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Let $D$ be a Ramsey ultrafilter on $\omega$. Suppose that $X_0 \supseteq X_1 \supseteq \ldots$ are sets in $D$. Since $D$ is a Ramsey ultrafilter, and therefore a $p$-point, there exists $Y \in D$ such that $Y \setminus X_n$ is finite for all $n \in \omega$. Define a sequence as follows:

$y_0 =$ the least $y_0 \in Y$ such that $\{ y \in Y \mid y > y_0\} \subseteq X_0$,

$y_1 =$ the least $y_1 \in Y$ such thaty_1 > y_0 and $\{ y \in Y \mid y > y_0\} \subseteq X_{y_0}$,

$\ldots$

$y_n =$ the least $y_n \in Y$ such that $y_n > y_{n-1}$ and $\{ y \in Y \mid y > y_n \} \subseteq X_{y_{n-1}}$.

Define $A_n := \{y \in Y \mid y_n < y \le y_{n+1} \}$.

Now my question is, why is the following statement true: "Since $D$ is Ramsey, there exists a set $\{ z_n \}_{n=0}^{\infty} \in D$ such that $z_n \in A_n$ for all $n$" ?

It seems to me that to this, we need to show that the $A_n$'s are elements of some partition where each partition element is not in $D$. So my idea was to define $\displaystyle\mathcal{A} = \left\{\left( \omega \setminus \bigcup_{n\in \omega} A_n \right), A_1, A_2, \ldots \right\}$, a partition of $\omega$. But I am not sure how to show each of these sets is not in $D$. Is there a better partition to take?

Any help would be appreciated, thank you. [This is from Jech, 9.2]

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Note that $\{ A_n \}_{n \in \omega}$ is a partition of $Y_0 = \{ y \in Y : y > y_0 \} \in D$.

The "Ramsey property" of Ramsey ultrafilters also holds for partitions of any set in the ultrafilter by "small" sets. (If $\{ B_n \}_{n \in \omega}$ is a partition of $Y \in D$ by $D$-small sets, then $\{ B_n \}_{n \in \omega} \cup \{ \omega \setminus Y \}$ is a partition of $\omega$ by $D$-small sets, so there is a choice set $Z$ for this partition in $D$. Consider $Z \cap Y \in D$.)

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    @Paul: Oh, yes. Of course.2012-03-14