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 that$y_1 > y_0$ and $\{ y \in Y \mid y > y_0\} \subseteq X_{y_0}$,


$y_n =$ 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]


1 Answer 1


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 $Z \cap Y \in D$.)

  • $\begingroup$ Why is $Y_0 \in D$ ? $\endgroup$ Mar 14, 2012 at 13:43
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    $\begingroup$ @Paul: Note that $\{ y \in \omega : y > y_0 \} \in D$ since $D$ is a nonprincipal ultrafilter, and take the intersection of this set with $Y$. $\endgroup$
    – user642796
    Mar 14, 2012 at 13:47
  • $\begingroup$ Hmm... I feel like I'm being a bit dense. $D$ being nonprincipal means it does not have a least element, so how does that tell me that $\{ y \in \omega \mid y > y_0 \}$ is in the filter? Thanks $\endgroup$ Mar 14, 2012 at 14:12
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    $\begingroup$ @Paul: Recall that a principal ultrafilter on $\omega$ is one of the form $D = \{ a \subseteq \omega : n \in \omega \}$ for some $n$; all other ultrafilters are nonprincipal. It is a quick to prove that an ultrafilter $D$ on $\omega$ is principal iff it contains a finite set. (If $D$ is nonprincipal then given $\{ n_1 , \ldots , n_k \} \subseteq \omega$ we know that $\omega \setminus \{ n_i \} \in D$ for all $i$ and so $\bigcap_{i=1}^k ( \omega \setminus \{ n_i \} ) = \omega \setminus \{ n_1 , \ldots , n_k \} \in D$; if $D$ is principal, then by definition it contains a singleton.) $\endgroup$
    – user642796
    Mar 14, 2012 at 15:04
  • $\begingroup$ $D = \{a \subseteq \omega \mid n \in a\}$ in your comment right? $\endgroup$ Mar 14, 2012 at 16:03

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