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% Compact subsets of the Baire space
% A. Miller  Nov 2012
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\def\al{\alpha}
\def\be{\beta}
\def\es{\emptyset}
\def\forces{{\;\Vdash}}
\def\hposet{{\mathbb H}}
\def\la{\langle}
\def\name#1{\stackrel{\circ}{#1}}
\def\om{\omega}
\def\proof{\par\noindent Proof\par\noindent}
\def\poset{{\mathbb P}}
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\def\qed{\par\noindent QED\par\bigskip}
\def\ra{\rangle}
\def\res{\upharpoonright}
\def\rmand{\mbox{ and }}
\def\rmiff{\mbox{ iff }}
\def\rmor{\mbox{ or }}
\def\seq{\om^{<\om}}
\def\sm{{\setminus}}
\def\su{\subseteq}

\newtheorem{theorem}{Theorem}%[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{prop}[theorem]{Proposition}

\def\address{\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}}
 

\begin{document}

\begin{center}
{\large Compact subsets of the Baire space}
\end{center}

\begin{flushright}
Arnold W. Miller\\
Nov 2012
\end{flushright}

Results in this note were obtained in 1994 and reported on
at a meeting on Real Analysis in
{\L}odz, Poland, July 1994.

Let $\om^\om$ be the Baire space, infinite sequences
of natural numbers with the product topology.
In this topology a set $K\su\om^\om$ is compact iff
there exists a finite
branching tree $T\subseteq\seq$ such that
$$K=[T]=^{def}\{x\in\om^\om: \forall n\in\om\;\; x\res n\in T\}.$$

\begin{theorem}\label{thm1}
  If there exists a countable standard model of ZFC, then
  there exists $M$, a countable standard model of ZFC, $N\supseteq M$,
  a generic extension of $M$, and $T\in N$ a finite
  branching subtree of $\seq$ with the properties
  that
  \begin{enumerate}
    \item[(a)] $\forall f\in [T]\cap N \;\;\exists g\in M\cap\om^\om
    \;\; \forall^\infty n\;f(n)< g(n)$
    and
    \item[(b)] $\forall g\in M\cap\om^\om\;\;
    \exists f\in [T]\cap N\;\;\exists^\infty n\;g(n)<f(n)$.
  \end{enumerate}
\end{theorem}


I don't know how to do this over an arbitrary ground model $M$, for
example a model of $V=L$.

I don't know if we can have (a) and the stronger
condition:

\bigskip

\noindent $\;\;\;(b)^\pr\;\;\forall g\in M\cap\om^\om\;
    \exists f\in [T]\cap N\; \;\forall^\infty n\;g(n)<f(n)$.


\bigskip

This Theorem is related to Michael's problem \cite{M} of 
whether there must be
a Lindel\"{o}f space $X$ such that $X\times\om^\om$ is not
Lindel\"{o}f and M.E.Rudin's characterization of that problem \cite{R}.
See also Alster \cite{A} and Lawrence \cite{L}.

\begin{center}
Proof of the Theorem.
\end{center}

Let $\poset$ be the natural forcing for producing a finite
splitting tree using finite conditions.  Namely, $p\in\poset$
iff $p\su\om^{<\om}$ is a finite subtree and $p\leq q$ iff
$p\supseteq q$ is an end extension of $q$.  End extension means
if $s\in p\sm q$ then $s\supseteq t$ for some $t\in q$ which is terminal
in $q$, i.e., has no extensions in $q$.   This order is countable
and hence forcing equivalent to adding a single Cohen real.
The union of $\poset$-generic filter is a tree $T\su\om^{<\om}$
which determines a compact set $K\su\om^\om$ by
$K=[T]$.

Given $X\su\om^\om$ define $\hposet_X$ to be a version
of Hechler forcing restricted to $X$:
$$\hposet_X=\{(s,F)\;:\; s\in\om^{<\om}\rmand F\in [X]^{<\om}\}$$
and $(s,F)\leq (t,H)$ iff $F\supseteq H$, $s\supseteq t$, and
$$\forall f\in H\;\;\forall n\;\; 
[\;|t|\leq n<|s| \;\to\;  f(n)\leq s(n)\;].$$
Forcing with this determines a $g\in\om^\om$ such that
$$\forall f\in X\;\;\forall^\infty n\;\; f(n)\leq g(n).$$

For $V$ a countable transitive model of
set theory let $K=[T]$ be
$\poset$-generic over $V$ and $g_\al$
$\hposet_{X_\al}$-generic over
$V[K]\la g_\be\;:\;\be<\al\ra$ where $X_\al=H_\al\cup K_\al$;
$$H_\al=\om^\om\cap V\la g_\be:\be<\al\ra
\;\;\rmand\;\;
K_\al = K  \cap  V[K]\la g_\be\;:\;\be<\al\ra.$$
We use finite supports and so this is a ccc.

