% Cardinal characteristic for relative $\ga$-sets, A.Miller
% LaTeX2e main result obtained 4-23-2003 last revised 11-21-2008
\documentclass[12pt]{article}
\usepackage{amssymb}
\def\appendix{\par} % to see appendix
%\def\appendix{\end{document}} %% to hide appendix
\def\om{\omega}
\def\al{\alpha}
\def\ga{\gamma}
\def\uu{{\mathcal U}}
\def\vv{{\mathcal V}}
\def\ff{{\mathcal F}}
\def\gg{{\mathcal G}}
\def\rr{{\mathbb R}}
\def\infsets{[\om]^\om}
\def\sq{\subseteq}
\def\sm{\setminus}
\def\bb{{\mathfrak b}}
\def\dd{{\mathfrak d}}
\def\pp{{\mathfrak p}}
\def\cc{{\mathfrak c}}
\def\rmand{{\mbox{ and }}}
\def\rmiff{{\mbox{ iff }}}
\def\sigc{{\bf \Sigma}^0_2}
\def\siga{{\bf \Sigma}^1_1}
\def\poset{{\mathbb P}}
\def\forces{{| \kern -2pt \vdash}}
\def\force{\forces}
\def\sqbdd{{\sqsubseteq^{{\rm bounded}}}}
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{question}[theorem]{Question}
\newtheorem{prop}[theorem]{Proposition}
\def\proof{\par\noindent Proof\par\noindent}
\def\qed{\par\noindent QED\par}
\begin{document}
\begin{center}
The cardinal characteristic for relative $\ga$-sets
\end{center}
\begin{flushright}
Arnold W. Miller \footnote{
Thanks for partial support from the BEST conference 2003,
support from Boise State University, and to
Tomek Bartoszynski, Justin Moore, and Marion Scheepers, for their
hospitality during the time the main result in this paper was
obtained.
\par Mathematics Subject Classification 2000: 03E35 54D20 03E50
\par Keywords: pseudo-intersection cardinal, relative $\ga$-set,
continuum, covering property, filter on $\om$.}
\end{flushright}
\begin{quote}
Abstract: For $X$ a separable metric space define
$\pp(X)$ to be the smallest cardinality of
a subset $Z$ of $X$ which is not a relative $\ga$-set in $X$, i.e.,
there exists an $\om$-cover of $X$ with no $\ga$-subcover of
$Z$. We give a characterization of $\pp(2^\om)$ and
$\pp(\om^\om)$ in terms of definable free filters on $\om$
which is related to the pseudo-intersection number $\pp$.
We show that for every uncountable standard analytic space $X$ that either
$\pp(X)=\pp(2^\om)$ or $\pp(X)=\pp(\om^\om)$. We show that
the following statements are each relatively consistent with ZFC:
(a) $\pp=\pp(\om^\om) < \pp(2^\om)$ and
(b) $\pp < \pp(\om^\om) =\pp(2^\om)$
\end{quote}
First we remind the reader of
the definition of a $\ga$-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
$F\sq U$. The space $X$ is a $\ga$-set iff for every $\om$-cover $\uu$
of $X$ there
exists a sequence $(U_n\in\uu:n<\om)$ such that for every $x\in X$ for all
but finitely many $n$ we have $x\in U_n$, equivalently
$$X=\bigcup_{m<\om}\bigcap_{n>m}\;U_n \;\;\;\mbox{ or }\;\;\;
\forall x\in X\;\forall^{\infty} n\in\om\;\; x\in U_n.$$
We refer to the sequence $(U_n:n<\om)$ as a
$\ga$-cover of $X$, although technically we are supposed to assume
that the $U_n$ are distinct.
In this paper all our spaces are
separable metric spaces, so we may assume that all $\om$-covers are
countable. This is because we can replace an arbitrary $\om$-cover with a
refinement consisting of finite unions of basic open sets.
The $\ga$-sets were first considered by Gerlits and Nagy \cite{gn}.
One of the things that they showed was the following.
