\documentclass{amsart} \usepackage{amssymb} \usepackage{amsmath} \usepackage{mathabx} \usepackage{stmaryrd} \usepackage{mathbbol} \theoremstyle{plain} \newtheorem{thm}{Theorem}[section] \newtheorem{cor}[thm]{Corollary} \newtheorem{lem}[thm]{Lemma} \theoremstyle{definition} \newtheorem{defn}[thm]{Definition} \newtheorem{exer}{Exercise} \newtheorem{exam}{Example} \theoremstyle{remark} \newtheorem{rem}[thm]{Remark} \newtheorem{fact}[thm]{Fact} \newtheorem{hyp}[thm]{Hypothesis} \newcommand{\Q}{\mathbb{Q}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\R}{\mathbb{R}} \newcommand{\N}{\mathbb{N}} \newcommand{\bP}{\mathbb{P}} \newcommand{\bQ}{\mathbb{Q}} \newcommand{\bone}{\mathbb{1}} \newcommand{\res}{\restriction} \newcommand{\dom}{\operatorname{dom}} \newcommand{\ran}{\operatorname{ran}} \newcommand{\cof}{\operatorname{cof}} \newcommand{\ot}{\text{o.t.}} \newcommand{\on}{\text{ON}} \newcommand{\card}{\text{CARD}} \newcommand{\ac}{\text{AC}} \newcommand{\zf}{\text{ZF}} \newcommand{\zfc}{\text{ZFC}} \newcommand{\sP}{\mathcal{P}} \newcommand{\sL}{\mathcal{L}} \newcommand{\sA}{\mathcal{A}} \newcommand{\sF}{\mathcal{F}} \newcommand{\sB}{\mathcal{B}} \newcommand{\sU}{\mathcal{U}} \newcommand{\sC}{\mathcal{C}} \newcommand{\sI}{\mathcal{I}} \newcommand{\sM}{\mathcal{M}} \newcommand{\ch}{\text{CH}} \newcommand{\gch}{\text{GCH}} \newcommand{\sh}{\text{SH}} \newcommand{\vb}{\text{var}} \newcommand{\pa}{\text{PA}} \newcommand{\fa}{\text{F}} \newcommand{\con}{\text{CON}} \newcommand{\bz}{\boldsymbol{0}} \newcommand{\lD}{{\Delta}} \newcommand{\lS}{{\Sigma}} \newcommand{\lP}{{\Pi}} \newcommand{\models}{\vDash} \newcommand{\proves}{\vdash} \newcommand{\sg}{\text{sg}} \newcommand{\seq}{\text{Seq}} \newcommand{\lh}{\text{lh}} \newcommand{\pred}{\text{pred}} \newcommand{\hod}{\text{HOD}} \newcommand{\od}{\text{OD}} \newcommand{\trcl}{\text{tr\,cl}} \newcommand{\cl}{\text{cl}} \newcommand{\rank}[1]{|#1|} \newcommand{\llex}{<_{\text{lex}}} \newcommand{\comp}{\parallel} \newcommand{\forces}{\Vdash} \newcommand{\fn}{\text{FN}} \newcommand{\ccc}{\text{c.c.c.}} \newcommand{\cc}{\text{c.c.}} \newcommand{\sat}{\text{sat}} \newcommand{\coll}{\text{coll}} \newcommand{\cub}{\text{c.u.b.}} \newcommand{\conc}{{}^\smallfrown} \newcommand{\dq}{\dot{Q}} \newcommand{\ts}{\bP * \dq} \newcommand{\tss}{\bP_1 * \dot{\bP}_2} \newcommand{\lb}{\llbracket} \newcommand{\rb}{\rrbracket} \newcommand{\supp}{\text{supp}} \newcommand{\itf}{\langle \bP_\beta, \dq_\beta \rangle_{\beta< \alpha}} \newcommand{\ma}{\text{MA}} \newcommand{\inter}{\text{int}} \begin{document} \begin{center} \bf Applications of Martin's Axiom \end{center} \bigskip \section{Products of \ccc{} Spaces} We consider the question of when the product of two \ccc{} topological spaces is \ccc. The next lemma shows the question is the same for partial orders, topological spaces, and compact Hausdorff spaces. Recall that for any partial order $\bP= \langle P, \leq \rangle$ there is an associated topological space $X_P$ on $P$, where a neighborhood base at $p \in P$ consists of the single open set $N_p=\{ q \in P \colon q \leq p\}$. Clearly this space is not (except for very trivial partial orders) even $T_1$. \begin{lem} $(\zfc)$ Let $\kappa$ be an infinite regular cardinal. Then the following are equivalent. \begin{enumerate} \item \label{pr1} There are $\kappa$-c.c.{} partial orders $\bP$, $\bQ$ such that $\bP \times \bQ$ is not $\kappa$-c.c. \item \label{pr2} There $\kappa$-c.c.{} topological spaces $X$, $Y$ such that the product $X \times Y$ is not $\kappa$-c.c. \item \label{pr3} There are $\kappa$-c.c.\ compact Hasdorff spaces $X$, $Y$, such that the product $X \times Y$ is not $\kappa$-c.c.