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I am trying to understand this article. This article states that $Y$ is $T_1$ space iff there exists a regular space $X$ (having at least two points), such that every continuous map from $X$ to $Y$ is constant.

I do not understand the construction of space $X$. Please anyone help me to show what regular space $X$ that satisfies that condition if $Y=\mathbb{R}$.

Antonio Maria Di Mauro
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flourence
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  • If $Y = \Bbb{R}$ what topology are you equipping $Y$ with? The standard topology is not just $T_1$ – graydad Dec 01 '14 at 06:54
  • I consider the standard topology – flourence Dec 01 '14 at 07:01
  • I edited my answer to reflect some corrections pointed out to me by @BrianM.Scott (and a link to a related answer of his), and it also has more references now. – Mirko Dec 01 '14 at 23:06
  • It is worth mentioning that the paper you are linking to is translation of the paper Horst Herrlich: Wann sind alle stetigen Abbildungen in $Y$ konstant? I have made the translation based on the request of the poster of this question. – Martin Sleziak Sep 18 '15 at 10:58

2 Answers2

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Eric van Douwen’s paper ‘A regular space on which every continuous real-valued function is constant’, Nieuw Arch. Wisk. $30$ $(1972)$, $143$-$145$, actually gives a ‘machine’ for starting with a $T_3$ space having two points that cannot be separated by a continuous real-valued function and producing from it a $T_3$ space on which all continuous real-valued functions are constant. I no longer have a copy of the paper or easy access to it, so I’m working from memory, but the construction below is either the same as or very similar to Eric’s.

Let $Z$ be any $T_3$ space with points $p$ and $q$ that cannot be separated by a continuous real-valued function. Let $Y=Z\setminus\{p,q\}$, and let $\kappa=|Y|$. Give $\kappa$ the discrete topology, and let $X=\kappa\times Z$ with the resulting product topology. For notational convenience let $X_\xi=\{\xi\}\times Z$, $p_\xi=\langle\xi,p\rangle$, and $q_\xi=\langle\xi,q\rangle$ for $\xi<\kappa$. Clearly $\kappa$ is infinite, so there is a bijection $\varphi:\kappa\to\kappa\times Y\subseteq X$. We may (and do) further assume that if $\varphi(\xi)=\langle\eta,y\rangle$, then $\eta\ne\xi$.

Define an equivalence relation $\sim$ on $X$ as follows.

  • Of course $x\sim x$ for all $x\in X$.
  • We identify all copies of $p$ to a single point: $p_\xi\sim p_\eta$ for all $\xi,\eta<\kappa$.
  • For all $\xi<\kappa$, $q_\xi\sim\varphi(\xi)$; this identifies each point of $\kappa\times Y$ with a unique $q_\xi$.

Let $X'=X/\!\sim$; we hope to show that $X'$ is $T_3$, and that every real-valued continuous function on $X'$ is constant.

Think of $Z$ as a string with endpoints $p$ and $q$. The idea of $\sim$ is to glue all of the copies of $p$ together into one point $p^*$, and then to run a separate string from $p^*$ to each other point, gluing the $q$ end of that string to the point. (I’ve always thought of this construction as Eric’s Spaghetti Machine.) This ought to ensure that if $x$ and $y$ are any points of $X'$, and $f:X'\to\Bbb R$ is continuous, the string from $p^*$ to $x$ ensures that $f(x)=f(p^*)$, while the string from $p^*$ to $y$ ensures that $f(y)=f(p^*)$, so that $f(x)=f(y)$.

And in fact it does. Let $\pi:X\to X'$ be the canonical quotient map, and let $p^*=\pi(p_0)$ (which is of course the same as $\pi(p_\xi)$ for each $\xi<\kappa$). If $f:X'\to\Bbb R$ is continuous, then $f\circ\pi:X\to\Bbb R$ is continuous. Let $g=f\circ\pi$; then $g(p_\xi)=g(q_\xi)$ for all $\xi<\kappa$, since $g\upharpoonright X_\xi$ is a continuous real-valued function on $X_\xi$, which is homeomorphic to $Z$. Let $x'\in X'\setminus \{p^*\}$ be arbitrary. Then $x'=\pi\big(\langle\eta,y\rangle\big)$ for some $\langle\eta,y\rangle\in\kappa\times Y$, and there is a $\xi\in\kappa\setminus\{\eta\}$ such that $\varphi(\xi)=\langle\eta,y\rangle$. Then $\pi(q_\xi)=\pi\big(\varphi(\xi)\big)=\pi\big(\langle\eta,y\rangle\big)$, so $f(x')=g\big(\langle\eta,y\rangle\big)=g(q_\xi)=g(p_\xi)=f(p^*)$, and $f$ is indeed constant on $X'$.

It’s actually harder to show that $X'$ is $T_3$.

