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% The gamma - Borel conjecture
% A. Miller Mar 2003 last revised July 8, 2004
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\def\diamondb{\Diamond(\bb)}
\def\om{\omega}
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\begin{document}
\begin{center}
{\large The $\gamma$-Borel conjecture}
\end{center}
\begin{flushright}
Arnold W. Miller\footnote{
Thanks to Boise State University
for support during the time this paper
was written and to Alan Dow for some helpful discussions and to
Boaz Tsaban for some suggestions to improve an earlier version.
\par Mathematics Subject Classification 2000: 03E35; 03E17
\par Keywords: $\gamma$-set, Hechler forcing,
Laver forcing, strong measure zero set.
}
\end{flushright}
\begin{center}
Abstract
\end{center}
\begin{quote}
In this paper we prove that it is consistent that every $\gamma$-set
is countable while not every strong measure zero set is countable.
We also show that it is consistent that every strong $\gamma$-set is
countable while not every $\gamma$-set is countable. On the other
hand we show that every strong measure zero set is countable iff
every set with the Rothberger property is countable.
\end{quote}
A set of reals $X$ has strong
measure zero iff for any sequence $(\epsilon_n:n<\om)$ of
positive reals there exists a sequence of intervals $(I_n:n<\om)$
covering $X$ with each $I_n$ of length less
than $\epsilon_n$. Laver \cite{laver} showed that it is relatively
consistent with ZFC that the Borel conjecture is true, i.e.,
every strong measure zero set is countable.
Sets of reals called $\gamma$-sets were first considered by Gerlits and Nagy
\cite{gn}. They showed that every $\gamma$ set has strong measure zero and
that Martin's Axiom implies every set of reals of size smaller than the
continuum is a $\gamma$-set. A $\gamma$-set of size continuum is constructed
in Galvin and Miller \cite{gm} using MA.
Next we define $\gamma$-set. An open cover $\uu$ of a topological space $X$
is an $\om$-cover iff for every finite $F\sq X$ there exists $U\in \uu$ with
$F\sq U$ and $X\notin\uu$. An open cover $\uu$ of $X$ is a $\gamma$-cover iff
$\uu$ is infinite and each $x\in X$ is in all but finitely many $U\in\uu$.
Finally, $X$ is a $\gamma$-set iff $X$ is a separable metric space in which
every $\om$-cover contains a $\gamma$-subcover.
\bigskip
Paul Szeptycki asked if it was possible to have a sort of weak Borel
conjecture be true, i.e., every $\gamma$-set countable, while the Borel
conjecture is false. We answer his question positively. We use Hechler
\cite{hechler} forcing, $\hh$, for adding a dominating real, an analysis of it
due to Baumgartner and Dordal \cite{bd}, properties of
Laver forcing $\ll$, and a characterization
of $\hh$ due to Truss \cite{truss}.
\bigskip
\begin{theorem} \label{mainthm}
If $\hh$ is iterated $\om_2$ times with finite support, $\hh_{\om_2}$,
over a model of CH, then in the resulting model every $\gamma$-set
is countable but every set of reals of cardinality $\om_1$ has
strong measure zero.
\end{theorem}
\proof
For $f\in \om^\om$, define $\uu_f$ to be the following family of
clopen subsets of $2^\om$.
$$\uu_f=\{C_F:\exists n\;\; F \sq 2^{f(n)}, |F|\leq n\} \mbox{ where }
C_F=\{x\in 2^\om: x\res f(n)\in F\}\}$$
Note that for any finite $A\sq 2^\om$ there exists $C\in \uu_f$ with
$A\sq C$. Also $2^\om\notin U_f$ provided that $2^{f(n)}>n$ all $n$.
