% 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} \usepackage{amssymb} \def\om{\omega} \def\sq{\subseteq} \def\reals{{\mathbb R}} \def\dom{{\rm dom}} \def\min{{\rm min}} \def\proof{\par\noindent Proof\par\noindent} \def\poset{{\mathbb P}} \def\qposet{{\mathbb Q}} \def\eposet{{\mathbb E}} \def\forces{{| \kern -2pt \vdash}} \def\force{\forces} \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