Question

Hi, I know that there are models in ZFC where don't exist p-points, but I can't find (neither on internet) a proof that I understand, some of you could help me please?

Another question: I know that assuming the continuum hypothesis then there exist Ramsey ultrafilters, but I don't know how to prove it. I found a proof that if we assume the continuum hypothesis then there exist p-points (http://www.logic.univie.ac.at/~wolfgang/Diplomarbeit_Wolfgang_Wohofsky.pdf).

Answer 1

I agree with the comments that you'll have to learn iterated forcing in order to understand the construction of models of ZFC without P-points. Your second question, though, is much easier. To prove the existence of Ramsey ultrafilters assuming CH, proceed as in the proof of Theorem 1.13 in the Diplomarbeit you cited, but make the following changes. Note first that the number of partitions of the set $[\omega]^2$ of pairs into two pieces is the cardinal of the continuum, i.e., $\aleph_1$ since we're assuming CH. So fix an enumeration of these partitions in a sequence of length $\omega_1$. The set that is called $X_{\alpha+1}$ (in formula (1.1) at the bottom of page 8) should instead be given a temporary name, say $Y_\alpha$. Now apply Ramsey's theorem to find an infinite subset of $Y_\alpha$ that is homogeneous for the $\alpha$th partition in your fixed enumeration. Define $X_{\alpha+1}$ to be that infinite homogeneous set. The proof that you get an ultrafilter (an in fact a P-point) still works, but the extra steps, shrinking $Y_\alpha$ to $X_{\alpha+1}$ using Ramsey's theorem, guarantee that your ultrafilter will be a Ramsey ultrafilter.

An alternative characterization of Ramsey ultrafilters is that every function $\omega\to\omega$ becomes either constant or one-to-one when restricted to a suitable set in the ultrafilter. If you're willing to use this characterization, then the extra step in the construction can be simplified. Begin by enumerating all functions $\omega\to\omega$. Then produce $X_{\alpha+1}$ as an infinite subset of $Y_\alpha$ on which the $\alpha$th function in your enumeration is constant or one-to-one. The existence of such a subset is obvious, so you avoid the need for Ramsey's theorem; on the other hand, the proof that this alternative characterization is equivalent to Ramseyness of the ultrafilter is a non-trivial result of Kunen.