\bigskip\noindent 
Our models are $M=V[\la g_\al\;:\;\al<\om_1\ra]$ and $N=M[K]$.

\bigskip
Condition (a): In $N$, $K=\cup_{\al<\om_1}K_\al$ and $g_\al$
eventually dominates each element of $K_\al$.

\bigskip
Condition (b): It suffices to show that for all $\al<\om_1$
there exists $f\in [T]$ such that $\exists^\infty n\; g_\al(n)<f(n)$.
This is sufficient because $\{g_\al\;:\;\al<\om_1\}$ is a
dominating family in $M$.

\bigskip\noindent
{\bf Claim}. For any $\al<\om_1$ and $s\in T$ there exists
$t\supseteq s$ with $t\in T$ such that $t(n)>g_\al(n)$ for some
$n$ with $|s|<n<|t|$.

\bigskip
Assuming the Claim it is easy to produce $f\in [T]$ which is
infinitely often larger than $g_\al$.  The rest of the proof is
to prove the Claim.

\bigskip
We now describe the posets.

\begin{enumerate}
\item For each $\al\leq\om_1$ define $\poset_\al$:
\begin{enumerate}
\item $(p,(s_i,K_i,H_i)_{i<\al})\in \poset_\al$ iff
\item $p\in\poset$ and
\item 
$s_i=K_i=H_i=\es$ for all but finitely many $i$,
\item $s_i\in\om^{<\om}$, $K_i$ is finite set of 
$\poset_i$-names for
elements of $\om^\om$ and for all $\tau\in K_i$
$$(p,(s_j,K_j,H_j)_{j<i})\forces_{\poset_i} \tau\in [T]$$
\item $H_i$ is a finite set of $\hposet_i$-names 
for elements
of $\om^\om$.
\item \label{shadow} Let 
$N=max\{s(n)\;: s\in p \rmand n<s|\}$ and let
$f\in\om^\om$ be the constant function $N$.  Then $\check{f}\in H_i$
for every $i<\al$ with $K_i$ nonempty.
\end{enumerate}
\item 
$\hposet_\al=\{(p,(s_i,H_i,K_i):i<\al)\in \poset_\al\;:\;
p=\es \rmand \forall i<\al\;\; K_i=\es \}$
\item \label{name} $\tau$ is $\hposet_i$-name
for an element
of $\om^\om$ iff
$\tau\su\hposet_i\times \{\check{\la n,m\ra}: n,m<\om\}$ and
$${1}\forces_{\poset_i}\forall n\;\exists !m \;\la n,m\ra\in \tau.$$  
\item Define the ordering on $\poset_\al$ by:
$(p^\pr,(s_i^\pr,K_i^\pr,H^\pr_i)_{i<\al})\leq (p,(s_i,K_i,H_i)_{i<\al})$
iff 
\begin{enumerate}
\item $p^\pr\leq p$ and for each $i<\al$
\item $H_i\su H_i^\pr$ and $K_i\su K_i^\pr$
\item \label{forcing} $\forall \tau\in H_i\cup K_i\;\;
\forall n\;\; |s_i|\leq n<|s_i^\pr|\;\; \to$
$$(p,(s_j,K_j,H_j)_{j<i}) \forces_{\poset_i} s^\pr(n)\geq \tau(n).$$
\end{enumerate}
\end{enumerate}

There are two possible orderings on $\hposet_\al$.
They only differ on the forcing relation in the
conclusion of clause \ref{forcing}, i.e., use
$\forces_{\hposet_i}$ instead of $\forces_{\poset_i}$.
It will be necessary to show that these are in fact the same.
It is the reason
behind the innocuous condition \ref{shadow}. 
We will need to drop $p$
but retain some information about $p$ in the $H_i$'s.

