% On squares of spaces and Fsigma-sets, A.Miller % result obtained Fall 2002, latest version 6-28-2004 % latex2e - in topproc syle \documentclass[11pt]{amsart} \setcounter{page}{1} %%%%%%% \setlength{\textwidth}{4.4in} %%%%%%% \setlength{\textheight}{7.0in} %%%%%%% \setlength{\evensidemargin}{1in} %%%%%%% \setlength{\oddsidemargin}{1in} %%%%%%% \setlength{\topmargin}{.8in} %%%%%%% \newtheorem{theorem}{Theorem} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{question}[theorem]{Question} \newtheorem{prop}[theorem]{Proposition} \def\appendix{\par} % to see appendix %\def\appendix{\end{document}} %% to hide appendix \def\cc{{\mathfrak c}} \def\tt{{\mathfrak t}} \def\reals{{\mathbb R}} \def\proof{\par\noindent Proof.\par\noindent} \def\res{\upharpoonright} \def\qed{\par\noindent QED\par} \def\rmor{\mbox{ or }} \def\rmiff{\mbox{ iff }} \def\rmand{\mbox{ and }} \def\om{\omega} \def\al{\alpha} \def\be{\beta} \def\ka{\kappa} \def\del{\delta} \def\diag{\Delta} \def\infsets{{[\om]^\om}} \def\sq{\subseteq} \def\sm{\setminus} \def\sig{\sigma} \def\uu{{\mathcal U}} \def\vv{{\mathcal V}} \def\ww{{\mathcal W}} \def\bb{{\mathcal B}} \def\ff{{\mathcal F}} \def\CC{{\mathcal C}} \begin{document} \noindent %%%%%% \begin{picture}(150,36) %%%%%% \put(5,20){\tiny{To appear in}} %%%%%% \put(5,7){\textbf{Topology Proceedings}} %%%%%% \put(0,0){\framebox(140,34){}} %%%%%% \put(2,2){\framebox(136,30){}} %%%%%% \end{picture} %%%%%% \vspace{0.5in} \title[squares]% {On squares of spaces and $F_\sig$-sets} \author{Arnold W. Miller} \address{University of Wisconsin-Madison, Department of Mathematics, Van Vleck Hall, 480 Lincoln Drive, Madison, Wisconsin 53706-1388} \email{miller@math.wisc.edu} \thanks{Thanks to the Fields Institute for Research in Mathematical Sciences at the University of Toronto for their support during the time these results were obtained, and to Juris Steprans who directed the special program in set theory and analysis. Also we thank Gary Gruenhage for his translation of the topology problem into properties of sets of reals.} \subjclass[2000]{03E35 54B10 54E35} \begin{abstract} We show that the continuum hypothesis implies there exists a Lindel\"of space $X$ such that $X^2$ is the union of two metrizable subspaces but $X$ is not metrizable. This gives a consistent solution to a problem of Balogh, Gruenhage, and Tkachuk. The main lemma is that assuming the continuum hypothesis there exist disjoint sets of reals $X$ and $Y$ such that $X$ is Borel concentrated on $Y$, i.e., for any Borel set $B$ if $Y\sq B$ then $X\sm B$ is countable, but $X^2\sm\diag$ is relatively $F_\sig$ in $X^2\cup Y^2$. \end{abstract} \maketitle In Balogh, Gruenhage, and Tkachuk \cite{bgt} the following question is asked: \bigskip Question 4.1. Let $X$ be a regular paracompact space $X$ such that $X\times X$ is the union of two metrizable subspaces. Must $X$ be metrizable? What if $X$ is Lindel\"of? \bigskip \begin{theorem}\label{one} Assume the continuum hypothesis. Then there exists a nonmetrizable regular Lindel\"of space $X$ such that $X^2$ is the union of two metrizable subspaces. \end{theorem} We first prove the