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\begin{document}
\begin{center}
{\Huge Special sets of reals}
\footnote{in {\bf Set Theory of the Reals},
ed Haim Judah, Israel Mathematical Conference Proceedings,
vol 6 (1993), 415-432, American Math Society.}
\end{center}
\begin{center}
Arnold W. Miller
\end{center}
\section*{Introduction}
This is a survey paper which will update Miller [1984b].
It is concerned with totally imperfect subsets of the reals. A set
of reals is totally imperfect iff
it does not contain an uncountable closed set.
Examples of such sets are:
Luzin sets, Sierpi\'{n}ski sets, concentrated sets, strong measure zero sets,
universal measure zero sets, perfectly meager sets,
strong first category sets, $\sigma$-sets, $\lambda$-sets, $Q$-sets,
$\gamma$-sets, $s_0$-sets, and $C^\prime$-sets.
We have concentrated this survey on answering the questions which appeared
in the earlier one.
\section*{Forcing Axiom}
\ques Is $\neg CH$ consistent with either of:
(P1) For every ccc poset $\Bbb P$ of cardinality less than or
equal to the continuum $\goth c$,
there exists a Luzin set of filters, i.e.,
$\langle G_\alpha:\alpha<\omega_1\rangle$ such that each
$G_\alpha$ is a $\Bbb P$-filter and for every dense set $D\subseteq
\Bbb P$ for all but countably many $\alpha$ we have
$G_\alpha\intersect D\not=\emptyset$.
(P2) For all ccc $\sigma$-ideals $I$ in the Borel subsets of $\reals$
there exists an $I$-Luzin set, i.e., an uncountable set $X\subseteq\reals$
such that for every $A\in I$ we have $A\intersect X$ is countable.
\bigskip
These two properties seem related because of the following result
of Martin and Solovay [1970].
\begin{theorem} (Martin, Solovay)
MA is equivalent to the statement that for every
ccc $\sigma$-ideal $I$ in the Borel subsets of $\reals$
and for every $J\in [I]^{<\continuum}$ we have $\Union J\not=\reals$.
\end{theorem}
\begin{corollary} \label{corms}
MA is equivalent to the statement that $\continuum$ is regular and
for every
ccc $\sigma$-ideals $I$ in the Borel subsets of $\reals$ there exists
a $\continuum$-$I$-Luzin set, i.e., $X\in[\reals]^{\continuum}$ such
that for every $A\in I$ we have $|A\intersect X|<\continuum$.
\end{corollary}
In Miller-Prikry [1984a] it is shown that (P2) is consistent with $\neg CH$.
It is also shown that the regularity of $\continuum$ is necessary
in Corollary \ref{corms}. Surprisingly Todorcevic [1991]
has shown that (P1) implies CH.
\begin{theorem} (Todorcevic) \label{tod}
The continuum hypothesis is equivalent to the proposition that every
ccc poset $\Bbb P$ of cardinality less than or equal
to the continuum $\goth c$ has a Luzin set of filters.
\end{theorem}
\proof
That this is true assuming CH is easy since such a poset contains only
$\omega_1$ maximal antichains.
To prove the other direction we will use the following four lemmas.
First define for $f,g\in\bairespace$,
$f\leq g$ iff for all $n\in\omega$
we have $f(n)\leq g(n)$, and
$f\leq^* g$ iff for all but finitely many $n\in\omega$
we have $f(n)\leq g(n)$.
\begin{lemma}\label{tod1}
Suppose $ F\subseteq\bairespace$ are increasing functions well-ordered
by $\leq^*$ and unbounded. Then there exists distinct $f,g\in {F}$ such
that $f\leq g$.
\end{lemma}
\proof
Since $F$ is unbounded its order type under $\leq^*$ cannot have
countable cofinality. Consequently, we can find $g_0\in { F}$ such
that $\{f\in { F}: f\leq^* g_0\}$ is dense in $ F$.
Let ${ F}^\prime\subseteq F$ be unbounded such that there
exists $n_0$ such that for every $f\in F^\prime$ and $n\geq n_0$
we have $g_0(n)\leq f(n)$. Let
$$T=\{s\in\seq : \{f\in F^\prime :f\supseteq s\}\; {\rm unbounded }\}.$$
It is easy to see that $T$ cannot be finitely branching. Consequently,
there exists $s\in T$ with $|s|\geq n_0$ and for which there exists
infinitely many $n$ such that $s\concat n\in T$. Let $f\supset s$ with
$f\leq^*g_0$ and $f\in F$. For some $k>n_0$ we have that
$g_0(l)\geq f(l)$ for all $l\geq k$. Let $g\in F^\prime$ be such
that $g\supseteq s\concat n$ where $n>f(k)$. Then $f\leq g$ because:
\begin{itemize}
\item $f(l)=g(l)=s(l)$ for $l\leq |s|$,
\item $f(l)\leq f(k)\leq g(|s|)=n\leq g(l)$ for $|s|\leq l\leq k$ (by choice
of $n$ and since they are increasing),
\item $f(l)\leq g_0(l)\leq g(l)$ for $l>k$ (by choice of $k$ and definition
of $F^\prime$).