Note that any condition not satisfying condition \ref{shadow}
can be extended to one 
that does simply by putting the required $\check{f}$ into $H_i$.
Define 
$$H_i^+=\left\{
\begin{array}{ll}
H_i\cup\{\check{f}\} & \mbox{ if } K_i\neq\es \\
H_i & \mbox{ otherwise }
\end{array}\right.$$
Then
$(p,(s_i,K_i,H^+_i)_{i<\al})\leq (p,(s_i,K_i,H_i)_{i<\al})$.
So conditions satisfying (1a-1f) are dense in those
satisfying (1a-1e).

\begin{lemma}\label{generic}
(a) Suppose that 
$\tau\su\hposet_\al\times \{\check{\la n,m\ra}: n,m<\om\}$
and
$\theta(x)$ is a Borel predicate with parameters
from $V$.  For any 
$(p,(s_i,K_i,H_i)_{i<\al})\in \poset_\al$
$$(\es,(s_i,\es,H_i)_{i<\al})\forces_{\hposet_\al} \theta(\tau)\rmiff
(p,(s_i,K_i,H_i)_{i<\al})\forces_{\poset_\al} \theta(\tau).$$

\noindent (b) Every $D\su \hposet_\al$ in $V$ which is dense
in $\hposet_\al$ 
is predense in $\poset_\al$.

\noindent (c) For every $G$ which is
$\poset_\al$-generic over $V$ if $G^\pr=G\cap \hposet_\al$,
then $G^\pr$ is $\hposet_\al$-generic over $V$.
\end{lemma}
\proof
This is by induction on $\al$.  $\hposet_0=\{\es\}$ is the trivial partial
order and so $\tau$ evaluates to a ground model real.  Hence
the trivial condition in $\poset_0=\poset*\{\es\}$
decides $\theta(\tau)$.
Suppose that the Lemma is true for all $i<\al$.
This implies that the forcing in clause \ref{forcing} coincides
for the posets $\hposet_{i}$ and $\poset_{i}$. Also the forcing
in the notion of $\hposet_i$-name (\ref{name})
for an element of $\om^\om$.

Proof of (b). Suppose $D\su\hposet_\al$ is dense in $\hposet_\al$.  We
claim that it is predense in $\poset_{\al}$.
Given an arbitrary $(p,(s_i,K_i,H_i)_{i<\al}$.

Let $(\es,(s_i^\pr,\es,H_i^\pr)_{i<\al})\leq (\es,(s_i,\es,H_i)_{i<\al})$
be any element of $D$.  We need to show that for some $p^+\leq p$
that
$$(p^+,(s_i^\pr,K_i,H_i^\pr)_{i<\al}\leq (p,(s_i,K_i,H_i^+)_{i<\al}$$
The only problem is clause \ref{forcing} for 
the case that $\tau\in K_i$.  

To choose $p^+$ let
$n_1=\max\{|s_i^\pr|:i<\al\}$ and let $p^+$ be obtained by adding
$n_1+1$ or more zeros to each terminal node of $p$.
Note that the $f\in H_i$ given by clause \ref{shadow} dominates
all $s\in p^+$.  For any $i<\al$ and $\tau\in K_i$ since
$f\in H_i$ we know that for any $n$ with $|s_i|\leq n <|s^\pr_i|$
that $s_i^\pr(n)\geq f(n)\geq r(n)$ for any $r\in p^+$ and
since $\tau\res n+1$ is forced to be in $p^+$ 
(since it is forced that $\tau\in [T]$), we have verified
clause \ref{forcing}.

Proof of (c). Since the suborder is the same, condition (b) implies
that if $G$
$\poset_\al$-generic over $V$, then $G\cap \hposet_\al$
is $\hposet_\al$-generic over $V$.

Proof of (a).
Suppose that $(\es(s_i,\es,H_i)_{i<\al})\forces_{\hposet_\al} \theta(\tau)$.  
Then given any $G$ $\poset_\al$-generic over $V$ with
$(p,(s_i,K_i,H_i)_{i<\al})\in G$ we have that
$$(\es,(s_i,\es,H_i)_{i<\al})\in G^\pr=^{def}G\cap \hposet_\al.$$
By the definition of forcing
we have that $V[G^\pr]\models\theta(\tau^{G^\pr})$.  But
$\tau^{G^\pr}=\tau^{G}$ and by Borel absoluteness 
$V[G]\models\theta(\tau^{G})$.  Consequently by the definition
of forcing
$$(p,(s_i,K_i,H_i)_{i<\al})\forces_{\poset_\al} \theta(\tau)$$