Let $\ll$ denote Laver forcing \cite{laver}.
\begin{lemma}\label{one} Suppose $M$ is a model of set theory, $f$ is
$\ll$-generic over $M$, and $X\sq 2^\om$ is in $M$. Then
$$M[f]\models\forall \cc\in [\uu_f]^\om\;\;
|\bigcap{\cc}\cap X |\leq \om $$
\end{lemma}
\proof
For a tree $p\sq \om^{<\om}$ and $s\in p$ we define
$$p_s=\{t\in p: t\sq s \rmor s\sq t\}$$
A Laver condition (or Laver tree) is a tree $p\sq \om^{<\om}$
with a special node $s\in p$, called the stem of $p$, which has
the properties that
\begin{enumerate}
\item $p_s=p$ and
\item for every $t\in p$ with $|t|\geq |s|$ there exist infinitely many
$n<\om$ with $tn\in p$.
\end{enumerate}
The order is $p\leq q$ iff $p\sq q$.
As usual we define $p\leq_0 q$ iff $p\leq q$ and $\stem(p)=\stem(q)$.
Somewhat nonstandardly let us write
$$\leaves(p)=\{r\in p: \stem(p)\sq r\}$$
and for each $s\in\leaves(p)$ define
$$\split(p,s)=\{n\in\om: sn\in p\} $$
Suppose that the lemma is false.
Let $p^*$ be a Laver condition such that
$$p^*\forces\lq
\bigcap\name{\cc}\cap X =\name{Y}\mbox{ is uncountable }
\rmand \name{\cc}\sq \uu_f \mbox{ is infinite}\rq$$
By cutting $\cc$ down (if necessary) we may suppose that
$\cc=\{C_{F_n}: n\in Q\}$ where $F_n\sq 2^{f(n)}$ with $|F_n|\leq n$
and $Q\in \infsub$.
Working in $M$ using standard arguments
of Laver forcing \cite{laver} we can prove the following
Claims.
\bigskip
{\bf Claim.} Suppose that $p$ is an arbitrary condition such that
$$p\forces \{\name{s}_i:i\alpha$.
\begin{lemma} (Baumgartner and Dordal \cite{bd}) Suppose $N$ is
a model of set theory and
$$N\models (a_\alpha\in \infsub:\alpha<\om_1)
\mbox{ is eventually narrow. }$$
Then for any $G_{\om_2}$ which is $\hh_{\om_2}$-generic
over $N$, we have that
$$N[G_{\om_2}]\models (a_\alpha\in \infsub:\alpha<\om_1)
\mbox{ is eventually narrow. }$$
\end{lemma}
Now we prove that every $\gamma$-set in $M[G_{\om_2}]$ countable.
Since $\gamma$-sets are zero dimensional we need only worry about
uncountable $Y\sq 2^\om$. Let $X\sq Y$ be a subset
of size $\om_1$. Construct $g:\om\to\om$ so that for every
$n<\om$ if $m=g(n)$, then
$$|\{x\res m:x\in X\}|>n$$
By the usual ccc finite support iteration arguments we can find
$\alpha<\om_2$ so that $X,g\in M[G_\alpha]$ and letting $h=h_\alpha$ be
the next Hechler real added we
have that $h(n)>g(n)$ for all $n$.
From Lemma \ref{main} and the remark following it we see that in
$N=M[G_{\alpha+1}]$
the set
$\bigcap\cc\cap X$ is countable for every infinite $\cc\sq \uu_h$.
Now since $h(n)>g(n)$ there is no $U\in \uu_h$ which covers $X$, however
$\uu_h$ is an $\om$-cover of $2^\om$ and hence of $Y$.
Now let $X=\{x_\alpha:\alpha<\om_1\}$ and $\uu_h=\{U_n:n<\om\}$.
In the model $N=M[G_{\alpha+1}]$
define $a_\alpha=\{n<\om: x_\alpha\in U_n\}$. Note that
$$N\models (a_\alpha\in \infsub:\alpha<\om_1)
\mbox{ is eventually narrow. }$$
Otherwise if $b\sq^* a_\alpha$ for uncountably many $\alpha$, then
for some infinite $c\sq b$
$$Z=\{x_\alpha: c\sq a_\alpha\}$$
is uncountable. But then $Z\sq \bigcap\{U_n:n\in c\}$
which contradicts Lemma \ref{five}.
