% LaTex2e
% On lambda'-sets, A. Miller
% Dec 2002 last revised Mar 30, 2004
% to appear in Topology Proceedings
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\def\om{\omega}
\def\sq{\subseteq}
\def\bb{{\mathfrak b}}
\def\dd{{\mathfrak d}}
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\def\dom{{\rm dom}}
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\def\res{\upharpoonright}
\def\qed{\par\noindent {\bf QED}\par}
\def\daglamb{$(\dagger)$-$\lambda'$-set }
\def\rmor{\mbox{ or }}
\def\rmiff{\mbox{ iff }}
\def\rmand{\mbox{ and }}
\def\gameone{{G_{{\cal O,P}}(X_F,\infty)}}
\def\gamefin{{G_{{\cal O,P}}^f(X_F,\infty)}}
\def\gameonegen{{G_{{\cal O,P}}(X,x)}}
\def\gamefingen{{G_{{\cal O,P}}^{fin}(X,x)}}
\def\gamegam{{G_{{\cal F,C}}^{\gamma}(X)}}
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\begin{document}
\begin{center}
{\large On $\lambda'$-sets}
\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 Steprans who directed the special program in set
theory and analysis.
\par Mathematics Subject Classification 2000:
03E17, 03E35}
\end{flushright}
\begin{center}
Abstract
\end{center}
\begin{quote}
A set $X\sq 2^\om$ is a $\lambda'$-set iff for every countable set
$Y\sq 2^\om$ there exists a $G_\delta$ set $G$ such that
$(X\cup Y)\cap G=Y$. In this paper we prove two forcing
results about $\lambda'$-sets.
First we show that it is consistent that every $\lambda'$-set is
a $\gamma$-set. Secondly we show that it is independent whether or
not every \daglamb is a $\lambda'$-set.
\end{quote}
\section{$\lambda'$-sets and $\gamma$-sets}
% result obtained 9-13-2002
A set $X\sq 2^\om$ is a $\lambda'$-set iff for all countable
$A\sq 2^\om$ there exists a $G_\delta$ set $G$ such that
$$(X\cup A)\cap G=A$$
An $\om$-cover of $X$ is a countable set of open sets such that
every finite subset of $X$ is contained in an element of the cover.
A $\gamma$-cover of $X$ is a countable sequence of open subsets of
$X$ such that every element of $X$ is in all but finitely many
elements of the sequence.
\medskip\noindent
Define $X$ to be a {\bf $\gamma$-set} iff any $\om$-cover of $X$ contains a
$\gamma$-cover of $X$.
\medskip In this section we answer a question of Gary Gruenhage
who asked if there is always a $\lambda'$-set which is not a $\gamma$-set. We
answer this in the negative.
It is well known (see Gerlits and Nagy \cite{GN}) that MA($\sigma$-centered)
implies that every set of reals of cardinality less than the continuum is a
$\gamma$-set. The standard model for MA($\sigma$-centered) (see Kunen and Tall
\cite{KT}) is obtained as follows:
Suppose that $M$ is a countable standard model of ZFC+CH and we iterate
$\sigma$-centered forcings of size $\om_1$ in $M$ with a finite support
iteration of length $\om_2$. In the final model $M_{\om_2}$, we have
that MA($\sigma$-centered) is true and the continuum is $\om_2$.
\begin{theorem}
In the standard model for MA($\sigma$-centered) every
$\lambda'$ set has cardinality $\leq \om_1$,
and (it follows
from MA($\sigma$-centered)) every set of size $\om_1$ is
a $\gamma$-set. Hence, in this model, every $\lambda'$-set is
a $\gamma$-set.
\end{theorem}
\proof
We will use the following Lemma in our proof.
\begin{lemma}
Suppose that $\poset$ is a $\sigma$-centered forcing
such that
$$\forces \tau\in {2^\om}$$
Then there
exists a countable set $A\sq 2^\om$ in the ground model
such that for every $p\in\poset$ and open set $U\supseteq A$
coded in the ground model
there exists $q\leq p$ such that $q\forces \tau\in U$.
\end{lemma}
\proof
To prove the Lemma we will use the following Claim.
