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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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1 Answer 1

up vote 4 down vote accepted

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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In the sense that if $A_k$ is a partition, we take $B_n:=A_n$ if $|A_n\cap X|=1$ and put all the other $A_n$ in the first of the $B_n$ (that are infinite). So we apply (2) to the partition $B_n$. But we don't get that $|A_n\cap X|=1$ for all $n$. –  Jacob Fox Apr 23 '11 at 21:14
    
Jacob, you are correct. I have fixed my answer. –  Asaf Karagila Apr 23 '11 at 21:35

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