% LaTeX \documentstyle[12pt]{article} \newcommand{\res}{\upharpoonright} \newcommand{\propsub}{\subsetneqq} \newcommand{\qed}{\nopagebreak\par\noindent\nopagebreak$\blacksquare$\par} \input amssym.def \input amssym %%%%%% if amssym.def and amssym.tex are not available %%%%%% delete the lines above and uncomment lines below %\def\res{\uparrow} %\def\propsub{\subset} %\def\qed{\nopagebreak\par\noindent\nopagebreak$\Box$\par} %\def\Bbb#1{{\bf #1}} %\def\goth#1{{\bf #1}} \def\intersect{\cap} \def\Union{\bigcup} \def\union{\cup} \def\rmiff{\mbox{ iff }} \def\rmand{\mbox{ and }} \def\dom{{\rm dom}} \def\blocking{{\cal B}} \def\famil{{\cal G}} \def\proof{{\sc Proof. }} \def\reals{{\Bbb R}} \def\rationals{{\Bbb Q}} \def\realfunc{{\reals^\reals}} \def\almost{{\cal A}} \def\aa{{{\rm A}(\almost)}} \def\darboux{{\cal D}} \def\ad{{{\rm A}(\darboux)}} \def\A{{\cal H}} \def\D{{\cal D}} \def\E{{\cal E}} \def\F{{\cal F}} \def\G{{\cal G}} \def\cf{{\rm cf}} %\def\qed{\hfill$\Box$} \def\proj{{{\rm pr}_x}} \def\continuum{{\goth c}} \def\poset{{\Bbb P}} \def\posetx{{\Bbb P}^*} \def\poS{{\Bbb S}} \def\ONE{{\Bbb I}} \def\su{\subseteq} \def\eq{{\goth e}} \def\la{\langle} \def\ra{\rangle} \def\implies{\longrightarrow} %% Theorems, etc. \newtheorem{theorem}{Theorem}[section] \newtheorem{corollary}[theorem]{Corollary} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{problem}{Problem}[section] \newtheorem{example}{Example}[section] \newtheorem{definition}{Definition} \newtheorem{remark}{Fact}[section] \newcommand{\thm}[2]{\begin{theorem}\label{#1}#2\end{theorem}} \newcommand{\cor}[2]{\begin{corollary}\label{#1}#2\end{corollary}} \newcommand{\prop}[2]{\begin{proposition}\label{#1}#2\end{proposition}} \newcommand{\lem}[2]{\begin{lemma}\label{#1}#2\end{lemma}} \newcommand{\pr}[2]{\begin{problem}\label{#1}#2\end{problem}} \newcommand{\ex}[2]{\begin{example}\label{#1}#2\end{example}} \newcommand{\defi}[2]{\begin{definition}\label{#1}#2\end{definition}} \newcommand{\rem}[2]{\begin{remark}\label{#1}#2\end{remark}} \pagestyle{myheadings} \title{Cardinal invariants concerning functions whose sum is almost continuous.} \author{} \date{} \markright{Functions whose sum is almost continuous.} \begin{document} \maketitle {\small\noindent Krzysztof Ciesielski\footnotemark[1], Department of Mathematics, West Virginia University, Morgantown, WV 26506-6310 (kcies@wvnvms.wvnet.edu)} \medskip {\small\noindent Arnold W. Miller\footnote[1]{ The results presented in this paper were initiated, and partially obtained, during the Joint US--Polish Workshop in Real Analysis, {\L}{\'o}d{\'z}, Poland, July 1994. The Workshop was partially supported by the NSF grant INT--9401673. \par We want to thank Juris Steprans for many helpful conversations. \par AMS Subject Classification. Primary: 26A15; Secondary: 03E35, 03E50. \par This is in Real Analysis Exchange, 20(1994-5), 657-672. }, York University, Department of Mathematics, North York, Ontario M3J 1P3, Canada, Permanent address: University of Wisconsin-Madison, Department of Mathematics, Van Vleck Hall, 480 Lincoln Drive, Madison, Wisconsin 53706-1388, USA (miller@math.wisc.edu)} \begin{abstract} Let $\almost$ stand for the class of all almost continuous functions from $\reals$ to $\reals$ and let $\aa$ be the smallest cardinality of a family $F\su\realfunc$ for which there is no $g\colon\reals\to\reals$ with the property that $f+g\in\almost$ for all $f\in F$. We define cardinal number $\ad$ for the class $\darboux$ of all real functions with the Darboux property similarly. It is known, that $\continuum < \aa \leq 2^{\continuum}$ \cite{Nat:AC1}. We will generalize this result by showing that the cofinality of $\aa$ is greater that $\continuum$. Moreover, we will show that it is pretty much all that can be said about $\aa$ in ZFC, by showing that $\aa$ can be equal to any regular cardinal between $\continuum^+$ and $2^{\continuum}$ and that it can be equal to $2^{\continuum}$ independently of the cofinality of $2^{\continuum}$. This solves a problem of T.