The pseudo-intersection number $\pp$ is defined as follows:
$$\pp=\min\{|\ff|\;:\;\ff\sq\infsets\mbox{ has the FIP and }\neg\exists
X\in\infsets\; \forall Y\in\ff\; X\sq^* Y\}$$
where FIP stands for the finite intersection property, i.e., every
finite subset of $\ff$ has infinite intersection, and $\sq^*$ denotes
inclusion mod finite. The set $X$ in this definition is called the
pseudo-intersection of the family $\ff$.
Gerlits and Nagy \cite{gn} showed that every $\ga$-set
has strong measure zero (in fact, the Rothberger property
$C^{\prime\prime}$)
and that Martin's Axiom implies every set of reals of size smaller than the
continuum is a $\ga$-set. Their arguments show that
$$\pp={\rm non}(\ga\mbox{-set })=^{def}\min\{|X|\;:\; X \mbox{ is not a
$\ga$-set }\}$$
where we only consider separable metric spaces $X$.
The property of being a $\ga$-set is not hereditary.
In fact, a $\ga$-set $X$ of size continuum is constructed
in Galvin and Miller \cite{gm} using MA, which has the property that
there exists a countable $F\sq X$ such that
$X\sm F$ is not a $\ga$-set. However, any closed subspace of a $\ga$-set
is a $\ga$-set.
Babinkostova, Guido and Kocinac \cite{babink} have defined the notion of a
relative $\ga$-set.
This is also studied in Babinkostova, Kocinac, and Scheepers \cite{babink2}.
For $X\sq Y$ define $X$ to be a $\ga$-set relative to $Y$ iff
for every open $\om$-cover $\uu$ of $Y$ there exists
a sequence $(U_n\in\uu:n<\om)$ such that
$$X \sq \bigcup_{m<\om}\bigcap_{n>m}\;U_n.$$
Note that if $Z\sq X \sq Y$ and $X$ is a relative $\ga$-set in
$Y$, then $Z$ is also.
Define the following cardinal number:
$$\pp(Y)=\min\{|X|\;:\; X\sq Y \mbox{ is not a $\ga$-set relative to } Y\}.$$
Perhaps it should be written non($\ga$ relative to $Y$).
In Just, Scheepers, Szeptycki, and Miller \cite{jmss} many cardinal
characteristics for covering properties are shown to be equal to well-known
cardinals. Scheepers has noted that the cardinal
numbers of the relativized version of the Rothberger property
$C^{\prime\prime}$ work out to be
either cov(meager) (the cardinality of the smallest cover of the real line with
meager sets) or non(SMZ) (the cardinality of the smallest non strong measure
zero set of reals).
Scheepers has raised the question of what we can say about the
relativized versions for the $\ga$-property. We begin with the easy
\begin{prop}
$\pp\leq \pp(\om^\om)\leq \pp(2^\om)\leq \cc$
\end{prop}
\proof
If $X$ is a $\ga$-set, then it is a $\ga$-set relative to
any superspace. Let $|X|=\pp(\om^\om)$ be a subset of $\om^\om$
which is not a relative $\ga$-set. Then $X$ is not a $\ga$-set relative
to itself, and hence $\pp\leq |X|=\pp(\om^\om)$.
For the second inequality, suppose $X\sq 2^\om$ is not a
$\ga$-set relative to $2^\om$ with $|X|=\pp(2^\om)$.
Let $\uu$ be an $\om$-cover of $2^\om$ witnessing that $X$ is
not a relative $\ga$-set in $2^\om$.
Then
$$\{U\cup (\om^\om\sm 2^\om)\;:\; U\in\uu\}$$
is an $\om$-cover of $\om^\om$ witnessing that $X$
is not a $\ga$-set relative to $\om^\om$, and so
$\pp(\om^\om)\leq |X|=\pp(2^\om)$.
\qed
We give another characterization of $\pp(\om^\om)$ and
$\pp(2^\om)$. A filter is free iff it contains the cofinite sets.
