# Martin’s axiom is consistent

If $\kappa$ is an uncountable strong limit cardinal such that for any $\lambda<\kappa$, $\kappa^{\lambda}=\kappa$ then it is consistent  that $2^{\aleph_{0}}=\kappa$ and MA. This is shown by using finite support iterated forcing to construct a model of ZFC in which this is true. Historically, this proof was the motivation for developing iterated forcing.

## Outline

The proof uses the convenient fact that $MA_{\kappa}$ holds as long as it holds for all partial orders smaller than $\kappa$. Given the conditions on $\kappa$, there are at most $\kappa$ names for these partial orders. At each step in the forcing  , we force with one of these names. The result is that the actual generic subset we add intersects every dense subset of every partial order.

## Construction of $P_{\kappa}$

$\hat{Q}_{\alpha}$ will be constructed by induction  with three conditions: $|P_{\alpha}|\leq\kappa$ for all $\alpha\leq\kappa$, $\Vdash_{P_{\alpha}}\hat{Q}_{\alpha}\subseteq\mathcal{M}$, and $P_{\alpha}$ satisfies the ccc. Note that a partial ordering on a cardinal $\lambda<\kappa$ is a function  from $\lambda\times\lambda$ to $\{0,1\}$, so there are at most $2^{\lambda}<\kappa$ of them. Since a canonical name for a partial ordering of a cardinal is just a function from $P_{\alpha}$ to that cardinal, there are at most $\kappa^{2^{\lambda}}\leq\kappa$ of them.

At each of the $\kappa$ steps, we want to deal with one of these possible partial orderings, so we need to partition   the $\kappa$ steps in to $\kappa$ steps for each of the $\kappa$ cardinals less than $\kappa$. In addition, we need to include every $P_{\alpha}$ name for any level. Therefore, we partion $\kappa$ into $\langle S_{\gamma,\delta}\rangle_{\gamma,\delta<\kappa}$ for each cardinal $\delta$, with each $S_{\gamma,\delta}$ having cardinality $\kappa$ and the added condition that $\eta\in S_{\gamma,\delta}$ implies $\eta\geq\gamma$. Then each $P_{\gamma}$ name for a partial ordering of $\delta$ is assigned some index $\eta\in S_{\gamma,\delta}$, and that partial order will be dealt with at stage $Q_{\eta}$.

Formally, given $\hat{Q}_{\beta}$ for $\beta<\alpha$, $P_{\alpha}$ can be constructed and the $P_{\alpha}$ names for partial orderings of each cardinal $\delta$ enumerated by the elements of $S_{\alpha,\delta}$. $\alpha\in S_{\gamma,\delta}$ for some $\gamma_{\alpha}$ and $\delta_{\alpha}$, and $\alpha\geq\gamma_{\alpha}$ so some canonical $P_{\gamma_{\alpha}}$ name for a partial order $\hat{\leq}_{\alpha}$ of $\delta_{\alpha}$ has already been assigned to $\alpha$.

Since $\hat{\leq}_{\alpha}$ is a $P_{\gamma_{\alpha}}$ name, it is also a $P_{\alpha}$ name, so $\hat{Q}_{\alpha}$ can be defined as $\langle\delta_{\alpha},\hat{\leq}_{\alpha}\rangle$ if $\Vdash_{P_{\alpha}}\langle\delta_{\alpha},\hat{\leq}_{\alpha}\rangle$ satisfies the ccc and by the trivial partial order $\langle 1,\{\langle 1,1\rangle\}\rangle$ otherwise. Obviously this satisfies the ccc, and so $P_{\alpha+1}$ does as well. Since $\hat{Q}_{\alpha}$ is either trivial or a cardinal together with a canonical name, $\Vdash_{P_{\alpha}}\hat{Q}_{\alpha}\subseteq\mathcal{M}$. Finally, $|P_{\alpha+q}|\leq\sum_{n}|\alpha|^{n}\cdot(\operatorname{sup}_{i}|\hat{Q}_{i}% |)^{n}\leq\kappa$.