~Natkaniec \cite[Problem 6.1, p. 495]{Nat:AC1}. We will also show that $\ad=\aa$ and give a combinatorial characterization of this number. This solves another problem of Natkaniec. (Private communication.) \end{abstract} \section{$\!\!\!\!\!\!\!${\bf.} Preliminaries.} We will use the following terminology and notation. Functions will be identified with their graphs. The family of all functions from a set $X$ into $Y$ will be denoted by $Y^X$. Symbol $|X|$ will stand for the cardinality of a set $X$. The cardinality of the set $\reals$ of real numbers is denoted by $\continuum$. For a cardinal number $\kappa$ we will write $\cf(\kappa)$ for the cofinality of $\kappa$. A cardinal number $\kappa$ is regular, if $\kappa=\cf(\kappa)$. Recall also, that the Continuum Hypothesis (abbreviated as CH) stands for the statement $\continuum=\aleph_1$. A function $f\colon\reals\to\reals$ is {\em almost continuous} (in the sense of Stallings \cite{Stall}) if and only if for every open set $U\su\reals^2$ containing $f$ there exists a continuous function $g\su U$. So, every neighborhood of $f$ in the graph topology contains a continuous function. This concept was introduced by Stallings \cite{Stall} in connection with fixed points. We will use symbol $\almost$ to denote the family of almost continuous functions from $\reals$ to $\reals$. For $\F\su\reals^\reals$ define the cardinal ${\rm A}(\F)$ as follows: \begin{eqnarray*} {\rm A}(\F) & = & \min\{|F|\colon F\su\reals^\reals \&\ \neg\exists g\in\reals^\reals\ \forall f\in F\ f+g\in\F\}\\ & = & \min\{|F|\colon F\su\reals^\reals \&\ \forall g\in\reals^\reals\ \exists f\in F\ f+g\not\in\F\} \end{eqnarray*} For a generalization of the next theorem see Natkaniec \cite{Nat:AC1}. Fast \cite{fast} proved the same result for the family of Darboux functions. \thm{nat} {$\continuum < \aa \leq 2^{\continuum}$.}\qed At the Joint US--Polish Workshop in Real Analysis in {\L}{\'o}d{\'z}, Poland, in July 1994 A.~Maliszewski gave a talk mentioning several problems of his and T.~Natkaniec. Natkaniec asked whether or not anything more could be said about the cardinal $\aa$. (See also Natkaniec \cite[Problem 6.1, p. 495]{Nat:AC1} or \cite[Problem 1.7.1, p. 55]{Nat:AC2}.) In what follows we will show that pretty much nothing more can be said (in ZFC), except that the $\cf(\aa)>\continuum$. We will also study the family $\darboux\su\realfunc$ of Darboux functions. Recall that a function is {\em Darboux} if and only if it takes every connected set to a connected set, or (in the case of a real function) satisfies the intermediate value property. Note that $\almost\su\darboux$. This is because if for example $f(a)\continuum$ in Section 2. In Section 3 we will prove that some forcing axioms imply that $\aa$ can be any regular cardinal between $\continuum^+$ and $2^\continuum$ and that $\aa$ can be equal to $2^\continuum$ for any value of $2^\continuum$. The proof of the consistency of the forcing axioms used in Section 3 will be left for the Section 4. \section{$\!\!\!\!\!\!\!${\bf.} $\ad=\aa$ and its cofinality.} We will need the following definitions. For a cardinal number $\kappa\leq\continuum$ we define the family $$\darboux(\kappa)\su\reals^\reals$$ of {\em $\kappa$ strongly Darboux functions} as the family of all functions $f\colon \reals\to\reals$ such that for all $a,b\in \reals$, $a\continuum$.} \proof Pick a family $F\su \reals^\reals$ of cardinality continuum. We will find a function $g\in\reals^\reals$ such that $f+g\in\darboux(\continuum)$ for all $f\in F$. Let $$\la\la a_\xi,b_\xi,y_\xi,f_\xi\ra\colon \xi<\continuum\ra$$ be an enumeration of the set of all $$\la a,b,y,f\ra\in\reals\times\reals\times\reals\times F\mbox{ with $a\continuum.