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\begin{document}
\begin{center}
The onto mapping property of Sierpinski
\end{center}
\begin{flushright}
A. Miller \\
July 2014\\
revised May 2016
\end{flushright}
\bigskip
\noindent Define
(*) There exists $(\phi_n:\om_1\to \om_1:n<\om)$ such that
for every $I\in [\om_1]^{\om_1}$ there exists $n$ such
that $\phi_n(I)=\om_1$.
\bigskip
This is roughly what Sierpinski \cite{sier} refers to as $P_3$ but I think
he brings $\reals$ into it. I don't know French so I cannot say for sure
what he says but I think he proves that (*) follows from
the continuum hypothesis.
Here we show that the existence of a Luzin set implies (*) and
(*) implies that there exists a nonmeager set of reals of size $\om_1$.
We also show that it is relatively consistent that (*) holds
but there is no Luzin set. All the other properties in this paper,
(**), (S*), (S**), (B*) are shown to be equivalent to (*).
\bigskip
\begin{prop}(Sierpinski \cite{sier})
CH implies (*).
\end{prop}
\proof
Let $\om_1^\om=\bigcup_{\al<\om_1}\ff_\al$
where the $\ff_\al$ are countable and increasing.
For each $\al$ construct $(\phi_n(\al):n<\om)$ so that for
every $g\in \ff_\al$ there is a some $n$ such that $\phi_n(\al)=g(n)$.
Now suppose $I\su\om_1$. If no $\phi_n$ maps $I$ onto $\om_1$,
then there
exists $g\in\om_1^\om$ such that $g(n)\notin\phi_n(I)$ for every $n$.
If $g\in \ff_{\al_0}$, then $\al\notin I$ for every $\al\geq\al_0$.
This is because $g\in \ff_\al$ and so for some $n$ $g(n)=\phi_n(\al)$
and since $g(n)\notin \phi_n(I)$ we have $\al\notin I$.
\qed
\bigskip \noindent Define
(**) There exists $(g_\al:\om\to\om_1:\al<\om_1)$ such that for every
$g:\om\to\om_1$ for all but countably many $\al$ there are infinitely
many $n$ with $g(n)=g_\al(n)$.
\begin{prop}
(**) iff (*).
\end{prop}
\proof
To see (**) implies (*) let $\phi_n(\al)=g_\al(n)$. Then the proof
of the first proposition goes thru.
On the other hand suppose
$(\phi_n:\om_1\to \om_1:n<\om)$ witnesses (*). First note
that for any
$I\in [\om_1]^{\om_1}$ there are infinitely many $n$ such
that $\phi_n(I)=\om_1$. This is because if there are only
finitely many $n$ we could cut down $I$ in finitely many steps
so that there were no
$n$ with $\phi_n(I)=\om_1$.
Now define $g_\al\in\om_1^\om$ by
$g_\al(n)=\phi_n(\al)$. These witness (**).
Given any $g:\om\to\om_1$ if there is
an uncountable $I\su\om_1$ and $N<\om$ such that for every $\al\in I$
we have $g(n)\neq g_\al(n)$ for all $n>N$ then this means
that $g(n)\notin \phi_n(I)$ and for all $n>N$ and so (*) fails.
\qed
Obviously (**) is false if ${\goth b}>\om_1$
so (*) is not provable just from ZFC.
\begin{prop}
It is relatively consistent with any cardinal arithmetic that
(*) is true and ${\goth b}={\goth d}=\om_1$.
\end{prop}
\proof
Start with any $M$ a countable transitive model of ZFC.
Our final model is $M[g_\al,f_\al:\al<\om_1]$
where each $g_\al:\om\to\al$ is generic with respect to the poset of
finite partial functions from $\om$ to $\al$ and $f_\be\in\om^\om$ is
Hechler real over $M[g_\al,f_\al:\al<\be]$. The $\om_1$-sequence
is obtained by finite support ccc forcing.
By ccc for any $g\in \om_1^\om\cap M[g_\al,f_\al:\al<\om_1]$
there will be $\al_0<\om_1$ such
that $\al_0$ bounds the range of $g$ and
$g\in M[g_\al,f_\al:\al<\al_0]$. It follows by product genericity that
for every $\al\geq\al_0$ there are infinitely many $n$ such that
$g(n)=g_\al(n)$. The Hechler sequence $f_\al$ for $\al<\om_1$ shows
that ${\goth d}=\om_1$.
\qed
With a little more work we will prove that (*) follows
from the existence of a Luzin set (Prop \ref{luzin}).
