On the Existence of Ramsey Ultrafilters

Abstract

So far we have seen that $\mathfrak {p}=\mathfrak {c}$ implies the existence of Ramsey ultrafilters (see Proposition 10.9). In particular, if we assume CH, then Ramsey ultrafilters exist. Moreover, by Proposition 13.9 we know that MA(countable) implies the existence of $\mathfrak {2}^{\mathfrak {c}}$ mutually non-isomorphic Ramsey ultrafilters. Furthermore, by Theorem 21.5 we know that $\mathfrak {p}\le \mathit {cov}(\mathcal {M})$, and Chapter 13 ∣ Related Result 80 tells us that MA(countable) is equivalent to $\mathit {cov}(\mathcal {M})=\mathfrak {c}$. Hence, $\mathit {cov}(\mathcal {M})=\mathfrak {c}$ is a sufficient condition for the existence of Ramsey ultrafilters and it is natural to ask whether $\mathit {cov}(\mathcal {M})=\mathfrak {c}$ is necessary, too. In the first section of this chapter we shall give a negative answer to this question by constructing a model of $\mathsf {ZFC}+\mathit {cov}(\mathcal {M})<\mathfrak {c}$ in which there is a Ramsey ultrafilter. Since in that model we have $\mathfrak {h}=\mathfrak {c}$ and $\mathfrak {h}$ is related to the Ramsey property (cf. Chapter 9), one might think that perhaps $\mathfrak {h}=\mathfrak {c}$ implies the existence of a Ramsey ultrafilter; but this is not the case, as we shall see in the second section of this chapter.