Since the tail of a finite iteration of $\hh$ is itself a finite
support iteration of $\hh$, the
Baumgartner-Dordal Lemma applies and so,
$$N[G_{[\alpha+2,\om_2)}]= M[G_{\om_2}]$$
models that $(a_\alpha:\alpha<\om_1)$ is eventually
narrow.
But this implies that $Y$ is not a $\gamma$-set since if
$(U_n\in\uu:n\in b)$ is a $\gamma$-cover of $X\sq Y$, then
for some infinite $c\sq b$, we would have that
$X\cap \bigcap\{U_n:n\in c\}$ is uncountable, which implies that
for uncountably many $\alpha$ that $c\sq a_\alpha$. Contradicting
the fact the $a_\alpha$ are eventually narrow.
On the other hand, it is well known that forcing with $\hh$ adds
Cohen reals and adding Cohen reals makes
sets of reals of small cardinality into strong measure zero sets.
To see this suppose that
$(\epsilon_n>0:n<\om)\in M$ a model of set theory. In $M$ let
$(I_{nm}:m<\om)$ list all intervals with rational end points and
of length less than $\epsilon_n$. If
$x:\om\to \om$ is a Cohen real over $M$, then it is an easy density
argument to prove that
$$M\cap\reals \sq \bigcup_{n<\om} I_{n x(n)}$$
The usual arguments show that in the iteration every set of reals of
cardinality $\om_1$ has strong measure zero.
This proves Theorem \ref{mainthm}.
\qed
\bigskip Remark. It is also true in the Hechler real model that every set of
reals of size $\om_1$ is both in ${\sf S}_1(\Gamma,\Gamma)$ and
${\sf S}_1(\Omega,\Omega)$. For definitions, see Just, Miller, Scheepers,
and Szeptycki \cite{jmss}. This follows from the fact that $\bb>\om_1$
and $cov(\mathcal{M})>\om_1$, see Figure 4 \cite{jmss}.
\bigskip
Define. $X$ is $C''$ iff for every sequence
$(\uu_n:n<\om)$ of open covers of $X$ there exist
$(U_n\in\uu_n:n<\om)$ an open cover of $X$.
Equivalent terminology for $C''$ is the Rothberger property or
${\sf S}_1({\mathcal O},{\mathcal O})$.
Define. $C''$-BC to be the statement that every set of reals with the
property $C''$ is countable and let SMZ-BC denote the standard Borel
conjecture, every strong measure zero set is countable.
\begin{prop}
SMZ-BC is equivalent to $C''$-BC.
\end{prop}
\proof
It is only necessary to prove right to left.
If $\bb=\om_1$, then there exists an uncountable set of reals $Z$ concentrated
on a countable subset of itself, i.e., there exist countable $Q\sq Z$ with
the property that $Z\setminus U$ is countable for
every open set $U$ containing $Q$
(Besicovitch \cite{bes}, Rothberger \cite{roth}). Any
such set has property $C''$. To see these two results, let
$$X=\{f_\alpha\in\om^\om\;:\;\alpha<\om_1\}$$
be well ordered by $\leq^*$ and
unbounded. Let $h:\om^\om \to [0,1]$ be a homeomorphism with range the
irrationals in $[0,1]$. We claim that
$Y=h(X)$ is concentrated on $Q$ where $Q$ is the set of rationals in $[0,1]$.
Note that for any open $U\sq [0,1]$ containing $Q$ the set
$[0,1]\setminus U$ is compact and therefore
$C=h^{-1}([0,1]\setminus U)$ is a compact subset of $\om^\om$. Since
compact sets correspond to finitely branching trees, there exists
$f\in\om^\om$ such that $g\leq f$ for every $g\in C$. Since $X$ is
unbounded we have that $C \cap X$ is countable, hence
$Y\setminus U$ is countable. Hence $Z=Y\cup Q$ is
concentrated on $Q$. To see that concentrated sets have
property $C^{\prime\prime}$, suppose $Z$ is
concentrated on a countable subset of itself $Q$.