## Proof that $MA_{\lambda}$ holds for $\lambda<\kappa$

### Lemma: It suffices to show that $MA_{\lambda}$ holds for partial order with size $\leq\lambda$

###### Proof.

Suppose $P$ is a partial order with $|P|>\kappa$ and let $\langle D_{\alpha}\rangle_{\alpha<\lambda}$ be dense subsets of $P$. Define functions $f_{i}:P\rightarrow D_{\alpha}$ for $\alpha\kappa$ with $f_{\alpha}(p)\geq p$ (obviously such elements exist since $D_{\alpha}$ is dense). Let $g:P\times P\rightarrow P$ be a function such that $g(p,q)\geq p,q$ whenever $p$ and $q$ are compatible  . Then pick some element $q\in P$ and let $Q$ be the closure of $\{q\}$ under $f_{\alpha}$ and $g$ with the same ordering as $P$ (restricted to $Q$).

Since there are only $\kappa$ functions being used, it must be that $|Q|\leq\kappa$. If $p\in Q$ then $f_{\alpha}(p)\geq p$ and clearly $f_{\alpha}(p)\in Q\cap D_{\alpha}$, so each $D_{\alpha}\cap Q$ is dense in $Q$. In addition, $Q$ is ccc: if $A$ is an antichain  in $Q$ and $p_{1},p_{2}\in A$ then $p_{1},p_{2}$ are incompatible in $Q$. But if they were compatible in $P$ then $g(p_{1},p_{2})\geq p_{1},p_{2}$ would be an element of $Q$, so they must be incompatible in $P$. Therefore $A$ is an antichain in $P$, and therefore must have countable  cardinality, since $P$ satisfies the ccc.

By assumption  , there is a directed $G\subseteq Q$ such that $G\cap(D_{\alpha}\cap Q)\neq\emptyset$ for each $\alpha<\kappa$, and therefore $MA_{\lambda}$ holds in full. ∎

Now we must prove that, if $G$ is a generic subset of $P_{\kappa}$, $R$ some partial order with $|R|\leq\lambda$ and $\langle D_{\alpha}\rangle_{\alpha<\lambda}$ are dense subsets of $R$ then there is some directed subset of $R$ intersecting each $D_{\alpha}$.

If $|R|<\lambda$ then $\lambda$ additional elements can be added greater than any other element of $R$ to make $|R|=\lambda$, and then since there is an order isomorphism into some partial order of $\lambda$, assume $R$ is a partial ordering of $\lambda$. Then let $D=\{\langle\alpha,\beta\rangle\mid\alpha\in D_{\beta}\}$.

Take canonical names so that $R=\hat{R}[G]$, $D=\hat{D}[G]$ and $D_{i}=\hat{D}_{i}[G]$ for each $i<\lambda$ and:

 $\begin{array}[]{rl}\Vdash_{P_{\kappa}}&\hat{R}\texttt{ is a partial ordering % satisfying ccc and}\\ &\hat{D}\subseteq\lambda\times\lambda\texttt{ and }\\ &\hat{D_{\alpha}}\texttt{ is dense in }\hat{R}\end{array}$

For any $\alpha,\beta$ there is a maximal antichain $D_{\alpha,\beta}\subseteq P_{\kappa}$ such that if $p\in D_{\alpha,\beta}$ then either $p\Vdash_{P_{\kappa}}\alpha\leq_{\hat{R}}\beta$ or $p\Vdash_{P_{\kappa}}\alpha\nleq_{\hat{R}}\beta$ and another maximal antichain $E_{\alpha,\beta}\subseteq P_{\kappa}$ such that if $p\in E_{\alpha,\beta}$ then either $p\Vdash_{P_{\kappa}}\langle\alpha,\beta\rangle\in\hat{D}$ or $p\Vdash_{P_{\kappa}}\langle\alpha,\beta\rangle\not\in\hat{D}$. These antichains determine the value of those two formulas   .

Then, since $\kappa^{\operatorname{cf}\kappa}>\kappa$ and $\kappa^{\mu}=\kappa$ for $\mu<\kappa$, it must be that $\operatorname{cf}\kappa=\kappa$, so $\kappa$ is regular  . Then $\gamma=\operatorname{sup}(\{\alpha+1\mid\alpha\in\operatorname{dom}(p),p\in% \bigcup_{\alpha,\beta<\lambda}D_{\alpha,\beta}\cup E_{\alpha,\beta})<\kappa$, so $D_{\alpha,\beta},E_{\alpha,\beta}\subseteq P_{\gamma}$, and therefore the $P_{\kappa}$ names $\hat{R}$ and $\hat{D}$ are also $P_{\gamma}$ names.