\end{itemize}
\qed
\begin{lemma}\label{tod2}
Suppose $ F\subseteq\bairespace$ are increasing functions ordered
by $\leq^*$ in type $\omega_1$ and unbounded. Suppose $\Bbb P$ is the poset
of finite pairwise $\leq$-incomparable subsets of $ F$. Then
$\Bbb P$ has ccc.
\end{lemma}
\proof
First note that for every $A,B\subseteq F$ unbounded there
exists unbounded $A^\prime\subseteq A$ and $B^\prime\subseteq B$
such that for every $f\in A^\prime$ and $g\in B^\prime$ we have
$f$ and $g$ are $\leq$-incomparable. This is easy to see since
if we let
$$T_A=\{s\in\seq : \{f\in A :f\supseteq s\}\; {\rm unbounded }\}$$
$$T_B=\{s\in\seq : \{f\in B :f\supseteq s\}\; {\rm unbounded }\}$$
then these are both infinite branching trees and so we can choose
$s\in T_A$ and $t\in T_B$ with $s(k) h_\beta(\alpha)$ such that $f_\gamma\in p$.
\end{itemize}
Order
$\Bbb P$ by inclusion on each coordinate.
\bigskip\noindent
Claim. $\Bbb P$ has ccc.
\proof
If $p\union p^\prime\in \Bbb P_{F }$, then by Lemma \ref{tod3}
$$(p\union p^\prime\union \{f_\gamma\}, A \union A^\prime)\leq
(p,A),\; (p^\prime,A^\prime)$$
for all sufficiently large $\gamma$.\qed
Note that $\{(p,A):\gamma\in A\}$ is dense for each $\gamma\in\omega_2$.
If there is a Luzin set of filters, then one of them must
meet at least $\omega_2$ of these dense sets. Consequently
there exists $\Bbb P$-filter $G$ with $|G|=\omega_2$.
Now look at $X=\Union\{ A:\exists p\; (p,A)\in G\}$.
Since
$X$ has cardinality $\omega_2$, it has an $\omega_1^{th}$ element
and so the set $\Union\{p:\exists A\; (p,A)\in G\}$ has cardinality
$\omega_1$. But this contradicts Lemma \ref{tod1}.
This proves Theorem \ref{tod}.
\qed
Note that this proof also gives a counterexample to Theorem 2.3
of Miller and Prikry [1984a]. The mistake in the proof occurs
in the first sentence where
it is stated that ``without loss of generality we may assume $\Bbb P$ is
a boolean algebra''. To correct that theorem we must change it to read:
\begin{theorem} Suppose GCH and $\Bbb P$ is ccc of cardinality
$\omega_2$, then there exists a countably
closed poset
$\Bbb Q$ with $\omega_2$-cc and cardinality $\omega_2$ such that
in $V^{\Bbb Q}$ there exists $\langle G_\alpha:\alpha<\omega_1\rangle$
centered subsets of $\Bbb P$ such that for every dense $D\subseteq \Bbb P$
for all but countably many $\alpha$ we
have $G_\alpha\intersect D\not=\emptyset$.
\end{theorem}
Now, of course, in a boolean algebra centered subsets generate filters,
but in a partial order this may not be the case. This seems to be
one of the ingredients of Todorcevic's example. The following
example sheds more light on this.
\begin{theorem} (Juhasz-Kunen)
There exists a poset $\Bbb P$ which is $\sigma$-centered but
not $\sigma$-filtered.
\end{theorem}
\proof
Let $\Bbb P=\{(C,F): C\subseteq\cantorspace$ clopen,
$\emptyset\not= F\in [C]^{<\omega}, \mu(C)\leq {1\over |F|}\}$.
Order $\Bbb P$ by: $(C_0,F_0)\leq (C_1,F_1)$ iff $C_0\subseteq C_1$
and $F_0\supseteq F_1$.
To see that $\Bbb P$ is $\sigma$-centered, note that there are only
countably many clopen sets and any finite set of conditions with the same
clopen set have a lower bound. To see that it cannot be written
as the union of countably many filters, for any filter $G$
define
$$F_G=\Union\{F:\exists C\; (F,C)\in G\}$$
and
$$C_G=\Intersection\{C:\exists F\; (F,C)\in G\}.$$
Note that
$F_G\subseteq C_G$ and either $F_G$ is finite or $C_G$ has measure
zero. So in either case $F_G$ has measure zero. Consequently,
given any family $\{G_n:n\in\omega\}$ of filters there exists
$x\in\cantorspace\setminus \Union_{n\in\omega} F_{G_n}$.