$$ We will show that $\kappa\geq\ad$. Let $F\su\reals^\reals$ be a family of cardinality $\kappa$ witnessing $\kappa={\rm A}(\darboux(\omega_1))$: \begin{equation}\label{eqA} \forall g\in\reals^\reals\ \exists f\in F\ f+g\not\in\darboux(\omega_1). \end{equation} It is enough to find family $F^*\su\reals^\reals$ of cardinality $\kappa$ such that \begin{equation}\label{eq2} \forall g\in\reals^\reals\ \exists f^*\in F^*\ f^*+g\not\in\darboux. \end{equation} Define $F^*=\{h\in\reals^\reals\colon \;\exists f\in F\; h=^*f\}$, where $h=^*f$ if and only if the set $\{x\colon h(x)\neq f(x)\}$ is at most countable. Since $\kappa>\continuum$ and for every $f\in\reals^\reals$ the set $\{h\in\reals^\reals\colon h=^*f\}$ has cardinality $\continuum$, we have $|F^*|=\kappa$. It is enough to show that $F^*$ satisfies (\ref{eq2}). So, choose $g\in\reals^\reals$. Then, by (\ref{eqA}), there exists $f\in F$ such that $f+g\not\in\darboux(\omega_1)$. This means, that there are $ay$, and \item $(f^*+g)(x)\neq y$ for every $x\in (a,b)$. \end{itemize} Thus, $f^*+g\not\in\darboux$. \qed Now, we are ready for one of our main theorems. \thm{equqtion}{ $\ad=\aa$.} \proof We already know that $\aa\leq\ad$. So, by Lemma \ref{lemStrDar1}, it is enough to prove that ${\rm A}(\darboux(\omega_1))\leq\aa$. So, let $\kappa=\aa$. Then, by Theorem \ref{nat}, $\kappa>\continuum$ and, by the definition of $\aa$, there exists a family $F\su\reals^\reals$ of cardinality $\kappa$ witnessing it, i.e., such that \[ \forall g\in\reals^\reals\ \exists f\in F\ f+g\not\in\almost. \] In particular, by the definition of the family $\blocking$ of blocking sets (from Theorem \ref{ThBlock}), \begin{equation}\label{eq3} \forall g\in\reals^\reals\ \exists f\in F\ \exists B\in\blocking\ (f+g)\cap B=\emptyset. \end{equation} It is enough to find a family $F^*\su\reals^\reals$ of cardinality $\kappa$ such that \begin{equation}\label{eq4} \forall g\in\reals^\reals\ \exists f^*\in F^*\ f^*+g\not\in\darboux(\omega_1). \end{equation} In order to do this, choose a function $h_B\in\reals^\reals$ for every $B\in\blocking$ such that \[ (x,h_B(x))\in B\ \mbox{ for every }\ x\in\proj(B). \] Let \[ F^*=\{f-h_B\colon f\in F\ \&\ B\in\blocking\}. \] Clearly $F^*$ has cardinality $\kappa$, since $|\blocking|\leq\continuum<\kappa$. We will show that $F^*$ satisfies (\ref{eq4}). Let $g\in\reals^\reals$. Then, by (\ref{eq3}), there exist $f\in F$ and $B\in\blocking$ such that $(f+g)\cap B=\emptyset$. In particular, \[ [(f-h_B)+g]\cap(B-h_B)=[(f+g)\cap B]-h_B=\emptyset, \] where we define $Z-h_B=\{(x,y-h_B(x))\colon (x,y)\in Z\}$ for any $Z\su\reals^2$. But $(B-h_B)\supset\proj(B)\times\{0\}$. Hence, $[(f-h_B)+g]\cap[\proj(B)\times\{0\}]=\emptyset$. In particular, $[(f-h_B)+g]^{-1}(0)\cap\proj(B)=\emptyset$. So, $f^*=f-h_B\in F^*$, while $(f-h_B)+g\not\in\darboux(\omega_1)$ since, by Theorem \ref{ThBlock}, $\proj(B)$ contains a non-degenerate interval. \qed To prove the next theorem we need a few more definitions. For a set $X\su\reals$ and a cardinal number $\kappa\leq\continuum$ we define the family $$\darboux(X,\kappa)\su\reals^X$$ as the family of all functions $f\colon X\to\reals$ such that for all $a,b\in X$, $a\continuum$.} \proof It is obvious that $\cf(\eq_\kappa)>\kappa$ since $\kappa$ can be split into $\kappa$ many sets of size $\kappa$. \qed \section{$\!\!\!\!\!\!\!${\bf.} Forcing axioms and the value of $\aa$.} In this section we will prove the following two theorems. \thm{thForMain}{ Let $\lambda\geq\kappa\geq\omega_2$ be cardinals such that $\cf(\lambda)>\omega_1$ and $\kappa$ is regular. Then it is relatively consistent with ZFC that the Continuum Hypothesis ($\continuum=\aleph_1$) is true, $2^{\continuum}=\lambda$, and $\aa=\kappa$. } So for example if $2\leq n\leq 17$, then it is consistent that $$\continuum=\aleph_1< \aa=\aleph_n\leq\aleph_{17}=2^\continuum.