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In many books I find these two definitions of Ramsey ultrafilters, extremely similar but different:

1)For every partition $\mathbb{N}=\bigsqcup A_k$ with $A_k\not\in\mathcal{U}$ there exists $X\in\mathcal{U}$ such that $|X\cap A_k|=1$.

2)For every partition $\mathbb{N}=\bigsqcup A_k$ with $A_k\not\in\mathcal{U}$ there exists $X\in\mathcal{U}$ such that $|X\cap A_k|\leq1$.

And in many proves these books pass from one of these definitions to the other without saying why they can do that. So I think they are equivalent, but I can't prove that despite I think it should be easy, otherwise the books should have spent some words about that.

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For 1 implies 2 is trivial.

For the other direction, for all $A_k$ which don't meet $X$ choose some $a_k\in A_k$ and take $X\cup\{a_k\mid A_k\cap X = \emptyset\}$. It is in the ultrafilter as needed.

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    Jacob, you are correct. I have fixed my answer.2011-04-23