Let $x'\in X'\setminus\{p^*\}$; there is some $\langle\eta,y\rangle\in\kappa\times Y$ such that $x'=\pi\big(\langle\eta,y\rangle\big)$. Let $U$ be an open nbhd of $x'$; the $\sim$-equivalence classes are closed in $X$, so $X'$ is $T_1$, and without loss of generality we may assume that $\pi(q_\eta)\notin U$. Let $V_0=X_\eta\cap\pi^{-1}[U]$; $V_0$ is an open nbhd of $y$ in $X$. Let

$$K_0=\{\xi<\kappa:\varphi(\xi)\in V_0\}=\{\xi<\kappa:\pi(q_\xi)\in\pi[V_0]\}\;,$$

and let $$V_1=\bigcup_{\xi\in K_0}\big(X_\xi\cap\pi^{-1}[U]\big)\;.$$

In general, given $V_n$ for some $n\in\omega$, let

$$K_n=\{\xi<\kappa:\varphi(\xi)\in V_n\}=\{\xi<\kappa:\pi(q_\xi)\in\pi[V_n]\}\;,$$

and let $$V_{n+1}=\bigcup_{\xi\in K_n}\big(X_\xi\cap\pi^{-1}[U]\big)\;.$$

Finally, let $V=\bigcup_{n\in\omega}V_n$; then $V=\pi^{-1}[U]$.

There is an open nbhd $W_0$ of $\langle\eta,y\rangle$ in $X$ such that $\operatorname{cl}_XW_0\subseteq V_0$. For each $\xi\in\bigcup_{n\in\omega}K_n$ there is an open nbhd $G_\xi$ of $q_\xi$ in $X$ such that $G_\xi\subseteq X_\xi$, and $\operatorname{cl}_XG_\xi\subseteq\pi^{-1}[U]$. Given $W_n$ for some $n\in\omega$, let $L_n=\{\xi<\kappa:\pi(q_\xi)\in\pi[W_n]\}$, and let

$$W_{n+1}=W_n\cup\bigcup_{\xi\in L_n}G_\xi\;.$$

Then $W=\bigcup_{n\in\omega}W_n$ is open in $X$, and $W=\pi^{-1}\big[\pi[W]\big]$, so $\pi[W]$ is an open nbhd of $x'$ in $X'$. Let $H_0=\operatorname{cl}_XW_0$. Given $H_n$ for some $n\in\omega$, let $M_n=\{\xi<\kappa:q_\xi\in\pi[H_n]\}$, and let

$$H_{n+1}=H_n\cup\bigcup_{\xi\in M_n}\operatorname{cl}_XG_\xi\;.$$

Then $H=\bigcup_{n\in\omega}H_n$ is closed in $X$, and $\pi^{-1}\big[\pi[H]\big]=H$, so $\pi[H]$ is closed in $X'$. Clearly $\pi[W]\subseteq\pi[H]\subseteq U$, so

$$x'\in\pi[W]\subseteq\operatorname{cl}_{X'}\pi[W]\subseteq\pi[H]\subseteq U\;.$$

I leave it to you to modify the argument slightly to show that $X'$ is $T_3$ at the point $p^*$ as well.

