Exercises in Set Theory: Iterated Forcing and Martin’s Axiom

01 Jun 2013

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

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  Chapter 16: Forcing

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. Booth actually proved in 1970 that this works assuming only Martin’s Axiom. 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 $\alpha$.