Let $(\uu_n:n<\om)$ be a sequence of
open covers of $Z$. Let $(U_{2n}\in \uu_{2n}:n<\om)$ cover $Q$ and
then choose $(U_{2n+1}\in \uu_{2n+1}:n<\om)$ to cover the countably many
elements of $Z\setminus \bigcup_{n<\om} U_{2n}$.
So assume $\bb>\om_1$.
Suppose there is an uncountable strong measure zero set. Then by
standard arguments there exists an
$X\sq 2^\om$ with $|X|=\om_1$ such that for every $f\in\om^\om$ there exists
$(s_n\in 2^{f(n)}:n<\om)$ such that for every $x\in X$ there are infinitely
many $n$ with $s_n\sq x$.
\bigskip\noindent
{\bf Claim.} $X$ has property $C''$.
\proof
Let $(\uu_n:n<\om)$ be open covers of $X$. Without loss we may assume
each element of each $\uu_n$ is of the form $[s]$ for some $s\in 2^{<\om}$.
Since $|X|<\bb$ we can find finite $A_n\sq 2^{<\om}$ so that
$s\in A_n$ implies $[s]\in \uu_n$ and for each $x\in X$ for all but
finitely many $n$ there exists $s\in A_n$ with $s\sq x$. Let
$f:\om\to\om$ be such that $f(n)>\max\{|s|:s\in A_n\}$. Using
strong measure zero of $X$ choose $s_n\in 2^{f(n)}$ so that
every element of $X$ is in infinitely many $[s_n]$. Define $t_n\in A_n$
as follows. If there exists $t\in A_n$ with $t\sq s_n$ then
let $t_n$ be such. If there isn't, choose $t_n$ arbitrarily. We
claim that $\{[t_n]:n<\om\}$ covers $X$. For any $x\in X$ for all
but finitely many $n$ we have that there exists $t\in A_n$ with
$t\sq x$. But for infinitely many $n$ we have that $s_n\sq x$.
Since $|s_n|>|t_n|$ it must be the case that $t_n\sq x$ for infinitely
many $t_n$.
This proves the Claim and the Proposition.
\qed
\bigskip
Define $X$ is a strong $\gamma$-set iff there exists an increasing sequence of
integers $(k_n:n<\om)$ so that for every sequence $(\uu_n:n<\om)$ where $\uu_n$
is a $k_n$-cover of $X$ (i.e., covers every $k_n$ element subset of $X$) there
exists a $\gamma$-cover of the form $(U_n\in\uu_n:n<\om)$. These were first
defined in Galvin and Miller \cite{gm}. Tsaban \cite{tsaban} has shown
that an equivalent definition results if we always require $k_n=n$.
\begin{theorem}
In the Cohen real model, i.e., $\om_2$ Cohen reals added to a model
of CH, every strong $\gamma$-set is countable but there is an uncountable
$\gamma$-set.
\end{theorem}
\proof
First we construct an uncountable $\gamma$-set.
This proof is a modification of the construction from Just, Miller,
Scheepers, and Szeptycki \cite{jmss} section 5.
Without loss of generality we may rearrange the generic set
of $\om_2$ Cohen reals to have order type $\om_2+\om_1$ and
assume that
$$N=M[x_\alpha\sq\om:\alpha<\om_1]$$
where $M$ is determined by the first
$\om_2$ Cohen reals. Note that $M$ fails to
satisfy CH. Construct $y_\alpha\in\infsub$ descending mod finite so that
$(y_\beta: \beta<\alpha)\in M[x_\beta:\beta<\alpha]$ as follows:
At stage $\alpha+1$ let
$$y_{\alpha+1}=x_{\alpha+1}\cap y_\alpha$$
This is infinite because $x_{\alpha+1}$ is Cohen generic over $y_\alpha$ At
limit stages choose $y_\alpha\in M[x_\beta:\beta<\alpha]$ so that
$y_\alpha\sq^*y_\beta$ all $\beta<\alpha$.
The choice of $y_\alpha$ can be made in some canonical way
so that the sequence of names $(\name{y_\alpha}:\alpha<\om_1)$
is in $M$. For $y\in [\om]^\om$ let
$[y]^{*\om}=\{x\in [\om]^\om\;:\; x\sq^* y\}$.
