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\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}