{} \item \label{pr4} There are complete Boolean algebras $\sB$, $\sC$ such that their product $\sB \times \sC$ is not $\kappa$-c.c. \end{enumerate} \end{lem} \begin{proof} Note first that for any partial order $P$, $P$ is $\kappa$-c.c.{} iff $X_P$ is $\kappa$-c.c.{} Also $X_{P\times Q}=X_P\times X_Q$. Suppose first that $P$, $Q$ are $\kappa$-c.c.{} partial orders, but $P \times Q$ is not $\kappa$-c.c.{} Then $X_P$, $X_Q$ are $\kappa$-c.c.{} topological spaces. Since $P \times Q$ is not $\kappa$-c.c.{}, neither is $X_{P \times Q}=X_P \times X_Q$. So (\ref{pr1}) implies (\ref{pr2}). To see (\ref{pr2}) implies (\ref{pr1}), let $X$, $Y$ be two topological spaces which are $\kappa$-c.c., but $X \times Y$ is not $\kappa$-c.c.\ Let $P$ be the collection of non-empty open sets in $X$ ordered by $U \leq V$ iff $U \subseteq V$. Likewise define $Q$. Clearly $P$, $Q$ are partial orders. $P$ and $Q$ are $\kappa$-c.c.\ partial orders since a $\kappa$ antichain in $P$ would be a $\kappa$ collection of pairwise disjoint open sets in $X$. Let $\{ W_\alpha\}_{\alpha< \kappa}$ be a $\kappa$ family of pairwise disjoint open sets in $X \times Y$. Without loss of generality we may assume that $W_\alpha=U_\alpha \times V_\alpha$ is a product of basic open sets. Let $p_\alpha=(U_\alpha, V_\alpha) \in P\times Q$. The $p_\alpha$ form an antichain in $P \times Q$ (since if $(U, V) \leq (U_\alpha,V_\alpha), (U_\beta, V_\beta)$ then $U \subseteq U_\alpha \cap U_\beta$, $V \subseteq V_\alpha \cap V_\beta$, so $U \times V \subseteq W_\alpha \cap W_\beta$). So, $P \times Q$ is not $\kappa$-c.c. (\ref{pr4}) implies (\ref{pr1}) trivially, since Boolean algebras are partial orders. For the other direction, let $P$, $Q$ be as in (\ref{pr1}). Let $\sB$ be the regular open algebra of $P$, and likwise define $\sC$. Thus, $\sB$ and $\sC$ are complete Boolean algebras. Since every element of $\sB$ (i.e., every regular open subset of $X_P$) contains a basic open subset of $X_P$, it is clear that $\sB$ is $\kappa$-c.c., and likewise for $Q$. For $p \in P$ let $B_p=\inter \cl (N_p) \in \sB$ and likewise define $C_q \in \sC$ for $q \in Q$. Let $\{ (p_\alpha,q_\alpha)\}_{\alpha< \kappa}$ be an antichain in $P \times Q$. Let $B_\alpha=B_{p_\alpha}$, and likewise for $C_\alpha$. Then the $(B_\alpha, C_\alpha)$ form an antichain in $\sB \times \sC$. For suppose $(B,C) \leq (B_\alpha,C_\alpha), (B_\beta,C_\beta)$. We may assume $B=B_p$, $C=C_q$. So, $B_p \subseteq B_{p_\alpha} \cap B_{p_\beta}$. This says $N_{p_\alpha}$ and $N_{p_\beta}$ are dense below $p$. Thus, there is an $r \in P$ with $r \leq p_\alpha, p_\beta$. Similarly, $q_\alpha$ and $q_\beta$ are compatible. Hence $(p_\alpha, q_\alpha)$ and $(p_\beta, q_\beta)$ are compatible, a contradiction. (\ref{pr3}) trivially implies (\ref{pr2}). To finish we show that (\ref{pr1}) implies (\ref{pr3}). Given $P$ and $Q$, let $\sB$ and $\sC$ again be their completions. Let $X'$ be the Stone space of the Boolean algebra $\sB$, that is, $X'$ is the collection of ultrafilters on $\sB$. $X'$ is a topological space with basic open sets of the form $N_V=\{ \sU \in X' \colon V \in \sU \}$ for $V$ a regular open set in $X_P$. $X'$ is clearly Hausdorff, and it is a standard fact (see the following exercise) that $X'$ is compact. $X'$ is also $\kappa$-c.c., for if $N_{V_\alpha}$ were pairwisw disjoint open sets in $X'$, then the $V_\alpha$ would be pairwise disjoint open sets in $\sB$, contradicting $\sB$ being $\kappa$-c.c.