% On squares of spaces and Fsigma-sets, A.Miller
% result obtained Fall 2002, latest version 6-28-2004
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\begin{document}
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\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}