New solutions to exercises from Thomas Jech’s book “Set Theory”:

Last November, I tried to provide some details of the proof given in chapter 7, regarding the fact that the continuum hypothesis implies the existence of a Ramsey ultrafilter. Peter Krautzberger pointed out that the proof could probably work assuming only Martin’s Axiom. This was indeed proved by Booth in 1970 and the missing argument is actually given in exercise 16.16. For completeness, I copy the details on this blog post.

Remember that the proof involves contructing a sequence ${(X_{\alpha})}_{{\alpha<2^{{\aleph_{0}}}}}$ of infinite subsets of $\omega$. The induction hypothesis is that at step $\alpha<2^{{\aleph_{0}}}$, for all $\beta_{1},\beta_{2}<\alpha$ we have $\beta_{1}<\beta_{2}\implies X_{{\beta_{2}}}\setminus X_{{\beta_{1}}}$ is finite. It is then easy to show the result for the successor step, since the construction satisfies $X_{{\alpha+1}}\subseteq X_{\alpha}$. However at limit step, to ensure that $X_{\alpha}\setminus X_{\beta}$ is finite for all $\beta<\alpha$, the proof relies on the continunum hypothesis. This is the only place where it is used.

Assume instead Martin’s Axiom and consider a limit step $\alpha<2^{{\aleph_{0}}}$. Define the forcing notion $P_{\alpha}=\{(s,F):s\in{[\omega]}^{{<\omega}},F\in{[\alpha]}^{{<\omega}}\}$ and $(s^{{\prime}},F^{{\prime}})\leq(s,F)$ iff $s\subseteq s^{{\prime}}$, $F\subseteq F^{{\prime}}$ and $s^{{\prime}}\setminus s\subseteq X_{\beta}$ for all $\beta\in F$. It is clear that the relation is reflexive and antisymmetric. The transitivity is almost obvious, just note that if $(s_{3},F_{3})\leq(s_{2},F_{2})$ and $(s_{2},F_{2})\leq(s_{1},F_{1})$ then for all $\beta\in F_{1}\subseteq F_{2}$ we have $s_{3}\setminus s_{1}\subseteq s_{3}\setminus s_{2}\cup s_{2}\setminus s_{1}% \subseteq X_{\beta}$.

The forcing notion satisfies ccc or even property (K): since ${[\omega]}^{{<\omega}}$ is countable, for any uncountable subset $W$ there is $t\in{[\omega]}^{{<\omega}}$ such that $Z=\{(s,F)\in W:s=t\}$ is uncountable. Then any $(t,F_{1}),(t,F_{2})\in Z$ have a common refinement $(t,F_{1}\cup F_{2})$.

For all $n<\omega$, define $D_{n}=\{(s,F):|s|\geq n\}$. Let $\beta_{1}>\beta_{2}>...>\beta_{k}$ the elements of $F$. We show by induction on $1\leq m\leq k$ that $\bigcap_{{i=1}}^{m}X_{{\beta_{i}}}$ is infinite. This is true for $m=1$ by assumption. If it is true for $m-1$ then

$\bigcap_{{i=1}}^{{m-1}}X_{{\beta_{i}}}=\bigcap_{{i=1}}^{m}X_{{\beta_{i}}}\cup% \bigcap_{{i=1}}^{m}X_{{\beta_{i}}}\setminus X_{{\beta_{m}}}$ |

The left hand side is infinite by induction hypothesis. The second term of the right hand side is included in $X_{{\beta_{1}}}\setminus X_{{\beta_{m}}}$ and thus is finite. Hence the first term is infinite and the result is true for $m$. Finally, for $m=k$, we get that $\bigcap_{{\beta\in F}}X_{\beta}$ is infinite. Pick $x_{1},x_{2},...,x_{n}$ distinct elements from that set and define $(s^{{\prime}},F^{{\prime}})=(s\cup\{x_{1},...x_{n}\},F)$. We have $(s^{{\prime}},F^{{\prime}})\in D_{n}$, $s\subseteq s^{{\prime}}$, $F\subseteq F^{{\prime}}$ and for all $\beta\in F$, $s^{{\prime}}\setminus s=\{x_{1},...x_{n}\}\subseteq X_{\beta}$. This shows that $D_{n}$ is dense. For each $\beta<\alpha$, the set $E_{\beta}=\{(s,F):\beta\in F\}$ is also dense: for any $(s,F)$ consider $(s^{{\prime}},F^{{\prime}})=(s,F\cup\{\beta\})$.

By Martin’s Axiom there is a generic filter $G$ for the family $\{D_{n}:n<\omega\}\cup\{E_{\beta}:\beta<\alpha\}$ of size $|\alpha|<2^{{\aleph_{0}}}$. Let $X_{\alpha}=\{n<\omega:\exists(s,F)\in G,n\in s\}$. For all $n<\omega$, there is $(s,F)\in G\cap D_{n}$ and so $|X_{\alpha}|\geq|s|\geq n$. Hence $X_{\alpha}$ is infinite. Let $\beta<\alpha$ and $(s_{1},F_{1})\in G\cap E_{\beta}$. For any $x\in X_{\alpha}$, there is $(s_{2},F_{2})\in G$ such that $x\in s_{2}$. Hence there is $(s_{3},F_{3})\in G$ a refinement of $(s_{1},F_{2}),(s_{2},F_{2})$. We have $x\in s_{2}\subseteq s_{3}$ and $s_{3}\subseteq s_{1}\subseteq X_{\beta}$. Hence $X_{\alpha}\setminus X_{\beta}\subseteq s_{1}$ is finite and the induction hypothesis is true at step $\alpha$.