Since $(\cantorspace,\{x\})\in \Bbb P$ we have
$\Bbb P\not=\Union_{n\in\omega}G_n$.
\qed
\section*{Strong first category}
\ques (Galvin) Does every Sierpinski set have strong first category?
\bigskip
An uncountable set of reals is a Sierpinski set iff it meets every
measure zero set in a countable set. A set has strong first category
iff it can be translated away from every measure zero set.
This problem remains open although there are some partial results.
\begin{theorem} (Bartoszynski-Judah [1990])
(CH) Every Sierpinski set is the union of two sets of sets of
strong first category.
\end{theorem}
\proof
Let $X=\{x_\alpha:\alpha<\omega_1\}$ be any Sierpinski set.
Construct $X_i^\alpha$ countable for $\alpha<\omega_1$ and $i<2$
so that
\begin{itemize}
\item $X_i^\alpha\subseteq X_i^\beta$ if $\alpha<\beta$,
\item $ X_0^\alpha\intersect X_1^\alpha =\emptyset$,
\item $ X_0^\alpha\union X_1^\alpha \subseteq X$, and
\item $x_\alpha\in X_0^{\alpha+1}\union X_1^{\alpha+1}$.
\end{itemize}
Afterwards $X_i=\Union_{\alpha<\omega_1}X_i^\alpha$ will
give us our partition $X=X_0\union X_1$ into sets of strong
first category.
Let $\langle G_\alpha,i_\alpha\rangle$ for $\alpha<\omega_1$ list
all pairs $\langle G,i\rangle$ for $G$ a $G_\delta$-set of measure
zero and $i\in \{0,1\}$.
Let $X_i^0=\emptyset$ and for
limit ordinals $\alpha$ let $X_i^\alpha=\Union_{\beta<\alpha}X_i^\beta$.
At stage $\alpha+1$ proceed as follows. Let $i=i_\alpha$ and
$G=G_\alpha$.
First find $z$ such that
$$(z+G)\intersect X^\alpha_{i}=\emptyset.$$
This is easy to do because the set $X^\alpha_{i}$ is countable,
so any $z$ not in the measure zero set $X^\alpha_{i}- G$
will do. Now let
\smallskip
$X^{\alpha+1}_{i}= X^{\alpha}_{i}$ and let
\smallskip
$X^{\alpha+1}_{1-i}= X^{\alpha}_{1-i}\union ((z+G)\intersect
X)\union\{x_\alpha\}$ (where $\{x_\alpha\}$ is added only
if it is not in $X^{\alpha+1}_{i}$.
\smallskip
\noindent So $(z+G)\intersect X\subseteq X_{1-i}$ and therefore
$(z+G)\intersect X_i=\emptyset$.
\qed
\ques It is also an open question of Galvin if the union of two strong
first category sets must have strong first category.
\bigskip
\begin{theorem} (Bartoszynski-Judah [1990]) \label{BJ90}
If ZFC is consistent, then so is ZFC + there exists a Sierpinski set +
every Sierpinski set has strong first category.
\end{theorem}
This model is obtained by starting with a model of MA+$\neg$CH and adding
$\omega_1$ random reals.
Note that
by a Lowenheim-Skolem argument, if there exists a
Sierpinski set which fails to have strong first category, then
there exists an inner model of CH in which there exists
a Sierpinski set which fails to have strong first category.
Woodin [to appear] has shown that
$\Sigma^2_1$ statements are absolute between models of CH,
assuming there is a Woodin cardinal.
He has noted that the existence of a
Sierpinski set which fails to have strong first category is a
$\Sigma^2_1$ statement, consequently, it is unlikely to be independent
of CH. Note that in the model of Theorem \ref{BJ90} CH fails.
\begin{theorem} (Jasinski-Weiss [1991]) \label{JW99}
If $S$ is a Sierpinski set and
$\Union_{n<\omega}C_n$ is a union of compact
sets of measure zero, then there exists $x$ such that
$(x+\Union_{n<\omega}C_n)\intersect S=\emptyset$.
\end{theorem}
\proof
In fact,
$\{x: (x+\Union_{n<\omega}C_n)\intersect S=\emptyset\}$ is comeager.
The following is copied from Miller [1990/1].
Let the $C_n$ be increasing and compact.
Let $\Intersect_{n<\omega}U_n$ be decreasing, open, $\mu(U_n)< {1\over n}$,
and $\rationals + \Union_{n<\omega}C_n\subseteq \Intersect_{n<\omega}U_n$.
\medskip
\noindent Claim. $Y=\{x:x+\Union_{n<\omega}C_n\subseteq
\Intersect_{n<\omega}U_n\}$ is a
dense $G_\delta$ and hence comeager.