\bigskip
Claim. Suppose $\Sigma\sq \poset$ is a centered subset. Then
there exists $x\in 2^\om$ such that for every $p\in\Sigma$ and
for every $n<\om$ there exists $q\leq p$ such that
$$q\forces \check{x}\res n= \tau\res n.$$
\noindent pf: Otherwise by the compactness of $2^\om$ there exists a finite
set $$\{p_m:mn \;\;a_n < f(n)\}$$
Definition. A set $X\sq 2^\om$ is
a {\bf \daglamb} iff
for every $f\in\om^\om$ we have $X\cap G_f$
is a $\lambda'$-set.
\begin{theorem}\label{dagger1}
Suppose that the continuum hypothesis is true or even just
$\bb=\dd$. Then there exists a
\daglamb which is not a $\lambda'$-set.
\end{theorem}
\begin{theorem}\label{dagger2}
In the Cohen real model (Cohen's original model for not CH) every
\daglamb is a $\lambda'$-set.
\end{theorem}
\bigskip
\noindent Proof of Theorem \ref{dagger1}
Assume CH. Let $\{f_\alpha\in\om^\om:\alpha<\om_1\}$
be a scale. That is, for $\alpha<\beta$ we have that
$f_\alpha<^* f_\beta$ and for all $g\in\om^\om$ there exists
$\alpha<\om_1$ such that $g<^* f_\alpha$.
We may also assume that the $f_\alpha$ are strictly increasing.
Let $X\sq [\om]^\om$
be the set of ranges of the elements of the scale. Then for
any $g\in\om^\om$ we have that $G_g\cap X$ is countable and hence
a $\lambda'$-set. On the other hand $X$ is not a $\lambda'$-set because
of the countable set $[\om]^{<\om}$. If $U\sq P(\om)$ is
an open set containing $[\om]^{<\om}$, then
$K=P(\om)\setminus U$ is a compact subset of $[\om]^{\om}$.
If we identify $[\om]^\om$ with the strictly increasing
elements of $\om^\om$ (via the homeomorphism $a\mapsto\{a_0,a_1,\ldots\})$),
then there exists $f\in\om^\om$ such that
for all $g\in K$ we have $\forall n\;\; g(n)g_\alpha(n)$$
Note that for
every $g\in\om^\om\cap N$ there exists $\alpha<\om_1$ such that
$$\forall\beta>\alpha\;\; \exists^\infty n \;\;f_\beta(n)>g(n).$$
Suppose by way of contradiction that for some $g\in N[G]\cap\om^\om$
and all $\alpha<\om_1$ we have that $f_\alpha\leq^* g$. Then for
some $\Sigma\in[\om_1]^{\om_1}$ and $n<\om$ we have
that
$$\forall m>n\;\;\forall \alpha\in\Sigma\;\; f_\alpha(m)\leq g(m)$$
Let $q\in G$ force this fact.
Now since $\poset$ is a countable poset, there exists some $p\in G$
with $p\leq q$ such that
$$\Gamma=\{\alpha<\om_1: p\forces \alpha\in\dot{\Sigma}\}$$
is uncountable (and by definability of forcing it is in $N$). But note that
$\{f_\alpha:\alpha\in\Gamma\}$ is unbounded and so for some $m>n$
the set $\{f_\alpha(m):\alpha\in\Gamma\}$ is unbounded in $\om$.
Let $r\leq p$ decide $g(m)$, i.e., for some $k<\om$ suppose
$$r\force \dot{g}(m)=k.$$
Choose $\alpha\in\Gamma$ such that $f_\alpha(m)>k$, then $r$ forces a
contradiction and the Lemma is proved.
\qed
\begin{lemma}\label{lemdagg2}
Suppose $N$ is a countable standard model of ZFC+CH, $\poset$ is a countable
poset in $N$, and
$$N\models Y\sq 2^\om \mbox{ is not a } \lambda'\mbox{ - set}$$
Then for $G$ $\poset$-generic over $N$ we have that
$$N[G]\models Y \mbox{ is not a } \lambda'\mbox{ - set}$$
\end{lemma}
\proof
Let $D\sq 2^\om$ be countable in $N$ and witness that
$Y$ is not a $\lambda'$-set, i.e.,. there is no $G_\delta$ set $\bigcap_nU_n$
coded in $N$ with
$$\bigcap_nU_n\cap (Y\cup D)=D$$
Working in $N$ let $D=\{x_n:n<\om\}$ and
let $Z=Y\setminus D$ and for each $z\in Z$ define
$f_z\in\om^\om$ such that $f_z(n)$ is the least $m$ such that
$x_n\res m\not= z\res m$.
