$MA_{\aleph_0}$ always holds

Given a countable collection of dense subsets of a partial order, we can selected a set $\langle p_n\rangle_{n<\omega}$ such that $p_n$ is in the $n$ -th dense subset, and $p_{n+1}\leq p_n$ for each $n$ . Therefore $CH$ implies $MA$ .

If $MA_\kappa$ then $2^{\aleph_0}>\kappa$ , and in fact $2^\kappa=2^{\aleph_0}$

$\kappa\geq\aleph_0$ , so $2^\kappa\geq 2^{\aleph_0}$ , hence it will suffice to find an surjective function from $\operatorname{P}(\aleph_0)$ to $\operatorname{P}(\kappa)$ .

Let $A=\langle A_\alpha\rangle_{\alpha<\kappa}$ , a sequence of infinite subsets of $\omega$ such that for any $\alpha\neq\beta$ , $A_\alpha\cap A_\beta$ is finite.

Given any subset $S\subseteq\kappa$ we will construct a function $f:\omega\rightarrow\{0,1\}$ such that a unique $S$ can be recovered from each $f$ . $f$ will have the property that if $i\in S$ then $f(a)=0$ for finitely many elements $a\in A_i$ , and if $i\notin S$ then $f(a)=0$ for infinitely many elements of $A_i$ .

Let $P$ be the partial order (under inclusion) such that each element $p\in P$ satisfies:

• $p$ is a partial function from $\omega$ to $\{0,1\}$
• There exist $i_1,\ldots,i_n\in S$ such that for each $j<n$ , $A_{i_j}\subseteq \operatorname{dom}(p)$
• There is a finite subset of $\omega$ , $w_p$ , such that $w_p=\operatorname{dom}(p)-\bigcup_{j<n} A_{i_j}$
• For each $j<n$ , $p(a)=0$ for finitely many elements of $A_{i_j}$

This satisfies ccc. To see this, consider any uncountable sequence $S=\langle p_\alpha\rangle_{\alpha<\omega_1}$ of elements of $P$ . There are only countably many finite subsets of $\omega$ , so there is some $w\subseteq\omega$ such that $w=w_p$ for uncountably many $p\in S$ and $p\upharpoonright w$ is the same for each such element. Since each of these function's domain covers only a finite number of the $A_\alpha$ , and is $1$ on all but a finite number of elements in each, there are only a countable number of different combinations available, and therefore two of them are compatible.

Consider the following groups of dense subsets:

• $D_n=\{p\in P\mid n\in\operatorname{dom}(p)\}$ for $n<\omega$ . This is obviously dense since any $p$ not already in $D_n$ can be extended to one which is by adding $\langle n,1\rangle$
• $D_\alpha=\{p\in P\mid \operatorname{dom}(p)\supseteq A_\alpha\}$ for $\alpha\in S$ . This is dense since if $p\notin D_\alpha$ then $p\cup\{\langle a,1\rangle\mid a\in A_\alpha\setminus\operatorname{dom}(p)\}$ is.
• For each $\alpha\notin S$ , $n<\omega$ , $D_{n,\alpha}=\{p\in P\mid m\geq n \wedge p(m)=0\}$ for some $m<\omega$ . This is dense since if $p\notin D_{n,\alpha}$ then $\operatorname{dom}(p)\cap A_\alpha=A_\alpha\cap\left(w_p\cup \bigcup_{j} A_{i_j}\right)$ . But $w_p$ is finite, and the intersection of $A_\alpha$ with any other $A_i$ is finite, so this intersection is finite, and hence bounded by some $m$ . $A_\alpha$ is infinite, so there is some $m\leq x\in A_\alpha$ . So $p\cup\{\langle x,0\rangle\}\in D_{n,\alpha}$ .

By $MA_\kappa$ , given any set of $\kappa$ dense subsets of $P$ , there is a generic $G$ which intersects all of them. There are a total of $\aleph_0+|S|+(\kappa-|S|)\cdot\aleph_0=\kappa$ dense subsets in these three groups, and hence some generic $G$ intersecting all of them. Since $G$ is directed, $g=\bigcup G$ is a partial function from $\omega$ to $\{0,1\}$ . Since for each $n<\omega$ , $G\cap D_n$ is non-empty, $n\in\operatorname{dom}(g)$ , so $g$ is a total function. Since $G\cap D_\alpha$ for $\alpha\in S$ is non-empty, there is some element of $G$ whose domain contains all of $A_\alpha$ and is $0$ on a finite number of them, hence $g(a)=0$ for a finite number of $a\in A_\alpha$ . Finally, since $G\cap D_{n,\alpha}$ for each $n<\omega$ , $\alpha\notin S$ , the set of $n\in A_\alpha$ such that $g(n)=0$ is unbounded, and hence infinite. So $g$ is as promised, and $2^\kappa=2^{\aleph_0}$ .