right\rangle$, and let $$T_{n,k}=\left\{\left\langle n\pm t,1-t-\frac1k\right\rangle:0<t\le1-\frac1k\right\}\;.$$ $T_{n,k}$ consists of the legs of an isosceles right triangle with its apex at $p_{n,k}$; the base of the triangle is not part of the space, but it’s the segment of the $x$-axis from $\left\langle n-1+\frac1k,0\right\rangle$ to $\left\langle n+1-\frac1k,0\right\rangle$, inclusive. Let $$Y_0=\bigcup_{n\in\Bbb Z}L_{2n}\;,$$ $$Y_1=\{p_{2n+1,k}:n\in\Bbb Z\text{ and }k\ge 2\}\;,$$ and $$Y_2=\bigcup\{T_{2n+1,k}:n\in\Bbb Z\text{ and }k\ge 2\}\;.$$ Let $Y=Y_0\cup Y_1\cup Y_2$.

Points of $Y_2$ are isolated. A nbhd of $p_{n,k}$ must contain all but at most finitely many points of the ‘tent’ $T_{n,k}$ whose peak it is. And a nbhd of a point $\langle n,y\rangle\in L_n$ must contain all but at most finitely many points of the line segment $(n-1,n+1)\times\{y\}$. (The points on this segment are the points at height $y$ on the righthand sides of the tents $T_{n-1,k}$ and the lefthand sides of the tents $T_{n+1,k}$.

Now let $X=Y\cup\{p^-,p^+\}$, where $p^-$ and $p^+$ are distinct new points. For each $\alpha\in\Bbb R$, $\{p^-\}\cup\{\langle x,y\rangle\in Y:x<\alpha\}$ is a nbhd of $p^-$, and $\{p^+\}\cup\{\langle x,y\rangle\in Y:x>\alpha\}$ is a nbhd of $p^+$. You’ll need to check, but the nbhd assigments really do specify a topology on $X$. (This is essentially Thomas’s original presentation of the space in A Regular Space, Not Completely Regular, The American Mathematical Monthly, Vol. 76, No. 2 (Feb., 1969), pp. 181-182. S&S give a different presentation of it as their example $94$.

It’s not too hard to show that if $f:X\to\Bbb R$ is continuous, then $f(p^-)=f(p^+)$. The key step is observing that if $f(p_{n,k})=\alpha$, then there are at most countably many points $z\in T_{n,k}$ such that $f(z)\ne\alpha$.

Answer 2

I haven't looked at this in too much detail, but it looks like the following's going on.

You start with a $\mathbb{Z}$-indexed stack of copies of $P$, which I'll write $P(i)$. Each $P(i)$ is slit along the positive $A_\Omega$-axis $\ell(i)$. What this presumably means is that you add a second copy of $\ell(i)$ -- let's call them $\ell(i)_+$ and $\ell(i)_-$ -- and adjoin extra open sets around the points in $\ell(i)_+$ that only intersect the upper half-plane, and likewise for $\ell(i)_-$. So what you have now is a 'plane' missing its origin and with its right half-plane split in two along the $A_\Omega$-axis, with two copies of that axis. You then glue each $\ell(i)_-$ to $\ell(i-1)_+$, which is what the picture is showing. This gives you the desired corkscrew. (I'm not quite sure what your confusion about this gluing is -- if you were gluing fourth quadrants to fourth quadrants, you wouldn't really get a screw, right?) The space $X$ they describe is this corkscrew with two points $a_+$ and $a_-$ added, which I'm guessing you understand.

As Steen and Seebach explain in points 1-3, this is regular but not completely regular. It's clearly locally regular around every point but $a_+$ and $a_-$; meanwhile, every closed set not containing $a_+$ must lie below some $P(i)$, and removing $P(i)$ splits that closed set from $a_+$, proving regularity at $a_+$ (likewise for $a_-$). On the other hand, if you have a map $f:X \to \mathbb{R}$, then since $\Omega$ is uncountable, $f$ is constant outside some countable interval $(-\alpha_i,\alpha_i)$ on the $i$th $A_\Omega$-axis. The supremum (or infimum, if you're using their weird backwards indexing) of the countably many $\alpha_i$ is a countable ordinal $\alpha$ with $f$ constant on each $A_\Omega$-axis outside $(-\alpha,\alpha)$. For $\beta > \alpha$, the copies of the point $(\beta, 0)$ on each $P(i)$ converge to $a_+$ as $i \to \infty$ and to $a_-$ as $i \to -\infty$, and they have the same value under $f$, so $f(a_+) = f(a_-)$, disproving complete regularity. If you want to get rid of $T_1$ you just double some points.