$$ \thm{thForMain2}{ Let $\lambda$ be a cardinal such that $\cf(\lambda)>\omega_1$. Then it is relatively consistent with ZFC that the Continuum Hypothesis ($\continuum=\aleph_1$) holds and $\aa=\lambda=2^{\continuum}$. } It follows from Theorem \ref{thForMain2} that $\aa$ can be a singular cardinal, e.g. $\aa=\aleph_{\omega_2}$ where $\continuum^+=\omega_2$. We do not know how to get $\aa$ strictly smaller than $2^\continuum$ and singular. The technique of proof is a variation on the idea of a Generalized Martin's Axiom (GMA). In this section we will formulate the forcing axioms and show that they imply the results. The proof of the consistency of these axioms will be left for Section 4. For a partially ordered set $(\poset,\leq)$ we say that $G\su\poset$ is a $\poset$-filter if and only if \begin{itemize} \item for all $p,q\in G$ there exists $r\in G$ with $r\leq p$ and $r\leq q$, and \item for all $p,q\in\poset$ if $p\in G$ and $q\geq p$, then $q\in G$. \end{itemize} Define $D\su \poset$ to be dense if and only if for every $p\in\poset$ there exists $q\in D$ with $q\leq p$. For any cardinal $\kappa$ and poset $\poset$ define MA$_\kappa(\poset)$ (Martin's Axiom for $\poset$) to be the statement that for any family $\D$ of dense subsets of $\poset$ with $|\D|<\kappa$ there exists a $\poset$-filter $G$ such that $D\cap G\not=\emptyset$ for every $D\in \D$. >From now on, let $\poset$ be the following partial order $$\poset= \{p\;| \; p:X\to\reals, X\su\reals, \rmand |X|<\continuum\}$$ i.e., the partial function from $\reals$ to $\reals$ of cardinality less than $\continuum$. Define $p\leq q$ if and only if $q\su p$, i.e., $p$ extends $q$ as a partial function. \lem{geq}{MA$_\kappa(\poset)$ implies $\aa\geq\kappa$.} \proof We know by Theorem \ref{Comb} that $\aa=\eq_\continuum>\continuum$. Thus, it is enough to prove that MA$_\kappa(\poset)$ implies $\eq_\continuum\geq\kappa$ for $\kappa>\continuum$. Note that for any $\poset$-filter $G$ since any two conditions in $G$ must have a common extension, $\Union G$ is a partial function from $\reals$ to $\reals$. Moreover, it is easy to see that for any $x\in\reals$ the set $$D_x=\{p\in\poset: x\in\dom(p)\}$$ is dense in $\poset$ and that $\Union G\colon\reals\to\reals$ for any $\poset$-filter $G$ intersecting all sets $D_x$. Let $\la S_\alpha:\alpha<\continuum \ra$ be a partition of $\reals$ into pairwise disjoint sets of size $\continuum$. Also for any $f\in{\reals^\reals}$ and $\alpha<\continuum$ the set $$D_{f,\alpha}=\{p\in\poset: \exists x\in(\dom(p)\cap S_\alpha)\;\; p(x)=f(x)\}$$ is dense in $\poset$. Given any $F\subseteq {\reals^\reals}$ with $|F|<\kappa$ let $$\D=\{D_x:x\in\reals\}\cup\{D_{f,\alpha}: f\in F,\alpha<\continuum\}.$$ Notice that $|\D|=\continuum<\kappa$. Applying MA$_\kappa(\poset)$ we can find a $\poset$-filter $G$ such that $G$ meets every $D\in\D$. Letting $g=\bigcup G\colon\reals\to\reals$ we see that $|f\cap g|=\continuum$ for every $f\in\F$. \qed The proof of Lemma \ref{geq} is a kind of forcing extension of the inductive argument used in the proof of Theorem \ref{nat}. Notice also, that Theorem \ref{thForMain2} follows immediately from Lemma \ref{geq}, Theorem \ref{nat} and the following theorem. \thm{thForMain2Proof}{ Let $\lambda$ be a cardinal such that $\cf(\lambda)>\omega_1$. Then it is relatively consistent with ZFC+CH that $2^{\continuum}=\lambda$ and that MA$_\lambda(\poset)$ holds. } Thus, we have proved Theorem \ref{thForMain2} modulo Theorem \ref{thForMain2Proof}. Theorem \ref{thForMain2Proof} will be proved in Section 4. Lemma \ref{geq} shows also one inequality of Theorem \ref{thForMain}. To prove the reverse inequality we will use a different partial order $(\posetx,\leq)$. It is similar to $\poset$ but in addition has some side conditions. \[ \posetx=\{(p,E):p\in\poset \rmand E\su\realfunc \mbox{ with } |E|<\continuum\}. \] Define the ordering on $\posetx$ by \begin{eqnarray*} (p,E)\leq (q,F) & \rmiff & p\leq q\ \rmand \ E\supseteq F\\ & \rmand & \forall x\in\dom(p)\setminus\dom(q)\;\; \forall f\in F\; p(x)\not=f(x). \end{eqnarray*} The idea of the last condition is that we wish to create a generic function $g\in \realfunc$ with the property that for many $f$ we have $g(x)\not=f(x)$ for almost all $x$. Thus, the condition $(q,F)$ `promises' that for all new $x$ and old $f\in F$ it should be that $g(x)\not=f(x)$. For a cardinal number $\kappa$ define Lus$_\kappa(\posetx)$ to be the statement: \begin{quote} There exists a sequence $\la G_\alpha:\alpha<\kappa\ra$ of $\posetx$-filters, called a $\kappa$-Lusin sequence, such that for every dense set $D\su\posetx$ \[ |\{\alpha<\kappa\colon \;G_\alpha\cap D=\emptyset\}|<\kappa. \] \end{quote} Thus we have a Lusin sequence of $\posetx$-filters. This is also known as a kind of Anti-Martin's Axiom. See vanDouwen and Fleissner \cite{df}, Miller and Prikry \cite{mp}, Todorcevic \cite{tod}, and Miller \cite{sur} for a similar axiom. \lem{leq}{ Suppose $\continuum<\kappa$, $\kappa$ is regular, and Lus$_\kappa(\posetx)$. Then $\aa\leq\kappa$. } \proof Let $\la G_\alpha:\alpha<\kappa\ra$ be a $\kappa$-Lusin sequence of $\posetx$-filters and let \[ g_\alpha=\Union \{p:\exists F\; (p,F)\in G_\alpha\}. \] Then $g_\alpha$ is a partial function from $\reals$ into $\reals$. Similarly to the last proof, let $$D_x=\{(p,F)\in\posetx\colon \; x\in\dom(p)\}.$$ To see that $D_x$ is dense let $(q,F)$ be an arbitrary element of $\posetx$ and suppose it is not already an element of $D_x$. The set $Q=\{f(x):f\in F\}$ has cardinality less than $\continuum$ so there exists $y\in\reals\setminus Q$. Let $p=q\union\{(x,y)\}$. Then $(p,F)\leq (q,F)$ and $(p,F)\in D_x$. Thus, each $D_x$ is dense in $\posetx$. Hence, since $\continuum<\kappa$ and $\kappa$ is regular, we may assume the each $g_\alpha$ is a total function. For each $f\in\realfunc$ define $$D(f)=\{(p,E)\in\posetx: \; f\in E\}.$$ Note that for any $(p,F)$ if we let $E=F\union\{f\}$, then $(p,F)\leq (p,E)$. Hence $D(f)$ is dense. Next, note that by the nature of definition of $\leq$ in $\posetx$, if $(p,F)\in G$, where $G$ is a $\posetx$-filter, and $g=\Union \{p\colon\exists F\;(p,F)\in G\}$, then for any $f\in F$ we have $g(x)\not=f(x)$ except possibly for the $x$ in the domain of $p$. Therefore for any $f\in \realfunc$ there exists $\alpha<\kappa$ such that $|g_\alpha\cap f|<\continuum$. Thus, the family $\{g_\alpha\colon\alpha<\kappa\}$ shows that $\aa=\eq_\continuum\leq\kappa$ as was to be shown. \qed \lem{equiv}{ For any regular $\kappa$ we have Lus$_\kappa(\posetx)\implies $MA$_\kappa(\posetx)\implies$MA$_\kappa(\poset)$. } \proof This first implication needs that $\kappa$ is regular but is true for any partial order. Given a family $\D$ of dense subsets of $\posetx$ of cardinality less than $\kappa$ and $\la G_\alpha:\alpha<\kappa\ra$ a Lusin sequence for $\posetx$ it must be that for some $\alpha<\kappa$ that $G_\alpha$ meets every element of $\D$. The second implication follows from the fact that in some sense $\poset$ is `living inside' of $\posetx$. Let $r:\reals\to\reals$ be a map with of $|r^{-1}(y)|=\continuum$ for every $y\in\reals$. Define $$\pi:\posetx\to\poset \mbox{ by } \pi(p,F)=r\circ p.$$ Notice that if $(p,E)\leq(q,F)$ then $\pi(p,E)\leq\pi(q,F)$. This implies that $\pi(G)$ is a $\poset$-filter for any $\posetx$-filter $G$. Furthermore, we claim that if $D\su\poset$ is dense, then $\pi^{-1}(D)$ is dense in $\posetx$. To see this, let $(p,F)\in\posetx$ be arbitrary. Since $D$ is dense, there exists $q\leq \pi(p,F)$ with $q\in D$. Now, find $s\in\poset$ extending $p$ such that $r\circ s= q\supseteq r\circ p$ and $s(x)\neq f(x)$ for every $x\in\dom(s)\setminus\dom(p)$ and $f\in F$. This can be done by choosing \[ s(x)\in r^{-1}(q(x))\setminus\{f(x)\colon f\in F\} \] for every $x\in\dom(q)\setminus\dom(p)$. Then, $(s,F)\leq (p,F)$ and $(s,F)\in\pi^{-1}(q)\su\pi^{-1}(D)$. This gives us the second implication, since if $\D$ is a family of dense subsets of $\poset$ with $|\D|<\kappa$ and $G$ is a $\posetx$-filter meeting each element of $\{\pi^{-1}(D):D\in\D\}$, then $\pi(G)$ is a $\poset$-filter meeting each element of $\D$. \qed It follows from Lemmas \ref{geq}, \ref{leq}, and \ref{equiv} that Lus$_\kappa(\posetx)$ implies $\aa=\kappa$. In particular, Theorem \ref{thForMain} follows from the following theorem. \thm{thForMainProof}{ Let $\lambda\geq\kappa\geq\omega_2$ be cardinals such that $\cf(\lambda)>\omega_1$ and $\kappa$ is regular. Then it is relatively consistent with ZFC+CH that $2^{\continuum}=\lambda$ and Lus$_\kappa(\posetx)$ holds. } Theorem \ref{thForMainProof} will be proved in Section 4. \section{$\!\!\!\!\!\!\!${\bf.} Consistency of our forcing axioms.} In this section we will prove Theorems \ref{thForMain2Proof} and \ref{thForMainProof}. For Theorem \ref{thForMain2Proof}, start with a model of GCH and extend it by forcing with the countable partial functions from $\lambda$ to $\omega_1$. For Theorem \ref{thForMainProof} start with a model of $$2^\omega=\omega_1+2^{\omega_1}=\lambda$$ and do a countable support iteration of $\posetx$ of length $\kappa$. $\posetx$ is isomorphic to the eventual dominating partial order. For the expert this should suffice. The rest of this section is included for our readers who are not set theorists. For similar proofs see for example Kamo \cite{kam} and Uchida \cite{uc}. We begin with some basic forcing terminology and facts. (See Kunen \cite{Kun}.) For a model $M$ of set theory ZFC and a partial order set $(\poS,\leq)$ a filter $G\su\poS$ is {\em $\poS$-generic over $M$} if $G$ intersects every dense $D\su\poS$ belonging to $M$. The fundamental theorem of forcing states that for every model $M$ of ZFC and every partial order $\poS$ from $M$ there exists model $M[G]$ of ZFC (called an {\em $\poS$-generic extension of $M$}) such that $G$ is $\poS$-generic over $M$ and $M[G]$ is the smallest model of ZFC such that $M\su M[G]$ and $G\in M[G]$. Thus, the simplistic idea for getting MA$_\kappa(\poset)$ is to start with model $M$ of $ZFC$, take $\poset$ from $M$ and look at the model $M[G]$, where $G$ is $\poset$-generic over $M$. Then, $G$ intersects ``all'' dense subsets of $\poset$ and we are done. There are, however, two problems with this simple approach. First, ``all'' dense subsets of $\poset$ means ``all dense subsets from $M$'' and we like to be able to talk about all dense subsets from our universe, i.e., from $M[G]$. Second, our partial order is a set described by some formula as the set having some properties. There is no reason, in general, that the same description will give us the same objects in $M$ and in $M[G]$. The second problem will not give us much trouble. For the generic extensions we will consider, the definition of $\poset$ will give us the same objects in all models we will consider. In the case of the partial order $\posetx$ this will not be the case, but the new orders $\posetx$ will be close enough to the old so that it will not bother us. To take care of the first of the mentioned problems, we will be constructing a Lusin sequence $\la G_\alpha\colon\alpha<\kappa\ra$ by some kind of induction on $\alpha<\kappa$: our final model can be imagined as $N=M[G_0][G_1]\ldots[G_\alpha]\ldots$ and we will make sure that every dense subset $D\in N$ of $\posetx$ is taken care of from some stage $\alpha<\kappa$. We need some more definitions and facts. Given a partial order we say that $p,q$ are {\em compatible} if there exists $r$ such that $r\leq p$ and $r\leq q$. A partial order is {\em well-met} provided for any two elements $p,q$ if $p$ and $q$ are compatible, then they have a greatest lower bound, i.e., there exists $r$ such that $r\leq p$ and $r\leq q$ and for any $s$ if $s\leq p$ and $s\leq q$, then $s\leq r$. Notice that both partial orders $\poset$ and $\posetx$ used in Lemmas \ref{geq} and \ref{leq} are well-met. For the case of $\posetx$ if $(p,E)$ and $(q,F)$ are compatible, then $(p\union q,E\union F)$ is there greatest lower bound. A subset $L$ of a partial order is {\em linked} if any two elements of $L$ are compatible. A partial order is {\em $\omega_1$-linked} provided it is a union of $\omega_1$ linked subsets. Assuming the Continuum Hypothesis note that the poset $\poset$ used in the proof of Lemma \ref{geq} has cardinality $\omega_1$ hence it is $\omega_1$-linked. Note that for any $p\in\poset$ if we define $$L_p=\{(q,F)\in\posetx: q=p\},$$ then $L_p$ is a linked subset of $\posetx$, hence $\posetx$ is also $\omega_1$-linked. A subset $A$ of a partial order is an {\em antichain} if any two elements of $A$ are incompatible. We say that a partial order has the {\em $\omega_2$-chain condition} ({\em $\omega_2$-cc}) if every its antichain has cardinality less than $\omega_2$. Clearly $\omega_1$-linked implies the $\omega_2$-chain condition. Finally we say a partial order is countably closed if any descending $\omega$-sequence $\la p_n:n\in\omega\ra$ (i.e., $p_{n+1}\leq p_n$ all $n$) has a lower bound. Notice that both of our partial orders are countably closed. All partial orders we are going to consider here will be countably closed and will satisfy $\omega_2$-chain condition. In particular, it is known that if the generic extension $M[G]$ of $M$ is obtained with such partial order, then $M[G]$ and $M$ have the same cardinal numbers, the same real numbers, the same countable subsets of real numbers and the same sets $\reals^X$ for any countable set $X\in M$. In particular, $\poset$ will be the same in $M[G]$ as in $M$. Let us also notice that every dense set contains a maximal antichain and if $A$ is a maximal antichain, then $D=\{p:\exists q\in A\;\; p\leq q\}$ is a dense set. Thus a filter $G$ is $\poS$-generic over a model $M$ if and only if it meets every maximal antichain in $M$. \bigskip {\sc Proof of Theorem \ref{thForMain2Proof}.} Take a model $M$ of ZFC+GCH. For a set $X$ in $M$ let \[ \poS_X= \{p\in\poset^X\colon p(x)=\emptyset \mbox{ for all but countably many $x\in X$}\}. \] Define an ordering on $\poS_X$ by $p\leq q$ if and only if $p(x)\leq q(x)$ for every $x\in X$. Now, let $\lambda$ be as in Theorem \ref{thForMain2Proof} and let $G$ be a $\poS_\lambda$ generic over $M$. We will show that MA$_\lambda(\poset)$ holds in $M[G]$. It is easy to see that $\poS_\lambda$ is countably closed. It is also known that $\poS_\lambda$ satisfies $\omega_2$-cc and that $2^{\omega_1}=\lambda$ in $M[G]$. (See Kunen \cite[Ch. VII, Lemma 6.10 and Thm. 6.17]{Kun}.) Now, for $\alpha<\lambda$ let $G_\alpha=\{p(\alpha)\colon p\in G\}$. Then, each $G_\alpha$ is a filter in $\poset$. We will show that for every family $\D$ of dense subsets of $\poset$ with $|\D|<\lambda$ there exists $\alpha<\kappa$ such that $G_\alpha$ intersects every $D$ from $\D$. In order to argue for it we need two more facts about forcing $\poS_X$. (See Kunen \cite[Ch. VII]{Kun}: Thm. 1.4 and 2.1 for (A) and Lemma 5.6 for (B).) \begin{description} \item{(A)} If $X,Y\in M$ are disjoint and $G$ is $\poS_{X\cup Y}$-generic over $M$, then $G_X=G\cap\poS_X$ is $\poS_X$-generic over $M$, $G_Y$ is $\poS_Y$-generic over $M[G_X]$, and $$M[G_X][G_Y]=M[G].