\ Likewise define $Y'$. To see $X' \times Y'$ is not $\kappa$-c.c., let $(p_\alpha,q_\alpha)$ again be a $\kappa$ size antichain in $P \times Q$. Let $B_\alpha$, $C_\alpha$ be as before, and consider the basic open sets $N_{B_\alpha} \times N_{C_\alpha}$ in $X' \times Y'$. These are pairwise disjoint since if $N_{B_\alpha} \cap N_{B_\beta} \neq \emptyset$ then $B_\alpha \cap B_\beta \neq \emptyset$, and then (as argued above) $p_\alpha$ and $p_\beta$ are compatible in $P$ (and likewise for $Q$). \end{proof} We showed before that if there is a Suslin tree $T$, then $T$ gives a partial order which is \ccc{} (i.e., $T$ with the reverse of the tree ordering) but for which $T \times T$ is not \ccc{} We now show that $\ch$ also implies the non-productivity of \ccc{} \begin{thm} [Galvin, Laver] Assume $\zfc+\ch$. Then there are two $\ccc$ partial orders $P$, $Q$ whose product $P \times Q$ is not \ccc{} \end{thm} \begin{cor} Assuming $\ch$, there are two \ccc{} compact Hausdorff spaces whose product is not \ccc{} \end{cor} \begin{proof} Suppose $f \colon (\omega_1)^2 \to \{0, 1\}$ is a partition of the pairs of countable ordinals. For $i \in \{0,1\}$ we define the partial order $P_i$ to consist of all finite $p \subseteq \omega_1$ such that $f''p^2 = \{ i \}$ (that is, $P_i$ consists of finite sets which are homogeneous for color $i$). We order $P_i$ by: $p \leq q$ iff $p \supseteq q$. No matter how we choose $f$, the product $P_0 \times P_1$ will not be \ccc, since the elements $(\alpha, \alpha)$ for $\alpha < \omega_1$ form an antichain. It remains to show that we can choose $f$ so that $P_0$ and $P_1$ are \ccc{} Let (using $\ch$) $\{ F_\alpha \}_{\alpha < \omega_1}$ enumerate all $\omega$ sequences $F_\alpha=(F_\alpha^0, F_\alpha^1, \dots, F_\alpha^n, \dots)$ of pairwise disjoint finite subsets of $\omega_1$. Suppose we have defined $f(\alpha,\beta)$ when $\max \{ \alpha, \beta \} < \gamma$. We define $f(\alpha, \gamma)$ for $\alpha < \gamma$ as follows. Let $S_n$, $n < \omega$, enumerate all objects of the form $S_n= \langle F_{\alpha_n}, X_n ,i_n \rangle$ such that \begin{enumerate} \item $\alpha_n < \gamma$, $i \in \{ 0, 1\}$, and $X_n$ is a finite subset of $\gamma$. \item $X_n \cap \bigcup_k F_{\alpha_n}^k= \emptyset$ and there are infinitely many $k \in \omega$ such that $f'' (F_{\alpha_n}^k \cup X_n)=\{ i\}$. \end{enumerate} By diagonalizing, there is a sequence $\{ Y_l \}_{l \in \omega}$ with each $Y_l$ of the form $F_{\alpha_n}^k$ for some $n=n(l)$, $k=k(l)$, such that the $Y_l$ are pairwise disjoint, for each $n$ there are infinitely many $l$ such that $n(l)=n$, and $f'' (Y_l \cup X_{n(l)})=\{i_{n(l)})\}$. For $\alpha \in Y_l$, define $f(\alpha,\gamma)=i_{n(l)}$. The definition of $f$ gives us the following property. (*): for any $\alpha < \gamma$, finite $X \subseteq \gamma$, and $i \in \{ 0,1\}$, if $\sup(\cup F_\alpha)) < \gamma$, $X \cap (\cup F_\alpha) = \emptyset$, and there are infiniely many $n$ such that $f'' (F_\alpha^n \cup X)^2=\{ i\}$, then there are infinitely many $n$ such that $f'' (F_\alpha^n \cup X)^2=\{ i\}$ and $f''(F_\alpha^n \cup \{ \gamma\})^2 =\{ i \}$. To see this works, suppose that $\{ p_\alpha \}_{\alpha < \omega_1}$ was an $\omega_1$-antichain for $P_0$ (the other case being similar). Thinning to $\lD$-system, we may assume that $p_\alpha= r \cup G_\alpha$, where $G_\alpha \cap G_\beta= \emptyset$ for $\alpha \neq \beta$. To get a contradiction it suffices to find $\alpha \neq \beta$ such that $f''(G_\alpha \cup G_\beta)=\{ 0 \}$. Consider the first $\omega$ many elements, and fix $\alpha< \omega_1$ such that $F_\alpha=(G_0,G_1,\dots, G_n, \dots)$. Let $\delta >\alpha$ and $\delta > \sup \bigcup_n G_n$. Fix one of the $G_\beta$ with $G_\beta \cap \delta= \emptyset$ (which we can do as the $G_\beta$ are pairwise disjoint). Say $G_\beta=\{ \gamma_1,\dots, \gamma_m\}$. Using (*) $m$ times (at step $p$ we use $X=\{ \gamma_1,\dots, \gamma_{p-1}\}$, $\gamma=\gamma_p$) we thin the $G_n$ sequence to an infinite subsequence $(G'_0,G'_1,\dots)$ such that $f''(G'_n \cup G_\beta)^2=\{ 0 \}$. This contradicts the $p_\alpha$ being an antichain. \end{proof} We now show that $\ma + \neg \ch$ implies that the product of two \ccc\ partial orders is \ccc\ \begin{defn} A partial order $\bP$ is strongly \ccc\ if whenever $A \subseteq P$ with $|A|=\omega_1$, then there is a $B \subseteq A$ with $|B|=\omega_1$ such that for any finite $\{ p_1,\dots, p_n \} \subseteq B$ there is a $p \in P$ with $p\leq p_1,\dots, p_n$. \end{defn} \begin{lem} \label{ccct1} Assume $\zfc+ \ma +\neg \ch$. Then every \ccc\ partial order is strongly \ccc{} \end{lem} \begin{proof} Let $P$ be \ccc, and $A =\{ p_\alpha\}_{\alpha< \omega_1}$. First we claim that there is a $p \in A$ such that any $q \leq p$ is compatible with $p_\alpha$ for uncountably many $\alpha$. For if not, then we could get $q_{\alpha_0} \leq p_{\alpha_0}$ (where $\alpha_0=0$) and $\alpha_1$ such that $q_{\alpha_0}$ is incompatible with all $p_\beta$ for $\beta \geq \alpha_1$. We could then get $q_{\alpha_1} \leq p_{\alpha_1}$ and $\alpha_2$ so that $q_{\alpha_1}$ is incompatible with all $p_\beta$ for $\beta \geq \alpha_2$. Continuing we define an $\omega_1$ antichain $\{ q_{\alpha_i} \}_{i < \omega_1}$. We may assume $p_0$ is such that any $q \leq p_0$ is compatible with $\omega_1$ many $p_\alpha$. Let $Q$ be the partial order consisting of all finite subsets $u=\{ q_1,\dots,q_n\} \subseteq P$ which contain $p_0$ and which are compatible, that is, there is a $q \in P$ with $q \leq q_1,\dots, q_n$. We order $Q$ by $u \leq v$ iff $u \supseteq v$. We claim that $Q$ is \ccc\ For suppose $\{ u_\alpha \}_{\alpha< \omega_1}$ were an uncountable antichain in $Q$. For each $\alpha < \omega_1$, let $r_\alpha \in P$ with $r_\alpha$ extending all of the elements of $u_\alpha$. Since $P$ is \ccc, there are $\alpha \neq \beta$ such that $r_\alpha \parallel r_\beta$. Let $r \leq r_\alpha, r_\beta$. Then $r$ extends all elements of $u_\alpha$ and $u_\beta$, so $u_\alpha \cup u_\beta \in Q$, a contradiction. Thus, $Q$ is \ccc\ For each $\alpha < \omega_1$, let $D_\alpha \subseteq Q$ be those $u \in Q$ which contain a $p_\beta$ for some $\beta > \alpha$. Then $D_\alpha$ is dense in $Q$. For given $u \in Q$, let $q \in P$ extend all of the elements of $u$. Since $p_0 \in u$, $q$ is compatiblw with $\omega_1$ many of the $p_\beta$. Let $\beta > \alpha$ with $p_\beta \parallel q$. Then $u \cup\{ p_\beta \} \in D_\alpha$. By $\ma$ (and the fact that $\omega_1 < 2^\omega$ from $\neg \ch$) there is a filter $G$ on $Q$ meeting all of the $D_\alpha$. Let $B= \cup G$, so $B \subseteq A$. Since $G$ meets all of the $D_\alpha$, $|G|=\omega_1$. Since $G$ is a filter, for any finite $\{ u_1,\dots,u_n \} \subseteq G$, there is a $p \in G$ extending all of the elements in $\bigcup_{i=1}^n u_i$. It follws that finite number of elements from $B$ have a common extension in $P$. \end{proof} \begin{lem} \label{ccct2} If $P$ is strongly \ccc and $\Q$ is \ccc, then $P \times Q$ is \ccc\ \end{lem} \begin{proof} Suppose $(p_\alpha,q_\alpha)$ were an uncountable antichain in $P \times Q$. Since $P$ is strongly \ccc, we may thin the sequence so that for any finite subset $u$ of $\{ p_\alpha \}_{\alpha< \omega_1}$ the elements of $u$ have a common extension (actually, all we need is that any two are compatible). Since $Q$ is \ccc, get now $\alpha < \beta$ such that $q_\alpha \parallel q_\beta$. Then $p_\alpha \parallel p_\beta$ and $q_\alpha \parallel