5
$\begingroup$

Consider Prikry's forcing for changing the cofinality of a measurable cardinal into $\omega.$ The forcing has the Prikry property and one can prove this either directly or using Rowbottom's theorem which is a Ramsey type theorem.

Now consider Magidor's forcing for changing the cofinality of a large cardinal into $\omega_1$ (or any other uncountable regular cardinal). The proof of Prikry property for it (that I have seen) is direct.

My question is:

Question. Is there a suitable Ramsey type theorem such that one can use it to present a simplified proof of the Prikry property for Magidor's forcing?

Any references, if there are, are appreciated.

$\endgroup$
3
  • $\begingroup$ The Ramsey argument only works for Prikry with normal filters, right? Not for the generalized tree based Prikry? $\endgroup$
    – Asaf Karagila
    Commented May 4, 2017 at 7:17
  • $\begingroup$ There is such a Ramsey type theorem for the tree version, see Bukovsky's paper Iterated ultrapowers and Prikry forcing . $\endgroup$ Commented May 4, 2017 at 7:35
  • $\begingroup$ I may also mention that Prikry's original paper includes a Ramsey type theorem for the forcing notion defined in Section 1.3 of Gitik's paper prikry type forcings, namely "One-Element Prikry Forcing which adds a Prikry Sequence to a Singular Cardina which is a limit of $\omega$-many measurable cardinals". $\endgroup$ Commented May 4, 2017 at 7:36

1 Answer 1

4
$\begingroup$

There is an analog of Rowbottom's theorem which can be extracted from the proof of the Magidor forcing:

Suppose that $u$ is a sequence of ultrafilters on $\kappa$ of length $\lambda<\kappa$ and $f\colon [\kappa]^{<\omega}\to 2$. Then there are sets $A_\nu\in u(\nu)$ for $\nu<\lambda$ such that whenever $\langle \nu_0,\dots,\nu_{k-1}\rangle$ is a finite sequence of ordinals less than $\lambda$, the function $f$ is constant on the set of increasing sequences $\langle \alpha_0,\dots,\alpha_{k-1}\rangle\in A_{\nu_0}\times\dots\times A_{\nu_{k-1}}$.

This can be extended to Radin forcing on sequences of ultrafilters of length less than $\kappa^+$; I'm not sure what could be done for longer sequences.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .