%\documentclass{article}
\documentclass{amsart}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{amssymb}
\usepackage{amsfonts}
\usepackage{hyperref}
\usepackage{paralist}
\usepackage{amsfonts}
\usepackage{mathrsfs}
\usepackage{graphicx}
\usepackage{url}
\usepackage{comment}
\usepackage{xspace}
%\usepackage[nolabel]{showlabels}
%\usepackage{showlabels}
%\usepackage{slovak}

\numberwithin{equation}{section}



%\excludecomment{mine}
\includecomment{mine}
%\excludecomment{detail}
\newenvironment{detail}{\begin{footnotesize}(}{)\end{footnotesize}}


%theorems
\newtheorem*{THM}{Theorem}
\newtheorem*{REM}{Remark}


\newcommand{\powerset}[1]{\mc P(#1)}
\newcommand{\Zobr}[3]{\ensuremath{#1\colon #2\to #3}}
\newcommand{\Zobrto}[3]{\ensuremath{#1\colon #2\mapsto #3}}
\newcommand{\RR}{\ensuremath{\mathbb R}}
\newcommand{\NN}{\ensuremath{\mathbb N}}
\newcommand{\vp}{\ensuremath{\varphi}}
\newcommand{\ve}{\ensuremath{\varepsilon}}
\newcommand{\sm}{\ensuremath{\setminus}}
\newcommand{\emps}{\ensuremath{\emptyset}}

\newcommand{\limti}[1]{\lim\limits_{#1\to\infty}}
\newcommand{\limtol}[2]{\lim\limits_{{#1} \to {#2}^-}}
\newcommand{\dcc}[1]{\ensuremath{\lfloor #1 \rfloor}} %dolna cela cast
\newcommand{\hcc}[1]{\ensuremath{\lceil #1 \rceil}} %horna cela cast
\newcommand{\limsupti}[1]{\limsup\limits_{#1\to\infty}}
\newcommand{\liminfti}[1]{\liminf\limits_{#1\to\infty}}
\newcommand{\Ra}{\ensuremath{\Rightarrow}}
\newcommand{\Lra}{\ensuremath{\Leftrightarrow}}
\newcommand{\abs}[1]{\lvert#1\rvert}
\newcommand{\absl}[1]{\left\lvert#1\right\rvert}
\newcommand{\norm}[1]{\lVert#1\rVert}
\newcommand{\ol}[1]{\ensuremath{\overline{#1}}}
\newcommand{\ul}[1]{\ensuremath{\underline{#1}}}
\newcommand{\inv}[1]{#1^{-1}}
\newcommand{\Invobr}[2]{\inv{#1}(#2)}
\newcommand{\Obr}[2]{#1[#2]}
%\newcommand{\intrv}[2]{\ensuremath{\langle #1,#2 \rangle}}
%\newcommand{\intrvl}[2]{\ensuremath{\langle #1,#2 )}}
\newcommand{\intrv}[2]{\ensuremath{[#1,#2]}}
\newcommand{\intrvr}[2]{(#1,#2]}
\newcommand{\intrvl}[2]{\ensuremath{[ #1,#2 )}}
\newcommand{\seqq}[3]{({#1}_{#2})_{#3=1}^\infty}
\newcommand{\seq}[2]{\seqq{#1}{#2}{#2}}


\newcommand{\mc}[1]{\ensuremath{\mathcal{#1}}}
\newcommand{\mfr}[1]{\ensuremath{\mathfrak{#1}}}

\newcommand{\Dots}[3]{\ensuremath{#1_{#2}\ldots#1_{#3}}}
\newcommand{\Pdots}[3]{\ensuremath{#1_{#2}+\dots+#1_{#3}}}
\newcommand{\Cdots}[3]{\ensuremath{#1_{#2}\cdot\ldots\cdot#1_{#3}}}
\newcommand{\Ldots}[3]{\ensuremath{#1_{#2},\ldots,#1_{#3}}}