For $\ff\sq P(\om)$ a free filter on $\om$, define
$$\pp_\ff=\min\{|X|\;:\; X\sq\ff\;\;\rmand\;
\neg\exists a\in\infsets\;\; \forall b\in X\;\; a\sq^* b\}.$$
Note that $\pp$ is the minimum of $\pp_\ff$ for
$\ff\sq P(\om)$ a free filter, since every family with the FIP generates
a filter.
We have the following characterizations:
\begin{theorem}\label{characterization}
\par (a) $\pp(\om^\om)$ is the minimum
of $\pp_\ff$ such that $\ff\subseteq P(\om)$ is a ${\bf \Sigma}^1_1$
free filter.
\par (b) $\pp(2^\om)$ is the minimum of
$\pp_\ff$ such that $\ff\sq P(\om)$ is a $\sigc$ free filter.
\end{theorem}
\proof
Suppose $X\sq \om^\om$ with $|X|=\pp(\om^\om)$ and
$\uu$ is an open $\om$-cover of $\om^\om$ witnessing that $X$ is
not a relative $\ga$-set.
Without loss of generality we may assume that $\uu$ is a countable
family of clopen sets, say $\uu=\{U_n:n\in\om\}$. Let $f:\om^\om\to P(\om)$
be the Marczewski \cite{marc} characteristic function of sequence
$$f(x)=\{n:x\in U_n\}.$$
This is a continuous mapping so
its image $\gg=f(\om^\om)$ is ${\bf \Sigma}^1_1$.
Since $\uu$ was an $\om$-cover the
image $\gg$ has
the FIP and note that the filter $\ff$ generated by a
${\bf \Sigma}^1_1$
family $\gg$ with the FIP is ${\bf \Sigma}^1_1$, i.e.,
$$X\in\ff \rmiff \exists F\in [\gg]^{<\om}\;\; \cap F\sq X.$$
Now assume $|X| < \pp_\ff$ and hence
$|f(X)|<\pp_\ff$. Then
there exists $a\in[\om]^\om$ such that for each $b\in X$ we have that
$a\sq^* f(b)$. It follows that $(U_n\;:\;n\in a)$ is a $\ga$-cover of
$X$ which is a contradiction. Hence $\pp(\om^\om)=|X|\geq \pp_\ff$ and so
$$\pp(\om^\om)\geq \min\{\pp_\ff:\ff \mbox{ is a $\siga$ free filter }\}.$$
To see the other inequality, suppose $\ff\sq P(\om)$ is a
${\bf \Sigma}^1_1$
filter and $X\sq \ff$ has no pseudo-intersection with
$|X|=\pp_\ff$. Let $f:\om^\om\to\ff$ be a continuous onto map. For each
$n\in\om$ define $U_n=f^{-1}(\{x\in\ff\;:\; n\in x\})$. Define
$\uu=\{U_n\;:\;n\in\om\}$. Then $\uu$ is an $\om$-cover of $\om^\om$.
Choose $Y\sq\om^\om$ with $f(Y)=X$ and $|Y|=|X|$.
If $Y$ is relative $\ga$ in $\om^\om$,
then there exists $a\in\infsets$
such that $(U_n\;:\;n\in a)$ is a
$\ga$-cover of $Y$. For each $b\in X$ we have $c\in Y$ with $f(c)=b$.
For each $n$ if $c\in U_n$, then
$f(c)\in f(U_n)$ and so $n\in b$. It follows that $a\sq^*c$
for all $c\in X$. Since we are assuming that there is no such $a$, we
must have that $Y$ is not a $\ga$-set relative to $\om^\om$ and therefore
$$\pp(\om^\om)\leq |Y|=|X|=\pp_\ff$$
and therefor
$$\pp(\om^\om)\leq \min\{\pp_\ff:\ff \mbox{ is a
$\siga$ free filter }\}.$$
The proof for $\pp(2^\om)$ is similar. To see that
$$\pp(2^\om)\geq \min\{\pp_\ff:\ff \mbox{ is a
$\sigc$ free filter }\}$$
choose $X\sq 2^\om$ with $|X|=\pp(2^\om)$ and $\uu$ a countable
clopen $\om$-cover
of $2^\om$ with no $\ga$-subcover of $X$.
