% Accepted to RESEARCH section by Krzysztof Chris Ciesielski
% Received May 1, 2003. Final, accepted version received on 8/4/03.
% LaTeX2e
% A non-hereditary Borel-cover gamma-set
% A. Miller Dec 2002 last revised August 4, 2003
\documentclass[12pt]{article}
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\def\min{{\rm min}}
\def\proof{\par\noindent Proof\par\noindent}
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\def\res{\upharpoonright}
\def\qed{\par\noindent QED\par}
\def\bb{{\mathfrak b}}
\def\dd{{\mathfrak d}}
\def\cc{{\mathfrak c}}
\def\pp{{\mathfrak p}}
\def\Tau{{\mathrm{T}}}
\def\rmor{\mbox{ or }}
\def\rmiff{\mbox{ iff }}
\def\dual#1{\widetilde{#1}}
\def\rmand{\mbox{ and }}
\def\aron{Aronszajn }
\def\ss{\mathsf{S}}
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{question}[theorem]{Question}
\newtheorem{conjecture}[theorem]{Conjecture}
\begin{document}
\begin{center}
{\large A Nonhereditary Borel-cover $\gamma$-set}
\end{center}
\begin{flushright}
Arnold W. Miller\footnote{
Thanks to the Fields Institute for Research in Mathematical Sciences
at the University of Toronto for their support during the time this paper
was written and to Juris Stepr\={a}ns who
directed the special program in set
theory and analysis.
\par Mathematics Subject Classification 2000: 03E50; 03E17
}
\end{flushright}
\begin{center}
Abstract
\end{center}
\begin{quote}
In this paper we prove that if there is a Borel-cover $\gamma$-set of
cardinality the continuum, then there is one which is not hereditary.
\end{quote}
% Result obtained 9-18-02
In this paper we answer some of the questions raised by Bartoszy\'{n}ski and
Tsaban \cite{BT} concerning hereditary properties of sets defined by certain
Borel covering properties.
Define. An $\om$-cover of a set $X$ is a family of sets such that
every finite subset of $X$ is included in an element of the cover
but $X$ itself is not in the family.
Define. A $\gamma$-cover of a set $X$ is an infinite family of sets such
that every element of $X$ is in all but finitely many elements of the
family.
Define. A set $X$ is called a Borel-cover $\gamma$-set iff every countable
$\om$-cover of $X$ by Borel sets contains a $\gamma$-cover.
These concepts were introduced by Gerlits and Nagy \cite{GN} for open covers.
Being a Borel-cover $\gamma$-set is equivalent to saying that for any
$\om$-sequence of countable Borel $\om$-covers of $X$ we can choose one
element from each and get a $\gamma$-cover of $X$ -- this is denoted
$\ss_1({\cal B}_\Omega, {\cal B}_\Gamma)$. The equivalence was proved by
Gerlitz and Nagy \cite{GN} for open covers but the proof works also for Borel
covers as was noted in Scheepers and Tsaban \cite{ST}:
Let ${\cal B}_n$ be Borel $\om$-covers of $X$. Since
$$\{U\cap V: U\in{\cal U}, V\in{\cal V}\}$$
is an $\om$-cover if ${\cal U}$ and ${\cal V}$ are, we may assume that
${\cal B}_{n+1}$ refines ${\cal B}_{n}$. Let $x_n$ for $n<\om$ be distinct
elements of $X$ and let
$${\cal B}=\{A\setminus \{x_n\}: n<\om, A\in {\cal B}_{n}\}$$
It is easy
to check that ${\cal B}$ is an $\om$-cover of $X$. Now let
${\cal C}$ be a
$\gamma$-subcover of ${\cal B}$. Note that for any fixed $n$ at
most
finitely many of the elements of ${\cal C}$ can be of the form
$A\setminus
\{x_n\}$. By refining $\cal C$ we may assume at most one thing is
taken
from each ${\cal B}_n$ and since they are refining we can fatten up
${\cal
C}$ to take exactly one element of each ${\cal B}_n$.
Define. A family of
subsets of $X$, $\;{\cal U}$ is a $\tau$ cover of
$X$ iff every element of
$X$ is in infinitely many elements of ${\cal U}$ and
for every $x,y\in X$
at least one of the sets
$$\{U\in{\cal U}: x\in U,\; y\notin U\} \rmor
\{U\in{\cal U}: x\notin U,\; y\in U\}$$
is finite.
