\documentclass[12pt]{article} \usepackage{amssymb} \def\aa{{\mathcal A}} \def\al{\alpha} \def\be{\beta} \def\ff{{\mathcal F}} \def\forces{{\;\Vdash\;}} \def\ga{\gamma} \def\goth{\mathfrak} \def\om{\omega} \def\poset{{\mathbb P}} \def\prodaa{\prod_{n<\om}\aa_n} \def\proof{\par\noindent proof:\par} \def\qed{\nopagebreak\par\noindent\nopagebreak$\Box$\par} \def\reals{{\mathbb R}} \def\res{\mathord{\upharpoonright}} \def\rmand{\mbox{ and }} \def\rmor{\mbox{ or }} \def\si{\sigma} \def\sm{\setminus} \def\su{\subseteq} \newtheorem{theorem}{Theorem} % [section] \newtheorem{prop}[theorem]{Proposition} \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