### Lemma: For any $\gamma$, $G_{\gamma}=\{p\upharpoonright\gamma\mid p\in G\}$ is a generic subset of $P_{\gamma}$

###### Proof.

First, it is directed, since if $p_{1}\upharpoonright\gamma,p_{2}\upharpoonright\gamma\in G_{\gamma}$ then there is some $p\in G$ such that $p\leq p_{1},p_{2}$, and therefore $p\upharpoonright\gamma\in G_{\gamma}$ and $p\leq p_{1}\upharpoonright\gamma,p_{2}\upharpoonright\gamma$.

Also, it is generic. If $D$ is a dense subset of $P_{\gamma}$ then $D_{\kappa}=\{p\in P_{\kappa}\mid p\leq q\in D\}$ is dense in $P_{\kappa}$, since if $p\in P_{\kappa}$ then there is some $d\leq p\upharpoonright$, but then $d$ is compatible with $p$, so $d\cup p\in D_{\kappa}$. Therefore there is some $p\in D_{\kappa}\cap G_{\kappa}$, and so $p\upharpoonright\in D\cap G_{\gamma}$. ∎

Since $\hat{R}$ and $\hat{D}$ are $P_{\gamma}$ names, $\hat{R}[G]=\hat{R}[G_{\gamma}]=R$ and $\hat{D}[G]=\hat{D}[G_{\gamma}]=D$, so

 $\begin{array}[]{rl}V[G_{\gamma}]\vDash&\hat{R}\texttt{ is a partial ordering % of }\lambda\texttt{ satisfying the ccc and}\\ &\hat{D_{\alpha}}\texttt{ is dense in }\hat{R}\end{array}$

Then there must be some $p\in G_{\gamma}$ such that

 $p\Vdash_{P_{\gamma}}\hat{R}\texttt{ is a partial ordering of }\lambda\texttt{ % satisfying the ccc}$

Let $A_{p}$ be a maximal antichain of $P_{\gamma}$ such that $p\in A_{p}$, and define $\hat{\leq}^{*}$ as a $P_{\gamma}$ name with $(p,m)\in\hat{\leq}^{*}$ for each $m\in\hat{R}$ and $(a,n)\in\hat{\leq}^{*}$ if $n=(\alpha,\beta)$ where $\alpha<\beta<\lambda$ and $p\neq a\in A_{p}$. That is, $\hat{\leq}^{*}[G]=R$ when $p\in G$ and $\hat{\leq}^{*}[G]=\in\upharpoonright\lambda$ otherwise. Then this is the name for a partial ordering of $\lambda$, and therefore there is some $\eta\in S_{\gamma,\lambda}$ such that $\hat{\leq}^{*}=\hat{\leq}_{\eta}$, and $\eta\geq\gamma$. Since $p\in G_{\gamma}\subseteq G_{\eta}$, $\hat{Q}_{\eta}[G_{\eta}]=\hat{\leq}_{\eta}[G_{\eta}]=R$.

Since $P_{\eta+1}=P_{\eta}*Q_{\eta}$, we know that $G_{Q_{\eta}}\subseteq Q_{\eta}$ is generic since http://planetmath.org/node/3258forcing with the composition  is equivalent     to successive forcing. Since $D_{i}\in V[G_{\gamma}]\subseteq V[G_{\eta}]$ and is dense, it follows that $D_{i}\cap G_{Q_{\eta}}\neq\emptyset$ and since $G_{Q_{\eta}}$ is a subset of $R$ in $P_{\kappa}$, $MA_{\lambda}$ holds.

## Proof that $2^{\aleph_{0}}=\kappa$

The relationship between Martin’s axiom and the continuum hypothesis  tells us that $2^{\aleph_{0}}\geq\kappa$. Since $2^{\aleph_{0}}$ was less than $\kappa$ in $V$, and since $|P_{\kappa}|=\kappa$ adds at most $\kappa$ elements, it must be that $2^{\aleph_{0}}=\kappa$.

Title Martin’s axiom is consistent MartinsAxiomIsConsistent 2013-03-22 13:21:59 2013-03-22 13:21:59 mathcam (2727) mathcam (2727) 7 mathcam (2727) Result msc 03E50