\bigskip\noindent
{\bf Claim.} Suppose $(\uu_n:n<\om)\in M[x_\beta:\beta\leq \alpha]$
is a family of
$\om$-covers of
$$\finsub\cup \{y_\beta:\beta\leq \alpha\}$$
Then there exists a sequence $(U_n\in\uu_n:n<\om)$ in
$M[x_\beta:\beta\leq \alpha+1]$
which is a $\gamma$-cover of
$$\finsub\cup \{y_\beta:\beta\leq \alpha\} \cup
[y_{\alpha+1}]^{*\om}$$
\bigskip
\proof
Let $\bigcup_n F_n=\finsub\cup \{y_\beta:\beta\leq \alpha\}$
be an increasing union of finite sets and define
$\vv_n=\{U\in\uu_n : F_n\sq U\}$
and note that they are $\om$-covers.
Next inductively define $\ww_n$ by $\ww_0=\vv_0$ and
$$\ww_{n+1}=\{U\cap V: U\in\vv_n, V\in\ww_n\}$$
and note that they are $\om$-covers which refine each other.
Working in the ground model
$M[x_\beta:\beta\leq \alpha]$ construct an increasing sequence
$k_n$ and $U_n\in \ww_n$ so that
$$\{x\sq\om:x\cap [k_n,k_{n+1})=\emptyset\}\sq U_n$$
this can be done since $\ww_n$ is an $\om$-cover
of $\finsub$. Now since $x_{\alpha+1}$ is Cohen
real the following set will be infinite:
$$A=\{n<\om: x_{\alpha+1}\cap [k_n,k_{n+1})=\emptyset\}$$
The same or larger set will work for $y_{\alpha+1}$
and so $(U_n:n\in A)$ will be a $\gamma$-cover of
$[y_{\alpha+1}]^{*\om}$. The refining conditions on $\ww_n$ means we can
fill it in on the complement of $A$ and the choice of $\vv_n$ means
it is a $\gamma$-cover of the rest.
\qed
The Claim shows that
$\finsub\cup\{y_\alpha:\alpha<\om_1\}$ is
a $\gamma$-set.
\bigskip Next we show that there are no uncountable strong $\gamma$-sets.
Suppose for contradiction that $X\sq 2^\om$ is an uncountable strong
$\gamma$-set witnessed by $(k_n:n<\om)$ in the model $N$. By the usual ccc
arguments we may suppose that $X,(k_n:n<\om)\in M$ where $M\sq N$ is some model
of CH. Let $u\in N\cap \om^\om$ be Cohen generic over $M$ and $v\in N\cap
\om^\om$ Cohen generic over $M[u]$ so that if we let
$$\uu_n=\{[s]:s\in 2^{u(n)}\}$$
then (using that $N$ thinks $X$ is strong $\gamma$) there exists
$$(\vv_n\in [\uu_n]^{\leq k_n}:n<\om)\in M[u,v]$$
so that
$\forall x\in X\forall^\infty n\;\; x\in\cup\vv_n$.
Let $\poset$ denote Cohen forcing and since
it is countable there must be some $(p,q)\in\poset\times\poset$ and
$N<\om$ such that
$$(p,q)\forces (\name{\vv}_n\in [\name{\uu}_n]^{\leq k_n}:n<\om)$$
and
$$Y=\{x\in X:(p,q)\forces \forall n>N \;\; x\in\cup\name{\vv}_n\}$$
is uncountable. Fix $n>N,|p|$. Now since $Y$ is uncountable there
exist some level $l<\om$ with
$$|\{x\res l: x\in Y\}|> k_{n}$$
Let $r\supseteq p$ be an extension with $r(n)=l$. But this is a
contradiction since
\begin{itemize}
\item $(r,q)\forces \lq\vv_n\sq 2^{l} \rmand |\vv_n|\leq k_{n}\rq$, and
\item $(r,q)\forces \lq x\in\vv_{n}\rq$ for every $x\in Y$ and
so $(r,q)\forces \lq \{x\res l: x\in Y\}\sq \vv_n\rq$
\end{itemize}
\qed
Remark. T. Bartoszy\'nski has shown that in the iterated superperfect real
model every strong $\gamma$-set is countable. Superperfect forcing is also
called rational perfect set forcing, see Miller \cite{super}. The principle
$\diamondb$ (see D\v zamonja, Hru\v s\'ak, and Moore \cite{dhm}) implies that
there is an uncountable $\gamma$-set. Since $\diamondb$ holds in the iterated
superperfect real model, we get another model for the consistency of strong
$\gamma$-BC but not $\gamma$-BC.
