2
$\begingroup$

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 $Y=\mathbb{R}$.

$\endgroup$
4
  • $\begingroup$ If $Y = \Bbb{R}$ what topology are you equipping $Y$ with? The standard topology is not just $T_1$ $\endgroup$
    – graydad
    Dec 1, 2014 at 6:54
  • $\begingroup$ I consider the standard topology $\endgroup$
    – flourence
    Dec 1, 2014 at 7:01
  • $\begingroup$ 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. $\endgroup$
    – Mirko
    Dec 1, 2014 at 23:06
  • $\begingroup$ 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. $\endgroup$ Sep 18, 2015 at 10:58

2 Answers 2

5
$\begingroup$

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 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 $x'\in X'$ 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 $x'\in X'\setminus\{p^*\}$, and let $U$ be an open nbhd of $x'$. There is some $\langle\eta,y\rangle\in\kappa\times Y$ such that $x'=\pi\big(\langle\eta,y\rangle\big)$. 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.

$\endgroup$
3
  • $\begingroup$ 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? $\endgroup$
    – flourence
    Dec 3, 2014 at 4:30
  • $\begingroup$ @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. $\endgroup$ Dec 3, 2014 at 4:38
  • $\begingroup$ 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. $\endgroup$
    – Alex M.
    May 6, 2018 at 16:51
1
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$ Dec 1, 2014 at 21:45
  • $\begingroup$ @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. $\endgroup$
    – Mirko
    Dec 1, 2014 at 22:51
  • $\begingroup$ You’re welcome. $\endgroup$ Dec 1, 2014 at 22:53

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.