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Let ${\cal U}$ be a non-principal ultrafilter on $\omega$. If $\kappa>0$ is a cardinal, we say that a function $c:\omega \to \kappa$ is a coloring for ${\cal U}$ if for all $U\in{\cal U}$ the restriction $c|_U$ is not constant. The coloring number of ${\cal U}$ is the least cardinal $\mu$ such that there is a coloring $c: \omega\to\mu$ for ${\cal U}$.

Is there a non-principal ultrafilter ${\cal U}$ on $\omega$ with finite coloring number?

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    $\begingroup$ On odd days I don't believe in AC, so that my answer would be no, there does not exist a non-principal ultrafilter on $\omega$. $\endgroup$
    – Benoît Kloeckner
    Commented Jul 13, 2017 at 13:16
  • $\begingroup$ @BenoîtKloeckner, get back to us tomorrow quatorze juillet ($2\times 7$). $\endgroup$
    – Mikhail Katz
    Commented Jul 13, 2017 at 13:44
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    $\begingroup$ If $\kappa$ is finite and $c:\omega \to \kappa$ then there is $i \in \kappa$ such that $c^{-1}(i) \in \mathcal{U}$, so $c$ is not a coloring. It makes no difference if $\mathcal{U}$ is principal or not. $\endgroup$
    – Ramiro de la Vega
    Commented Jul 13, 2017 at 13:53
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    $\begingroup$ @RamirodelaVega Not only does "non-principal" not matter, neither does the assumption that the ultrafilter's underlying set (and the domain of the coloring) is $\omega$. $\endgroup$
    – Andreas Blass
    Commented Jul 13, 2017 at 15:12

1 Answer 1

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If a function $c$ has finite image then $\omega$ decomposes into a disjoint union of finitely many subsets according to the value of $c$. By the property of the ultrafilter exactly one of those sets will be in $\mathcal U$. Thus $c$ is constant on a member of the ultrafilter. Therefore there is no ultrafilter with finite coloring number.

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