\newcommand{\enu}{\renewcommand{\theenumi}{\roman{enumi}}\renewcommand{\labelenumi}{{\rm (\theenumi)}}}
%\newcommand{\enu}{\renewcommand{\theenumi}{\roman{enumi}}\renewcommand{\labelenumi}{\textrm{(\theenumi)}}}
%\newcommand{\enu}{\renewcommand{\theenumi}{\roman{enumi}}\renewcommand{\labelenumi}{\mathrm{(\theenumi)}}}

\begin{document}


\author{Horst Herrlich}

\title{When are all continuous functions to $Y$ constant?}

\maketitle

Translation of \cite{HERKONST} (H. Herrlich: Wann sind alle stetigen Abbildungen in $Y$ konstant?).

Urysohn \cite{URYSOHNMACHTIGKEIT} asked whether for every regular space $X$ (having at least two points) there is a non-constant continuous map from $X$ do the space $Y$ of real numbers. This question was negatively answered by Hewitt \cite{HEWITT2URYSOHN}, Nov\'ak \cite{NOVAKREGULARCONSTANT} and Van Est-Freudenthal \cite{VANESTFREUDENTHAL}. The methods used by these authors (which go back to Tychonoff \cite{TYCHONOFFERWEITERUNG}) let us show relatively easy the following result:

\begin{THM}
Let $Y$ be a topological space. The following conditions are equivalent:\\
\emph{(a)} $Y$ is a $T_1$-space,\\
\emph{(b)} there exists a regular space $X$ (having at least two points), such that every continuous map from $X$ to $Y$ is constant.
\end{THM}

\begin{proof}
(b) \Ra (a). If $Y$ is not a $T_1$-space, then there are two different points in $Y$ such that $a$ is contained in every neighborhood of $b$. If $X$ is a regular space with at least two points, $B$ an open subset of $X$ such that $B\ne\emps$, $B\ne X$, then the function from $X$ to $Y$ defined as
$$f(x)=
  \begin{cases}
    a & \text{for} x\in B, \\
    b & \text{for} x\notin B
  \end{cases}
$$
is continuous and not constant.

(a) \Ra (b). Let $Y$ be a $T_1$-space. The construction of the space $X$ with the desired properties will be done in several steps.
First we construct for $i=1,2$ the spaces $R_{i}$ with a distinguished point $r_i$ in a such way that every continuous map from $R_i$ to $Y$ is constant on some neighborhood of $R_i$. Then we show that the space $T$, which is obtained from the product $R_1\times R_2$ by removing the point $(r_1;r_2)$,
has the property, that for every continuous map $f$ from $T$ to $Y$ there exist neighborhoods $U_i$ of $r_i$ such that $f$ is constant on $U_1\times U_2 - \{(r_1; r_2)\}$. Through gluing together countably many homeomorphic copies of the space $T$ and addition of two points $a$, $b$ we obtain a space $Q$ with the property, that every continuous map from $Q$ to $Y$ has the same value at the point $a$ as at the point $b$.
With the help of $Q$ we will show that every (regular) space $Z$ is embeddable into a (regular) space $Q(Z)$ such that every continuous map from $Q(Z)$ to $Y$ is constant on $Z$. This easily implies the existence of a space $Z$ with the properties required in (b).

1. Construction of spaces $R_i$. Let $\abs{Y}\le\aleph_\alpha$ and for $i=1,2$ let $R_i$ be a set with $\abs{R_i}=\aleph_{\alpha+i}$ and let $r_i$ be a fixed point from $R_i$. We declare a subset of $R_i$ to be open if and only if $\abs{R_i-B}<\aleph_{\alpha+1}$ whenever $r_i\in B$. Then $R_i$ is a regular space with the property that the intersection of $\aleph_{\alpha+i-1}$ neighborhoods of $r_i$ is again a neighborhood of $r_i$. If $f$ is a continuous map from $R_i$ to $Y$, then $\inv f(y_0) = \bigcap \{\Invobr f{Y-\{y\}}; y\in Y-\{y_0\}\}$ again a neighborhood of $r_i$, since it is an intersection of $\aleph_\alpha$ neighborhoods of $r_i$. Therefore $f$ is constant on a neighborhood of $r_i$.