q_\beta$, so $(p_\alpha,q_\alpha) \parallel (p_\beta, q_\beta)$, a contradiction. \end{proof} \begin{thm} \label{mapr} Assume $\zfc+\ma+\neg \ch$. Then the product of two \ccc{} partial orders is \ccc\ Likewise the product of two \ccc\ topological spaces is \ccc\ \end{thm} \begin{proof} Immediate from lemmas~\ref{ccct1} and \ref{ccct2}. \end{proof} As a consequence, we have the following topological result. \begin{thm} Assume $\zfc+\ma+\neg \ch$. Then if $(X_\alpha, \tau_\alpha)_{\alpha \in I}$ are \ccc\ topological spaces, then their product $X=\prod_{\alpha \in I} X_\alpha$ is also \ccc\ \end{thm} \begin{proof} Suppose $\{ V_\alpha \}_{\alpha < \omega_1}$ is an uncountable antichain (i.e., pairwise disjoint open sets) in $X$. Without loss of generality we many assume the $V_\alpha$ are basic open, say $V_\alpha= U_{\beta^\alpha_1} \times \dots \times U_{\beta^\alpha_n}$, where $U_{\beta^\alpha_i}$ is open in $X_{\beta_i}$ and $n$ depends on $\alpha$. We call $\{ \beta_1^\alpha, \dots, \beta_n^\alpha \}$ the support of $V_\alpha$. We may thin the antichain so that the supports form a $\lD$-system with say root $r=\{ \beta_1,\dots, \beta_m\}$. then clearly $\{ W_\alpha \}$ also must form an antichain, where $W_\alpha$ is the product of the $U_{\beta_i^\alpha}$ for $\beta_i^\alpha \in r$. This is a contradiction, as theorem~\ref{mapr} gives that $X_{\beta_1} \times \dots \times X_{\beta_m}$ is \ccc \end{proof} \section{Almost Disjoint Sets} If $x, y \subseteq \omega$, we say $x$ and $y$ are {\em almost disjoint} if $x \cap y$ is finite. \begin{lem} There is a $2^\omega$ sequence $\{ x_\alpha\}_{\alpha < 2^\omega}$ of infinite subsets of $\omega$ such that if $\alpha \neq \beta$ then $x_\alpha$, $x_\beta$ are pairwise disjoint. \end{lem} \begin{proof} Let $\{ y_\alpha \}_{\alpha<2^\omega}$ be distinct infinite subsets of $\omega$. Let $f \colon \omega^{< \omega} \to \omega$ be one-to-one. Define $x_\alpha =\{ f(y_\alpha \cap n)\colon n \in \omega \} \subseteq \omega$. Let $n \in y_\alpha - y_\beta$, say Then for any $m > \max \{ f(y_\alpha \cap 1), \dots, f(y_\alpha \cap n) \}$, if $m \in x_\alpha$ then $m \notin x_\beta$. \end{proof} \begin{thm} Assume $\zfc + \ma$. Then $\forall \kappa < 2^\omega\ (2^\kappa=2^\omega)$. In particular, $2^\omega$ is regular. \end{thm} \begin{proof} We use the almost disjoint sets forcing. Fix a sequence $\{ x_\alpha \}_{\alpha < 2^\omega}$ of infinite, almost disjoint subsets of $\omega$. Fix $\kappa< 2^\omega$, and we use this sequence to code subsets of $\kappa$. Let $A \subseteq \kappa$. We claim that there is an $x \subseteq \omega$ such that for all $\alpha < \kappa$ if $\alpha \in A$ then $x \cap x_\alpha$ is finite, and if $\alpha \notin A$ then $x \cap x_\alpha$ is infinite. This clearly suffices to show that $2^\kappa \leq 2^\omega$. To see the claim, consider the partial order $\bP$ which consists of all pairs $p=( s, F)$ where $s \in 2^{< \omega}$ and $F \subseteq A$ is finite. We regard $s$ as the characteristic function of the subset we are trying to build. If $p'=(s',F')$, then we define $p' \leq p$ iff $s'$ extends $s$, $F \subseteq F'$, and for all $n \in \dom(s')-\dom(s)$, if $s'(n)=1$ then for all $\alpha \in F$ we have $x_\alpha(n)=0$. To see $\bP$ is \ccc\ simply note that $(p,F)$ is compatible with $(p',F')$ if $s=s'$, and there are only countably many choices for $s$. For $\alpha \in A$, let $D_\alpha =\{ (s,F) \colon x_\alpha \in F \}$. Clearly $D_\alpha$ is dense as we may extend any $(s,F)$ to $(s, F\cup \{ x_\alpha \})$. For $\alpha < \kappa$, $\alpha \notin A$, and $k\in \omega$, let $D_{\alpha,k}=\{ (s,F) \colon \exists m \geq k\ (m \in