Now the family $X=\{f_z:z\in Z\}$ must be unbounded in $\leq^*$ in
$N$. Suppose not, then there exists $g\in\om^\om\cap N$ which
eventually dominates each element of $X$. It follows that if we
let
$$U_n=\bigcup_{m<\om} [x_m\res \max\{n,g(m)\}]$$
then
$$(\bigcap_{n<\om}U_n)\cap (Y\cup D)=D$$
which is a contradiction.
It follows from Lemma \ref{lemdagg1} that $X$ is unbounded in
$N[G]$. I claim that $D$ cannot be $G_\delta$ in $Y\cup D$ in
the model $N[G]$. Suppose it is, and let $\bigcap_{n<\om}U_n$
be a $G_\delta$ in $N[G]$ such that
$$\bigcap_{n<\om}U_n\cap(Y\cup D)=D$$
For each $n$ let $g_n\in\om^\om$ be such that for every $m$ we have that
$$[x_m\res g_n(m)]\sq U_n.$$
Now for any $z\in Z$ there exist a $n$ such that $z\notin U_n$.
But this means that $f_z(m)\leq g_n(m)$ for every $m$ since otherwise
$$x_m\res g_n(m)= z\res g_n(m)$$
and then $z\in U_n$. This proves the Lemma.
\qed
Now we prove Theorem \ref{dagger2}. Suppose that $X\sq 2^\om$ is
in $M[G]$ where $G$ is $\poset_{\om_2}$-generic over $M$ and
$$M[G]\models X\mbox{ is not a } \lambda' \mbox{-set}$$
By Lowenheim-Skolem arguments there exists $\alpha<\om_2$
such that
$$X_\alpha=^{def}X\cap M[G_\alpha],\;\;
X_\alpha\in M[G_\alpha], \rmand
M[G_\alpha]\models X_\alpha\mbox{ is not a } \lambda' \mbox{-set}$$
Since being a $\lambda'$-set only depends on codes for $G_\delta$-sets
and reals are added by countable suborders of
$\poset_{[\alpha,\om_2)}$ it follows from Lemma \ref{lemdagg2} that
$$M[G]\models X_\alpha\mbox{ is not a } \lambda' \mbox{-set}$$
But if $f\in\om^\om\in M[G]$ is $\om^{<\om}$-generic over
$M[G_\alpha]$ then $X_\alpha\sq G_f$. It follows that
$$M[G]\models X \mbox{ is not \daglamb}$$
as was to be proved.
\qed
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\end{thebibliography}
\begin{flushleft}
Arnold W. Miller \\
miller@math.wisc.edu \\
http://www.math.wisc.edu/$\sim$miller\\
University of Wisconsin-Madison \\
Department of Mathematics, Van Vleck Hall \\
480 Lincoln Drive \\
Madison, Wisconsin 53706-1388 \\
\end{flushleft}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\appendix
\newpage
The appendices are not intended for final publication but for
the electronic version only.
\begin{center}
Appendix A
\end{center}
Here is a proof of
\bigskip
{\it {\bf Theorem } (Nyikos, Gruenhage, Sharma)
If $X\sq 2^\om$ is a
$\lambda'$-set which is dense in $2^\om$, then
${\cal P}$ has no winning strategy in $\gameone$.}
\proof
For contradiction let $\tau$ be a winning strategy for ${\cal P}$.