On the other hand, if it is not the case that
$(\es,(s_i,\es,H_i)_{i<\al})\forces_{\hposet_\al} \theta(\tau)$, 
then there exists
$(\es,(s_i^\pr,\es,H_i^\pr)_{i<\al})\leq (\es,(s_i,\es,H_i)_{i<\al})$ 
such that 
$$(\es,(s_i^\pr,\es,H_i^\pr)_{i<\al})\forces_{\hposet_\al} \neg\theta(\tau)$$
as we saw in the proof of (b) we may construct $p^+\leq p$ such that
$$(p^+,(s_i^\pr,K_i,H_i^\pr)_{i<\al})\leq (p,(s_i,K_i,H_i)_{i<\al})$$
$$(p^+,(s_i^\pr,K_i,H_i^\pr)_{i<\al})\forces_{\poset_\al} \neg\theta(\tau)$$
and therefor
it is not the case that
$(p,(s_i,K_i,H_i)_{i<\al})\forces_{\poset_\al} \theta(\tau)$.
\qed

Our next lemma finds a node $t$ in $T$ below $s$ which has
nothing to do with the currently mentioned elements of
$K$.  This will allow us to extend $t$ without having to
worry about extensions of the $s_i$ dominating these
elements of $K$.

\begin{lemma}\label{hide}
For any $\be$, $(p,(s_i,K_i,H_i)_{i<\be})\in \poset_\be$
and $s\in p$
there exists 
$$(p^\pr,(s_i^\pr,K_i^\pr,H_i^\pr)_{i<\be})\leq (p,(s_i,K_i,H_i)_{i<\be})$$ 
and $t\in p^\pr$ with $s\su t$
with the property 
that for every $j < \be$ and every $\tau\in K_j^{\pr}$
that
$$(p^\pr,(s_i^\pr,K_i^\pr,H_i^\pr)_{i<j})\forces t\not\su\tau$$
\end{lemma}
\proof
The proof is by induction on $\be$.  For $\be$ limit
it is trivial.  For the successor case $\be+1$ suppose we are
given $(p,(s_i,K_i,H_i)_{i\leq \be}) $ and $s\in p$.
Let $n=|K_\be|$.  Extend p so as to have at least $n+1$ nodes
$t_1,t_2,\ldots, t_{n+1}$ below $s$ and having the
same length, say $|t_i|=k>|s|$.  Extend $(p,(s_i,K_i,H_i)_{i<\be})$
to $(\hat{p},(\hat{s}_i,\hat{K}_i,\hat{H}_i)_{i<\be})$ which
decides $\tau\res k$ for each $\tau\in K_\be$.  
Some $\hat{t}=t_i$ is not ruled out.  Apply the induction hypothesis
to $(\hat{p},(\hat{s}_i,\hat{K}_i,\hat{H}_i)_{i<\be})$ and $\hat{t}$.
\qed

We are now ready to prove the Claim.
Let $T_s=\{t\in T\;:\; s\su t\}$ and suppose for contradiction
that for some 
$(p,(s_i,K_i,H_i)_{i\leq \al})\in \poset_{\al+1}$ with $s\in p$
that
$$ (p,(s_i,K_i,H_i)_{i\leq \al})\forces \forall t\in T_s\;\;
\forall n\;\; (|s|<n<|t|\;\to\; t(n)<g_\al(n)).$$
Without loss by Lemma \ref{hide} we may assume
that there exists $t\in p$ a terminal node of $p$ extending $s$ with 
the property that all $\tau \in \bigcup_{i\leq\al}K_i$
are being forced incompatible with $t$.
Extend $t$ by concatenating zeros to
it so that $|t|=n_0>|s_\al|$.
By Lemma \ref{generic} we can find
$$(\es,(s_i^\pr,\es,H_i^\pr)_{i<\al})\leq
(\es,(s_i,\es,H_i)_{i<\al})$$ and $N<\om$
such that for
each $\tau\in H_\al$
and for each $n$ with $|s_\al|\leq n\leq n_0$
$$(\es,(s_i^\pr,\es,H_i^\pr)_{i<\al})\forces_{\hposet_\al} \tau(n)<N.$$

In addition we may assume that $N>s(n)$ for all $s\in p$ and $n<|s|$.
Now we define $p^+$ as follows.  We extend $t$ by adding
$N+1$ to it's end, i.e., $t^\pr\supseteq t$ with $t^\pr(n_0)=N+1$.  
For all other terminal nodes of $p$ we extend by adding zeros until they
are longer than any of the lengths of the $s_i^\pr$.  We define
$s_\al^\pr$ to be the extension of $s_\al$ of length $n_0+1$ gotten
by adding the constant sequence $N$.

Then 
$$(p^+,(s_i^\pr,K_i,H_i^\pr)_{i\leq \al})\leq (p,(s_i,K_i,H_i)_{i\leq \al})$$
because none
of the elements of any $K_i$ go thru the node $t$ and all the other nodes
are extended by zeros.  But this is a contradiction, $t^\pr(n_0)=N+1$
and this condition forces that $g_\al(n_0)=N$ since
$s_\al^\pr(n_0)=N$.

