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Let $D$ be a Ramsey ultrafilter on $\omega$, and let $F: [\omega]^{n+1} \to \{1,\ldots, k\}$ be a function. Define the function $F_a: [\omega\setminus \{a \}]^n \to \{1, \ldots, k\}$ for each $a\in \omega$, given by $F_a (x) = F( x \cup \{a \})$. Let's suppose that for each $a$, we can find some $H_a \in D$ such that $F_a:[H_a]^{n}\to \{1,\ldots, k\}$ is constant.

Now I'm having some difficulty proving the following statement: "There exists $X \in D$ such that the constant value of $F_a$ is the same for all $a \in X$." This is from Jech(9.2).

Does anyone have any advice about how to prove this?

Thanks very much.

$\bf{NOTE}$: $[X]^n$ means subsets of $X$ of size $n$.

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up vote 2 down vote accepted

Partition $\omega$ into $k$ sets $A_i$, $1\le i\le k$, by setting $a\in A_i$ iff the constant value of $F_a$ on $[H_a]^n$ is $i$.

Since $D$ is an ultrafilter (you do not need that it is Ramsey here, but you need that it is ultra), exactly one of the sets $A_i$ is in $D$. Of course, this is the set $X$ in the statement.

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There are only $k$ possible values for the constant value of $F_a$ on $[H_a]^n$. This gives you a partition of $\omega$ into $k$ parts. $D$ is an ultrafilter, so one of the parts must belong to $D$. Note that this does not require that $D$ be Ramsey.

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