Clearly any
$\gamma$-cover is a $\tau$-cover. These covers were introduced
in Tsaban
\cite{tsaban}.
\begin{theorem}
Suppose there is a Borel-cover
$\gamma$-set of size the continuum. Then
there is a Borel-cover
$\gamma$-set $X$ and subset $Y$ of $X$ which is not a
Borel-cover
$\gamma$-set. In fact, there is an open $\om$-cover of $Y$
with no
$\tau$-subcover.
\end{theorem}
\proof
For $X\subset P(\om)$ let
$$\dual{X}=\{\om\setminus a: a\in X\}$$
be the dual of $X$, i.e., the set
of complements of elements of $X$.
Let $P\sq [\om]^\om$ be a perfect set
of
independent subsets of $\om$. This means that for every disjoint
pair
$F_1,F_2$ of finite subsets of $P$ the set
$$\bigcap
F_1\cap\bigcap\dual{F_2}\mbox{ is infinite.}$$
Such a set was first
constructed by
Fichtenholtz, Kantorovich, and Hausdorff, see Kunen
\cite{kunen}.
To construct
one, let $Q=\{(n,s):n\in\om, s\sq P(n)\}$.
Define
$A_x=\{(n,s): x\cap n\in s\}$ for each $x\sq \om$ and
$$P=\{A_x:x\sq
\om\}\sq P(Q)=2^Q.$$
Let $Z\sq P$ be a Borel-cover $\gamma$-set
of
cardinality the continuum.
\bigskip
\noindent Claim. $Z\cup\dual{Z}$ is
a Borel-cover $\gamma$-set.
proof: Let $\{B_n:n<\om\}$ be a Borel
$\om$-cover of
$Z\cup\dual{Z}$. Then it is easy to see that
$\{(B_n\cap
\dual{B_n}):n<\om\}$ is an $\om$-cover of
$Z$. This is because if
$(F\cup\dual{F})\sq B_n$ then
$F\sq (B_n\cap\dual{B_n})$.
Since $Z$ is a
Borel-cover $\gamma$-set
there exists an $a\in [\om]^\om$ such
that
$$\{(B_n\cap \dual{B_n}):n\in a\}$$ is a $\gamma$-cover
of $Z$. But
then it is also a $\gamma$-cover of $\dual{Z}$.
This proves the
Claim.
\bigskip
Let $X=Z\cup\dual{Z}$ and to pick $Y\sq X$ as required
we
will choose for
each $a\in Z$ to put either $a\in Y$ or $(\om\setminus
a)\in Y$
(but not both).
Since $Z$ was a subset of $P$ and $P$ was
independent we will have
that the intersection of any finite subset of $Y$
is infinite. In
particular,
$$ {\cal U}=\{U_n:n\in\om\}\mbox{ where }
U_n=\{a\sq \om:n\in a\} $$
is an $\om$-cover of $Y$. But $\{U_n:n\in b\}$
is a
$\gamma$-cover of $Y$ iff $b\sq^* a$ for every $a\in Y$.
But this is
easy to defeat. Using that $Z$ has cardinality the
continuum let
$Z=\{a_\alpha:\alpha<\cc\}$ and let
$[\om]^\om=\{b_\alpha:\alpha<\cc\}$.
For each $\alpha$
if $b_\alpha\sq^* a_\alpha$ put $(\om\setminus a_\alpha)$
into
$Y$ and otherwise put $a_\alpha$ into $Y$.
To construct $Y$ so that
${\cal U}$ has no $\tau$-subcovers
can be done
by using two elements
$a_0,a_1$ of $Z$ for each $b\in [\om]^\om$.
First note that the set
$\{U_n:n\in b\}$ is a $\tau$-cover of $Y$ iff
$b$ meets every element of
$Y$ in an infinite set and for
every two elements $a_0,a_1$ of $Y$
either
$(a_0\cap b)\sq^* a_1$ or $(a_1\cap b)\sq^* a_0$.
Notation:
$a^{(0)}=a$ and $a^{(1)}=\om\setminus a$.
\bigskip\noindent
Claim. There
exists $i,j$ in $\{0,1\}$ such that
\par (a) $b\cap a_i^{(j)}$ is finite
or
\par (b) both $b\cap a_0^{(i)}\cap a_1^{(1-j)}$ and
$b\cap
a_0^{(1-i)}\cap a_1^{(j)}$ are infinite.
proof: Assume case (a) fails for all $i,j$ in $\{0,1\}$.