$$ \item{(B)} If $A\su M$ then there exists $X\in M$ with $|X|\leq |A|+\omega_1$ such that $A\in M[G_X]$. \end{description} Now, let $G_\lambda$ be $\poS_\lambda$ generic over $M$ and let $\D\in M[G_\lambda]$ be a family of dense subsets of $\poset$ with $|\D|<\lambda$. Let $\A$ be a family of maximal antichains, one contained in each element of $\D$. Then, $|A|\leq\omega_1$ for each $A\in \A$, since $\poset$ satisfies $\omega_2$-cc. So, by (B), there is $X\su\lambda$ from $M$ of cardinality $|\A| \cdot \omega_1<\lambda$ such that $\A\in M[G_X]$. Choose $\alpha\in\lambda\setminus X$. Then since $G_\alpha$ is $\poset$-generic over $M[G_X]$ it follows that $G$ meets each element of $\A$ hence of $\D$. \qed \bigskip Next we prepare to prove Theorem \ref{thForMainProof}. As mentioned in the beginning of the section, we will try to prove it by defining some sequence $\la \poS_\alpha\colon\alpha\leq\kappa\ra$ of partial orders and try to obtain our final model as $N_\kappa=M[G_\kappa]$ where every $G_\alpha$ is an $\poS_\alpha$-generic over an appropriate initial model. This technique is called iterated forcing and needs a few words of introduction. We can define in $M$ an iterated forcing $\la \poS_\alpha:\alpha<\kappa\ra$ by induction on $\alpha$. At successor stages we define $$\poS_{\alpha+1}=\poS_\alpha\times{\posetx}^{M[G_\alpha]}.$$ where ${\posetx}^{M[G_\alpha]}$ is $\posetx$ in the sense of $M[G_\alpha]$. (Since we add new elements of $\reals^\reals$ the partial order $\posetx$ changes as our models increase.) We can't really do it precisely this way, because $\poset_{\alpha+1}$ must be in $M$. However, it is possible to find its approximation, $\hat{\poset}_\alpha$, in $M$, called a name for $\poset_\alpha$, and use this instead. (See Kunen \cite[Ch. VII sec. 5]{Kun}). For limit ordinals $\lambda<\kappa$, define $\poS_\lambda$ to a set of functions $f$ with domain $\lambda$ such that $f|_{\alpha}\in \poS_\alpha$ for each $\alpha<\lambda$ and $f(\alpha)=\ONE$ for all but countable many $\alpha$. Here we use $\ONE$ to denote the largest element of any partial order. Countable support iterations originated with Laver \cite{lav}. For details see Baumgartner \cite{Baum} or Kunen \cite[Ch. VII sec. 7]{Kun}. The proof that follows will involve a basic lemma used to show various generalizations of Martin's Axiom hold for one cardinal up. (See Baumgartner \cite{Baum} and Shelah \cite{sh}). In particular, we will need the following theorem. \thm{thBaum}{ (Baumgartner) Assume the Continuum Hypothesis. Suppose $\la \poS_\alpha:\;\alpha<\kappa\ra$ is a countable support iteration of countably closed well-met $\omega_1$-linked partial orders. Then for every $\alpha\leq\kappa$ we have that $\poS_\alpha$ is countably closed and satisfies the $\omega_2$-chain condition. } Actually we need only a very weak version of this theorem, for example, something analogous to \cite[Theorem VII, 7.3]{Kun} of Kunen. Now, we are ready for the proof of Theorem \ref{thForMainProof}. {\sc Proof of Theorem \ref{thForMainProof}.} Take a model $M$ of ZFC+CH in which $2^{\continuum}=\lambda$, and $\kappa$ is a regular cardinal with $\omega_2\leq\kappa\leq\lambda$. Let $\poS_\alpha$ be a countable support iteration $\{\poset_\alpha\colon\alpha<\kappa\}$, where $\poset_\alpha={\posetx}^{M[G^\alpha]}$ for all $\alpha<\kappa$. Here for $\alpha<\kappa$ let $G^\alpha=G^\kappa|_\alpha$. Then $G^\alpha$ is $\poS_\alpha$-generic filter over $M$. Let $G^\kappa$ be an $\poS_\kappa$-generic filter over $M$. We will show that Lus$_\kappa(\posetx)$ holds in $M[G^\kappa]$. In the model $M[G^\alpha]$ the partial order ${\posetx}^{M[G^\alpha]}$ can be decoded from $\poS_\alpha$ and we can also decode a filter $G_\alpha$ which is ${\posetx}^{M[G^\alpha]}$-generic over $M[G^\alpha]$. We claim that the sequence $\la G_\alpha\colon\alpha\in \kappa\ra$ is a Lusin sequence for $\posetx$ in $M[G^\kappa]$. So, let $D\in M[G^\kappa]$ be a dense subset of $\posetx$ and let $A\in M[G]$ be a maximal antichain contained in $D\su\posetx$. Then, $|A|\leq\omega_1$, since $\posetx$ satisfies $\omega_2$-cc. So, by the fact similar to (B) above, there is $\beta<\kappa$ such that $A\in M[G^\beta]$. Then, for every $\alpha\geq\beta$, the filter $G_\alpha$ is generic over $M[G^\alpha]\supseteq M[G^\beta]$ and so, $G_\alpha$ intersects both $A$ and $D$. 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