Let $f:2^\om\to P(\om)$
be defined by $f(x)=\{n:x\in U_n\}$. Then $f$ is continuous and so
its range is a compact set $f(2^\om)=C\sq P(\om)$ which has the FIP.
Note that the filter $\ff$ generated by $C$ is $\sigc$ in
$P(\om)$. To see this note that
for
each $n<\om$ the map $h:C^n\to P(\om)$ defined by
$$h(X_1,\ldots,X_n)=X_1\cap X_2\cap\cdots\cap X_n$$
is continuous and so its range $C_n$ is compact. For each $n$
let $D_n$ be the compact set
$$D_n=\{(x,y)\;:\;x\in C_n \rmand x\sq y\}.$$
and let $\pi(x,y)=y$ be the projection
onto the second coordinate. Then
$$\ff=\cup_{n<\om}\pi(D_n)$$
and so $\ff$ is $\sigc$.
Hence, if $|X|<\pp_\ff$, then
$|f(X)|<\pp_\ff$ and therefor
there exists $a\in \infsets$ with $a\sq^* f(x)$ for each $x\in X$
and therefor $x\in U_n$ for all but finitely many $n\in a$ and
$(U_n:n\in a)$ is a $\ga$-cover of $X$, which is a contradiction.
To see that
$$\pp(2^\om)\leq \min\{\pp_\ff:\ff \mbox{ is a
$\sigc$ free filter }\}$$
suppose that $\ff$ is a $\sigc$ free filter in $P(\om)$ and
for contradiction $\pp_\ff<\pp(2^\om)$.
First note that there exists a compact $C\sq\ff$ such that for every $x\in\ff$
there exists a $y\in C$ with $x=^*y$. To see this, suppose that
$\ff=\cup_{n<\om}C_n$. For each $n<\om$ let
$$C_n^*=\{x\sq\om : n\sq x\rmand
\exists y\in C_n\;\; \forall i\geq n (i\in y\rmiff i\in x)\}$$
then $C=\cup_{n<\om} C_n^*$ does the trick. Now suppose
$X$ is a subset of $\ff$ with no pseudo-intersection
and $|X|=\pp_\ff<\pp(2^\om)$. Choose a map $f:2^\om\to C$ which is
continuous and onto
and select $Y\sq 2^\om$ with $|Y|=|X|$ such that
for each $x\in X$ there exists $y\in Y$ with $f(y)=^*x$.
Let
$$U_n=f^{-1}(\{x\in C\;:\; n\in x\}).$$
Then $\uu=\{U_n:n<\om\}$
is an $\om$-cover of $2^\om$ and since $Y$ is a relative $\ga$-set
there exists
$a\in \infsets$ such that for every $y\in Y$ we have that
$y\in U_n$ for all but finitely many $n\in a$. Hence
for each $x\in X$ there is $y\in Y$ with $a\sq^* f(y)=^*x$
which means that $X$ does have a pseudo-intersection which is
contrary to what we assumed.
\qed
For another paper studying the connection between $\ga$-sets and
free filters, see LaFlamme and Scheepers \cite{ls}.
\begin{lemma}\label{lem1}
\par (a) Suppose that $X$ is homeomorphic to a closed subspace of
$Y$, then $\pp(Y)\leq \pp(X)$.
\par (b) Suppose that $f:X\to Y$ is continuous and onto, then
$\pp(X)\leq \pp(Y)$.
\end{lemma}
\proof
\medskip\noindent (a) Suppose $Z\sq X$ with $|Z|=\pp(X)$ is not relatively
$\ga$ in $X$ and this is witnessed by a family $\uu$ of open sets of $Y$ which
is an $\om$-cover of $X$.
Then
$$\{U\cup(Y\sm X)\;:\; U\in\uu\}$$
is an $\om$-cover of $Y$ which shows that $Z$ is not relatively $\ga$ in
$Y$. Hence $\pp(Y)\leq |Z|=\pp(X)$.
\medskip\noindent (b) Suppose $Z\sq Y$ with $|Z|=\pp(Y)$ is not relatively
$\ga$ in $Y$ and this is witnessed by an $\om$-cover
$\uu$. Choose $W\sq X$ with $|W|=|Z|$ and $f(W)=Z$.