following Lemma. \begin{lemma} (CH) There are uncountable disjoint sets $X,Y\sq 2^\om$ such that \begin{enumerate} \item $X$ is Borel concentrated on $Y$, i.e. every Borel set in $2^\om$ containing $Y$ contains all but countably many elements of $X$, \item $Y^2\sm\diag$ is $F_\sig$ in $X^2\cup Y^2$, and \item $X^2\sm\diag$ is $F_\sig$ in $X^2\cup Y^2$. \end{enumerate} Here $\diag=\{(x,x): x\in 2^\om\}$. \end{lemma} \proof We identify the Cantor space $2^\om$ with the power set $P(\om)$ of $\om$. We use $\infsets$ to stand for the infinite subsets of $\om$. Define for $y\in \infsets$ $$[y]^{*\om}=\{x\in[\om]^\om: x\sq^* y\}$$ where $\sq^*$ stands for inclusion mod finite. Let $\langle B_\al:\al<\om_1\rangle$ be all Borel subsets of $[\om]^\om$. We construct $y_\al$ for $\al<\om_1$ so that \begin{enumerate} \item $\al<\beta$ implies $y_\beta\sq^* y_\al$ and $y_\beta\not=^* y_\al$ and \item either $y_{\al}\notin B_\al$ or $[y_{\al}]^{*\om}\sq B_\al$. \end{enumerate} These conditions are easy to get. Given $y_\beta$ for $\beta<\al$ and $B_\al$ let $y\in \infsets$ be arbitrary with $y\sq^* y_\beta$ but $y_\beta\not=^* y$ for each $\beta<\al$. If $[y]^{*\om}$ is not a subset of $B_\al$, then simply take $y_{\al}\in [y]^{*\om}\sm B_\al$, otherwise take $y_{\al}=y$. Let $$X=\{y_\al\sm y_{\al+1}:\al<\om_1\} \rmand Y=\{y_\al:\al<\om_1\}$$ Iff $B$ is any Borel set containing $Y$, then choose $\al$ so that $B=B_\al$. At stage $\al$ of the construction it must have been that $[y_{\al}]^{*\om}\sq B_\al$. But this means that $x_\beta\in B_\al$ for all $\beta\geq\al$. So $X$ is Borel concentrated on $Y$. If we define $$F=\{(u,v)\in P(\om)\times P(\om): (u\sq^*v \rmor v\sq^*u)\rmand u\not=v\}$$ then $F$ is an $F_\sig$ set and $$F\cap (X^2\cup Y^2) = (Y^2\sm\diag)$$ Also if we define $$H=\{(u,v)\in P(\om)\times P(\om): u\cap v=^*\emptyset\}$$ then $H$ is an $F_\sig$ set and $$H\cap (X^2\cup Y^2) = (X^2\sm\diag)$$ This finishes the proof of the Lemma. \qed Now define the following Michael-line like topology. Suppose that $M$ is a topological space and $X\sq M$. Then $M(X)$ is the topological space on the same set but with the following topology. For $x\in X$ we make $x$ an isolated point, i.e., add $\{x\}$ to the topology of $M(X)$. For any point $y\in M\sm X$ neighborhoods in $M$ form a neighborhood basis for $y$ in $M(X)$. It is easy to see that $M(X)$ is regular for any regular space $M$ and subset $X\sq M$. The following is Exercise 5.5.2 from Engelking \cite{engel}: \begin{prop}\label{prop} Suppose $M$ is a metric space and $X\sq M$. Then $M(X)$ is metrizable iff $X$ is an $F_\sig$ set in $M$. \end{prop} Our example is $M(X)$ where $X$ and $Y$ are from the Lemma and $M=X\cup Y$ has its usual (separable metric) topology as a subspace of $2^\om$. It follows from the Proposition that $M(X)$ is not metrizable. \bigskip\noindent Claim 1. $M(X)$ is a Lindel\"of space. Take any open cover $\uu$ of $M(X)$. Open sets in $M(X)$ have the form $U \cup Z$ where $U$ is open in $M$ and $Z\sq X$ is arbitrary. Then since $Y$ has its standard topology, countably many elements of $\uu$ will cover $Y$, say $$\{(U_n\cup