2. Proof of the properties of $T$. Let $f$ be a continuous map from $T$ to $Y$. Then for every $x\in R_1-\{r_1\}$ there exists a neighborhood $U_x$ of $r_2$ such that $f$ is constant on $\{x\}\times U_x$. $V_2=\bigcap \{U_x; x\in R_1-\{r_1\}\}$ is a neighborhood or $r_2$.
If $y$ is an arbitrary element of $V_2-\{r_2\}$, then there is a neighborhood $V_1$ of $r_1$, such that $f$ is constant on  $V_1\times\{y\}$. Therefore $f$ is constant on $(V_1\times V_2)-\{(r_1;r_2)\}$.

3. Construction of the space $Q$. Let assign to every number $n$ a space $T_n$, which is homeomorphic to $T$, and whose elements will be denoted by $(x;y;n)$. We add to the topological sum of the spaces $T_n$ two points $a$, $b$ and define the neighborhoods of $a$ (resp. $b$) as all sets for which there exists an integer $n$ with $\{\bigcup T_m; m\ge n\}\cup \{a\}\subseteq B$ (resp. $\{\bigcup T_m; m\ge n\}\cup \{b\}\subseteq B$), then we get a regular space $R$.

If we identify in $R$

1. for every even $n$ and every $x\in R_1-\{r_1\}$ the two points $(x;r_2;n)$ and $(x;r_2;n+1)$,

2. for every odd $n$ and every $x\in R_2-\{r_2\}$ the two points $(r_1;x;n)$ and $(r_1;x;n+1)$,

then we obtain a regular space $Q$ with the property such that $f(a)=f(b)$ holds for every continuous map from $Q$ to $Y$.

4. Construction of $Q(Z)$ for an arbitrary space $Z$. A subset $B$ of $Z\times Q$ is open in $Z\times Q$ if and only if the following holds:
\begin{enumerate}
  \item If $(z;x)$ is an element of $B$, then there is a neighborhood $U$ of $x$ in $Q$ with $\{z\}\times U\subset B$.
  \item If $(z;a)$ is an element of $B$, then there is a neighborhood $U$ of $z$ in $Z$ with $U\times\{a\}\subset B$.
\end{enumerate}
If we identify in the above space $Z\times Q$ all points of the set $Z\times\{b\}$, then we obtain a space $Q(Z)$. The assignment $h(z)=(z;a)$ defines a homeomorphic map $h$ from $Z$ into $Q(Z)$. So we can assume that $Z$ is canonically embedded into $Q(Z)$. If $f$ is a continuous map from $Q(Z)$ to $Y$, then $f$ is constant on $Z$. If $Z$ is regular, then so is $Q(Z)$.

5. Definition of $X$. Let $X_0$ be a singleton. By induction we define $X_{n+1}=Q(X_n)$. Then $X_0\subset X_1\subset X_2\subset \dots $ are regular spaces. Let a subset of $X=\bigcup\{X_n; n=0,1,\dots\}$ be open if and only if $B\cap X_n$ is open for every $n$. The space $X$ is a regular space. Every continuous map from $X$ to $Y$ is constant.
\end{proof}

\begin{REM}
The above results has a trivial analogue:\\
Let $X$ be a topological space. The following conditions are equivalent:\\
\emph{(a)} $X$ is connected,\\
\emph{(b)} there is a regular space $Y$ (having at least 2 points), such that every continuous map from $X$ to $Y$ is constant.
\end{REM}


\bibliographystyle{shorter}
%\bibliographystyle{plain}
\bibliography{martin}


\end{document}