Let $\vv=\{f^{-1}(U)\;:\; U\in\uu\}$. Since $f$ is onto, $\vv$ is
an $\om$-cover of $X$. We claim that there is no sequence
$(U_n\in\uu: n<\om)$ such that
$(f^{-1}(U_n): n<\om)$ is a $\ga$-cover of $W$.
This is
because $x\in f^{-1}(U_n)$ implies $f(x)\in U_n$ and since
$f(W)=Z$, then
$(U_n:n<\om)$ would
be a $\ga$-cover of $Z$.
It follows that $\pp(X)\leq |W|=|Z|=\pp(Y)$.
\qed
\begin{theorem}
Suppose $X$ is an uncountable $\siga$ set in a Polish space,
i.e., a nontrivial standard analytic space, then
\par (a) if $X$ is not $\sigma$-compact, then $\pp(X)=\pp(\om^\om)$ and
\par (b) if $X$ is $\sigma$-compact, then $\pp(X)=\pp(2^\om)$.
\end{theorem}
\proof
Every $\siga$ set is the continuous image of $\om^\om$ and
every uncountable $\siga$ set
contains a homeomorphic copy of $2^\om$.
It follows from Lemma \ref{lem1} that
$$\pp(\om^\om)\leq \pp(X)\leq \pp(2^\om).$$
\medskip\noindent (a) If $X$ is not $\sigma$-compact, then Hurewicz
\cite{hurewicz} (see Kechris \cite{kechris} 21.18) proved that there exists a
closed subspace of $X$ which is homeomorphic to $\om^\om$. Hence by Lemma
\ref{lem1} we have $\pp(X)\leq\pp(\om^\om)$.
\medskip\noindent (b) Suppose $X$ is $\sigma$-compact. We need to
show that $\pp(2^\om)\leq \pp(X)$.
We first consider the special case that $X=\om \times 2^\om$.
Choose $Y\sq \om \times 2^\om$ to be
non relatively $\ga$ in $\om \times 2^\om$ with $|Y|=\pp(\om\times 2^\om)$.
Since $\om\times 2^\om$ is zero dimensional
we can assume that there exists an $\om$-cover $\uu=\{C_n:n<\om\}$
of clopen sets in $\om\times 2^\om$ with no $\ga$-subcover of $Y$.
As in the proof of Theorem \ref{characterization}
we consider $f:\om \times 2^\om\to P(\om)$
defined by
$$f(x)=\{n<\om: x\in C_n\}.$$
The function $f$
is continuous since the $C_n$ are clopen.
The image
$$f(\om\times 2^\om)\sq P(\om)$$
is a $\sigma$-compact family of subsets of $\om$
with the finite intersection property.
Hence $f(\om\times 2^\om)$ generates a $\sigma$-compact filter $\ff$
as in the proof of Theorem \ref{characterization}.
Note that $f(Y)$ is a subset of $\ff$ without a pseudo-intersection.
Hence $\pp_\ff\leq |f(Y)|\leq |Y|=\pp(\om \times 2^\om)$ and so we have
$\pp(2^\om)\leq \pp(\om \times 2^\om)$ and hence
$\pp(2^\om)= \pp(\om \times 2^\om)$.
Now suppose that $X$ is any $\sigma$-compact metric space.
Note that there is a continuous onto mapping $f:\om \times 2^\om \to X$
and so by Lemma \ref{lem1} we have that
$$\pp(X)\geq \pp(\om \times 2^\om)=\pp(2^\om).$$
\qed
The main result of this paper is the following theorem:
\begin{theorem}\label{main}
The following statements are each relatively consistent with ZFC:
\par (a) $\pp=\pp(\om^\om) < \pp(2^\om)$ and
\par (b) $\pp < \pp(\om^\om) =\pp(2^\om)$
\end{theorem}
\proof
\medskip\noindent Part(a).