We will also show that (*) implies there is a nonmeager set
of reals of size $\om_1$ (Prop \ref{nonmeag}) and so in the
random real model (*) fails and ${\goth b}={\goth d}=\om_1$.
\bigskip
\bigskip
\bigskip\noindent
Actually I think Sierpinski
considers what appears to be a stronger version:
\bigskip\noindent Define
(S*) There exists $(\phi_n:\om_1\to \om_1:n<\om)$ such that
for every $I\in [\om_1]^{\om_1}$ for all but finitely many $n$
$\;\;\;\phi_n(I)=\om_1$.
\bigskip
Surprisingly (S*) is equivalent to (*).
\begin{prop}\label{sstar}
(S*) iff (*).
\end{prop}
\proof
We show (**) implies (S*).
Let $a_0=1$ and $a_{n+1}=1+\sum_{i\leq n} a_i$.
Let $$\aa_n=\{u \;\;|\;\; \exists D\in [\om_1]^{a_n}\;\; u:D\to\om_1\}
\rmand \prodaa=\{g \;\;|\;\; \forall n \; g(n)\in \aa_n\}$$
Since each $\aa_n$ has cardinality $\om_1$ from (**) we
get $(g_\al\in \prodaa:\al<\om_1)$ such that for every
$g\in\prodaa$ for all but countably many $\al$ there are
infinitely many $n$ such that $g(n)=g_\al(n)$. For each
$\al<\om_1$ define $h_\al:\om\to\om_1$ so that if $g_\al(n)=u_n:A_n\to\om_1$
for every $n$ then
$$h_\al\res (A_n\sm \cup_{i\be$ for all $\al\ge\al_0$
there will be infinitely
many $n$ with $g_\al(n)=\hat{k}(n)$. This means
that $g_\al(n)=f_\be(k(n))$. Since $k$ is one-to-one, there
will be infinitely many such $n$ where $f_\be(k(n))=f_\al(k(n))$.
But $g_\al(n)=f_\al(k(n))$ implies $\hat{g}_\al(n)=k(n)$.
To get rid of the requirement that $k$ be one-to-one,
let $j:\om_1\times\om \to \om_1$ be a bijection and
$\pi:\om_1\to \om_1$ be projection onto first coordinate, i.e.,
$\pi(j(\al,n))=\al$. Define $h_\al(n)=\pi(\hat{g}_\al(n))$.
Given any $k:\om\to\om_1$ define $\hat{k}(n)=j(k(n),n)$.
Then since $\hat{k}$ is one-to-one for all but countably
many $\al$ there will be infinitely many $n$ with
$\hat{g}_\al(n)=\hat{k}(n)$. But this implies
$$h_\al(n)=\pi(\hat{g}_\al(n))=\pi(\hat{k}(n))=k(n)$$
Hence $(h_\al:\al<\om_1)$ satisfies (**).
\qed
\begin{prop}\label{nonmeag}
Suppose (*), then there exists $(x_{\al,\be}\in 2^\om:\al,\be<\om_1)$
such that for every dense open $D\su 2^\om$ there exists
$\al_0<\om_1$ such that for every $\al\geq\al_0$ there is
a $\be_\al<\om_1$ such that $x_{\al,\be}\in D$ for every $\be\geq\be_\al$.
\end{prop}
\proof
We use that there are
$\{h_\al:\om\to\om\;:\;\al<\om_1\}$ with the property that
for every $X\in[\om]^\om$ and $h:\om\to\om$ for all but countably
many $\al$ there are infinitely many $n\in X$ with $h(n)=h_\al(n)$
(see (S**) in the proof of Prop \ref{sstar}).
This implies that there exists
$(X_\al\in [\om]^\om:\al<\om_1)$ such that for every $Y\in [\om]^\om$
for all but countably many $\al$ there are infinitely many
$x\in X_\al$ such that $|Y\cap [x,x^+)|\geq 2$ where $x^+$ is
the least element of $X_\al$ greater than $x$. Fix $\al$ and
enumerate $X_\al=\{k_n:n<\om\}$ in strict increasing order.
Define
$$P_\al=\{g:\om\to FIN(\om,2)\;:\;\forall n\;\; g(n)\in 2^{[k_n,k_{n+1})}\}$$
By (S**) there exists $g_{\al,\be}\in P_\al$ for $\be<\om_1$ with the
property that for any $h$ in $P_\al$ and infinite $Y\su\om$ for
all but countably many $\be$ there are infinitely many $n\in Y$ with
$h(n)=g_{\al,\be}(n)$. Define $x_{\al,\be}\in 2^\om$ by
$x_{\al,\be}(m)=g_{\al,\be}(n)(m)$ where $n$ is the unique integer
with $k_n\leq m