x_\alpha \wedge s(m)=1) \}$. To see this is dense, note that $x_\alpha$ is almost disjoint from all the $x_\beta$ for $\beta \in F$. Hence there is am $m \geq k$ such that $m \in x_\alpha - \bigcup_{\beta \in F} x_\beta$. Extend $s$ to $s'$ where $s'(m)=1$ and $s'(l)=0$ for all other $l \in \dom(s')-\dom(s)$. Then $(s',F) \leq (s,F)$ and $(s',F) \in D_{\alpha,k}$. By $\ma$, let $G$ be a filter on $\bP$ meeting all of the $D_\alpha$ for $\alpha \in A$ and all of the $D_{\alpha,k}$ for $\alpha \notin A$. Let $x= \bigcup \{ s \colon \exists F \ (s,F) \in G \}$ be the real determined from $G$. Suppose first that $\alpha \in A$. Since $D_\alpha$ is dense, let $(s,F) \in G \cap D_\alpha$. Suppose $(s',F') \in G$ as well. Since $G$ is a filter, there is some $(s'', F'') \in G$ with $(s'',F'') \leq (s,F), (s',F')$. Thus, for all $n \in \dom(s'')-\dom(s)$ we have $s''(n)=1 \rightarrow (n \notin x_\alpha)$. Hence $x \cap x_\alpha \subseteq \{ n \colon s(n)=1 \}$, so $x$ is almost disjoint from $x_\alpha$. Suppose now that $\alpha \notin A$. For any $k \in \omega$ there is an $(s,F) \in G \cap D_{\alpha,k}$. Thus there is an $m \in x \cap x_\alpha$ with $m \geq k$. This shows $x \cap x_\alpha$ is infinite. \end{proof} \section{Measure and Category} We first show that $\ma + \neg \ch$ implies the $< 2^\omega$ additivity of measure and category. First we consider category. Recall a subset of a Polish space is meager if it is contained in the countable union of closed nowhere dense sets. A set $A$ is comeager if its complement is meager, equivalently if $A$ contains a countable intersection of dense open sets. The Baire category theorem (for complete metric spaces) says that every comeager set is dense, in particular non-empty. It is immediate that a countable union of meager sets is meager (equivalently, a countable intersection of comeager sets is comeager). The following result shows that with $\ma$ we may extend this to $< 2^\omega$ size unions. \begin{thm} Assume $\ma$. Then the union of $< 2^\omega$ meager sets in a Polish space is meager. \end{thm} \begin{proof} Let $\kappa < 2^\omega$ and fix dense open sets $U_\alpha$, $\alpha < \kappa$. We show that there are dense open sets $V_n$ such that $\bigcap_n V_n \subseteq \bigcap_{\alpha< \kappa} U_\alpha$. Fix a base $\{ B_n \}_{n \in \omega}$ for the Polish space. Let $\bP$ consist of all finite sequences $p= \langle (W_0,F_0),\dots,(W_n,F_n) \rangle$ where each $F_i \subseteq \kappa$ is finite, $W_i$ is a finite union of basic open sets, and $W_i \subseteq \bigcap_{\alpha \in F_i} U_\alpha$. If $p'=\langle (W''_0,F_0),\dots,(W'_m,F'_m) \rangle$ then we define $p' \leq p$ iff $m \geq n$ and for all $i \leq n$ we have $F'_i \supseteq F_i$ and $W'_i \supseteq W_i$. $\bP$ is \ccc{} since if $m=n$ and $W_i=W'_i$, then $p$ is compatible with $p'$. For each $\alpha < \kappa$, let $D_\alpha$ be those $p$ such that $\alpha \in F_i$ for some $i$. Clearly each $D_\alpha$ is dense. For each $i$ and basic open set $B_j$, let $E_{i,j}$ be those $p$ such that $W_n \cap B_j \neq \emptyset$. Easily each $E_{i,j}$ is also dense. From $\ma$, let $G$ be a filter on $\bP$ meeting all of these dense sets. Define $V_i$ to be the union of all the $W_i$ such that $(W_i,F_i)$ is the $i^{\text{th}}$ coordinate of a $p \in G$. Since $G$ meets each $E_{i,j}$, each $V_i$ is dense open. Let $x \in \bigcap_i V_i$. Fix $\alpha < \kappa$ and we show that $x \in W_\alpha$. Let $p \in G \cap D_\alpha$, say $p= \langle (W_0,F_0),\dots,(W_n,F_n) \rangle$ with $\alpha \in F_i$. If $q=\langle (W'_0,F'_0),\dots,(W'_m,F'_m) \rangle$ is also in $G$, then (if $m \geq i$) $W'_i \subseteq U_\alpha$ as