\proof
Since $Y$ contains the rationals $\rationals$ it is enough to see that
it is a $G_\delta$. But note that for each $n$
$$\{x:x+C_n\subseteq U_n\}$$
is open, since $C_n$ is compact and $U_n$ is open.
Since
$$Y=\{x:x+\Union_{n<\omega}C_n\subseteq \Intersect_{n<\omega}U_n\}=
\Intersect_{n<\omega}\{x:x+C_n\subseteq U_n\}$$
it is $G_\delta$ and the Claim is proved.
To prove the theorem let $X=\Intersect_{n<\omega}U_n\intersect S$
and let $Z=Y\setminus(X-\Union_{n<\omega}C_n)$. Since
$X$ is countable and so $X-\Union_{n<\omega}C_n$ is meager, it
is enough to see that $(z+\Union_{n<\omega}C_n)\intersect S=\emptyset$
for every $z\in Z$.
But $z\in Y$ implies
$z+\Union_{n<\omega}C_n\subseteq\Intersect_{n<\omega}U_n$ implies
$(z+\Union_{n<\omega}C_n)\intersect S\subseteq X$ implies (if nonempty) that
$z+c=x$ where $c\in \Union_{n<\omega}C_n$ and $x\in X$ so finally
$z\in X-\Union_{n<\omega}C_n$.
\qed
The referee has informed me that Theorem \ref{JW99} was also
proved by Tim Carlson.
The following is an open question.
\ques Is it consistent to have every
strong measure zero set countable (Borel conjecture) and every
first category zero set countable (Dual Borel conjecture)?
\bigskip
Carlson's theorem that in the Cohen real model the Dual Borel
conjecture holds has been extended by Judah-Shelah [1989] to show
that it is consistent with MA($\sigma$-centered) that the
Dual Borel conjecture holds. This argument has been improved
by Pawlikowski [1990] to show that any model obtained by the
finite support iteration of $\sigma$-centered forcing the dual
Borel conjecture holds. Judah-Shelah-Woodin [1990] have shown
that the Borel conjecture is consistent with the continuum arbitrarily
large. See Bartoszynski-Shelah [to appear] or Bartoszynski-Judah [to appear]
for a correction to that argument.
\section*{Q-sets and strong measure zero}
A set of reals is a Q-set iff every subset is a relative $G_\delta$.
A set of reals $X$ has strong measure zero iff for any sequence
of positive reals $\langle \epsilon_n:n\in\omega\rangle$ the set
$X$ can be covered by sequence of sets
$\langle I_n:n\in\omega\rangle$ where each $I_n$ has diameter
less than $\epsilon_n$.
\ques (Fleissner) Does every Q-set have strong measure zero?
\bigskip
This question was answered by the following theorem.
\begin{theorem}
(Judah-Shelah [1988]) If we add one Mathias or Laver real $f$
to a model $M$ of MA+$\neg$CH, then in $M[f]$ every uncountable
set of reals $X\in M$ remains a Q-set and furthermore does
not have strong measure zero.
\end{theorem}
Judah-Shelah [to appear] constructed a Q-set in a model
$V[G]$ where $V$ is a model of GCH and the extension satisfies
the Sack's property: for every $f\in\bairespace\intersect V[G]$
there exists $\langle H_n:n\in\omega\rangle\in V$ such that
for every $n$ $f(n)\in H_n$ and $|H_n|\leq 2^n$.
Strong measure zero sets are also denoted as $C$-sets.
Rothberger defined $C^\prime$ and $C^{\prime \prime}$ as follows.
A set of reals $X$ is $C^{\prime \prime}$ iff for any
sequence $\langle {\cal U}_n : n\in\omega\rangle$ of open covers
of $X$ there exists $\langle {U}_n : n\in\omega\rangle$ with
each $U_n\in {\cal U}_n$ such that
$X\subseteq\Union_{n\in\omega}{U}_n$. The definition of
$C^\prime$ is the same except the open covers ${\cal U}_n$ are
also required to be finite.
\ques (Rothberger) Does every set with property
$C^\prime$ have property $C^{\prime \prime}$?
\bigskip
This question is answered in the negative in Miller-Fremlin
[1988] where it is shown that assuming CH that there is a
$C^\prime$ set which is not $C^{\prime \prime}$.
Another question about Q-sets was answered by Knight [to appear]
where it is shown to be consistent to a have a $\Delta$-set
which is not a Q-set. A $\Delta$-set is a set of reals
$X$ with the property that for every sequence $X_n\subseteq X$
with the property that $\Intersection_{n\in\omega}X_n=\emptyset$
there exists open sets $U_n$ with $X_n\subseteq U_n$ and
$(\Intersection_{n\in\omega} U_n) \intersect X=\emptyset$.
For more about $\Delta$-sets see Tanaka [1986], [1990].