The four sets $a_0^{(i)}\cap a_1^{(j)}$ partition
$\om$ into infinite sets since $a_0$ and $a_1$ are independent.
If all four meet $b$ in an infinite set then we are done. So
assume that $b\cap a_0^{(i)}\cap a_1^{(j)}$ is finite for
some $i,j$. But
since $b\cap a_0^{(i)}$ is infinite it must be that
$b\cap a_0^{(i)}\cap a_1^{(1-j)}$ is infinite. A similar
argument shows
$b\cap a_0^{(1-i)}\cap a_1^{(j)}$ is infinite.
This proves the Claim.
\bigskip
To kill off the possibility of $b$ giving a $\tau$-subcover we
put $a_i^{(j)}$ into $Y$ in case (a) or put
both $a_0^{(i)}, a_1^{(j)}$ into $Y$ in case (b).
This proves the Theorem.
\qed
\bigskip Remark. Tsaban points out the following corollary of our result. In
Problem 7.9 of Bukovsk\'y, Rec{\l}aw, and Repick\'y \cite{brr} it is asked
whether
every $\gamma$-set of reals which is also a $\sigma$-set is a hereditary
$\gamma$-set. It is shown in Scheepers and Tsaban \cite{ST} that every
Borel-cover $\gamma$-set (more generally $\ss_1({\cal B}_\Gamma, {\cal
B}_\Gamma)$-set) is a $\sigma$-set. Hence the answer to the problem is no.
\bigskip
The following result is due to Brendle \cite{brendle}. Our proof is a slight
modification of a result of Todor\v{c}evi\'{c} -- see Theorem 4.1 of Galvin
and Miller \cite{GM} and is perhaps simpler.
\begin{theorem} \label{ch}(Brendle)
Assume CH. Then there exists a Borel-cover $\gamma$-set of size
$\om_1$.
\end{theorem}
\proof
The idea is to construct an \aron tree
$T\sq \om^{<\om_1}$ labeled by perfect sets.
We construct perfect subtrees $p_s\sq 2^{<\om}$ for
$s\in T$ and $X_s\sq [p_s]$ countable dense sets such that
\begin{enumerate}
\item if $s\sq t$ then $p_s \supseteq p_t$,
\item if $s$ and $t$ are incomparable then $[p_s]\cap [p_t]=\emptyset$,
\item if $\alpha<\beta<\om_1$ and $n<\om$ then for
every $s\in T_\alpha$ there exists $t\in T_\beta$ with $s\sq t$
and $p_s\cap 2^n=p_t\cap 2^n$, and
\item for every sequence $(B_n:n<\om)$ of Borel subsets of
$2^\om$ there exists $\alpha<\om_1$ such that either
for some finite $F\sq \cup \{X_s:s\in T_{\leq \alpha+1}\}$
no $B_n$ covers $F$ or there exists an $a\in [\om]^\om$
such that $\{B_n:n\in a\}$ is a $\gamma$-cover of
$$\cup \{[p_s]:s\in T_{\leq \alpha+1}\}
\;\cup\; \bigcup \{X_s:s\in T_{\leq \alpha}\}$$
\end{enumerate}
After the construction is completed we will let $X=\cup\{X_s:s\in T\}$.
The
last item guarantees that $X$ will be a Borel-cover $\gamma$-set.
The first
three items are simply to guarantee that our construction
can continue at
limit levels. To do the last item we use the following
Lemma.
Define for
any perfect tree $p$ and $s\in p$,
$$p\langle s\rangle =\{t\in p: t\sq s
\rmor s\sq t\}$$
\begin{lemma}
Suppose $\langle p_n:n<\om\rangle $ are
perfect trees and
$({\cal B}_n:n<\om)$
is a sequence of countable Borel
$\om$-covers of
$\cup_{n<\om}[p_n]$. Then
there exists
perfect pairwise
disjoint subtrees $q_n\sq p_n$ and
$\{B_n\in {\cal B}_n:n<\om\}$ which is
a $\gamma$-cover
of $\cup_{n<\om}[q_n]$.
\end{lemma}
\proof
We can begin
by refining the $p_n$'s so that $[p_n]$ are pairwise disjoint.
So we may as
well assume this to begin with.
Also since Borel sets have the (relative)
property of Baire with
respect to each perfect set, by passing to perfect
subsets we may
assume that each of our Borel covers is an open cover.
Note that for finite sequences $(k_i:i