# Why is this set a Ramsey Ultrafilter?

I am reading Jech's book on Set Theory just now and I had a question about Ramsey ultrafilters.

Jech defines a Ramsey ultrafilter to be: "A non-principal ultrafilter $D$ on $\omega$ is a Ramsey ultrafilter if for every partition $\{A_n | n\in \omega \}$ of $\omega$ into $\aleph_0$ pieces such that $A_n \not\in D$ for all $n$, there exists $X \in D$ such that $X \cap A_n$ has one element for all $n \in \omega$."

Now if $2^{\aleph_0} = \aleph_1$, then we can enumerate all the partitions of $\omega$ by $\mathcal{A}_{\alpha}$ for $\alpha < \omega_1$, and then construct an $\omega_1$-sequence of infinite subsets of $\omega$ as follows: 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 $\beta < \alpha$.

Now I understand the above, but he defines the set $D = \{ X | X \supseteq X_{\alpha} \text{ for some } \alpha \in \omega_1\}$ and says it is a Ramsey ultrafilter. But I cannot see how it is nonprincipal, an ultrafilter, or Ramsey.

I would appreciate any help with those three points.

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## 1 Answer

Because the $X_\alpha$ are almost decreasing, we have that $D$ is closed under intersections and it is clearly closed under enlargements, and all members are infinite.
We can assume that all finite partitions of $\omega$ were included in our enumeration as well.

To see it is Ramsey (i.e. the partition property part):

Let $\{ A_n \mid n \in \omega \}$ be a partition of $\omega$, and assume no $A_n$ is in $D$. This partition occurs in the enumeration as some $\cal{A}_{\alpha_0}$. If $X_{\alpha_0+1}$ would be a subset of one of the $A_n (\in \cal{A}_{\alpha_0})$, then this $A_n$ would be in $D$ by definition, so this cannot happen. By construction, we have that $X_{\alpha_0 + 1} \in D$ meets every $A_n$ in at most one point, which is what needed to be shown for the partition property.

To see it is an ultrafilter consider for a subset A, the partition $\cal{A}_\alpha = \{ A , \omega\setminus A \}$, and note that the second option cannot hold for $X_{\alpha+1}$, as this makes the set finite, so one of the others hold and makes one of them a member of $D$.

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There I was slaving over an answer, and just as I was having this understand that the construction only chooses infinite sets you posted this answer which is quite cleaner than what I intended to post. :-) –  Asaf Karagila Feb 1 '12 at 13:04
Thanks for your response. But why can we assume that we can include finite partitions? –  Paul Slevin Feb 1 '12 at 14:07
And how do I know that the sequence is almost decreasing? –  Paul Slevin Feb 1 '12 at 14:18
@PaulSlevin this is by construction, isn't it? Almost decreasing means that $A_\alpha \setminus A_\beta$ is finite for all $\alpha > \beta$, and for successors we have just standard decreasing, and for limits we construct a predecessor. And Jech enumerates all partitions, so why not the finite ones too? It helps the proof later, and makes no real difference in the construction... –  Henno Brandsma Feb 1 '12 at 14:53
I looked in Jech, and he doesn't seem to say he just enumerates the infinite partitions, just "partitions". So I think my assumption is justified. I looked into other books that construct Ramsey ultrafilters under CH (or $\frak{t} = 2^{\aleph_0}$, or MA) and the others also take care of the requirement by considering finite partitions (or partitions with just 2 sets) as well, one way or another. –  Henno Brandsma Feb 4 '12 at 7:24