\section*{Hierarchy order}
If $X$ is a separable metric space, then define $ord(X)$
(Borel rank or Baire order) to be the
smallest $\alpha<\omega_1$ such that every Borel set in $X$
is $\Sigma^0_\alpha$ in $X$. A set $A$ is analytic in $X$ iff
$A=\Union_{f\in\bairespace}\Intersect_{n\in\omega}B_{f\res n}$ for
some Borel sets $B_{s}$ in $X$ for $s\in\seq$.
\ques (Mauldin) Is it consistent to have a space $X$ where every
$ord(X)<\omega_1$ but not every analytic set is Borel?
\bigskip
This question remains open. A related result is proved in Miller [1990a]:
assuming CH there exists a separable
metric space $X$ such that every analytic in $X$ set is Borel in $X$ but there
exists a set analytic in ${X^2}$ set which is not Borel in $X^2$.
This doesn't answer the question since the space $X^2$
has Borel subsets of arbitrarily large rank while $X$ has bounded Borel
rank.
The following is another open question
along similar lines.
\ques Is it consistent to have a space $X$ where $ord(X)>3$ but the
difference hierarchy inside the $\Delta_3^0$ sets is bounded?
\bigskip
The paper Miller [1990a] contains some related results about projective
hierarchies.
\section*{Universal measurable sets}
A set of reals is universally measurable iff it is measurable
with respect to any measure on the real line.
\ques (Mauldin) Is it consistent to have only $\continuum$ many
universally measurable sets?
\bigskip
This question remains open.
It is theorem of Laver that it is consistent
that there are only $\continuum$ universal measurable zero sets.
This holds, for example, in the random real model (see Miller [1983])
The following theorem of Grzegorek and Ryll-Nardzewski seems relevant.
\begin{theorem} (Grzegorek, Ryll-Nardzewski [1980])
There
are universally measurable sets which are not equal to a Borel set
modulo a universal measure zero set.
\end{theorem}
\proof
Let $W$ be the $\Pi^1_1$ set of well-orderings. Then $W$ is
universally measurable. On the other hand if $B$ is
any Borel set then $W\Delta B$ cannot be universally measurable.
$$W\Delta B= (W\setminus B)\union (B\setminus W)$$
The set $B\setminus W$ is $\Sigma^1_1$, hence if it is uncountable it would
contain a perfect set and not be universal measure zero. So without
loss we may assume $B\subseteq W$. But then by the boundedness theorem
it is easy to get a perfect subset of $W\setminus B$ and
so it doesn't have universal measure zero.
\qed
\section*{Perfectly meager}
A set of reals $P$ is perfect iff it is homeomorphic to
$\cantorspace$.
A set of reals $X$ is perfectly meager iff for every perfect set
$P$ the set $X\intersect P$ is meager in $P$.
\ques (Marczewski) Is the product of perfectly meager sets
perfectly meager?
\bigskip
This was answered by Reclaw.
\begin{theorem} \label{rec3}
(Reclaw [to appear]) If there exists a Luzin set, for example
assuming CH, then there exists $X$ and $Y$ are perfectly meager
such that $X\cross Y$ is not perfectly meager.
\end{theorem}
\proof
Let $\langle C_s:s\in\seq\rangle$ be a family
of perfect subsets of $\cantorspace$ such that
\begin{itemize}
\item $C_{\langle\rangle}=\cantorspace$,
\item for each $s\in\seq$
$\langle C_{s\concat n}:n\in\omega\rangle$ is a disjoint family of
nowhere dense subsets of $C_s$,
\item for each $s\in\seq$
every $t\in\binseq$ if $C_s\intersect [t]\not=\emptyset$, then
for some $n\in\omega$ we have $C_{s\concat n}\subseteq [t]$.
\end{itemize}
The Luzin function $f:\bairespace\rightarrow \cantorspace$ is
defined by $\{f(x)\}=\Intersect_{n\in\omega}C_{x\res n}$.
We will use the following forcing Lemma about $f$.
Let $\Bbb P=\seq$, that is Cohen forcing for $\bairespace$.
\begin{lemma} \label{rec2}
Let $(x,y)$ be the pair of Cohen reals added by forcing with
$\Bbb P^2$. For every perfect $P\subseteq(\cantorspace)^2$ and
$p\in\Bbb P$ there exists $q\leq p$ and $C\subseteq P$ closed
nowhere dense in $P$ such that
$q\forces$``$(x,f(y))\in C$ or $(x,f(y))\notin P$".
\end{lemma}
\proof
Let $p=\langle p_o,p_1\rangle$.
\medskip\noindent Case 1. $\{x\}\cross C_{p_1}\not\subseteq P$.
Since $P$ is closed there exists $t\in\binseq$ such that
$[t]\intersect C_{p_1}\not=\emptyset$ and $q_0\leq p_0$ such that
$q_0\forces$``$(\{x\}\cross [t])\intersect P=\emptyset$''.
For some $m\in\omega$ we have that $C_{p_1\concat m}\subseteq [t]$.
Let $q_1=p_1\concat m$. Then $q\forces$``$(x,f(y))\notin P$''.
\medskip\noindent Case 2. $\{x\}\cross C_{p_1}\subseteq P$.
Let $q_0\subseteq x$ with $q_0\leq p_0$ such that
$q_0\forces$``$\{x\}\cross C_{p_1}\subseteq P$''. Let
$$H=\{u\in\cantorspace:\{u\}\cross C_{p_1}\subseteq P\}.$$
It is easy
to see that $H$ is closed. Since there is a comeager in $[q_0]$ set
of Cohen reals in $H\intersect [q_0]$ it must be that
$[q_0]\subseteq H$ and thus $[q_0]\cross C_{p_1}\subseteq P$.
If we let $q_1=p_1\concat n$ for an arbitrary $n$ and $C=[q_0]\cross C_{q_1}$,
then $C\subseteq P$ is nowhere dense in $P$ and
$q\forces$``$(x,f(y))\in C$''.
\qed
Finally we show how the lemma proves Theorem \ref{rec3}.
Suppose
$$K=\{(x_\alpha,y_\alpha) : \alpha<\kappa\}$$
is a Luzin set. This means that $\kappa$ is uncountable, but for every
countable transitive model $M$ of a fragment of set theory,
all but countably many of the $(x_\alpha,y_\alpha)$ are
$\Bbb P^2$ generic over $M$. It follows from Lemma \ref{rec2}
that
$$K_1=\{(x_\alpha,f(y_\alpha)):\alpha<\kappa\}$$
is perfectly
meager. To see this, let $P\subseteq(\cantorspace)^2$ be any
perfect set. Let $M$ be a countable transitive model of a partial fragment
of set theory which contains a code for $P$. Let $\{C_n:n\in\omega\}$
be all closed nowhere dense in $P$ sets which are coded in $M$.
By Lemma \ref{rec2} for all but countably many $\alpha$ we
have that $(x_\alpha,f(y_\alpha))\notin P$ or $(x_\alpha,f(y_\alpha))\in
\Union_{n\in\omega}C_n$. Hence $K_1\intersect P$ is meager
in $P$. Similarly
$$K_2= \{(f(x_\alpha),y_\alpha):\alpha<\kappa\}$$
is perfectly meager. However consider
$$H=\{\langle (x_\alpha,f(y_\alpha)),(y_\alpha,f(x_\alpha))\rangle
:\alpha<\kappa\}.$$
$H\subseteq K_1\cross K_2$ and $H$ is homeomorphic to the Luzin set
$K$ since it is essentially just the graph of the function
$f\cross f:K\rightarrow\cantorspace$. It follows that $H$ is Luzin
in its closure and so $K_1\cross K_2$ is not perfectly meager.
\qed
A generalization of this result appears in Pawlikowski [1989].
\section*{$\sigma$-sets}
A set of reals is a $\sigma$-set iff every relative Borel set
is a relative $F_\sigma$-set, i.e. $ord(X)=2$.
\ques Can you map a $\sigma$-set continuously onto the reals?
\bigskip
This was answered by Reclaw.
\begin{theorem} \label{recl2}
(Reclaw) If a set of reals $X$ can be mapped continuously
onto $\cantorspace$, then $ord(X)=\omega_1$.
\end{theorem}
\proof
For a countable $H\subset P(Y)$ define the Borel hierarchy generated
by $H$ as follows. Let
$$\Sigma_0(H)=H$$
and for $\alpha<\omega_1$ let
$$\Sigma_{\alpha}(H)=\{\union_{n\in\omega}(Y\setminus A_n):
A_n\in\Sigma_{\beta_n}(H),\beta_n<\alpha\}.$$
Let Borel($H$) be the union of all $\Sigma_{\alpha}(H)$ for
$\alpha<\omega_1$.
We will need the following theorem which generalizes
the classical theorem of Lebesgue.
\begin{theorem} (Bing, Bledsoe, Mauldin [1974]) \label{BBM}
Suppose $H\subseteq P(\cantorspace)$ is a countable family such
that every clopen set is in Borel($H$). Then $ord(H)=\omega_1$.
\end{theorem}
\proof
To prove this theorem we will need the following two lemmas.
Given a countable $H\subseteq P(\cantorspace)$ let
$$R=\{C\cross A:A\in H, C\subseteq \cantorspace\;\mbox{ is clopen}\}.$$
\begin{lemma} (Universal sets) \label{univ}
For each $\alpha$ with $1\leq\alpha<\omega_1$ there exists
$U\in\Sigma_\alpha(R)$ which is universal for $\Sigma_\alpha(H)$ sets,
i.e., for every $A\in\Sigma_\alpha(H)$, there exists $x\in\cantorspace$
such that $A=\{y: (x,y)\in U\}$.
