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\begin{center}
{\langlerge Remark 3.4\\ A Dedekind Finite Borel Set}
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\begin{flushright}
Arnold W. Miller\\
August 2012
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\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}}

Asaf Karagila pointed out that Remark 3.4 in my paper,
A Dedekind finite Borel set \cite{ded},
directly contradicts Theorem 3.3 (c) in
Does GCH imply AC locally? by A. Kanamori and D. Pincus \cite{kanamori}.
Neither of the papers provides a
proof.\footnote{
Aki Kanomori tells me that
Theorem 3.3 (c) was incorrectly stated.
Instead of ${\mathcal N}$ it
should be the Solovay-type inner model ${\mathcal N}_0\subseteqbseteq {\mathcal N}$.
This is the model
$HOD(^\omegaega ON)$ of ${\mathcal N}$.
See page 155 of {\bf Consequences of the
Axiom of Choice} by Rubin and Howard who call
${\mathcal N}_0$ Truss's model.}
Here we give a proof of Remark 3.4 \cite{ded}.

The Feferman-Levy model ${\mathcal N}$ is described in
Cohen \cite{cohen} p.143, Jech
\cite{jech} p.142,
and in the on-line only appendix to Miller \cite{longbor}.
We use the notation from \cite{longbor}.
${\mathbb Col}$ is the Levy collapse of $\aleph_\omega$ , ${\mathbb Col}_n$
the collapse of $\aleph_k$ for $k<n$, $G_n={\mathbb Col}_n\cap G$
where $G$ ${\mathbb Col}$-generic
over $M$. The model ${\mathcal N}$ is the symmetry model
$M\subseteq{\mathcal N}\subseteq M[G]$ determined by the group ${\mathcal H}$ of
automorphism of ${\mathbb Col}$ which finitely permute finitely
many of the collapsing
maps domains and the filter of subgroups generated by
$(H_n\;:\;n<\omega)$ where
$$H_n=\{\rho\in {\mathcal H}\;:\; \forall p\in{\mathbb Col}_n\;\;
\rho(p)=p\}.$$

\begin{define}
Let $\langle,\rangle:\omega\times\omega\to\omega$ be a fixed bijection,
i.e., a pairing function.  For each $n\in\omega$
define the map $\pi_n:2^\omega\to 2^\omega$ by:
$$\pi_n(x)=y \mbox { iff } \forall m\in\omega\;\;y(m)=x(\langle n,m\rangle).$$
\end{define}

\begin{define}
In the Feferman-Levy model ${\mathcal N}$, take $F_n=M[G_n]\cap 2^\omega$.
Define
$$B=\{x\in 2^\omega\;:\;\forall n\;\; \pi_n(x)\in F_{n+1}{\setminus} F_n
\mbox{ or } [\pi_n(x)\in F_n \mbox{ and } \pi_n(x)=\pi_{n+1}(x)]
\}.$$
\end{define}

The set $B$ is uncountable because there is a map $h$ from $B$ onto $2^\omega$.
Define $h$ by $h(x)=\pi_n(x)$ iff $\pi_n(x)=\pi_m(x)$ for all $m>n$.  Such an
$n$ must exist because for any $x$ there exists $n$ such that $x\in F_n$ and
hence $\pi_m(x)\in F_n$ for all $m$.  It is easy to check that $h$ maps $B$ onto
$2^\omega$.

\begin{prop}
In the Feferman-Levy model ${\mathcal N}$ the set $B$ has the
property that there is no one-to-one map
taking $2^\omega$ into $B$.
\end{prop}

\par\noindent Proof\par\noindent
Suppose for contradiction that
there is in ${\mathcal N}$ a one-to-one map  $f:2^\omega\to B$.
Define
$$C_n=\{x\in B\;:\; \forall m>n \;\;\pi_m(x)=\pi_n(x)\}$$
and note that $B=\bigcup_nC_n$ and each $C_n$ is countable.

Let
$$p{\;\Vdash}s \name{f}:2^\omega\to \name{B} \mbox{ is one-to-one }$$
and suppose that $p\in{\mathbb Col}_n$ and $H_n$
(the subgroup of ${\mathcal H}$ which
is the identity on ${\mathbb Col}_n$)
fixes $\name{f}$.  Since
$C_n$ is countable and $f$ is one-to-one, we may choose
$N>>n$ and $q\leq p$ such that
$$q{\;\Vdash}s {f}^{-1}(\name{C}_n)\subseteq \name{F}_N.$$
Now let $\name{x_N}$ be a canonical name for a real which codes
the generic collapse of $\aleph_N$.  The important property
it has is that $x_N\in F_{N+1}{\setminus} F_{N}$ and its values are
completely determined by $N^{th}$ coordinate of the generic
filter and not any earlier
ones, e.g., the $\leq n$ coordinates. Take $r\leq q$ such that
$$r{\;\Vdash}s \name{f}(\name{x}_N)=\name{y}, \pi_n(\name{y})=\tau, \mbox{ and }
\tau\in \name{F}_{n+1}{\setminus} \name{F}_n$$
where $\tau$ is a ${\mathbb Col}_{n+1}$-name for an element of $2^\omega$.
For $G$ ${\mathbb Col}$-generic over $M$ and containing
$r$ note that $\tau^{G_{n+1}}\in M[G_{n+1}]{\setminus} M[G_{n}]$.  So we can
find $k$ and $r_0,r_1\leq r$ such that
\begin{itemize}
\item $r_0$ and $r_1$ are the same on every coordinate except $n$,
\item $r_0{\;\Vdash}s \tau(k)=0$, and
\item $r_1{\;\Vdash}s \tau(k)=1$.
\end{itemize}

Now let $\rho\in{\mathcal H}$ be an automorphism of ${\mathbb Col}$
which fixes every coordinate except
$n$ and makes $\rho(r_0)$ and $r_1$ compatible.
To do this move the domain of $r_0\upharpoonright(\{n\}\times\omega)$ disjoint from the
domain of $r_1$.
Since $\rho\in H_n$ it fixes $\name{f}$, i.e.,
$\rho(\name{f})=\name{f}$.   Since $\rho$ fixes the $N^{th}$
coordinate $\rho(\name{x}_N)=\name{x}_N$.
But this means that
$$\rho(r_0){\;\Vdash}s \pi_n(\name{f}(\name{x}_N))(k)=0$$
and
$$r_1{\;\Vdash}s \pi_n(\name{f}(\name{x}_N))(k)=1$$
which contradicts that $\rho(r_0)$ and $r_1$ are compatible.
\par\noindent QED\par

\address

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