\bigskip
Claim. For any play $(F_1,\ldots,F_n)$ by ${\cal O}$
there exists $y\in 2^\om$ so that for all finite
$F\sq X$ and $k<\om$ there exists $F_{n+1}\supseteq F$
a finite subset of $X$ such that
$$ y\res k\sq \tau(F_1,F_2,\dots, F_n,F_{n+1})$$
proof: Let $U$ be a regular ultrafilter on the $[X]^{<\om}$,
ie.,
$$\{F\in [X]^{<\om}: x\in F\}\in U \mbox{ for all }x\in X.$$
Define $y\in 2^\om$ by
$$y(k)=1 \rmiff \{F\in [X]^{<\om}:\tau(F_1,F_2,\dots, F_n,F)(k)=1\}\in U$$
Note that the length of $\tau(F_1,F_2,\dots, F_n,F)$ is greater
than $k$ for $U$ almost all $F$ because $X$ is dense.
Also, for any $k<\om$ and finite $F\sq X$ we have that
$$\{F_{n+1}: F\sq F_{n+1},\;
y\res k\sq \tau(F_1,F_2,\dots, F_n,F_{n+1})\}\in U$$
Hence the Claim is proved.
\bigskip
Let $\om^{<\om}=\{s_i:i<\om\}$.
Using the Claim construct $\langle F_s,y_s :s\in \om^{<\om}\rangle$
so that
\begin{enumerate}
\item $F_{s}$ contains all $y_{s_i}$ which are in $X$ for $i<|s|$ and
\item $y_s\res k \sq \tau(F_{s\res 1},F_{s\res 2},\dots, F_s,F_{sk})$
\end{enumerate}
Let $Y=\{y_s:s\in \om^{<\om}\}$ and suppose $U_n$ is a descending
sequence of open sets such that
$$\bigcap_{n<\om}U_n\cap (X\cup Y)=Y$$
Construct $f\in \om^\om$ as follows. Since $y_{f\res n}\in U_n$ there
exists $k<\om$ such that $[y_{f\res n}\res k]\sq U_n$.
Let $f(n)=k$ and
thus we have that for each $n$
$$\tau(F_{f\res 1},F_{f\res 1},\ldots, F_{f\res (n+1)})=^{def}s_n
\rmand [s_n]\sq U_n$$
But this is a losing play for ${\cal P}$.
Suppose $x\in X$ and $\exists^\infty n \;\;s_n\sq x$.
Then $x\in \bigcap_{n<\om}U_n$ and so
$x\in Y$. But by our construction each element of $Y$ is
in $F_{f\res n}$ for
all but finitely many $n$ and since $s_n\not\sq x$
for $x\in F_{f\res n}$ we have a contradiction.
\qed
\bigskip
Remark.
For a kind of weak converse suppose that there exists a countable
$$A\sq 2^\om\setminus X$$
which is not $G_\delta$ in $X\cup A$, then ${\cal P}$
has a winning strategy. Let
$$A=\{a_n:n<\om\}$$
be listed with infinitely many repetitions. Let
${\cal P}$ play some $\sigma_n\sq a_n$ such that
$[\sigma_n]\cap F_n=\emptyset$. To see that it is winning let
$$G=\{z\in 2^\om:\exists^\infty n \;\;\sigma_n\sq z\}$$
Since $A\sq G$ and is not relatively $G_\delta$ there exists $x\in X$
such that $\sigma_n\sq x$ for infinitely many $n$. Hence $\sigma_n$
does not converge to $\infty$.
Player ${\cal P}$ also has a winning strategy
if $X$ contains a perfect subset $Q$. Just let player
${\cal P}$ play a sequence $\sigma_n$ such that
$$G=\{x\in Q:\exists^\infty n\; \sigma_n\sq x\}$$
is comeager in $Q$.
\newpage
\begin{center}
Appendix B
\end{center}
If
$X=\{f_\alpha\in \om^\om: \alpha<\bb\}$
is well-ordered by eventual dominance and unbounded. Then Rothberger
showed the set $X$ is a $\lambda'$-set with respect to
$\om^\om$.
This follows from the following lemma.
\begin{lemma}(Rothberger)
Suppose $Z_\beta=\{f_\alpha:\alpha<\beta\}\sq \om^\om$ is
well-ordered by eventual dominance,
and $A\sq \om^\om$ is countable and for
every $g\in A$ there exists $\alpha<\beta$ such that
$\exists^\infty n\; g(n)