\end{lemma}
\proof
For $\alpha=1$: Let $H=\{A_n:n\in\omega\}$ let
$$U=\Union_{n\in\omega}\{x: x(n)=1\}\cross (\cantorspace\setminus A_n).$$
For $\alpha>1$: Let $x\mapsto \langle x_n:n\in\omega\rangle$ be
a nice recursive coding taking
$\cantorspace\rightarrow(\cantorspace)^\omega$. Let $\beta_n$ for
$n\in\omega$ be cofinal in $\alpha$, and $U_n\in\Sigma_{\beta_n}(R)$
be universal for $\Sigma_{\beta_n}(H)$ sets. Define
$U_n^{\prime}$ by $(x,y)\in U_n^{\prime}$ iff $(x_n,y)\in U_n$.
It is easy to check that $U_n^\prime$ is also
$\Sigma_{\beta_n}(R)$ and universal for $\Sigma_{\beta_n}(H)$.
But now taking
$$U=\Union_{n\in\omega}(\cantorspace\setminus U_n^\prime)$$
gives us a set in
$\Sigma_\alpha(R)$ which is universal for $\Sigma_\alpha(H)$ sets.
\qed
\begin{lemma} (Diagonalization) \label{diag}
Suppose
that every clopen set is in Borel($H$).
Then for every $B\in$Borel($R$) the set $\{x:(x,x)\in B\}$
is in Borel($H$).
\end{lemma}
\proof
For $B=C\cross A$ where $A\in H, C\subseteq \cantorspace$ is clopen.
Note that $\{x:(x,x)\in B\}=C\intersect A$. Since the by assumption
$C\in$Borel($H$), we have the result for elements of $R$.
To do Borel($R$) is an easy induction.
\qed
Now we give a proof of Theorem \ref{BBM}. Suppose
Borel($H)=\Sigma_\alpha(H)$. By Lemma \ref{univ} there exist
$U$ in Borel($R$) which is universal for $\Sigma_\alpha(H)$
and hence Borel($H$). By Lemma \ref{diag} the set
$$D=\{x:(x,x)\notin U\}$$
is in Borel($H$). But this means that
for some $x$ that $D=\{y:(x,y)\in U\}$. But then
$x\in D$ iff $x\notin D$.
\qed
Now we give a proof of Theorem \ref{recl2}.
Suppose that $f:X\rightarrow \cantorspace$ is onto and continuous.
Let $\cal C$ be a a countable clopen basis for $X$ and let
$H=\{f^{\prime\prime}C: C\in\cal C\}$. Since it is clear
that $H$ contains all clopen sets, by Theorem \ref{BBM}, the
$ord(H)=\omega_1$. But the map $f$ takes the Borel hierarchy
of $X$ directly over the hierarchy on Borel($H$), so $ord(X)=\omega_1$.
\qed
An extension of Reclaw's result to Souslin (operation A) sets
appears in Miller [1990a].
\section*{$\lambda$-sets and $\lambda^\prime$-sets}
A set of reals $X$ is a $\lambda$-set iff every countable
subset of $X$ is a relative $G_\delta$ in $X$. A set
of reals $X\subseteq Y$ is a $\lambda^\prime$-set with respect to $Y$
iff for every countable $F\subseteq Y$ the set $X\union F$ is
a $\lambda$-set. It is a theorem of ZFC that there are $\lambda^\prime$
sets of cardinality $\omega_1$.
This section will clear up some confusion about
a remark (p 266) in Reed [1983] which is attributed to me.
The cardinal $\goth b$ is the minimum cardinality of
a set $F\subseteq \bairespace$ such that $F$ is unbounded, i.e.
there does not exist $g\in\bairespace$ such that $f\leq^* g$.
In Reed [1983] this is referred to as the least $\alpha$ for which
$M(\alpha)$ holds.
The cardinal $\goth d$ is the minimum cardinality of
a set $F\subseteq \bairespace$ such that $F$ is a dominating
family, i.e., for every $g\in\bairespace$ there exists
a $f\in F$ such that $g\leq^* f$. Obviously
$\goth b\leq \goth d\leq \goth c$.
\begin{theorem} (Rothberger [1939])
There exists a $\lambda$-set $X$ of cardinality $\goth b$,
in fact, $X$ is a $\lambda^\prime$-set with respect
to the irrationals.
\end{theorem}
\proof
Let $F=\{f_\alpha:\alpha<\goth b\}\subseteq\bairespace$ be an unbounded family
such that $\alpha\leq\beta$ implies $f_\alpha\leq^*f_\beta$. We claim
that for every countable $H\subseteq\bairespace$ the set
$F\union H$ is a $\lambda$-set. It is enough to see that for every
countable $H\subseteq\bairespace$
$H$ is a relative $G_\delta$ in $F\union H$. Since $F$ is unbounded
there exists $\alpha<\goth b$ such that for every $h\in H$ we have
$f_{\alpha}\not\leq^* h$.