well, since $p$ is compatible with $q$ (any common extension will have $i^{\text{th}}$ coordinate $(W, F)$ where $\alpha \in F$ and $W \supseteq W_i \cup W'_i$). Thus, $x \in V_i \subseteq U_\alpha$. \end{proof} We now prove the analogous result for measure. \begin{thm} \label{addm} Assume $\ma$. Let $\mu$ be a $\sigma$-finite Borel measure on a separable metric space $X$. Then the union of $< 2^\omega$ subsets of $X$ of $\mu$ measure $0$ has $\mu$ measure $0$. \end{thm} \begin{proof} We may assume that $\mu$ is finite, that is, a probability measure. Recall that any Borel probability measure on a metric space is regular, that is, for every Borel set $B \subseteq X$ and every $\epsilon >0$, there is a closed set $F$ and an open set $U$ with $F \subseteq B \subseteq U$ and $\mu(U-F)< \epsilon$. Recall also that a set $A \subseteq X$ is defined to have measure $0$ if it is contained in a Borel set of measure $0$, and a set is measurable if is equal to a Borel set modulo a set of measure $0$. The measurable sets form a $\sigma$-algebra containing the Borel sets, and $\mu$ extends to a countably additive measure on this algebra. Fix $\kappa< 2^\omega$, and suppose $A_\alpha$, for $\alpha<\kappa$, are measure zero sets. Fix $\epsilon >0$, and we show that there is an open set $U$ containing $A=\bigcup_{\alpha<\kappa}A_\alpha$ of measure $< \epsilon$. Let $\bP$ consist of all open sets in $X$ of measure $< \epsilon$. Define $p \leq q$ iff $p \supseteq q$. We show that $\bP$ is \ccc{} Suppose $\{ p_\alpha \}_{\alpha < \omega_1}$ were an uncountable antichain in $\bP$. By thinning, we may assume that for some $\epsilon_1 < \epsilon$ that $\mu(p_\alpha)<\epsilon_1$ for all $\alpha$. For each $\alpha$ there is a finite union of basic open sets (with respect to a fixed countable base for $X$) $U_\alpha \subseteq p_\alpha$ such that $\mu(p_\alpha -U_\alpha)< (\epsilon-\epsilon_1)/2$. Fix $\alpha \neq \beta$ such that $U_\alpha=U_\beta$. Then $\mu(p_\alpha \cup p_\beta) \leq \mu(U_\alpha)+ \mu(p_\alpha-U_\alpha)+\mu(p_\beta-U_\alpha) < \epsilon$. For each $\alpha< \kappa$, let $D_\alpha$ be those $p_\alpha$ containing $A_\alpha$. Easily each $D_\alpha$ is dense. By $\ma$, let $G$ be a filter on $\bP$ meeting all of the $D_\alpha$. Let $V$ be the union of all the open sets $p \in G$. Clearly $V$ contains each $A_\alpha$. We claim that $\mu(V) \leq \epsilon$. Since $G$ is a filter, any finite union of open sets $p \in G$ has measure $< \epsilon$. By countable additivity, any countable union of sets $p \in G$ has measure $\leq \epsilon$. However, in any second countable space the union of a family of open sets is given by a union of a countable subfamily. Hence, $\mu(V)\leq \epsilon$. \end{proof} As a corollary we get a little more: \begin{cor} Assume $\ma$. Let $\mu$ be a $\sigma$-finite Borel measure on a separable metric space $X$. Then the algebra $\sM$ of measurable sets is closed under $\kappa < 2^\omega$ unions and intersections for all $\kappa < 2^\omega$. Also $\mu$ is $\kappa$-additive for all $\kappa < 2^\omega$. \end{cor} \begin{proof} We prove this by induction on $\kappa <2^\omega$. Suppose $\{A_\alpha \}_{\alpha< \kappa}$ are given with $A_\alpha \in \sM$. We show that $A=\bigcup_{\alpha< \kappa} A_\alpha$ is in $\sM$. Let $B_\alpha=A_\alpha- \bigcup_{\beta < \alpha} A_\beta$ be the disjointification of the $A_\alpha$. By induction, each $B_\alpha$ is in $\sM$. From $\sigma$-finiteness, only countably many of the $B_\alpha$ can have non-zero $\mu$ measure. The union of the rest has $\mu$ measure $0$ by theorem~\ref{addm}. Thus, $A$ is the union of countably many sets in $\sM$ together with a measure $0$ set (which is by definition measurable). Thus $A \in \sM$. This show $\sM$ is $< 2^\omega$-additive. If the $A_\alpha$ are pairwise disjoint, then again only countably many of them can have non-zero $\mu$ measure, and the union of the rest has measure $0$. The countable additivity of $\mu$ then gives that $\mu(A)= \sum \mu(A_\alpha)$. \end{proof} \section{Random and Generic Reals} We introduce two new forcings for adding a real, one corresponding to measure and the other to category, $\bP_m$ and $\bP_c$. The category version turns out to be isomorphic (actually, have an isomorphic completion) to ordinary Cohen forcing, so this is not really a new forcing but just a different point of view. In the measure case, $\bP_m$ will be a new forcing, which we call the random real forcing. Let $X$ be a Polish space (e.g., $X=\R$), and $\sI$ an ideal on $X$. For example, $\sI$ could be the ideal $\sI_m$ of Lebesgue measure $0$ sets, or the ideal $\sI_c$ of meager sets. We define an equivalence relation $\sim_I$ of the Borel subsets of $X$ by $A \sim_I B$ iff $A \triangle B \in \sI$. For $B \subseteq X$ a Borel set, we let $[B]_{\sI}$, or just $[B]$ if $\sI$ is understood, denote the equivalence class of $B$. We let $\bP_{\sI}$ be the partial order whose elements are the equivalence classes $[B]$ ($B$ a Borel subset of $X$) with $B \notin \sI$, ordered by $[A] \leq [B]$ iff $A-B \in \sI$. This is easily well-defined and a partial order. In fact, $\bP_{\sI}$ is actually a Boolean algebra under the obvious operations (for example $[A]+[B]=[A \cup B]$, $-[A]=[X-A]$) provided we add $0=[\emptyset]$ back in. We say $\sI$ is \ccc{} if the Boolean algebra $\bP_{\sI}$ is \ccc{} In other words, there is no $\omega_1$ sequence $A_\alpha$ of Borel subsets of $X$ such that for $\alpha \neq \beta$ we have $A_\alpha \triangle A_\beta \in \sI$. For a general ideal, this will not be a complete Boolean algebra, but we have the following. \begin{lem} If $\sI$ is \ccc{} and countably-additive then $\bP_{\sI}$ is a complete Boolean algebra. \end{lem} \begin{proof} Consider a collection $\{ [A_\alpha] \}_{\alpha < \kappa} \}$ of elements of the Boolean algegra. We show that the least upper bound $\sum [A_\alpha]$ exists. Let $\sB$ be a maximal collection subject to being an antichain in $\sP_{\sI}$ and for each $b \in \sB$ there is an $\alpha$ such that $b \leq [A_\alpha]$. Since $\bP_{\sI}$ is \ccc{} we may write $\sB= \{ [B_n]\}_{n \in \omega}$. Let $B= \bigcup_n B_n$ (we have implicitly chosen representatives for the equivalence classes). We first claim that $[B]=\sum [B_n]$. Clearly $[B_n] \leq [B]$ for each $n$. On the other hand, if $[B']$ is also such that $[B_n] \leq [B']$ for each $n$, then by countable additivity of $\sI$ we get that $B=\bigcup_n B_n \subseteq B' \mod \sI$. Hence $[B]$ is the least upper bound of the $[B_n]$. Next we show that $[B]$ is the least upper bound of the $[A_\alpha]$. Sice for each $n$, $[B_n] \leq [A_\alpha]$ for some $\alpha$, clearly any upper bound for the $[A_\alpha]$ is an upper bound for the $[B_n]$. So, it suffices to show that $B$ is an upper bound for the $[A_\alpha]$. Fix $\alpha$, and we show that $[A_\alpha] \leq [B]$. If not, then $A_\alpha -B \notin \sI$. Then $[A_\alpha -B]$ cound be added to the $[B_n]$, contradicting the maximality of $\sB$. \end{proof} \begin{exer} Show that for any Polish space $X$, the category forcing $\bP_c$ is \ccc{} show that for any $\sigma$-finite Borel measure $\mu$ on a separable metric space $X$, the corresponding measure forcing $\bP_m$ is \ccc{} \end{exer} \begin{cor} $\bP_c$ and $\bP_m$ are complete Boolean albegras. \end{cor} \end{document}