Given an $\om$-cover $\uu$ of $2^\om$
define the poset $\poset(\uu)$ as follows:
\begin{enumerate}
\item $p\in \poset(\uu)$ iff $p=(F,(U_n\in\uu:nn$. Then $\forall x\in V\cap 2^\om\;\;\forall^\infty n\;\;
x\in U_n$.
\end{lemma}
\proof
$\sigma$-centered is clear, since if $(N^p_n:ng(n)\}.$$
\begin{lemma}\label{inequal}
$\pp(\om^\om)\leq \bb$
\end{lemma}
\proof
Suppose $X\sq \om^\om$ and $|X|<\pp(\om^\om)$. We need
to show that $X$ is eventually dominated. Without loss of
generality we may assume that the elements of $X$ are increasing
and $X$ is infinite.
For each $n<\om$ let
$$\uu_n=\{U_m^n:m<\om\}\mbox{ where } U_m^n=\{f\in\om^\om: f(n) n_0\; x(n)<\tau(n)$. By making $n_0$ larger
(if necessary) we may assume that
$$|F^p|=k_{n_0}\rmand (U_m^{n_0}:mn_0\;\; x(n)n_0$.
By our construction of $g$ we have that there exists $q\leq p$ such that
$q\forces \tau(l) n_0 \tau(n)>x(n)$.
This proves the Claim and the Lemma.
\qed
It follows (see Bartoszynski and Judah \cite{barto} Theorem 6.5.4)
that the finite support iteration using $\poset(\uu_\al)$ at
stage $\al$
does not add a dominating real and so over a ground model which
satisfies CH we have that
$V[G_{\om_2}]$ satisfies that $\bb=\om_1$ and hence $\pp(\om^\om)=\om_1$
by Lemma \ref{inequal}.
This proves Theorem \ref{main} part (a), the consistency of
$\pp=\pp(\om^\om) < \pp(2^\om)$.
\bigskip
Part (b) (the consistency of $\pp < \pp(\om^\om) =\pp(2^\om)$) is simpler.
It is well known that $\pp>\om_1$ implies that $2^{\om_1}=2^\om$.
For example, see Rothberger \cite{roth}. Now starting with
a ground model $V$ which satisfies $2^\om=\om_2$ and $2^{\om_1}=\om_3$,
do a finite support iteration using
$\poset(\uu_\al)$ at stage $\al<\om_2$ where
$\uu_\al$ is an $\om$-cover of $V[G_\al]\cap \om^\om$.
Dovetail so that $\uu_\al$ for $\al<\om_2$ lists all countable $\om$-covers of
$\om^\om$ in the final model $V[G_{\om_2}]$.
This can be done since in
all these models the continuum is $\om_2$.
The analogue of Lemma \ref{routine} holds for $\om^\om$ in
place of $2^\om$ so
in the final model we
have that $\pp(\om^\om)=\om_2$.
Also we get $\pp=\om_1$ since $2^{\om_1}=\om_3>\om_2=2^\om$.
This finishes the proof of Theorem \ref{main}.
\qed
One obvious question is
\begin{question}
Is it consistent to have $\pp<\pp(\om^\om)<\pp(2^{\om})$?
\end{question}
\begin{question}(Scheepers)
Are either $\pp(\om^\om)$ or $\pp(2^{\om})$ the same
as some other well-known small cardinal?
See Vaughan \cite{vaughan} for a plethora of such cardinals.
\end{question}
In Laver's model \cite{laver} for the Borel conjecture, we have that
$\bb=\dd=\om_2$ and $\pp(2^\om)=\pp(\om^\om)=\om_1$.
In Laver's model there is a set of reals of size $\om_1$ which does not have
measure zero, i.e., non(measure)=$\om_1$, Judah and Shelah \cite{js}, see also
Bartoszynski and Judah \cite{barto} or Pawlikowski \cite{pawl}. But it is
easy to see that $\pp(2^\om)\leq$ non(measure), i.e.,
if $X\sq 2^\om$ and $|X|<\pp(2^\om)$ then $X$ has measure
zero. Let $\{x_n:n<\om\}\sq X$ be distinct and look at
$$\uu=\{C\sq 2^\om\;:\;\exists n\;\; x_n\notin C \mbox{ is clopen and }
\mu(C)<{1\over 2^n}\}.$$
This is an $\om$-cover of $2^\om$ and so there exists a sequence
$C_n\in\uu$ with $X\sq\cup_n\cap_{m>n}C_m$. For any $n$ at most finitely
many $C_n$ have measure $>{1\over 2^n}$
which shows that $X$ has measure zero.