This also holds for every $\beta>\alpha$ since $f_\beta\geq^* f_\alpha$.
So if
$$K=\{g\in\bairespace: f_\alpha(n)\not\leq^* g\}$$
then $K$ is a $G_\delta$ such that $H\subset K$ and
$|K\intersect (X\union H)|<\goth b$. So it now suffices to note the
following:
\medskip\noindent
Claim: Any set of reals $Y$ of cardinality less than $\goth b$
is a $\lambda$-set.
\medskip\noindent
Let $Y=\{x_n:n<\omega\}\union\{y_\alpha: \alpha<\kappa\}$. Let
$\{U_n(x_m):n\in\omega\}$ be a decreasing neighborhood base for
each $x_m$. For each $\alpha<\kappa$ define
$g_\alpha\in\bairespace$ so that $y_\alpha\notin U_{g_\alpha(n)}(x_n)$.
Since $\kappa<\goth b$ there exists $g\in\bairespace$ such
that $g_\alpha\leq^*g$ for all $\alpha<\kappa$.
Let $U=\Intersection_{n<\omega}(\Union_{m>n} U_{g(n)}(x_n)$. Then
$U$ is a $G_\delta$-set such that $U\intersect Y=\{x_n:n\in\omega\}$.
\qed
\begin{theorem} (Miller [1983])
In the Cohen real model there are no $\lambda$-sets of cardinality
greater than $\omega_1$.
\end{theorem}
\proof
It is shown that every set of reals of cardinality $\omega_2$ contains
the one-to-one continuous image of a Luzin set. But such
a set cannot be a $\lambda$-set. If $L$ is
Luzin and $D$ a countable dense subset of $L$, then $D$ cannot be relatively
$G_\delta$ in $L$. But if $f:L\rightarrow X$ is one-to-one and continuous,
then $f^{\prime\prime}D$ cannot be $G_\delta$ in $X$.
\qed
In the Cohen real model $\goth b=\omega_1$ and $\goth d=\goth c$.
\begin{theorem} \label{lav}
In the Laver model for the Borel conjecture
($\goth b=\goth d=\goth c=\omega_2$)
there does not exist a $\lambda^\prime$-set
with respect to the reals of cardinality $\omega_2$.
\end{theorem}
\proof
Lemma 14 of Laver [1976] implies the following lemma.
\begin{lemma}
Suppose $p\forces$``$\tau\in [0,1]$'', then there exists
$q\leq p$ and a finite set $U_s$ for each splitting node $s$ of
$q(0)$, such that for all $\epsilon >0$ and all but finitely many n
with $s\concat n\in q(0)$
$$q(0)_{s\concat n}\concat q\res[1,\omega_2)
\forces \exists u\in U_s |u-\tau|<\epsilon$$
\end{lemma}
\begin{lemma} \label{lav2}
Given the above $q,\tau,$ and $U=\Union_{s\in q(0)} U_s$,
for any $G_\delta$ set $G\supseteq U$ there exists $r\leq q$
such that $r\forces$``$\tau\in G$''.
\end{lemma}
\proof
This is an easy fusion argument on $q(0)$. If
$G=\Intersection_{n\in\omega}G_n$ where $G_n$ is open, then make
sure that everything in the $n^{th}$ forces that
$\tau\in G_n$.
\qed
Let $V[f_\beta:\beta<\omega_2]$ be the Laver model.
To prove Theorem \ref{lav} suppose that $X\subseteq [0,1]$ is
a $\lambda^\prime$-set in $V[f_\beta:\beta<\omega_2]$.
Via a Lowenheim-Skolem type argument
there exists $\alpha<\omega_2$ such that
$$(X\intersect V[f_\beta:\beta<\alpha]) \in V[f_\beta:\beta<\alpha]$$
and a function $f\in V[f_\beta:\beta<\alpha]$ such that
for every countable $D\subseteq [0,1]$ in $V[f_\beta:\beta<\alpha]$
we have $f(D)$ is a $G_\delta$ code for a set $G$ such that
$$G\intersect (X\union D)=D.$$
But now Lemma \ref{lav2} (applied with $V[f_\beta:\beta<\alpha]$
as the ground model) implies that $X\subseteq V[f_\beta:\beta<\alpha]$.
But since $V[f_\beta:\beta<\alpha]$ satisfies CH, the set $X$ has
cardinality $\omega_1$.
\qed
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\bigskip
\begin{flushright}
Department of Mathematics \\
University of Wisconsin \\
480 Lincoln Drive \\
Madison, WI 53706, USA \\
miller@math.wisc.edu. \\
\end{flushright}
\end{document}