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Assume CH. A tower is an almost decreasing family $(A_\alpha)_{\alpha\in\omega_1}$ with no pseudointersection. A selective (also called Ramsey) ultrafilter is one with the property that for every $f:[\omega]^{2}\rightarrow2$, there is some set $X$ in the ultrafilter with $|f''(X)|=1$.

As in the title, my question is whether or not every tower can be extended to a selective ultrafilter?

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The answer is no: for every ultrafilter $U$, there is a tower $T$ such that the only ultrafilter containing $T$ is $U$. (Not every tower has this property, of course.)

So it remains to show that there exist non-Ramsey ultrafilters, which we can do by diagonalizing against all possible colorings.

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  • $\begingroup$ Thanks for the answer. Could you point me to where I can see the proof of the first fact? $\endgroup$ – Horse Jan 18 '16 at 22:09
  • $\begingroup$ As an easier to see answer inspired by yours, note that every p point (under CH) is generated by a tower; take any p point which is not a q point, and the generating tower must not be extendable to a selective. Though this doesn't answer the p point part of my question and yours does, so I would still like to see how to prove the first fact in your answer. $\endgroup$ – Horse Jan 18 '16 at 22:22
  • $\begingroup$ For the existence of non-Ramsey ultrafilters you can also observe that if $p\in\omega^*$, then $p\cdot p$ is never Ramsey, where $\cdot$ is the Fubini (tensor) product. $\endgroup$ – Brian M. Scott Jan 18 '16 at 22:32
  • $\begingroup$ I should point out that Noah Schweber's argument is incorrect: Under CH, an ultrafilter $\mathcal{U}$ is generated by a tower if and only if $\mathcal{U}$ is a p point. There are ultrafilters which are not p points under CH. $\endgroup$ – Horse Jan 21 '16 at 1:23

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