It is also true that $\pp(2^\om)\leq$ non(SMZ), i.e., if $|X|<\pp(2^\om)$ then
$X$ has strong measure zero. The result of Gerlits and Nagy \cite{gn}, that
$\ga$-sets have the Rothberger property $C^{\prime\prime}$, relativizes to show
that if $X\sq 2^\om$ and $|X|<\pp(2^\om)$, then $X$ has the relative Rothberger
property and this implies that $X$ has strong measure zero.
\begin{question}\footnote{This has been answered. See appendix.}
Suppose that $Y=\cup_{n<\om}X_n$ is an increasing union where $Y$ is
a separable metric space. If each $X_n$ is relatively $\ga$ in $Y$, is
$Y$ a $\ga$-set? If not, suppose each $X_n$ is a $\ga$-set, then
is $Y$ a $\ga$-set?
\end{question}
Tsaban \cite{tsaban} Lemma 22 shows that the answer to this question in
the Borel cover case is yes. It is also connected to the existence of
a group which is a $\ga$-set, \cite{tsaban} Theorem 20.
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of sets and some of its applications. Fund. Math. 31(1938), 207-233.
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Pawlikowski, Janusz; Laver's forcing and outer measure.
in {\bf Set theory (Boise, ID,
1992--1994)}. 71--76, Contemp. Math., 192, Amer. Math. Soc., Providence, RI,
1996.
\bibitem{roth}
Rothberger, Fritz;
On some problems of Hausdorff and of Sierpi\'nski.
Fund. Math. 35, (1948). 29--46.
\bibitem{tsaban}
Tsaban, B; $o$-bounded
groups and other topological groups with strong combinatorial properties.
Proc. Amer. Math. Soc.
134 (2006), no. 3, 881--891.
\bibitem{vaughan}
Vaughan, Jerry E.; Small uncountable cardinals and topology. With an appendix
by S. Shelah. in {\bf Open problems in topology}.
195--218, North-Holland, Amsterdam,
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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
The appendix is not intended for final publication but for
the electronic version only.
\begin{center}
Appendix \\
Scheepers Remarks
\end{center}
\noindent Def. $X\sq Y$ is $C^{\prime\prime}$ (Rothberger) in $Y$ iff for every
sequence $(\uu_n:n<\om)$ of open covers of $Y$ there
exists a cover $(U_n\in\uu_n:n<\om)$ of $X$.
\bigskip\noindent
Prop. \par (a) non($C^{\prime\prime}$ in
$\om^\om$)=non($C^{\prime\prime}$)=cov(meager)=
non(SMZ in $\om^\om$)
\par (b) non($C^{\prime\prime}$ in $2^\om$) = non(SMZ in $2^\om$)=
non(SMZ in $\rr$)
\proof
Here we mean strong measure zero in the usual metric on
the reals and for $\om^\om$ or $2^\om$ the metric $d(x,y)={1\over n+1}$
where $n$ is minimal such that $x(n)\not=y(n)$.
\bigskip\noindent
(a) Fremlin and Miller
(1988) prove:
non($C^{\prime\prime}$)=cov(meager)=non(SMZ in $\om^\om$)
non($C^{\prime\prime})\leq $non($C^{\prime\prime}$ in $\om^\om$)
since if $X$ is not relatively $C^{\prime\prime}$ it is not $C^{\prime\prime}$.
non($C^{\prime\prime}$ in $\om^\om$) $\leq$ non(SMZ in $\om^\om$) since
$C^{\prime\prime}\sq$ SMZ.
\bigskip\noindent
(b) Suppose $X\sq 2^\om$ fails to
be relatively $C^{\prime\prime}$.