X_n:n<\om\}\sq \uu$$ where each $U_n$ open in $M$ and $X_n\sq X$. But since $X$ is Borel concentrated on $Y$ we have that $X\sm\cup\{U_n:n<\om\}$ is countable, so we need only add countably many more elements of $\uu$ to cover all of $M(X)$. \bigskip\noindent Claim 2. $M(X)^2$ is the union of two metrizable subspaces. Let \par $M_1=(X^2\sm \diag)\cup Y^2$ and \par $M_2= (X\times Y)\cup (Y\times X)\cup (X^2\cap \diag)$. \par\noindent Note that $M_1$ is $N(X^2\sm \diag)$ where $N=(X^2\sm \diag)\cup Y^2$ in its separable metric topology as a subspace of $2^\om\times 2^\om$. By the Lemma we have that $X^2\sm \diag$ is relatively $F_\sig$ in $N$ and so by Proposition \ref{prop} $M_1$ is metrizable. To see that $M_2$ is metrizable use the Bing Metrization Theorem: \begin{quote} A topological space is metrizable iff it is regular and has a $\sig$-discrete basis. \end{quote} A family $\bb$ of subsets of $X$ is discrete iff every point of $X$ has a neighborhood meeting at most one element of $\bb$. $\sig$-discrete means the countable union of discrete families. Note that for each $x\in X$ the sets $\{x\}\times Y$ and $Y\times\{x\}$ are open in $M_2$. Let $\bb$ be a countable open basis for $Y$. Then $$\CC=\{U\times\{x\},\;\;\{x\}\times U,\;\; \{(x,x)\}\;: \;\; x\in X, U\in\bb\}$$ is an open basis for $M_2$. It is $\sig$-discrete. The family $\{\{(x,x)\}:x\in X\}$ is discrete in $M_2$ since $X^2\cap \diag$ is closed in $M_2$. And for each fixed $U\in\bb$ the family $\{U\times\{x\}\;:\;x\in X\}$ is discrete in $M_2$. (For $(x,x)\in X$ use the neighborhood $\{x\}\times \{x\}$. For $(y,x)$ with $y\in Y$ and $x\in X$ use the neighborhood $Y\times\{x\}$ and for $(x,y)$ use the neighborhood $\{x\}\times Y$.) Similarly, for each $U\in\bb$ the family $\{\{x\}\times U\;:\;x\in X\}$ is discrete in $M_2$. Since $\bb$ is countable, $M_2$ has a $\sig$-discrete basis and is therefor metrizable. This proves Theorem \ref{one}. \qed The next Theorem is an easy generalization of Theorem \ref{one} using the tower cardinal $\tt$ which is defined as follows. $\tt$ is the minimum cardinality of a set $T\sq\infsets$ which is linearly ordered by $\sq^*$ but there does not exist $z\in\infsets$ with $z\sq^* y$ for every $y\in T$. Martin's axiom implies that $\tt=\cc$. \bigskip \begin{theorem} Suppose $\tt=\cc$. Then there exists a nonmetrizable regular paracompact space $X$ such that $X^2$ is the union of two metrizable subspaces. \end{theorem} \proof The main Lemma changes to: \begin{lemma} ($\tt=\cc$) There are disjoint sets $X,Y\sq 2^\om$ of cardinality $\cc$ such that \begin{enumerate} \item $X$ is Borel $\cc$-concentrated on $Y$, i.e., for every Borel set $B$ in $2^\om$, if $Y\sq B$ then $|X\sm B|<\cc$, \item $Y^2\sm\diag$ is $F_\sig$ in $X^2\cup Y^2$, and \item $X^2\sm\diag$ is $F_\sig$ in $X^2\cup Y^2$. \end{enumerate} \end{lemma} The proof is similar. The space $M=X\cup Y$ is the same. Since $X$ is not relatively Borel in $M$ we have by Proposition \ref{prop} that $M(X)$ is not metrizable. But $M(X)$ is regular and paracompact for any $X\sq M$ and metric $M$, see example 5.1.22 Engelking \cite{engel}. \qed \bigskip