# Construction of a Ramsey ultrafilter

I am having difficulties with the proof in Jech Set Theory concerning the existence of Ramsey filters in case the continuum hypothesis holds. A similar question about the same proof was asked here but i still have some problems.

In the proof we have a construction of an $\omega_1$ almost decreasing sequence of infinite subsets of $\omega$ (quote from the original post):

Given $X_{\alpha}$, let $X_{\alpha+1} \subseteq X_{\alpha}$ be such that either $X_{\alpha+1} \subseteq A$ for some $A \in \mathcal{A}_{\alpha}$, or that $|X_{\alpha+1} \cap A| \le 1$ for all $A \in \mathcal{A}_{\alpha}$. If $\alpha$ is a limit ordinal, let $X_{\alpha}$ be such that $X_{\alpha} \setminus X_{\beta}$ is finite for all $\beta < \alpha$.

My first question is regarding the construction itself, in the limit case. Why can i always find such a set $X_\alpha$. since $\alpha< \omega_1$ is countable i understand that it is enough to find an injection from

$\left\{ X \subseteq \omega : \left\vert{X}\right\vert =\aleph_0 \land \exists\beta<\alpha : \left\vert{X\setminus X_\beta}\right\vert = \aleph_0\right\}$ to $\alpha, \hspace{1mm}$ but i couldn't find one.

My second question is why the set $D = \{ X | X \supseteq X_{\alpha} \text{ for some } \alpha \in \omega_1\}$ is a Ramsey filter. This question was actually asked in the original post so i'll clarify what i didn't understand. In the answer (posted by Henno Brandsma) it is said that since $X_\alpha$ is almost decreasing, it is close under intersection. I cant see how that follows, i think i'm missing something basic here. Im ok with the rest (meaning i understand that if $D$ is a filter, it is a ramsey filter).

Thanks!

At a limit stage $\alpha$ let $\{Y_n:n\in\omega\}=\{X_\beta:\beta<\alpha\}$. Note that $\bigcap_{k\le n}Y_k$ is infinite for each $n\in\omega$. Now just choose $m_n\in\left(\bigcap_{k\le n}Y_k\right)\setminus\{m_k:k<n\}$ for each $n\in\omega$, and let $X_\alpha=\{m_n:n\in\omega\}$.
Suppose that $A,B\in D$. Then there are $\alpha,\beta<\omega_1$ such that $X_\alpha\subseteq A$ and $X_\beta\subseteq B$. Without loss of generality $\alpha\le\beta$, so $X_\beta\setminus X_\alpha$ is finite; let $F=X_\beta\setminus X_\alpha$, and observe that $X_\alpha\cap X_\beta=X_\beta\setminus F$. Now consider the partition $\{F,X_\beta\setminus F,\omega\setminus X_\beta\}$ of $\omega$. Say this is $\mathscr{A}_\gamma$; then $X_{\gamma+1}$ must be a subset of $X_\beta\setminus F$ or of $\omega\setminus X_\beta$, and the latter is clearly impossible, so $X_{\gamma+1}\subseteq X_\alpha\cap X_\beta\subseteq A\cap B$, and therefore $A\cap B\in D$.
• Thanks! nice use of the enumeration of the partitions and also clean construction of $X\alpha$ for limit $\alpha$. – Ariel Aug 12 '15 at 10:03