Brian M. Scott
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  • thank you for the proof. But can you so me what a continuous map define on space that is defined in A. Mysior's paper is not constant? – flourence Dec 03 '14 at 04:30
  • @flourence: Let $f(\langle x,1\rangle)=1$ for every $x\in\Bbb R$, and let $f$ send every other point of the space to $0$; then $f$ is continuous but not constant. – Brian M. Scott Dec 03 '14 at 04:38
  • This is funny: searching for the title of van Douwen article I stumbled upon one with the very same name, but by T.E. Gantner. It can be read here. – Alex M. May 06 '18 at 16:51
  • Can you explain why you write $K_0 = ... = {\xi < \kappa : \pi(q_\xi)\in \pi[V_0]}$? From what I understand, you're using that $q_\eta\notin V_0$. Why is that? – Jakobian Aug 04 '23 at 18:44
  • @Jakobian: For $\xi\in K_0$ we have $q_\xi\sim\varphi(\xi)\in V_0$, so $\pi(q_\xi)=\pi(\varphi(\xi))\subseteq\pi[V_0]$. – Brian M. Scott Aug 04 '23 at 19:33
  • It's true that $\xi\in K_0$ implies that $\pi(q_\xi)\in \pi[V_0]$, but I'm asking why the opposite implication is true – Jakobian Aug 04 '23 at 19:37
  • @Jakobian: If $\pi(q_\xi)\in\pi[V_0]$, then $q_\xi\sim\langle\eta,w\rangle$ for some $\langle\eta,w\rangle\in V_0$. But $\varphi(\xi)$ is the unique element of $\kappa\times Y$ equivalent to $q_\xi$, so $\varphi(x_\xi)=\langle\eta,w\rangle\in V_0$. – Brian M. Scott Aug 04 '23 at 21:35
  • @BrianM.Scott Wouldn't that only be true if $\langle \eta, w\rangle\in \kappa\times Y$ in the first place? I don't think that's a given. – Jakobian Aug 04 '23 at 21:47
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    @Jakobian: Okay, I think that I’ve finally figured out what the problem is: you’re worried about the possibility that $\pi(q_\eta)\in U$. Yes, that’s an oversight, but it’s easily fixed: the $\sim$-classes in $X$ are closed, so $X'$ is $T_1$, and without loss of generality we can assume that $\pi(q_\eta)\notin U$. I may not get to it until this evening (it’s 15:20 for me now), but I’ll make the necessary small modification at some point today. – Brian M. Scott Aug 05 '23 at 19:21
  • I also have doubts why the second equality after definition of $K_n$ holds. And also the claimed equality $V = \pi^{-1}[U]$. Could you explain those? – Jakobian Aug 10 '23 at 13:16
  • @Jakobian: Until this month I’d not thought about this example in nearly nine years; it would be helpful and save me some time if you could indicate exactly where you see a problem with each of those equations. – Brian M. Scott Aug 10 '23 at 19:36
  • The issue is similar as with $K_0$. We need $\pi(q_\xi)\in \pi[V_n]$ to imply $\varphi(\xi)\in V_n$. For this we need $q_\xi\notin V_n$. This was easy to fix for $n = 0$. For $n > 0$ we'd need something else, I don't know precisely, maybe a proof by induction, or whole modification of the proof, but we need to prevent $q_\xi$ for $\xi\in K_{n-1}$ to belong to $V_n$ somehow. I don't know if this is even salvageable, there is a lot of unproven claims here that follow from nowhere. I don't know if this is important, so maybe we could just delete all of them from $V_n$. – Jakobian Aug 11 '23 at 12:23
  • Since they belong to different $X_\xi$, the set ${q_\xi : \xi\in K_{n-1}}$ is still closed, so we could do that without modifying that $V_n$ is open, but I don't know if it doesn't break the proof somewhere along the way. – Jakobian Aug 11 '23 at 12:28
  • I had some help and I found this article by V. Tzannes. They say it's a modification of the argument by van Douwen, so it should help – Jakobian Aug 12 '23 at 21:05
  • @Jakobian: My apologies: I’ve been away for a while dealing with other more urgent things and haven’t had a chance to get back to this. I’ll try to get to it in the next few days. – Brian M. Scott Aug 31 '23 at 07:20
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Edit. My answer below are not precise:
Please read the comment posted by @BrianM.Scott after my answer to see what I really should have said,
as well as to a link to his detailed and self-contained answer to a related question (and for more references).

One such simple example is by Mysior,
A. Mysior, A regular space which is not completely regular
Proc. Amer. Math. Soc. 81 (1981), 652-653
http://www.ams.org/journals/proc/1981-081-04/S0002-9939-1981-0601748-4/
It is only a two-page paper with the example described on the first page, and a remark based on it on the second page showing the existence of a regular space such that every real-valued function defined on it is constant. Correction. The example by Mysior is of a regular space in which there are two points $a,b$ with the property that $f(a)=f(b)$ for every continuous real valued function $f$. This example could be used to construct a regular space on which every continuous real-valued function is constant. See the comment below by @BrianM.Scott. You may also see Exercises 2.7.17 and 2.7.18 in Engelking, General Topology (1989) (more references there).

Another one is the Tychonoff corkscrew, in Counterexamples in Topology, will try to find a link in a minute. This one used to be the standard example, but the example by Mysior is easier to follow.

Can't find the best link, but the Tychonoff corkscrew is based on the Tychonoff plank (it takes countably many Tychonoff planks and glues them together in a certain way). The Tychonoff plank is based on the product $[0,\omega]\times [0,\omega_1]$. These and some related examples (examples 86-91) are described in
Counterexamples in Topology (1970, 2nd ed. 1978) Lynn Steen and J. Arthur Seebach, Jr.

Mirko
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  • None of these is an example of a space on which all real-valued continuous functions are constant: they are examples of $T_3$ spaces with points $p$ and $q$ that cannot be separated by a continuous real-valued function. I describe another, due to Thomas, in this answer. Any of them can by used to construct the desired example, using the technique in the van Douwen paper cited by Mysior. – Brian M. Scott Dec 01 '14 at 21:45
  • @BrianM.Scott Of course you are right, thank you for correcting me, and thank you for providing the link to your answer with the example due to Thomas. – Mirko Dec 01 '14 at 22:51
  • You’re welcome. – Brian M. Scott Dec 01 '14 at 22:53