This proves the Claim and therefore Theorem \ref{thm1}.
\qed


By general forcing facts $N$ is a ccc generic extension of
$M$.  In fact,  $N$ is a generic extension of $M$ using 
a ccc suborder of $\poset_{\om_1}$, see Solovay \cite{solovay}
p.22 definition of $\Sigma$.   For a proof using
complete Boolean algebras see Grigorieff \cite{grig}.
However this factor forcing cannot be countable in $M$ even though
$K$ is added by a poset countable in $V$.

\begin{prop}
Suppose $M$ is a countable transitive model of ZFC, 
$G$ $\poset$-generic over $M$ where
$M\models \poset$ is countable.
Then $M$ and $N=M[G]$ fail to satisfy Theorem \ref{thm1}.
\end{prop}
\proof
Suppose not.  In $N$ let $T\su\om^{<\om}$ 
be a finitely branching tree satisfying:
  \begin{enumerate}
    \item[(a)] $\forall f\in [T]\cap N \;\;\exists g\in M\cap\om^\om
    \;\; \forall^\infty n\;f(n)< g(n)$
    and
    \item[(b)] $\forall g\in M\cap\om^\om\;\;
    \exists f\in [T]\cap N\;\;\exists^\infty n\;g(n)<f(n)$.
  \end{enumerate}
Define
$$B=\{s\in T\;:\; \exists g\in M\cap \om^\om\;\; \forall f\in [T_s]
\;\;\; f\leq^*g\}$$
to be the $M$-bounded nodes of $T$.
Without loss of generality we may assume that $B$ is empty.
To see this replace $T$ by $T_0=T\sm B$.  This will be an $M$-unbounded
tree for which the corresponding $B_0$ is empty.  This is true because
given any $H\su M\cap\om^\om$ such that $N\models H $ is countable,
there exists $g\in M\cap\om^\om$ with $h\leq^*g$ for all $h\in H$.

Working in $M$ suppose for some $p_0$ and name $T$
$$p_0\forces T\su\om^{<\om}\mbox{ is finitely branching tree and }B=\es.$$
We will prove that (a) fails.
Suppose $p\leq p_0$ and $p\forces s\in T$, then let
$$T(p)=\{t\in \om^{<\om}\;:\; s\su t \rmand \exists q\leq p\;\; 
q\forces t\in T\}.$$
Note that $T(p)$ is a subtree of $\om^\om$.  Clearly it cannot be finite
branching because then $s$ is $M$-bounded.  Hence for some
$k$ the set $\{t(k):t\in T(p)\}$ is infinite.
Using this observation we can build a name $\tau$ 
for an element of $[T]$ which is
infinitely often larger than any ground model real, so (a) fails.
Construct $(p_t, t_s)$ for $s\in \om^{<\om}$ such that
\begin{enumerate}
\item $p_{\la\ra}=p_0$ and $t_{\la\ra}={\la\ra}$.
\item $p_{si}$ for $i<\om$ is a maximal antichain beneath $p_s$.
\item $p_s\forces t_s\in T$.
\item $t_{si}\supseteq t_s$ and for \label{four}
some $k$ the set $\{t_{si}(k):i<\om\}$ is infinite.
\item For any $q\in \poset $ with $q\leq p_0$ there exists $s$ such that
$p_s\leq q$.
\end{enumerate}

For any generic filter $G$ 
containing $p_0$ there will be a unique $f\in\om^\om$
$p_{f\res n}\in G$ for all $n$.  Let $\tau$ be a name for 
$\cup_{n<\om} t_{f\res n}$.  We claim
$$p_0\forces \tau\in [T]\rmand 
\forall g\in M\;\exists^\infty\; k\;\; g(k)<\tau(k).$$
Suppose for contradiction that $q\leq p_0$ and $g\in M$ satisfy
$$q\forces \forall k \;\;\tau(k)\leq g(k).$$
Then some for some $s$ we have $p_s\leq q$.
By condition \ref{four}
we may find $k$ and $i$ 
so that $t_{si}(k)>g(k)$.
But then $p_{si}\forces \tau(k)>g(k)$ which is a
contradiction.
\qed


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\bibitem{grig}
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\bibitem{L}
B.Lawrence, The influence of a small cardinal on the
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\bibitem{M}
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\bibitem{R}
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\bibitem{solovay}
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\end{thebibliography}

\bigskip
\address
\end{document}