Note that by compactness of $2^\om$ we may assume
there is a sequence $(\uu_n:n<\om)$ of
finite clopen covers of $2^\om$ for which there is no
$(U_n\in\uu_n:n<\om)$ which covers $X$. Now choose $\epsilon_n>0$ so
that any interval $[s]$ with diameter less than $\epsilon_n$ is
a subset of some $U_n$. For the converse, suppose $X$ fails to
have SMZ in $2^\om$ which is witnessed by $(\epsilon_n:n<\om)$.
Then the sequence $\uu_n=\{C\sq 2^\om: {\rm diam}(C)<\epsilon_n\}$ witnesses
that it is not $C^{\prime\prime}$.
\par
non(SMZ in $2^\om$)=non(SMZ in $[0,1]$)=non(SMZ in $\rr$) is easy to prove.
\qed
These cardinals can be different, for example, in the iterated
modified Silver reals model, see Miller (1981),
cov(meager)$=\om_1$ while non(SMZ in $2^\om)=\om_2$.
\bigskip
\noindent Def. $X$ has the Menger property M iff for every sequence
$(\uu_n:n<\om)$ of open covers there exists $(\vv_n\in[\uu_n]^{<\om}:n<\om)$
such that $X\sq \cup_n\cup \vv_n$.
\noindent Def. $X$ has the Hurewicz property H iff for every sequence
$(\uu_n:n<\om)$ of open covers there exists $(\vv_n\in[\uu_n]^{<\om}:n<\om)$
such that $X\sq \cup_m\cap_{n>m}\cup\vv_n$.
\bigskip
Then it is known, see Miller and Fremlin (1988) that
$$\dd={\rm non}(M)\;\;\;\;\;\bb={\rm non}(H).$$
Relativizing H or M to $2^\om$ doesn't work since $2^\om$ has
property H and M. For $\om^\om$ it is easy to see:
\bigskip
Prop.
\par (a) non(M) $\leq$ non(M in $\om^\om$) $\leq\dd\leq $ non(M)
\par (b) non(H) $\leq$ non(H in $\om^\om$) $\leq\bb\leq $ non(H)
\bigskip\bigskip
\noindent Biblio
\bigskip
Miller, Arnold W.; Some properties of measure and category. Trans. Amer. Math.
Soc. 266 (1981), no. 1, 93--114.
\medskip
Miller, Arnold W.; Fremlin, David H.; On some properties of Hurewicz, Menger,
and Rothberger. Fund. Math. 129 (1988), no. 1, 17--33.
\begin{center}
Question 12
\end{center}
This was answered by Francis Jordan (There are no hereditary
productive $\ga$-spaces, eprint Spring 08). He proves that the increasing
countable union of $\ga$-sets is a $\ga$-set.
\begin{lemma}
For any $\ga$-set $Y$ and sequence $(\uu_n:n<\om)$ of $\om$-covers of $Y$
there exists $(\vv_n\in[\uu_n]^\om:n<\om)$ such that $\bigcup_{n<\om}\vv_n$
is a $\ga$-cover of $Y$.
\end{lemma}
\noindent The proof is left to the reader. (Gerlits-Nagy)
\bigskip
Suppose $X_n$ for $n<\om$ are $\ga$-sets, $X_n\subseteq X_{n+1}$
for each $n$
and $X=\bigcup_{n<\om}X_n$.
Given any $\om$-cover of $X$ we may extract from it
a sequence $\uu_n$ such that each $\uu_n$ is a $\ga$-cover of $X_n$.
Let $\uu_n^0=\uu_n$. Using the Lemma construct a sequence
$(\uu_n^p:p\leq n<\om)$ by induction on $p$ such that
$\uu_n^{p+1}\in [\uu_n^p]^\om$ and $\bigcup\{\uu_n^{p+1}:{p+1\leq n<\om}\}$
is a $\ga$-cover of $X_p$.
Any sequence $(U_n\in \uu_n^n:n<\om)$ is a $\ga$-cover of $X$.
\bigskip\bigskip\bigskip
TeXed on : \today
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