\medskip
Remark. Tsaban and Weiss \cite{tw} have shown that the following are
equivalent:
\begin{enumerate}
\item Every set of reals of strong measure zero is countable (Borel conjecture).
\item Every set of reals with the property $S_1(\Omega,\Omega)$ is countable.
\end{enumerate}
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numbers. Topology Appl. 17 (1984), no. 2, 145--155.
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\bibitem{laver}
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\end{thebibliography}
\begin{flushleft}
Arnold W. Miller \\
miller@math.wisc.edu \\
http://www.math.wisc.edu/$\sim$miller\\
University of Wisconsin-Madison \\
Department of Mathematics, Van Vleck Hall \\
480 Lincoln Drive \\
Madison, Wisconsin 53706-1388 \\
\end{flushleft}
\appendix
\newpage
\begin{center}
Appendix \\
\end{center}
This is not intended for publication but only for the electronic version.
\begin{theorem}(T. Bartoszynski)
In the iterated superperfect forcing model, every strong $\gamma$-set
is countable.
\end{theorem}
\proof
This model is obtained by the countable support iteration of length
$\om_2$ of superperfect forcing over a model of
CH.
First we consider one-step. Let
$f$ be superperfect generic over $M$ a model of set theory. Define
$(\uu_n:n<\om)$ by $$\uu_n=\{[s]: s\in 2^{f(n)}\}.$$
\bigskip\noindent{\bf Claim}. Let $g\in \om^\om\cap M$ and
$(\vv_n\in [\uu_n]^{m}\cup \name{\vv}_n)\sq
\cup\{K_s\;:\; s\in\splitnode(q)\}\rq$$
\qed
Now suppose for contradiction that $X$ is an uncountable
strong $\gamma$-set in the model $M[f_\alpha:\alpha<\om_2]$.
By the $\om_2$ chain condition and a Lowenheim-Skolem argument
there must be an $\alpha_0<\om_2$
with $X,(k_n:n<\om)\in M[f_\alpha:\alpha<\alpha_0]$ such that
$$M[f_\alpha:\alpha<\alpha_0]\models X \mbox{ is a strong $\gamma$-set
with witness } (k_n:n<\om)$$
Denote $M[f_\alpha:\alpha<\alpha_0]$ as $M_0$.
Now using $f_{\alpha_0}$ (the next superperfect real)
Let $\uu_n=\{[s]\;:\; s\in 2^{f_{\alpha_0}(n)}\}$. By the one step argument
for any $g\in M_0\cap \om^\om$
$$M_0[f_{\alpha_0}]\models\forall (\vv_n\in[\uu_n]^{g(n)}
|\{x\in X\cap 2^\om\;:\; \forall^\infty n\;\; x\in \cup\vv_n)|\leq\om.
$$
Denote $M_0[f_{\alpha_0}]$ as $M_1$. Our final model
$M_2 = M[f_\alpha:\alpha<\om_2]$ satisfies the Laver property over
the intermediate models. . This means
for any $f\in M_2\cap \om^\om$ such that there exists
$h\in M_1\cap \om^\om$ which bounds $f$, i.e., $f(n)h(n)$, then there must be infinitely many $i$ so that there exists
$nh(n)$.
Note that $x_\delta=\{k_{g(i)}:n<\om\}$. So as we
have remarked there are infinitely many $i$ (say $i\in C$)
so that there exists
elements of $n_i