Remark. The Michael line is the topological space $M(X)$ where $M$ is the unit interval, $[0,1]$, and $X$ the irrationals in $[0,1]$. Michael Granado in unpublished work has shown that the square of the Michael line is not the union of two metrizable subspaces. \begin{question} (Using just ZFC) Do there exist disjoint sets of reals $X$ and $Y$ such that $X$ is not $F_\sig$ in $X\cup Y$ but $X^2\sm\diag$ is $F_\sig$ in $X^2\cup Y^2$? \end{question} \bibliographystyle{amsplain} \begin{thebibliography}{99} \bibitem{bgt} Balogh, Zoltan; Gruenhage, Gary; Tkachuk, Vladimir; \textit{Additivity of metrizability and related properties}, Proceedings of the International Conference on Set-theoretic Topology and its Applications, Part 2 (Matsuyama, 1994). Topology Appl. \textbf{84} (1998), no. 1-3, 91--103. \bibitem{engel} Engelking, Ryszard;\textit{General topology}. Translated from the Polish by the author. Second edition. Sigma Series in Pure Mathematics, 6. Heldermann Verlag, Berlin, 1989. viii+529 pp. ISBN: 3-88538-006-4. \end{thebibliography} \appendix \newpage The appendix is not intended for final publication but for the electronic version only. \begin{center} Appendix \end{center} \begin{quote} Suppose $M$ is a metric space and $X\sq M$. Then $M(X)$ is metrizable iff $X$ is an $F_\sig$ in $M$. (Engelking 5.5.2) \end{quote} \proof Suppose $X$ is not $F_\sig$ in $M$, then $Y=M\sm X$ is closed in $M(X)$ ( since the points of $X$ are isolated, $X$ is open ). But $Y$ is not $G_\del$ in $M(X)$. To see, this suppose that $Y=\cap_{n\in\om} U_n$ were each $U_n$ is open in $M(X)$. Then there would exists $V_n$ open in $M$ and $X_n\sq X$ with $U_n=V_n\cup X_n$. But then $Y= \cap_{n\in\om} V_n$ which contradicts $Y$ is not $G_\del$ in $M$. For the converse, suppose $X$ is $F_\sig$ in $M$ and write it as the union of closed sets $X=\cup_{n<\om}C_n$. $M(X)$ is regular so it is enough by the Bing Metrization Theorem to check that it has a $\sig$-discrete base. Let $\bb$ be a $\sig$-discrete base for $M$. We claim that $$\bb \cup\{\{x\}:x\in X\}$$ which is a basis for $M(X)$ is $\sig$-discrete in $M(X)$. $\bb$ is $\sig$-discrete in $M$ so it is also $\sig$-discrete in $M(X)$. $$\{\{x\}:x\in X\}=\cup_{n<\om}\CC_n\mbox{ where } \CC_n=\{\{x\}:x\in C_n\}$$ shows that it is $\sig$-discrete, since for any $n$ if $x\notin C_n$ then $M\setminus C_n$ is a neighborhood of $x$ missing all elements of $\CC_n$. \bigskip \begin{quote} $M(X)$ is regular paracompact, whenever $M$ is metric. (Engelking 5.1.22) \end{quote} \proof Regular: Given $p\in M$ if $p\in X$ then it is has the clopen neighborhood $\{p\}$, if $p\notin X$, then the neighborhoods of $p$ in $M$ are also a neighborhood basis in $M(X)$. Paracompact: Let $\uu$ be an open cover of basic open sets in $M(X)$. We may assume it has the form: $$\uu=\vv\cup \{\{x\}:x\in Z\}$$ where $\vv$ is a family of basic open sets in $M$ and $Z=X\sm \cup\vv$. Since metric spaces are hereditarily paracompact, there exists a locally finite refinement $\ww$ of $\vv$ with $\cup\vv=\cup\ww$. But then $\ww\cup \{\{x\}:x\in Z\}$ is a locally finite refinement of $\uu$. \end{document}